[
    {
        "Problem": "The symbols $(a,b,\\ldots,g)$ and $[a,b,\\ldots, g]$ denote the greatest common divisor and least common multiple, respectively, of the positive integers $a,b,\\ldots, g$. For example, $(3,6,18)=3$ and $[6,15]=30$. Prove that",
        "Solution_1": "Consider an arbitrary prime $p$. Let $p^\\alpha$, $p^\\beta$, and $p^\\gamma$ be the greatest powers of $p$ that divide $a$, $b$, and $c$. \nWLOG let $\\alpha \\leq \\beta \\leq \\gamma$.\nExamining each factor in the equation, we see that the largest power of $p$ that divides the left hand side is $2\\gamma -(\\beta+\\gamma+\\gamma) = -\\beta$, and the largest power of $p$ that divides the right hand side is 2\\alpha -($\\alpha + \\beta + \\alpha) = -\\beta$. Since every prime has the same power in both expressions, the expressions are equal. $\\blacksquare$",
        "Solution_2": "Let $p = (a, b, c)$, $pq = (a, b)$, $pr = (b, c)$, and $ps = (c, a)$. Then it follows that $q, r, s$ are pairwise coprime and $a = pqsa'$, $b = pqrb'$, and $c = prsc'$, with $a', b', c'$ pairwise coprime as well. Then, we wish to show\n\nwhich can be checked fairly easily.",
        "Solution_3_.28very_long.2C_but_detailed.29": "Note: This solution is more of a \"bashy\" solution, and is easier to think of than the first 2 solutions.\nDividing both sides of the equation by $(a,b,c)^2$, and then multiplying both sides by $[a,b][b,c][c,a]$ gives us \nNow, we look at the prime factorisations of $a,b,c$. Let \n\n\n\nwhere all of the $x_1, y_1, z_1$ are nonnegative integers (they could be 0).\nThus, we have \nand \nNow, we see that $\\left(\\frac{[a,b,c]}{(a,b,c)}\\right) = 2^{\\max(x_1,y_1,z_1) - \\min(x_1,y_1,z_1)} \\cdot 3^{\\max(x_2,y_2,z_2) - \\min(x_2,y_2,z_2)} \\cdot 5^{\\max(x_3,y_3,z_3) - \\min(x_3,y_3,z_3)} \\cdots.$ Squaring the LHS will just double all the exponents on the RHS.\nAlso, we have $[a,b]=2^{\\max(x_1,y_1)} \\cdot 3^{\\max(x_2,y_2)} \\cdot 5^{\\max(x_3,y_3)} \\cdots$. We can also build similar equations for $[b,c],[c,a],(a,b),(b,c)(c,a)$, using 2 of $x,y,z$ and either the minimum of the exponents or the maximum.\nWe see that when we multiply $[a,b]$, $[b,c]$, and $[a,c]$ together, the exponents of the prime factorization will be $\\max(x_i,y_i)+ \\max(y_i,z_i)+ \\max(z_i,x_1)$, where $i$ is chosen for the $i$'th prime. \nWhen we multiply $(a,b)$, $(b,c)$, and $(a,c)$ together, the exponents of the prime factorization will be $\\min(x_i,y_i)+ \\min(y_i,z_i)+ \\min(z_i,x_1)$, where $i$ is chosen for the $i$'th prime.\nThus, when we divide the former by the latter, we have $\\max(x_i,y_i)+ \\max(y_i,z_i)+ \\max(x_i,z_i) - (\\min(x_i,y_i)+ \\min(y_i,z_i)+ \\min(x_i,z_i))$. We wish to prove that this is equal to the exponents we got from $([a,b,c]/(a,b,c))^2$.\nIn other words, this problem is now down to proving that  for any nonnegative integers $x_i, y_i, z_i$. WLOG, let $x_i \\le y_i \\le z_i$. Thus, computing maximums and minimums, this equation turns into\nFinally, canceling out the $y_i$'s and collecting like terms, we see the RHS is $2z_i - 2x_i$, which is equal to the LHS. Thus, this equation is true, and we are done. $\\square$",
        "Solution_4": "Let\nwhere $\\gcd(A,B,C)=\\gcd(x,y,z)=1$. (Existence proof missing.) Then, $\\operatorname{lcm}(a,b,c)^2=(dxyzABC)^2$ and $\\operatorname{lcm}(a,b)=dxyzAB$. Using this cyclically, we have\nNow, note that $\\gcd(a,b,c)^2=d^2$ and $\\gcd(a,b)=dx$. Using this cyclically, we have\n\n-brainiacmaniac31\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "A given tetrahedron $ABCD$ is isosceles, that is, $AB=CD, AC=BD, AD=BC$. Show that the faces of the tetrahedron are acute-angled triangles.",
        "Solution_1": "Suppose $\\triangle ABD$ is fixed.\nBy the equality conditions, it follows that the maximal possible value of $BC$ occurs when the four vertices are coplanar, with $C$ on the opposite side of $\\overline{AD}$ as $B$.\nIn this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.\nFor the sake of contradiction, suppose $\\angle ABD$ is non-acute.\nThen, $(AD)^2\\geq (AB)^2+(BD)^2$.\nIn our optimal case noted above, $ACDB$ is a parallelogram, so\n\nHowever, as stated, equality cannot be attained, so we get our desired contradiction.",
        "Solution_2": "It's not hard to see that the four faces are congruent from SSS Congruence. Without loss of generality, assume that $AB\\leq BC \\leq CA$. Now assume, for the sake of contradiction, that each face is non-acute; that is, right or obtuse. Consider triangles $\\triangle ABC$ and $\\triangle ABD$. They share side $AB$. Let $k$ and $l$ be the planes passing through $A$ and $B$, respectively, that are perpendicular to side $AB$. We have that triangles $ABC$ and $ABD$ are non-acute, so $C$ and $D$ are not strictly between planes $k$ and $l$. Therefore the length of $CD$ is at least the distance between the planes, which is $AB$. However, if $CD=AB$, then the four points $A$, $B$, $C$, and $D$ are coplanar, and the volume of $ABCD$ would be zero. Therefore $CD>AB$. However, we were given that $CD=AB$ in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.",
        "Solution_3": "Let $\\vec{a} = \\overrightarrow{DA}$, $\\vec{b} = \\overrightarrow{DB}$, and $\\vec{c} = \\overrightarrow{DC}$. The conditions given translate to\n\nWe wish to show that $\\vec{a}\\cdot\\vec{b}$, $\\vec{b}\\cdot\\vec{c}$, and $\\vec{c}\\cdot\\vec{a}$ are all positive. WLOG, $\\vec{a}\\cdot\\vec{a}\\geq \\vec{b}\\cdot\\vec{b}, \\vec{c}\\cdot\\vec{c} > 0$, so it immediately follows that $\\vec{a}\\cdot\\vec{b}$ and $\\vec{a}\\cdot\\vec{c}$ are positive. Adding all three equations,\n\nIn addition,\n\nEquality could only occur if $\\vec{a} = \\vec{b} + \\vec{c}$, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.",
        "Solution_4": "Suppose for the sake of contradiction that $\\angle BAC$ is not acute. Since all three sides of triangles $BAC$ and $CDB$ are congruent, those two triangles are congruent, meaning $\\angle BDC=\\angle BAC>90^{\\circ}$. Construct a sphere with diameter $BC$. Since angles $BAC$ and $BDC$ are both not acute, $A$ and $D$ both lie on or inside the sphere. We seek to make $AD=BC$ to satisfy the conditions of the problem. This can only occur when $AD$ is a diameter of the sphere, since both points lie on or inside the sphere. However, for $AD$ to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron $ABCD$ degenerate, creating a contradiction. Thus, all angles on a face of an isosceles tetrahedron are acute.",
        "Solution_5": "Proof by contradiction: Assume at least one of the tetrahedron's faces are obtuse. WLOG, assume $\\angle BAC$ is an obtuse angle. Using SSS congruence to prove that all four faces of the tetrahedron are congruent also shows that the angles surrounding point $A$ are congruent to angles in triangle $ABC$; namely, $\\angle BAD$ is congruent to $\\angle ABC$, and $\\angle CAD$ is congruent to $\\angle ACB$. Since the internal angles of triangle $ABC$ must add to 180 degrees, so do the angles surrounding point A. Now lay triangle $ABC$ on a flat surface. A diagram would make it clear that from the perspective of an aerial view, the \"apparent\" measures of $\\angle BAD$ and $\\angle CAD$ (which are most likely distorted visually, assuming these angles stick up vertically from the flat surface on which triangle $ABC$ lies) can never exceed the true measures of those angles (equality happens when these angles also lie flat on top of triangle $ABC$). This means that these two angles can never join to form side AD (because $\\angle BAC$ is more than the sum of $\\angle BAD$ and $\\angle CAD$ - a direct consequence of the facts that $\\angle BAC$ is obtuse and all three angles add up to 180 degrees), so the tetrahedron with obtuse triangle faces is impossible.",
        "Solution_6": "Lemma: given triangle $ABC$ and the midpoint of $BC$, which we will call $M$, we can say that if $AM > \\frac{BC}{2}$, then $\\angle A < 90$.\nProof: Since $M$ is the midpoint of $BC$, $BM = MC = \\frac{BC}{2}$. Since it is given that $AM > \\frac{BC}{2}$, we can substitute $\\frac{BC}{2}$ to get two inequalities:\n\nThe above inequalities imply that $\\angle C > \\angle CAM$ and $\\angle B > \\angle BAM$. Adding these inequalities and simplifying the RHS, we have that $\\angle C + \\angle B > \\angle A$. Adding $\\angle A$ to both sides, replacing the LHS with $180$ and dividing by $2$ gets us that $\\angle A < 90$. This is our desired inequality, so we are done.\nNote that all faces of this tetrahedron are congruent, by SSS.\nIn particular, we will use that $\\triangle ABD \\cong \\triangle BAC$. WLOG, assume that $\\angle ADB$ is the largest angle in triangle $ABD$.\nBecause $\\triangle ABD \\cong \\triangle BAC$, the median from $D$ to $AB$ is equal length to the median from $C$ to $AB$. These points meet at $E$, the midpoint of $AB$. By the triangle inequality,\nBy substituting $CD$ with $AB$ (this is a given in the problem) and $CE$ with $DE$, and then dividing by $2$, we get that\nBy the lemma we showed at the start, this implies that $\\angle ADB < 90$, and since we said that $\\angle ADB$ was the largest angle, triangle $ADB$ must be acute. Since all of the faces of this tetrahedron are congruent, then, all of the faces must be acute.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after $n$ selections ($n>1$), the product of the $n$ numbers selected will be divisible by 10.",
        "Solution": "For the product to be divisible by 10, there must be a factor of 2 and a factor of 5 in there.\nThe probability that there is no 5 is $\\left( \\frac{8}{9}\\right)^n$.\nThe probability that there is no 2 is $\\left( \\frac{5}{9}\\right)^n$.\nThe probability that there is neither a 2 nor 5 is $\\left( \\frac{4}{9}\\right)^n$, which is included in both previous cases.\nThe only possibility left is getting a 2 and a 5, making the product divisible by 10.\nBy complementarity and principle of inclusion-exclusion, the probability of that is $1- \\left( \\left( \\frac{8}{9}\\right)^n + \\left( \\frac{5}{9}\\right)^n - \\left( \\frac{4}{9}\\right)^n\\right)=\\boxed{1-(8/9)^n-(5/9)^n+(4/9)^n}$.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Let $R$ denote a non-negative rational number. Determine a fixed set of integers $a,b,c,d,e,f$, such that for every choice of $R$,",
        "Solution": "Note that when $R$ approaches $\\sqrt[3]{2}$, $\\frac{aR^2+bR+c}{dR^2+eR+f}$ must also approach $\\sqrt[3]{2}$ for the given inequality to hold. Therefore\n\nwhich happens if and only if\n\nWe cross multiply to get $a\\sqrt[3]{4}+b\\sqrt[3]{2}+c=2d+e\\sqrt[3]{4}+f\\sqrt[3]{2}$. It's not hard to show that, since $a$, $b$, $c$, $d$, $e$, and $f$ are integers, then $a=e$, $b=f$, and $c=2d$.\nNote, however, that this is a necessary but insufficient condition. For example, we must also have $a^2<2bc$ to ensure the function does not have any vertical asymptotes (which would violate the desired property). A simple search shows that $a=0$, $b=2$, and $c=2$ works."
    },
    {
        "Problem": "A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property.\n",
        "Solution": "Lemma: Convex pentagon $A_0A_1A_2A_3A_4$ has the property that $[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]$ if and only if $\\overline{A_{n - 1}A_{n + 1}}\\parallel\\overline{A_{n - 2}A_{n + 2}}$ for $n = 0, 1, 2, 3, 4$ (indices taken mod 5).\nProof: For the \"only if\" direction, since $[A_0A_1A_2] = [A_1A_2A_3]$, $A_0$ and $A_3$ are equidistant from $\\overline{A_1A_2}$, and since the pentagon is convex, $\\overline{A_0A_3}\\parallel\\overline{A_1A_2}$. The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed.\n\nLet $A'B'C'D'E'$ be the inner pentagon, labeled so that $A$ and $A'$ are opposite each other, and let $a, b, c, d, e$ be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon $A'B'C'D'E'$ is similar to pentagon $ABCDE$ with $AB = m(A'B')$ and parallelogram $ABCB'$ (and cyclic) has area 2. Supposing\n\nwe have\n\nSince $BCDC'$ and $CDED'$ are parallelograms, $BD' = EC' = (m - 1)a$. Triangle $EB'C'$ is similar to triangle $ECB$, so $me = BC = \\left(\\frac{2m - 1}{m - 1}\\right)e$, and with the requirement that $m > 1$,\n\nNow, we compute that $[AC'D'] = \\frac{1}{2m - 1} = \\sqrt{5} - 2$, and similar computation for the other four triangles gives $R = 5\\sqrt{5} - 10$. From the aforementioned pentagon similarity, $Q = P/m^2 = \\left(\\frac{7 - 3\\sqrt{5}}{2}\\right)P$. Solving for $P$, we have\nTo show that there are infinitely many pentagons with the given property, we start with triangle $A'B'C'$ and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result.\nAlternatively, one can shear a regular pentagon, a transformation which preserves areas and the property of a pair of lines being parallel."
    },
    {
        "Problem": "Two points $P$ and $Q$ lie in the interior of a regular tetrahedron $ABCD$. Prove that angle $PAQ < 60^\\circ$.",
        "Solution_1": "Let the side length of the regular tetrahedron be $a$. Link and extend $AP$ to meet the plane containing triangle $BCD$ at $E$; link $AQ$ and extend it to meet the same plane at $F$. We know that $E$ and $F$ are inside triangle $BCD$ and that $\\angle PAQ = \\angle EAF$\nNow let’s look at the plane containing triangle $BCD$ with points $E$ and $F$ inside the triangle. Link and extend $EF$ on both sides to meet the sides of the triangle $BCD$ at $I$ and $J$, $I$ on $BC$ and $J$ on $DC$. We have $\\angle EAF < \\angle IAJ$\nBut since $E$ and $F$ are interior of the tetrahedron, points $I$ and $J$ cannot be both at the vertices and $IJ < a$, $\\angle IAJ < \\angle BAD = 60$. Therefore, $\\angle PAQ < 60$.\nSolution with graphs posted at\nhttp://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\nhurdler: Remark on solution 1: This proof is not rigorous, in the very last step. The last step needs more justification."
    },
    {
        "Problem": "Let $\\{X_n\\}$ and $\\{Y_n\\}$ denote two sequences of integers defined as follows:\nThus, the first few terms of the sequences are:\nProve that, except for the \"1\", there is no term which occurs in both sequences.",
        "Solution": "We can look at each sequence $\\bmod{8}$:\nThe third and fourth terms are $3$ and $5$ $\\bmod{8}$. Plugging into the formula, we see that the next term is $11\\equiv 3\\bmod{8}$, and plugging $5$ and $3$, we get that the next term is $13\\equiv 5\\bmod{8}$. Thus the sequence $X$ repeats, and the pattern is $3,5,3,5,\\dots$.\nThe first and second terms are $1$ and $7$ $\\bmod{8}$. Plugging into the formula, we see that the next term is $17\\equiv 1\\bmod{8}$, and plugging $7$ and $1$, we get that the next term is $23\\equiv 7\\bmod{8}$. Thus the sequence $Y$ repeats, and the pattern is $1,7,1,7,\\dots$.\n\nCombining both results, we see that $X_i$ and $Y_j$ are not congruent $\\bmod{8}$ when $i\\geq 3$ and $j\\geq 2$. Thus after the \"1\", the terms of each sequence are not equal.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Three distinct vertices are chosen at random from the vertices of a given regular polygon of $(2n+1)$ sides. If all such choices are equally likely, what is the probability that the center of the given polygon lies in the interior of the triangle determined by the three chosen random points?",
        "Solution": "There are $\\binom{2n+1}{3}$ ways how to pick the three vertices. We will now count the ways where the interior does NOT contain the center. These are obviously exactly the ways where all three picked vertices lie among some $n+1$ consecutive vertices of the polygon.\nWe will count these as follows: We will go clockwise around the polygon. We can pick the first vertex arbitrarily ($2n+1$ possibilities). Once we pick it, we have to pick $2$ out of the next $n$ vertices ($\\binom{n}{2}$ possibilities).\nThen the probability that our triangle does NOT contain the center is\nAnd then the probability we seek is"
    },
    {
        "Problem": "Determine all the roots, real or complex, of the system of simultaneous equations",
        "Solution": "Let $x$, $y$, and $z$ be the roots of the cubic polynomial $t^3+at^2+bt+c$. Let $S_1=x+y+z=3$, $S_2=x^2+y^2+z^2=3$, and $S_3=x^3+y^3+z^3=3$. From this, $S_1+a=0$, $S_2+aS_1+2b=0$, and $S_3+aS_2+bS_1+3c=0$. Solving each of these, $a=-3$, $b=3$, and $c=-1$. Thus $x$, $y$, and $z$ are the roots of the polynomial $t^3-3t^2+3t-1=(t-1)^3$. Thus $x=y=z=1$, and there are no other solutions.\n",
        "Solution_2": "Let $P(t)=t^3-at^2+bt-c$ have roots x, y, and z. Then  using our system of equations, so $P(1)=0$. Thus, at least one of x, y, and z is equal to 1; without loss of generality, let $x=1$. Then we can use the system of equations to find that $y=z=1$ as well, and so $\\boxed{(1,1,1)}$ is the only solution to the system of equations.",
        "Solution_3": "Let $a=x-1,$ $b=y-1$ and $c=z-1.$ Then \n\n\n\nWe have\n\nThen one of $a, b$ and $c$ has to be 0, and easy to prove the other two are also 0. So $\\boxed{(1,1,1)}$ is the only solution to the system of equations.\nJ.Z.",
        "Solution_4": "We are going to use Intermediate Algebra Techniques to solve this equation.\nLet's start with the first one: $x+y+z=3$. This will be referred as the FIRST equation.\nWe are going to use the first equation to relate to the SECOND one ($x^2+y^2+z^2=3$) and the THIRD one ($x^3+y^3+z^3=3)$.\nSquaring this equation: $x^2+y^2+z^2+2xy+2yz+2xz=9$\nSubtracting this equation from the 2nd equation in the problem, we have $2xy+2yz+2xz=6$, so $xy+xz+yz=3$.\nNow we try the same idea with the cubed terms. Cube the first equation:\n$x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3yz^2+3zx^2+3zy^2=27$. Plug in $x^3+y^3+z^3=3$ and factor partially:\n$3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27$.\nNow here is the key step. Note that $z=3-x-y, y=3-x-z, x=3-y-z$. So we are going to substitute $y+z, x+z, x+y$ for each of the expressions and we get: $-x^3+3x^2-y^3+3y^2-z^3+3z^2=8-2xyz$ (I rearranged it a bit).\nResubstituting in the second and third equation: $-3+3(3)=8-2xyz$. So $xyz=1$.\nSo now we have three equations for the elementary symmetric sums of $x,y,z$:\nEquation 4: $x+y+z=3$ (this is also equation 1)\nEquation 5: $xy+yz+xz=3$\nEquation 6: $xyz=1$.\nIf we call the solutions of $t^3-3t^2+3t-1=0$ (Equation 7) $a,b,c$, then $x,y,z$ are the three roots $a,b,c$ but in some order. Notice that Equation 7 can be factored as $(t-1)^3=0$, which means that $t=1$. Therefore $(x,y,z)$ are permutations of $(1,1,1)$ in some order, which can be only $(1,1,1)$. (This step uses Vieta's formulas)\nTherefore, the only solution is $\\boxed{(x,y,z)=(1,1,1)}$.\n~hastapasta",
        "Solution_5": "From the question\nEquation 1 : $x+y+z=3$\nEquation 2 : $x^2+y^2+z^2=3$\nEquation 3 : $x^3+y^3+z^3=3$\nFrom Eq(1) and Eq(2),\nEquation 4 : $(x+y+z)^2=9 \\implies x^2+y^2+z^2+2\\sum xy = 9 \\implies \\sum xy = 3$\nFrom Eq(1), Eq(3), and the fact that $(x+y+z)^3=x^3+y^3+z^3+3\\prod(x+y)$,\nEquation 5 : $(x+y+z)^3=x^3+y^3+z^3+3\\prod(x+y)= 27 \\implies \\prod(x+y)=8$\nAs from Eq(1),  $\\ x+y=3-z$,$\\ y+z=3-x$,  $\\ z+x = 3- y$\nEquation 6 : $\\prod(3-x) =8$\nNow let there be a cubic function $f(t)$ with the roots $x,y,z$ so from the relation between roots and coefficients,\nEquation 7 : $f(t)= t^3-3t^2+3t-d$   (As we don't know the product let the constant be $d$)\nAs $f(t) = (t-x)(t-y)(t-z)$ then at $t=3$,\nEquation 8 : $f(3)=9-d=(3-x)(3-y)(3-z)$\nAnd from Eq(6),\nEquation 9 : $d= -1$\nSo finally,\nEquation 10 : $f(t)=t^3-3t^2+3t-1=(t-1)^{3}$\nSo, $\\ (x,y,z)=(1,1,1)$\n~Sparrow_1827\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.",
        "Solution": "Assume that the cube roots of three distinct prime numbers can be three terms of an arithmetic progression. Let the three distinct prime numbers be $p$, $q$, and $r.$ WLOG, let $p<q<r.$\nThen,\n\n\nwhere $m$, $n$ are distinct integers, and $d$ is the common difference in the progression. Then we have\n\nBecause $p$, $q$, $r$ are distinct primes, $pqr$ is not a perfect cube. Thus, the LHS is irrational but the RHS is rational, which is a contradiction. So, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression.",
        "Solution_2": "Assume the contrary and suppose that the cube roots of three distinct prime numbers, $p_1^\\frac{1}{3}$, $p_2^\\frac{1}{3}$, and $p_3^\\frac{1}{3}$, form an increasing arithmetic progression. Then we can write $p_2^\\frac{1}{3} = p_1^\\frac{1}{3} + md$ and $p_3^\\frac{1}{3} = p_1^\\frac{1}{3} + nd$ for some positive integral $d, m, n$ with $m < n$. Thus,\n\nfor some real $r$. Because $m$ and $n$ are integral, r is rational.\nThis equation rearranges to\nNote that if $a-b=x$, then  from the binomial theorem. As a result, cubing both sides of the equation gives\nLemma: The sum, product, and difference of two rational numbers are all rational, and that of a rational and irrational number are all irrational.\nProof: The first part is trivial if we write the fractions as $\\frac{m}{n}$ and $\\frac{p}{q}$, and find the common denominators or simplify fractions as\nneeded. The second part follows by contradiction: if the sum of a rational number and irrational number is rational, then the irrational number must be the difference of two rational numbers, which is rational. This is a contradiction; likewise for difference and product.\nThe LHS (left-hand side) of (*) is clearly rational (in fact, integral), and so the RHS must also be rational. This is the case for the first two terms $r^3p_2$ and $-(r-1)^3p_1$. But it is not the case for the third term. The number $(p_1p_2p_3)^\\frac{1}{3}$ is irrational because $p_1p_2p_3$, the product of three distinct primes, cannot be a perfect cube. Hence, $-3r(r-1)(p_1p_2p_3)^\\frac{1}{3}$, the product of a rational and an irrational number, is irrational, and the RHS is therefore the sum of two rational terms and one irrational term. This is irrational by the lemma, and hence we have a rational equalling an irrational, contradiction. It follows that the cube roots of three distinct prime numbers cannot form an arithmetic sequence."
    },
    {
        "Problem": "Let $a$, $b$, and $c$ denote three distinct integers, and let $P$ denote a polynomial having all integral coefficients.  Show that it is impossible that $P(a)=b$, $P(b)=c$, and $P(c)=a$.\n",
        "Solution": "It suffices to show that if $a,b,c$ are integers such that $P(a) = b$, $P(b)=c$, and $P(c)= a$, then $a=b=c$.\nWe note that\n\nso the quanitities $(a-b), (b-c), (c-a)$ must be equal in absolute value.  In fact, two of them, say $(a-b)$ and $(b-c)$, must be equal.  Then\n\nso $a=b= P(a)$, and $c= P(b) = P(a) = b$, so $a$, $b$, and $c$ are equal, as desired.  $\\blacksquare$",
        "Solution_1b": "Let $k$ be the value $(a-b), (b-c), (c-a)$ are equal to in absolute value. Assume $k$ is nonzero. Then each of $(a-b), (b-c), (c-a)$ is equal to $k$ or $-k$, so  where $m$ is one of -3, -1, 1, or 3. In particular, neither $m$ nor $k$ is zero, contradiction. Hence, $k = 0$, and $a, b, c$ are equal, yielding a final contradiction of the existence of these variables. $\\blacksquare$\nIn fact, this approach generalizes readily. Suppose that $n$ is an odd integer, and that there exist $n$ distinct integers $a_1, a_2, ..., a_n$ such that $P(a_i) = a_{i+1}$ and $P(a_n) = 1$, for $1 \\le i \\le n-1$. Then we have  so the differences are all equal in absolute value and thus equal to $k$ or $-k$ for some integer $k$. Adding the differences (in the order they are written) gives $0 = mk$ for some odd integer $m$, so $m$ is nonzero and hence $k = 0$. Thus, all the $a_i$ are equal, a contradiction of their distinctness, so the sequence $a_i$ cannot exist.",
        "Solution_2": "Consider the polynomial $Q(x) = (b-a)P(x) - (c-b)x - b^2 + ac.$ By using the facts that $P(a) = b$ and $P(b) = c$, we find that $Q(a) = Q(b) = 0.$ Thus, the polynomial $Q(x)$ has a and b as roots, and we can write $Q(x) = (x-a)(x-b)R(x)$ for some polynomial $R(x)$. Because $Q(x)$ and $(x-a)(x-b)$ are monic polynomials with integral coefficients, their quotient, $R(x)$, must also have integral coefficients, as can be demonstrated by simulating the long division. Thus, $Q(c)$, and hence $Q(c) - (x-a)(x-b)$, must be divisible by $(c-a)(c-b)$. But if $x=c-a$ and $y=c-b$, then we must have, after rearranging terms and substitution, that $(x-y)^2$ is divisible by $xy$. Equivalently, $S = x^2 + y^2$ is divisible by $xy$ (after canceling the $2xy$ which is clearly divisble by $xy$). Because S must be divisible by both x and y, we quickly deduce that x is divisible by y and y is divisible by x. Thus, x and y are equal in absolute value. A similar statement holds for a cyclic pemutation of the variables, and so x, y, and z are all equal in absolute value. The conclusion follows as in the first solution.",
        "Solution_3": "Assume for the sake of contradiction that is possible for unique integers $a,b,c$. Let $P(x)=d_1x^n+d_2x^{n-1}+\\cdots+d_n.$ Note that  \n\n\nSubtracting the second from the first, third from second, and first from third gives \n\n\n By the RHS, note that $a-b\\mid P(a)-P(b)$ so $\\lvert{a-b}\\rvert\\leq\\lvert{b-c}\\rvert.$  Similarly, $\\lvert{b-c}\\rvert\\leq\\lvert{c-a}\\rvert$ and $\\lvert{c-a}\\rvert\\leq\\lvert{a-b}\\rvert.$  Hence, $\\lvert{a-b}\\rvert\\leq\\lvert{b-c}\\rvert\\leq\\lvert{c-a}\\rvert\\leq\\lvert{a-b}\\rvert$ so $\\lvert{a-b}\\rvert=\\lvert{b-c}\\rvert=\\lvert{c-a}\\rvert.$  Assume WLOG that $a>b>c$ so $a-b=a-c$ and $b-c=c-a.$  From the first equation, we get $b=c$ and substituting this in the second gives $c=a.$  Hence, $a=b=c$, contradicting the uniqueness of $a,b,c.$ $\\blacksquare$",
        "Solution_4": "Assume that $a, b$ and $c$ are distinct integers which satisfy the given conditions. Then we have $P(c) = P(P(b)) = P(P(P(a))) = a$. Similarly $P(P(P(b))) = b$ and $P(P(P(c))) = c$. Since the polynomial has integer coefficients, this implies that the polynomial $P$ is $P(x) = x$. But this is a contradiction since then $P(a) = b$ can only be true if $a = b$ which contradicts the fact that $a, b$ and $c$ are distinct integers.",
        "Solution_5": "We get that $a-b|b-c, b-c|c-a, c-a|a-b$. Thus, we can say that $(b-c) = (a-b)k, (c-a) = (b-c)j, (a-b) = (c-a)m$, for non zero integers, not necesarily positive, k, j, and m. Multiplying these equations and canceling, we get that $kjm$=1. This means that either k=1, m=1, j=1, or two of the variables are equal to 1 and the other is equal to -1. If all of the variables are equal to zero, plugging in gives us that $a=b=c$, contradictory to the conditions given in the problem statement. If two of the variables are equal to -1, and the other is equal to 1, then WLOG let $k=m=-1, j=1$. But this implies that the distinct integer condition is not satisfied, and hence, this case is invalid. Thus, it is impossible. $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Prove that if $a$, $b$, and $c$ are positive real numbers, then",
        "Solution_1": "Consider the function $f(x)=x\\ln{x}$. $f''(x)=\\frac{1}{x}>0$ for $x>0$; therefore, it is a convex function and we can apply Jensen's Inequality:\nApply AM-GM to get\nwhich implies\nRearranging,\nBecause $f(x) = e^x$ is an increasing function, we can conclude that:\nwhich simplifies to the desired inequality.",
        "Solution_2": "Note that $(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\\left((abc)^{(a+b+c)/3}\\right)^3$.\nSo if we can prove that $a^ab^bc^c\\ge a^bb^cc^a$ and $a^ab^bc^c\\ge a^cb^ac^b$, then we are done.\nWLOG let $a\\ge b\\ge c$.\nNote that $(a^ab^bc^c)\\cdot \\left(\\dfrac{c}{a}\\right)^{a-b}\\cdot \\left(\\dfrac{c}{b}\\right)^{b-c}=a^bb^cc^a$. Since $\\dfrac{c}{a} \\le  1$, $\\dfrac{c}{b} \\le  1$, $a-b \\ge 0$, and $b-c \\ge 0$, it follows that $a^ab^bc^c \\ge a^bb^cc^a$.\nNote that $(a^ab^bc^c)\\cdot \\left(\\dfrac{b}{a}\\right)^{a-b}\\cdot \\left(\\dfrac{c}{a}\\right)^{b-c}=a^cb^ac^b$. Since $\\dfrac{b}{a} \\le  1$, $\\dfrac{c}{a} \\le  1$, $a-b \\ge 0$, and $b-c \\ge 0$, it follows that $a^ab^bc^c \\ge a^cb^ac^b$.\nThus, $(a^ab^bc^c)^3\\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\\left((abc)^{(a+b+c)/3}\\right)^3$, and cube-rooting both sides gives $a^ab^bc^c\\ge (abc)^{(a+b+c)/3}$ as desired.",
        "Solution_3": "WLOG let $a\\ge b\\ge c$. Let $b = ax$ and $c = ay$, where $x \\ge 1$ and $y \\ge 1$.\nWe want to prove that $(a)^{a}(ax)^{ax}(ay)^{ay} \\ge (a \\cdot ax \\cdot ay)^{\\frac{a + ax + ay}{3}}$.\nSimplifying and combining terms on each side, we get $a^{a + ax + ay}x^{ax}y^{ay} \\ge a^{a + ax + ay}(xy)^{\\frac{a + ax + ay}{3}}$.\nSince $a > 0$, we can divide out $a^{a + ax + ay}$ to get $x^{ax}y^{ay} \\ge (xy)^{\\frac{a + ax + ay}{3}}$.\nTake the $a$th root of each side and then cube both sides to get $x^{3x}y^{3y} \\ge (xy)^{1 + x + y}$.\nThis simplifies to $x^{2x-1}y^{2y-1} \\ge x^{y}y^{x}$.\nSince $2x - 1 \\ge x$ and $2y - 1 \\ge y$, we only need to prove $x^{x}y^{y} \\ge x^{y}y^{x}$ for our given $x, y$.\nWLOG, let $y \\ge x$ and $y =kx$ for $k \\ge 1$. Then our expression becomes\n$x^{x}(xk)^{xk} \\ge x^{xk}(xk)^{x}$\n$x^{x+xk}k^{xk} \\ge x^{x+xk}k^{x}$\n$k^{xk} \\ge k^{x}$\n$k^k \\ge k$\nThis is clearly true for $k \\ge 1$.",
        "Solution_4": "WLOG let $a\\ge b\\ge c$. Then sequence $(a,b,c)$ majorizes $(\\frac{a+b+c}{3},\\frac{a+b+c}{3},\\frac{a+b+c}{3})$. Thus by Muirhead's Inequality, we have $\\sum_{sym} a^ab^bc^c \\ge \\sum_{sym} a^{\\frac{a+b+c}{3}}b^{\\frac{a+b+c}{3}}c^{\\frac{a+b+c}{3}}$, so $a^ab^bc^c \\ge (abc)^{\\frac{a+b+c}{3}}$.",
        "Solution_5": "Let $x=\\frac{a}{\\sqrt[3]{abc}},$ $y=\\frac{b}{\\sqrt[3]{abc}}$ and $z=\\frac{c}{\\sqrt[3]{abc}}.$ Then $xyz=1$ and a straightforward calculation reduces the problem to \n\nWLOG, assume $x\\ge y\\ge z.$ Then $x\\ge 1,$ $z\\le 1$ and $xy=\\frac{1}{z} \\ge 1.$ Therefore,\nJ.Z.",
        "Solution_6": "Cubing both sides of the given inequality gives \nIf we take $a^{3a}b^{3b}c^{3c}$ as the product of $3a$ $a$'s, $3b$ $b$'s, and $3c$ $c's$, we get that\nby GM-HM, as desired.",
        "Solution_7": "Replacing $a$ with $ak$, $b$ with $bk$, and $c$ with $ck$, for some positive real $k$ we get:\n$a^{ak}k^{ak}b^{bk}k^{bk}c^{ck}k^{ck} = a^{ak}b^{bk}c^{ck}k^{ak+bk+ck} \\ge {k^3abc}^{{ak+bk+ck}/3} = k^{ak+bk+ck}{abc}^{{ak+bk+ck}/3}$\nThis means that this inequality is homogeneous since both sides have the same power of $k$ as a factor. Since the inequality is homogeneous, we can scale $abc$ so that their product is $1$, i.e. $abc = 1$. This makes the inequality turn into something much more nicer to deal with. Now we have to prove:\n$a^ab^bc^c \\ge 1$ given that $abc = 1$.\nNote that $a^a \\ge a$ even if $a \\le 1$. Therefore $a^a \\ge a$, $b^b \\ge b$, and $c^c \\ge c$. Multiplying these together we get:\n$a^ab^bc^c \\ge abc = 1$. This proves the desired result. Equality holds when $a = b = c = 1$.",
        "Solution_8_.28Rearrangement.29": "Let $\\log_2a=x$, $\\log_2b=y$ and $\\log_2c=z$, and WLOG $a\\ge b\\ge c >0$. \nThen we have both $a\\ge b\\ge c$ and $x\\ge y \\ge z$.\nBy the rearrangement inequality, \n$ax+by+cz\\ge ay + bz +cx$ and \n$ax+by +cz\\ge az+bx+cy$.\nSumming, \n$2(ax+by+cz)\\ge a(y+z)+b(x+z)+c(x+y)$\nAdding $ax+by+cz$,we get\n$ax+by+cz\\ge \\frac{(a+b+c)(x+y+z)}{3}$.\nNow we substitute back for $x,y,z$ to get:\n$\\log_2a^ab^bc^c\\ge \\log_2(abc)^{\\frac{a+b+c}{3}}$.\nRaising $2$ to the power of each side, we get\n$a^ab^bc^c\\ge (abc)^{\\frac{a+b+c}{3}}$",
        "Solution_9": "Assume without loss of generality that $a\\le b\\le c$. Then, we have\n\nIt follows that\n\nWe can \"distribute\" the exponents, which gives\n\nNotice that in the $b^{(c-b)/3}$ in the denominator of the third fraction can be brought out of the denominator by negating its exponent. This gives\n\nWe can move around terms on the left to get\n\nNow we combine the exponents in the numerators. This gives\n\nWe multiply both sides by $\\left(a^{(b-a)/3}\\right)\\left(a^{(c-a)/3}\\right),$ which gives\n\nWe then multiply both sides by $\\left(b^{(c-a)/3}\\right)\\left(c^{(b-a)/3}\\right):$\n\nMultiplying both sides by $\\left(b^{(b-a)/3}\\right)\\left(c^{(c-a)/3}\\right),$ then simplifying both sides with exponent laws, gives\n\nFinally, we multiply both sides by $(abc)^a.$ Combining exponents one last time gives the desired\n\n--MenuThreeOne\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Two boundary points of a ball of radius 1 are joined by a curve contained in the ball and having length less than 2. Prove that the curve is contained entirely within some hemisphere of the given ball.",
        "Solution": "Question: Why is the curve necessarily on a great circle? The curve is arbitrary–in space. Also, there is only one great circle through the two points, as three points determine a plane–a counterexample to this claim would then be readily found.\nDraw a Great Circle containing the two points and the curve. Then connect the two points with a chord in the circle. The circle has radius 1, so the circle has diameter 2, so the two points are all but diametrically opposite. Therefore we can draw a diameter $EF$ parallel to the chord but not on it.\nNow take the plane through EF perpendicular to AG. This creates two hemispheres, and the curve is in one and only one of them.",
        "Solution_2": "Let the curve intersect the sphere at points A and B. Let B' be the point diametrically opposite A, and let $Q$ be the plane through the center O of the sphere perpendicular to $BB'$ and passing through the midpoint of $BB'$. We claim that the curve must be in the hemisphere divided by $Q$ containing points A and B.\nAssume the contrary and suppose that the curve intersects $Q$ at point C. Reflect the portion of the curve from C to B to obtain a same-length curve from C to B'. Then the length of the curve equals the length of the new curve from A to B', which is at least the distance from A to B', or 2. This is a contradiction to the claim that the length of the curve is less than 2, so the curve must be contained in the hemisphere divided by $Q$ containing points A and B.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "A father, mother and son hold a family tournament, playing a two person board game with no ties. The tournament rules are:\n(i) The weakest player chooses the first two contestants.\n(ii) The winner of any game plays the next game against the person left out.\n(iii) The first person to win two games wins the tournament.\nThe father is the weakest player, the son the strongest, and it is assumed that any player's probability of winning an individual game from another player does not change during the tournament. Prove that the father's optimal strategy for winning the tournament is to play the first game with his wife.",
        "Solution": "There are three possible strategies for the father:\n1)  Sit out the first game\n2)  Play the first game against the weaker of the other two players (the mother).  This person is referred to throughout the solution as \"the weaker player\".\n3)  Play the first game against the stronger of the other two players (the son).  This person is referred to throughout the solution as \"the stronger player\".\nStrategy 1 is dominated by both strategy 2 and strategy 3.  If strategy 1 is used and the weaker player wins the first game, it becomes strategy 2, but with the father starting behind 0-1.  Likewise, if strategy 1 is used and the stronger player wins the first game, it becomes strategy 3, but with the father again starting behind 0-1.\nThus, the father must play the first game.\nBy the pigeonhole principle, there may be no more than 4 games; once the 4th game is finished, someone will have won two games.  Let $W$ denote a game the father wins, let $L$ denote a game the father loses, and let $x$ denote a game the father does not participate in.  Then, with the knowledge that the father plays the first game, the following ways for the father to win two games are possible:\n(A)  $WW$\n(B)  $WLxW$\n(C)  $LxWW$\nIf the father wins the first game, then he must either win the second game (A), or sit out the third game to win the fourth game (B).\nIf the father loses the first game, then he must sit out the second game and win the third and fourth games (C).\nNow we must determine which gives the largest probability of the father winning: playing the first game against the weaker player or the stronger player.  We define the following probabilities:\n$x$ is the chance of the father defeating the weaker player.  $(1-x)$ is the chance of the weaker player defeating the father.\n$y$ is the chance of the father defeating the stronger player.  $(1-y)$ is the chance of the stronger player defeating the father.\n$z$ is the chance of the weaker player defeating the stronger player.  $(1-z)$ is the chance of the stronger player defeating the weaker player.\nNote that $x > y$, and all of $x, y$ and $z$ are less than $0.5$.\nIf the father plays the weaker player first, we split the probability of winning the tournament in the three mutually exclusive cases above:\n$P(A) = xy$ because the father must defeat the weaker player, then the father must defeat the stronger player.\n$P(B) = x(1-y)zx$ because the father must defeat the weaker player, then stronger player must defeat the father.  Third, the weaker player must defeat the stronger player (because if the stronger player wins a second game, the tournament is over).  Fourth, the father plays the weaker player a second time and must win.\n$P(C) = (1-x)(1-z)yx$ because the weaker player must defeat the father, followed by the stronger player defeating the weaker player (because if the weaker player wins two in a row, he wins the tournament).  The father must then defeat the stronger player and the weaker player in the third and fourth game, respectively.\nFrom this, the total probability of the father winning by playing the weaker player first is:\n$P(TOTAL) = P(A) + P(B) + P(C) = xy + x(1-y)zx + (1-x)(1-z)yx$\nLikewise, we compute the probability of the father winning if he plays the stronger player first.  We do this by symmetry:\nAll $x$ will turn into $y$, because whenever the father defeated the weaker player before, he must defeat the stronger player now.\nAll $y$ will turn into $x$, because whenever the father defeated the stronger player before, he must defeat the weaker player now.\nAll $z$ will turn into $(1-z)$, because whenever the weaker player defeated the stronger player, now the stronger player must defeat the weaker player.\nAll $(1-z)$ will turn into $z$, because whenever the stronger player defeated the weaker player, now the weaker player must defeat the stronger player.\nApplying these transformations and calling the new probabilities $P'$, we find that:\n$P'(TOTAL) = P'(A) + P'(B) + P'(C) = yx + y(1-x)(1-z)y + (1-y)zxy$\nWe compare these probabilities.\n$P(A) = P'(A)$.  This makes sense, as the probabilitiy of the father winning the first two games is the same, regardless of the order in which they are played.\n$P(B) > P'(C)$:  Calculating $P(B) / P'(C)$, we get $x^2(1-y)z / xy(1-y)z$, which simplifies to $x/y$.\nSince $x > y$\n$x/y > 1$\n$P(B) / P'(C) > 1$\nso $P(B) > P'(C)$.\n$P(C) > P'(B)$:  Calculating $P(C) / P'(B)$, we get $(1-x)(1-z)yx / (1-x)(1-z)y^2$, which simpllifies again to $x/y$.  Thus, $P(C) > P'(B)$\nAdding the above equations and inequalities, we see that $P(TOTAL) > P'(TOTAL)$, and thus the father is better off playing the weaker of the two players (ie the mother) first.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Consider the two triangles $\\triangle ABC$ and $\\triangle PQR$ shown in Figure 1.  In $\\triangle ABC$, $\\angle ADB = \\angle BDC = \\angle CDA = 120^\\circ$.  Prove that $x=u+v+w$.",
        "Solution_1": "We rotate figure $PRQM$ by a clockwise angle of $\\pi/3$ about $Q$ to obtain figure $RR'QM'$:\nEvidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent.  Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations.  Then\n\nThen by symmetry,\nBut $ABC$ is composed of three smaller triangles.  The one with sides $w,v,a$ has area $\\tfrac{1}{2} wv \\sin 120^\\circ = \\frac{wv \\sqrt{3}}{4}$.  Therefore, the area of $ABC$ is\n\nAlso, by the Law of Cosines on that small triangle of $ABC$, $a^2 = w^2 + wv+ v^2$, so by symmetry,\n\nTherefore\n\nBut the area of triangle $PQR$ is $x^2 \\sqrt{3}/4$.  It follows that $u+v+w=x$, as desired.  $\\blacksquare$",
        "Solution_2": "Rotate $\\triangle ABC$ $60$ degrees clockwise about $A$ to get $\\triangle AB'C'$. Observe that $\\triangle ADD'$ is equilateral, which means $D'D=AD=u$. Also, $B',D',D,C$ are collinear because $\\angle B'D'A + \\angle AD'D = 120+60=180$ and $\\angle CDA + \\angle DD'A = 120+60=180$. The resulting $\\triangle B'AC$ has side lengths $b,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $A+60$.\nIf we perform the rotation about points $B$ and $C$, we get two triangles. One has side lengths $a,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $B+60$, and the other has side lengths $b,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $C+60$.\nThese three triangles fit together because $(A+60)+(B+60)+(C+60) = 360$. The result is an equilateral triangle of side length $u+v+w$.",
        "Solution_3": "\nAs in the first solution, we rotate and establish that $\\triangle MRM' \\cong \\triangle ABC$.\nLet $X$ and $Y$ be points on $\\overline{RQ}$ and $\\overline{PQ}$, respectively, such that $M$ lies on $\\overline{XY}$ and $\\overline{XY}\\parallel \\overline{PR}$.  We note that $m\\angle RXM = 120^\\circ$.  The rotation then takes $Y$ to $X$, so $m\\angle RXM'=m\\angle PYM = 120^\\circ$.\nIt follows that $RX=BD=v$, $MX=AD=u$, $M'X=CD=w$.\nSince $m\\angle MXM' = 120^\\circ$ and $m\\angle MQM' = 60^\\circ$, $MXM'Q$ is cyclic.\nBy Ptolemy's theorem,\n\nFinally, $RQ = RX+XQ = u+v+w$, as desired.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Let $A,B,C,D$ denote four points in space and $AB$ the distance between $A$ and $B$, and so on. Show that",
        "Solution_1": "\nIf we project points $A,B,C,D$ onto the plane parallel to $\\overline{AB}$ and $\\overline{CD}$, $AB$ and $CD$ stay the same but $BC, AC, AD, BD$ all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when $A,B,C,D$ are coplanar:\n\nLet $AD=a, AC=b, BC=c, BD=d, AB=m, CD=n$. We wish to prove that $a^2+b^2+c^2+d^2\\ge m^2+n^2$. Let us fix $\\triangle BCD$ and the length $AB$ and let $A$ vary on the circle centered at $B$ with radius $m$. If we find the minimum value of $a^2+b^2$, which is the only variable quantity, and prove that it is larger than $m^2+n^2-c^2-d^2$, we will be done.\nFirst, we express $a^2+b^2$ in terms of $c,d,m,\\theta,\\phi$, using the Law of Cosines:\n\n$a^2+b^2$ is a function of $\\theta$, so we take the derivative with respect to $\\theta$ and obtain that $a^2+b^2$ takes a minimum when\nDefine $p=a^2+b^2$ and $q=c^2+d^2$:\n",
        "Solution_2": "Let\n\nIt is clear that every other case can be reduced to this.\nThen, with the distance formula and expanding,\nwhich rearranges to the desired inequality.",
        "Solution_3": "Because the distances are all squared, we must only prove the result in one dimension, and then we can just add up the three individual inequalities for the $x$, $y$, and $z$ dimension to get the desired result. Let $x_a$, $x_b$, $x_c$, and $x_d$ be the positions of $A$, $B$, $C$, and $D$ respectively. Then we must show that,\nSo we are done.",
        "Solution_4_.28Vector_bash.29": "Let $a$, $b$, $c$, $d$ correspond to the position vectors of points A, B, C, and D, respectively, with respect to an arbitrary origin O. Let us also for simplicity define $a^2 = a \\cdot a = ||a||^2$, where $||a||$ is the magnitude of vector $a$. Because squares are non-negative, $a^2$ is non-negative for all vectors $a$. Thus,\n\nBecause dot product is linear, we expand to obtain\n\nfrom which we add $a^2 + b^2 + c^2 + d^2$ to both sides, rearrange, and complete the square to get\n\nAs $(a-b)^2 = ||a-b||^2 = ||AB||^2 = AB^2$ and likewise for the others,\n\nwhich is what we wanted to prove.\nNOTES:\n1. Equality holds when the vector equality $a + b = c + d$ holds, which occurs when A, B, C, and D are the vertices of a (planar) parallelogram, in that order.\n2. The algebra employed here is almost identical to the algebra used in Solution 3, which means that Solution 3 is just a simplification of this solution.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "If $P(x)$ denotes a polynomial of degree $n$ such that  for $k=0,1,2,\\ldots,n$, determine $P(n+1)$.",
        "Solution_1": "Let $Q(x) = (x+1)P(x) - x$, and clearly, $Q(x)$ has a degree of $n+1$.\nThen, for $k=0,1,2,\\ldots,n$, $Q(k) = (k+1)P(k) - k = (k+1)\\cdot \\dfrac{k}{k+1} - k = 0$.\nThus, $k=0,1,2,\\ldots,n$ are the roots of $Q(x)$.\nSince these are all $n+1$ of the roots of the $n+1^{\\text{th}}$ degree polynomial, by the Factor Theorem, we can write $Q(x)$ as  where $c$ is a constant.\nThus,\nWe plug in $x = -1$ to cancel the $(x+1)P(x)$ and find $c$:\n\nFinally, plugging in $x = n+1$ to find $P(n+1)$ gives:\n\nIf $n$ is even, this simplifies to $P(n+1) = \\dfrac{n}{n+2}$. If $n$ is odd, this simplifies to $P(n+1) = 1$. $\\Box$\n~Edits by BakedPotato66",
        "Solution_2": "It is fairly natural to use Lagrange's Interpolation Formula on this problem:\nthrough usage of the Binomial Theorem. $\\square$\n~lpieleanu (minor editing and reformatting)\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Two given circles intersect in two points $P$ and $Q$. Show how to construct a segment $AB$ passing through $P$ and terminating on the two circles such that $AP\\cdot PB$ is a maximum.\n\n",
        "Solution": "A maximum $AP \\cdot PB$ cannot be attained if $AB$ intersects segment $O_1O_2$ because a larger value can be attained by making one of $A$ or $B$ diametrically opposite $P$, which (as is easily checked) increases the value of both $AP$ and $PB$. Thus, assume $AB$ does not intersect $O_1O_2$.\nLet $E$ and $F$ be the centers of the small and big circles, respectively, and $r$ and $R$ be their respective radii.\nLet $M$ and $N$ be the feet of $E$ and $F$ to $AB$, and $\\alpha = \\angle APE$ and $\\epsilon = \\angle BPF$\nWe have:\n\n$AP\\times PB$ is maximum when the product $\\cos{\\alpha} \\cos{\\epsilon}$ is a maximum.\nWe have $\\cos{\\alpha} \\cos{\\epsilon}= \\frac{1}{2} [\\cos(\\alpha +\\epsilon) + \\cos(\\alpha -\\epsilon)]$\nBut $\\alpha +\\epsilon = 180^{\\circ} - \\angle EPF$ and is fixed, so is $\\cos(\\alpha +\\epsilon)$.\nSo its maximum depends on $cos(\\alpha -\\epsilon)$ which occurs when $\\alpha=\\epsilon$. To draw the line $AB$:\nDraw a circle with center $P$ and radius $PE$ to cut the radius $PF$ at $H$. Draw the line parallel to $EH$ passing through $P$. This line meets the small and big circles at $A$ and $B$, respectively.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "A deck of $n$ playing cards, which contains three aces, is shuffled at random (it is assumed that all possible card distributions are equally likely). The cards are then turned up one by one from the top until the second ace appears. Prove that the expected (average) number of cards to be turned up is $(n+1)/2$.",
        "Solution_1": "We begin by induction. Our base case is naturally when $n = 3$, as there can be no less than $3$ cards in the deck. The only way to turn up the second ace is to turn up the first, and then turn up the second, which requires $2$ moves. This indeed is equal to $(3 + 1)/2$. Next, we assume that the expected number of cards required to pull the second ace from a deck of n cards is in fact $(n + 1)/2$. Now, we see that to simulate the case with $n + 1$ cards, we simply must add a card to our deck. By our assumption, we can expect to turn up $(n + 1)/2$ cards to be turned up, including our second ace. Thus ahead of our second ace, we expect $(n - 1)/2$ cards , and after it $(n - 1)/2$ as well. When we add our card, it is ahead of the second ace half the time, and behind if half the time. Therefore we expect to add one card to the number of cards drawn half the time, and add nothing for half the time. Hence we expect to add half a card, and the expected value is now $((n + 1)/2) + 1/2)$, or $(n + 2)/2$, the desired outcome for the $n + 1$ case.",
        "Solution_2": "Although we can induct as in the situation above, we can use a more direct approach.\nConsider a deck.  The second ace is in the $k$th place.  If we draw the deck backwards (which has equal probability as drawing it forwards), then we will draw it in the $n+1-k$th place, since the $n$th place will become the first, the $n-1$th the second, etc.\nAlso, note that when second ace is still the second ace when we draw the deck backwards. (we need to note this, else this argument holds for the first ace and third ace as well)\nNow, this implies that the expected location of the second ace is in the $(k+n+1-k)/2=(n+1)/2$ place.\n",
        "Solution_3": "Like Solution 2, this solution is also a direct solution.\nThe three aces divide the deck into four piles with a non-negative number of cards. Let $a$, $b$, $c$, and $d$ be the number of cards in the first, second, third, and fourth piles, respectively, given a random deck $D$. Also, let $E(a), E(b), E(c), E(d)$ denote the expected value of $a$, $b$, $c$, and $d$ over all decks $D$. Then $a + b + c + d = n - 3$ (the deck minus the three aces).\nFor every deck $D$, there must exist other decks that permute the values of $a$, $b$, $c$, $d$. (i.e. if there is a deck with $a = 2, b = 3, c = 1, d = 4,$ there exists a deck with $a = 1, b = 2, c = 3, d = 4$.) Therefore, the value of $a$ over all decks $D$ is a permutation of the value of $b$ over all such decks, and as a result $E(a) = E(b)$. A similar result holds for $b$, $c$, and $d$, and so $E(a) = E(b) = E(c) = E(d)$. Because  $E(a) = E(b) = \\frac{n-3}{4}$. We want to compute the number of cards up to and including the second ace, which is just  as desired.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "If $A$ and $B$ are fixed points on a given circle and $XY$ is a variable diameter of the same circle, determine the locus of the point of intersection of lines $AX$ and $BY$. You may assume that $AB$ is not a diameter.\n",
        "Solution": "WLOG, assume that the circle is the unit circle centered at the origin. Then the points $A$ and $B$ have coordinates $(-a,b)$ and $(a,b)$ respectively and $X$ and $Y$ have coordinates $(r,s)$ and $(-r,-s)$. Note that these coordinates satisfy $a^2 + b^2 = 1$ and $r^2 + s^2 = 1$ since these points are on a unit circle. Now we can find equations for the lines:\n\nSolving these simultaneous equations gives coordinates for $P$ in terms of $a, b, r,$ and $s$:\n$P = \\left(\\frac{as}{b},\\frac{1 - ar}{b}\\right)$. These coordinates can be parametrized in Cartesian variables as follows:\n\nNow solve for $r$ and $s$ to get $r = \\frac{1-by}{a}$ and $s = \\frac{bx}{a}$ . Then since $r^2 + s^2 = 1, \\left(\\frac{bx}{a}\\right)^2 + \\left(\\frac{1-by}{a}\\right)^2 = 1$ which reduces to $x^2 + (y-1/b)^2 = \\frac{a^2}{b^2}.$ This equation defines a circle and is the locus of all intersection points $P$. In order to define this locus more generally, find the slope of this circle function using implicit differentiation:\n\nNow note that at points $A$ and $B$, this slope expression reduces to $y' = \\frac{-b}{a}$ and $y' = \\frac{b}{a}$ respectively, values which are identical to the slopes of lines $AO$ and $BO$. Thus we conclude that the complete locus of intersection points is the circle tangent to lines $AO$ and $BO$ at points $A$ and $B$ respectively.",
        "Solution_2": "Notice that $m\\angle AYP=\\frac{1}{2}m\\overarc{AB}$ (Inscried angle theorem) and that $m\\angle XAY=90$ since $XY$ is a diameter, and thus subtends an arc of $180$. This will hold for all $X$ and all $Y$, and so by AA similarity, the angle $APY$ will be constant for all P, thus implying that the points A, B, and all P will be concyclic.\nIf we assume that the center of this circle is $H$, we know that $m\\angle AHB=2m\\angle APB=2(90-\\frac{1}{2}\\overarc{AB})=180-\\overarc{AB}$. We can assume that $\\overline{AH}$ and $\\overline{BH}$ intersect the original circle at points $M$, and $N$ respectively. This will give us that $m\\angle AHB=\\frac{1}{2}|m\\overarc{MN}-m\\overarc{AB}|$ (Since $H$ lies on the perpendicular bisector of $AB$, we know that $M$, $N$ will be on the same side of $AB$.) Now we also know that $m\\overarc{MN}\\leq m\\overarc{AB}$ or $m\\overarc{MN}\\leq 360-m\\overarc{AB}$. The only case where $m\\overarc{MN}$ satisfies the measure of $\\angle AHB$, is when $M=A$ and $N=B$, implying that $AH$ and $BH$ are tangents, and so $H$ is the intersection of the tangents from $A$ and $B$ to the original circle.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$.",
        "Solution": "Either $a^2=0$ or $a^2>0$. If $a^2=0$, then $b^2=c^2=0$. Symmetry applies for $b$ as well. If $a^2,b^2\\neq 0$, then $c^2\\neq 0$. Now we look at $a^2\\bmod{4}$:\n$a^2\\equiv 0\\bmod{4}$: Since a square is either 1 or 0 mod 4, then all the other squares are 0 mod 4. Let $a=2a_1$, $b=2b_1$, and $c=2c_1$. Thus $a_1^2+b_1^2+c_1^2=4a_1^2b_1^2$. Since the LHS is divisible by four, all the variables are divisible by 4, and we must do this over and over again, and from infinite descent, there are no non-zero solutions when $a^2\\equiv 0\\bmod{4}$.\n$a^2\\equiv 1\\bmod{4}$: Since $b^2\\neq 0\\bmod{4}$, $b^2\\equiv 1\\bmod{4}$, and $2+c^2\\equiv 1\\bmod{4}$. But for this to be true, $c^2\\equiv 3\\bmod{4}$, which is an impossibility. Thus there are no non-zero solutions when $a^2\\equiv 1\\bmod{4}$.\nThus the only solution is the solution above: $(a,b,c)=0$.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\n"
    },
    {
        "Problem": "If the sum of the lengths of the six edges of a trirectangular tetrahedron $PABC$ (i.e., $\\angle APB=\\angle BPC=\\angle CPA=90^o$) is $S$, determine its maximum volume.",
        "Solution": "Let the side lengths of $AP$, $BP$, and $CP$ be $a$, $b$, and $c$, respectively. Therefore $S=a+b+c+\\sqrt{a^2+b^2}+\\sqrt{b^2+c^2}+\\sqrt{c^2+a^2}$. Let the volume of the tetrahedron be $V$. Therefore $V=\\frac{abc}{6}$.\nNote that $(a-b)^2\\geq 0$ implies $\\frac{a^2-2ab+b^2}{2}\\geq 0$, which means $\\frac{a^2+b^2}{2}\\geq ab$, which implies $a^2+b^2\\geq ab+\\frac{a^2+b^2}{2}$, which means $a^2+b^2\\geq \\frac{(a+b)^2}{2}$, which implies $\\sqrt{a^2+b^2}\\geq \\frac{1}{\\sqrt{2}} \\cdot (a+b)$. Equality holds only when $a=b$. Therefore\n$S\\geq a+b+c+\\frac{1}{\\sqrt{2}} \\cdot (a+b)+\\frac{1}{\\sqrt{2}} \\cdot (c+b)+\\frac{1}{\\sqrt{2}} \\cdot (a+c)$\n$=(a+b+c)(1+\\sqrt{2})$.\n$\\frac{a+b+c}{3}\\geq \\sqrt[3]{abc}$ is true from AM-GM, with equality only when $a=b=c$. So $S\\geq (a+b+c)(1+\\sqrt{2})\\geq 3(1+\\sqrt{2})\\sqrt[3]{abc}=3(1+\\sqrt{2})\\sqrt[3]{6V}$. This means that $\\frac{S}{3(1+\\sqrt{2})}=\\frac{S(\\sqrt{2}-1)}{3}\\geq \\sqrt[3]{6V}$, or $6V\\leq \\frac{S^3(\\sqrt{2}-1)^3}{27}$, or $V\\leq \\frac{S^3(\\sqrt{2}-1)^3}{162}$, with equality only when $a=b=c$. Therefore the maximum volume is $\\frac{S^3(\\sqrt{2}-1)^3}{162}$.",
        "Solution_2": "Note that by AM-GM \nso  Proceed as before.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "If $P(x)$, $Q(x)$, $R(x)$, and $S(x)$ are all polynomials such that\n\nprove that $x-1$ is a factor of $P(x)$.",
        "Solution_1": "In general we will show that if $m$ is an integer less than $n$ and $P_0, \\dotsc, P_{m-1}$ and $S$ are polynomials satisfying\n\nthen $P_k(1) = 0$, for all integers $0 \\le k \\le m-1$.  For the problem, we may set $n=5$, $m=3$, and then note that since $P(1)= 0$, $(x-1)$ is a factor of $P(x)$.\nIndeed, let $\\omega_1, \\dotsc, \\omega_{n-1}$ be the $n$th roots of unity other than 1.  Then for all integers $1 \\le i \\le n-1$,\n\nfor all integers $1 \\le i \\le n$.  This means that the $(m-1)$th degree polynomial\n\nhas $n-1 > m-1$ distinct roots.  Therefore all its coefficients must be zero, so $P_k = 0$ for all integers $0 \\le k \\le m-1$, as desired.  $\\blacksquare$",
        "Solution_2": "Let $\\zeta, \\xi, \\rho$ be three distinct primitive fifth roots of unity.\nSetting $x = \\zeta, \\xi$, we have\n\nThese equations imply that\n\nor\n\nBut by symmetry,\n\nSince $\\xi \\neq \\rho$, it follows that $Q(1) = R(1) = 0$.  Then, as noted above,\n\nso $(x-1)$ is a factor of $P(x)$, as desired.  $\\blacksquare$",
        "Solution_3": "Let $z, z^2, z^3$ be three of the 5th roots of unity not equal to one that satisfy $1 + z + z^2 + z^3 + z^4 = 0$ as a result. Plugging them into the equation gives the linear system of equations in $(A(1), B(1), C(1))$:\n\nDirect observation gives $(A(1), B(1), C(1)) = (0,0,0)$ as a solution, and there are no others because the system of equations is linear. Hence, $A(1) = 0$, and so $(x-1)$ is a factor of $A(x)$, as desired. $\\blacksquare$\nNote: We can generalize this approach to prove the claim in Solution 1 in an equally fast, concise, and readily understandable fashion."
    },
    {
        "Problem": "Determine all pairs of positive integers $(m,n)$ such that\n$(1+x^n+x^{2n}+\\cdots+x^{mn})$ is divisible by $(1+x+x^2+\\cdots+x^{m})$.",
        "Solution_1": "Denote the first and larger polynomial to be $f(x)$ and the second one to be $g(x)$. In order for $f(x)$ to be divisible by $g(x)$ they must have the same roots. The roots of $g(x)$ are the (m+1)th roots of unity, except for 1. When plugging into $f(x)$, the root of unity is a root of $f(x)$ if and only if the terms $x^n, x^{2n}, x^{3n}, \\cdots x^{mn}$ all represent a different (m+1)th root of unity not equal to 1.\nNote that if $\\\\gcd(m+1,n)=1$, the numbers $n, 2n, 3n, \\cdots, mn$ represent a complete set of residues minus 0 modulo $m+1$. However, if $gcd(m+1,n)=a$ not equal to 1, then $\\frac{(m+1)(n)}{a}$ is congruent to $0 \\pmod {m+1}$ and thus a complete set is not formed. Therefore, $f(x)$ divides $g(x)$ if and only if $\\boxed{\\\\gcd(m+1,n)=1}.$ $\\blacksquare$",
        "Solution_2": "We could instead consider $f(x)$ modulo $g(x)$. Notice that $x^{m+1} = 1 \\pmod {g(x)}$, and thus we can reduce the exponents of $f(x)$ to their equivalent modulo $m+1$. We want the resulting $h(x)$ with degree less than $m+1$ to be equal to $g(x)$ (of degree $m$), which implies that the exponents of $f(x)$ must be all different modulo $m+1$. This can only occur if and only if $gcd(m+1, n) = 1$, and this is our answer, as shown in Solution 1.",
        "Solution_3": "Notice that $1+x^n+x^{2n}+...+x^{mn}=\\frac{x^{mn+n}-1}{x^n-1}$ and $1+x+x^2+...+x^m=\\frac{x^{m+1}-1}{x-1}$, so it remains to prove that $(x^n-1)(x^{m+1}-1) \\mid (x^{mn+n}-1)(x-1)$. It is clear that $(x^n-1) \\mid (x^{mn+n}-1)$ and $(x^{m+1}-1) \\mid (x^{mn+n}-1)$. For any $k \\in \\mathbb{N}$, we can use the fact that $x^k-1=\\prod_{d \\mid k} \\Phi _d(x)$, where $\\Phi _d(x)$ is the dth cyclotomic polynomial. If $\\gcd(m+1,n) = d > 1$, then $x^{m+1}-1$ and $x^n-1$ share a common cyclotomic polynomial; namely, $\\Phi _d(x)$. But since all the factors of $x^{mn+n}-1$ are distinct, $(x^{mn+n}-1)(x-1)$ cannot be divisible by $(x^n-1)(x^{m+1}-1)$. We find that $\\gcd(m+1,n) = 1$ must be the solution, since the only shared polynomial is $\\Phi _1(x)=x-1$, and we are done."
    },
    {
        "Problem": "If $a$ and $b$ are two of the roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.",
        "Solution": "Given the roots $a,b,c,d$ of the equation $x^{4}+x^{3}-1=0$.\nFirst, Vieta's relations give $a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abc+abd+acd+bcd=0, abcd = -1$.\nThen $cd=-\\frac{1}{ab}$ and $c+d=-1-(a+b)$.\nThe other coefficients give $ab+(a+b)(c+d)+cd = 0$ or $ab+(a+b)[-1-(a+b)]-\\frac{1}{ab}=0$.\nLet $a+b=s$ and $ab=p$.\nThus, $0=ab+ac+ad+bc+bd+cd=p+s(-1-s)-\\frac{1}{p}$.   (1)\nAlso, $0=abc+abd+acd+bcd=p(-1-s)-s/p$.\nSolving this equation for $s$, $s= \\frac{-p^2}{p^2+1}$.\nSubstituting into (1): $\\frac{p^{6}+p^{4}+p^{3}-p^{2}-1}{p(p^2+1)^2}=0$.\nConclusion: $p =ab$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1=0$."
    },
    {
        "Problem": "Prove that if the opposite sides of a skew (non-planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.",
        "Solution": "We first prove that if the opposite sides are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals.\nLet the vertices of the quadrilateral be A, B, C, D, in that order. Thus, the opposite sides congruent condition translates to (after squaring):\n\nLet the position vectors of A, B, C, D be a, b, c, d with respect to an arbitrary origin $O$. Define the square of a vector $a^2 = a \\cdot a$. Thus, $(a-b)^2 = (c-d)^2$ and so  Similarly,  Subtracting the second equation from the first eventually results in  which factors as\nLet M and N be the midpoints of $AC$ and $BD$, with position vectors $m = \\frac{a+c}{2}$ and $n = \\frac{b+d}{2}$, respectively. Then notice that the perpendicularity condition for vectors gives $MN$ perpendicular to $AC$. Similarly (by adding the equations), we find that $MN$ is perpendicular to $BD$, completing the first part of the proof.\nProving the converse is straightforward also: indeed, $MN$ perpendicular to both diagonals gives\n\nAdding and subtracting the equations eventually simplifies to\n\nor $AB^2 = CD^2$ and $AD^2 = BC^2$. This reduces to $AB = CD$ and $AD = BC$, as desired. $\\blacksquare$",
        "Solution_2": "Let $A, B, C, D$ be position vectors rather than the specified points. We're given that $|A-B| = |C-D| \\rightarrow |A-B|^2=|C-D|^2 \\rightarrow (A-B) \\cdot (A-B) = (C-D) \\cdot (C-D) \\rightarrow \\\\ A \\cdot A - 2A \\cdot B + B \\cdot B - C \\cdot C + 2C \\cdot D - D \\cdot D=0$. We're also given that $|A-D| = |C-B| \\rightarrow |A-D|^2=|C-B|^2 \\rightarrow (A-D) \\cdot (A-D) = (C-B) \\cdot (C-B) \\rightarrow \\\\ A \\cdot A - 2A \\cdot D + D \\cdot D - C \\cdot C + 2C \\cdot B - B \\cdot B=0$. Adding the two equations gives $2A \\cdot A - 2A \\cdot B - 2A \\cdot D - 2 C \\cdot C + 2C \\cdot D + 2C \\cdot B = 0$. Dividing by 4 and rearranging gives $\\frac{A+C-B-D}{2} \\cdot (A-C) = 0$. On the left hand side of this equation, one of the two factors is the vector between the midpoints of the two diagonals, and the other factor is on of the two diagonals. Their dot product is $0$, so they are perpendicular. The same methods can be used to show that the vector between the two midpoints is perpendicular to the other diagonal. We can employ this process in reverse to get the converse (this isn't completely simple, but I'm lazy so it's left as an exercise i guess)."
    },
    {
        "Problem": "If $a,b,c,d,e$ are positive numbers bounded by $p$ and $q$, i.e, if they lie in $[p,q], 0 < p$, prove that\n\nand determine when there is equality.",
        "Solution": "Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but $x$. Then the expression on the LHS has the form $(r + x)(s + \\frac{1}{x}) = (rs + 1) + sx + \\frac{r}{x}$, where $r$ and $s$ are fixed. But this is convex. That is to say, as $x$ increases if first decreases, then increases. So its maximum must occur at $x = p$ or $x = q$. This is true for each variable.\nSuppose all five are $p$ or all five are $q$, then the LHS is 25, so the inequality is true and strict unless $p = q$. If four are $p$ and one is $q$, then the LHS is $17 + 4\\left(\\frac{p}{q} + \\frac{q}{p}\\right)$. Similarly if four are $q$ and one is $p$. If three are $p$ and two are $q$, then the LHS is $13 + 6\\left(\\frac{p}{q} + \\frac{q}{p}\\right)$. Similarly if three are $q$ and two are $p$.\n$\\frac{p}{q} + \\frac{q}{p} \\geq 2$ with equality iff $p = q$, so if $p < q$, then three of one and two of the other gives a larger LHS than four of one and one of the other. Finally, we note that the RHS is in fact $13 + 6\\left(\\frac{p}{q} + \\frac{q}{p}\\right)$, so the inequality is true with equality iff either (1) $p = q$ or (2) three of $v, w, x, y, z$ are $p$ and two are $q$ or vice versa."
    },
    {
        "Problem": "Given that $a,b,c,d,e$ are real numbers such that\n$a+b+c+d+e=8$,\n$a^2+b^2+c^2+d^2+e^2=16$.\nDetermine the maximum value of $e$.",
        "Solution_1": "By Cauchy Schwarz, we can see that $(1+1+1+1)(a^2+b^2+c^2+d^2)\\geq (a+b+c+d)^2$\nthus $4(16-e^2)\\geq (8-e)^2$ \nFinally, $e(5e-16) \\geq 0$ which means $\\frac{16}{5} \\geq e \\geq 0$\nso the maximum value of $e$ is $\\frac{16}{5}$.\nfrom: Image from Gon Mathcenter.net",
        "Solution_2": "Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem.\nWe get the following equations:\n$(1)\\hspace*{0.5cm} a+b+c+d+e=8\\\\ (2)\\hspace*{0.5cm} a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\\\\ (3)\\hspace*{0.5cm} 0=\\lambda+2a\\mu\\\\ (4)\\hspace*{0.5cm} 0=\\lambda+2b\\mu\\\\ (5)\\hspace*{0.5cm} 0=\\lambda+2c\\mu\\\\ (6)\\hspace*{0.5cm} 0=\\lambda+2d\\mu\\\\ (7)\\hspace*{0.5cm} 1=\\lambda+2e\\mu$\nIf $\\mu=0$, then $\\lambda=0$ according to $(6)$ and $\\lambda=1$ according to $(7)$, so $\\mu \\neq 0$. Setting the right sides of $(3)$ and $(4)$ equal yields $\\lambda+2a \\mu= \\lambda+2b \\mu \\implies 2a\\mu=2b \\mu \\implies a=b$. Similar steps yield that $a=b=c=d$. Thus, $(1)$ becomes $4d+e=8$ and $(2)$ becomes $4d^{2}+e^{2}=16$. Solving the system yields $e=0,\\frac{16}{5}$, so the maximum possible value of $e$ is $\\frac{16}{5}$.",
        "Solution_3": "A re-writing of Solution 1 to avoid the use of Cauchy Schwarz. We have\n and \n\nThe second equation times 4, then minus the first equation, \n\nThe rest follows.\nJ.Z."
    },
    {
        "Problem": "$ABCD$ and $A'B'C'D'$ are square maps of the same region, drawn to different scales and superimposed as shown in the figure. Prove that there is only one point $O$ on the small map that lies directly over point $O'$ of the large map such that $O$ and $O'$ each represent the same place of the country. Also, give a Euclidean construction (straight edge and compass) for $O$.\n",
        "Solution_1": "The point is obviously unique, because the two maps have different scales (but if P and Q where two fixed points the distance between them would be the same on both maps).\nLet the small map square be A'B'C'D' and the large be ABCD, where X and X' are corresponding points. We deal first with the special case where A'B' is parallel to AB. In this case let AA' and BB' meet at O. Then triangles OAB and OA'B' are similar, so O must represent the same point. So assume A'B' is not parallel to AB.\nLet the lines A'B' and AB meet at W, the lines B'C' and BC meet at X, the lines C'D' and CD meet at Y, and the lines D'A' and DA meet at Z. We claim that the segments WY and XZ meet at a point O inside the smaller square. W cannot lie between A' and B' (or one of the vertices A', B' of the smaller square would lie outside the larger square). If it lies on the opposite side of A' to B', then Y must lie on the opposite side of C' to D'. Thus the segment WY must cut the side A'D' at some point Z' and the side B'C' at some point X'. The same conclusion holds if W lies on the opposite side of B' to A', because then Y must lie on the opposite side of D' to C'. Similarly, the segment XZ must cut the side A'B' at some point W' and the side C'D' at some point Y'. But now the segments X'Z' and W'Y' join pairs of points on opposite sides of the small square and so they must meet at some point O inside the small square.\nNow the triangles WOW' and YOY' are similar (WW' and YY' are parallel). Hence OW/OY = OW'/OY'. So if we set up coordinate systems with AB as the x-axis and AD as the y-axis (for the large square) and A'B' as the x'-axis and A'D' as the y'-axis (for the small square) so that corresponding points have the same coordinates, then the y coordinate of O equals the y' coordinate of O. Similarly, XOX' and ZOZ' are similar, so OX/OZ = OX'/OZ', so the x-coordinate of O equals its x'-coordinate. In other words, O represents the same point on both maps.",
        "Solution_2": "The point is obviously unique because of the different scale and angle, which can also be seen as the point can and must be the unique center of spiral similarity taking ABCD to A'B'C'D', as under the spiral similarity the center does not move. (Recall that a spiral similarity is a dilation combined with a rotation, and that there is a unique spiral similarity taking any segment to another segment, provided the four points do not make a parallelogram.)\nTo construct this center of spiral similarity, reference EGMO Lemma 10.1. Draw line DD' and AA', and let them intersect at point X. Draw circles through points X,A,D and X,A',D' (using the perpendicular bisector constructions to find the center of the circle using two chords). The intersection point of the two circles other than X is the center of spiral similarity, and hence is O, which represents the same point on both maps."
    },
    {
        "Problem": "An integer $n$ will be called good if we can write\n$n=a_1+a_2+\\cdots+a_k$,\nwhere $a_1,a_2, \\ldots, a_k$ are positive integers (not necessarily distinct) satisfying\n$\\frac{1}{a_1}+\\frac{1}{a_2}+\\cdots+\\frac{1}{a_k}=1$.\nGiven the information that the integers 33 through 73 are good, prove that every integer $\\ge 33$ is good.",
        "Solution": "Lemma: If $g$ is a good integer, then so is $2g + 2$ and $2g + 9$.\nProof: Let $g = a_1 + a_2 + \\cdots + a_k$ such that  where the $a_i$ are positive integers. Then, notice that  so $g_1 = 2 + 2a_1 + 2a_2 + ... + 2a_k = 2 + 2g$ is also a good integer. Furthermore,  so $g_2 = 3 + 6 + 2a_1 + 2a_2 + ... + a_k = 9 + 2g$ is also good.\nThis completes the lemma. Now, we use induction to show that the integers from 33 to $n$ are all good, where $n \\ge 33$.\nBase case: Our given takes care of $33 \\le n \\le 73$.\nInductive step: Assume the integers between 33 and $k > 73$, inclusive, are all good. Now, we casework on parity of $k+1$.\nIf $k+1$ is odd, then write $k+1 = 2g + 9$ for some integer $g$.\nIt follows that because $k \\ge 74$, we must have\n\nThus, $33 \\le g < \\frac{k}{2}$, and so $g$ is good by the inductive hypothesis. Applying the lemma gives $2g + 9 = k+1$ as a good integer.\nIf $k+1$ is even, write $k+1 = 2g+2$ for some integer $g$. Therefore,  so that $33 < g < \\frac{k}{2}$, meaning that $g$ is a good integer by the inductive hypothesis. From the lemma, $k+1$ must be good as well.\nThis completes the inductive step and thus the induction. Therefore, all integers $n \\ge 33$ are good integers."
    },
    {
        "Problem": "(a) Prove that if the six dihedral (i.e. angles between pairs of faces) of a given tetrahedron are congruent, then the tetrahedron is regular.\n(b) Is a tetrahedron necessarily regular if five dihedral angles are congruent?",
        "Solution": "(a) Let $ABCD$ be the said tetrahedron, and let the inscribed sphere of $ABCD$ touch the faces at $W, X, Y, Z$. Then, $OW, OX, OY, OZ$ are normals to the respective faces. We know that the angle between any two normals is equal, so we have $|OW|=|OX|=|OY|=|OZ|$ at equal angles. Now, since  and similar for the other sides, we have that $WXYZ$ is a regular tetrahedron. Now, the faces of $ABCD$ are the tangent planes at $W$, $X$, $Y$, and $Z$. Then, consider a $120^{\\circ}$ rotation about $OW$. The rotation sends $X\\mapsto Y$, $Y\\mapsto Z$, and $Z\\mapsto X$. Thus we have $AB=AC$, $AC=AD$, $AD=AB$, and $BC=CD=DB$. Performing rotations about the other axes yields that $ABCD$ has equal edges, so it is regular.\n(b) Consider the normals $OW, OX, OY, OZ$. We can perform a transformation in which we slightly shift $X$, $Y$, and $Z$ closer such that $\\triangle XYZ$ is still equilateral, and such that all angles between every pair of normals is less that $180^{\\circ}$. Then, shift $W$ such that $WX=WY=XY$. Then five of the distances are equal but the sixth is not.\n~ tc1729"
    },
    {
        "Problem": "Nine mathematicians meet at an international conference and discover that among any three of them, at least two speak a common language. If each of the mathematicians speak at most three languages, prove that there are at least three of the mathematicians who can speak the same language.",
        "Solution": "Suppose no three mathematicians speak the same language. Then person A can share some language with at most 3 other delegates, because if he shared some language with 4 delegates there would be 3 with the same language. So there are 5 delegates who do not share a language with A. Let one of them be B. Using the same logic, one the remaining 4, let it be C, does not share a language with B. Now A, B, C do not have any common languages. Contradiction."
    },
    {
        "Problem": "Determine all non-negative integral solutions $(n_1,n_2,\\dots , n_{14})$ if any, apart from permutations, of the Diophantine Equation $n_1^4+n_2^4+\\cdots +n_{14}^4=1599$.",
        "Solution_1": "Recall that $n_i^4\\equiv 0,1\\bmod{16}$ for all integers $n_i$. Thus the sum we have is anything from 0 to 14 modulo 16. But $1599\\equiv 15\\bmod{16}$, and thus there are no integral solutions to the given Diophantine equation.",
        "Solution_2": "In base $16$, this equation would look like:\nWe notice that the unit digits of the LHS of this equation should equal to $F_{16}$. In base $16$, the only unit digits of fourth powers are $0$ and $1$. Thus, the maximum of these $14$ terms is 14 $1's$ or $E_{16}$. Since $E_{16}$ is less than $F_{16}$, there are no integral solutions for this equation."
    },
    {
        "Problem": "$N$ is the north pole. $A$ and $B$ are points on a great circle through $N$ equidistant from $N$. $C$ is a point on the equator. Show that the great circle through $C$ and $N$ bisects the angle $ACB$ in the spherical triangle $ABC$ (a spherical triangle has great circle arcs as sides).",
        "Solution": "\nSince $N$ is the north pole, we define the Earth with a sphere of radius one in space with $N=(0,0,1)$ and sphere center $O=(0,0,0)$\nWe then pick point $N$ on the sphere and define the $xz$-plane as the plane that contains great circle points $A$ , $B$, and $N$ with the $x$-axis perpendicular to the $z$-axis and in the direction of $A$.\nUsing this coordinate system and $x$, $y$, and $z$ axes $A=(cos(\\phi),0,sin(\\phi))$ where $\\phi$ is the angle from the $xy$-plane to $A$ or latitude on this sphere with $\\frac{-\\pi}{2} < \\phi < \\frac{\\pi}{2}$\nSince $A$ and $B$ are points on a great circle through $N$ equidistant from $N$, then $B=(-cos(\\phi),0,sin(\\phi))$\nSince $C$ is a point on the equator, then $C=(cos(\\theta),sin(\\theta),0)$ where $\\theta$ is the angle on the $xy$-plane from the origin to $C$ or longitude on this sphere with $-\\pi < \\phi \\le \\pi$\nWe note that vectors from the origin to points $N$, $A$, $B$, and $C$ are all unit vectors because all those points are on the unit sphere.\nSo, we're going to define points $N$, $A$, $B$, and $C$ as unit vectors with their coordinates.\nWe also define the following vectors as follows:\nVector $\\overrightarrow{V_{CN}}$ is the unit vector in the direction of arc $CN$ and tangent to the great circle of $CN$ at $C$\nVector $\\overrightarrow{V_{CA}}$ is the unit vector in the direction of arc $CA$ and tangent to the great circle of $CA$ at $C$\nVector $\\overrightarrow{V_{CB}}$ is the unit vector in the direction of arc $CB$ and tangent to the great circle of $CB$ at $C$\nTo calculate each of these vectors we shall use the cross product as follows:\n$\\overrightarrow{V_{CN}}=(\\overrightarrow{C}\\times\\overrightarrow{N})\\times\\overrightarrow{C}$\n$\\overrightarrow{V_{CN}}=(\\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\\times\\left\\langle 0,0,1 \\right\\rangle)\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle$\n$\\overrightarrow{V_{CN}}=\\left\\langle sin(\\theta),-cos(\\theta),0 \\right\\rangle\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle$\n$\\overrightarrow{V_{CN}}=\\left\\langle 0,0,sin^{2}(\\theta)+cos^{2}(\\theta) \\right\\rangle$\n$\\overrightarrow{V_{CN}}=\\left\\langle 0,0,1 \\right\\rangle$\nVector $\\overrightarrow{V_{CA}}$:\n$\\overrightarrow{V_{CA}}=(\\overrightarrow{C}\\times\\overrightarrow{A})\\times\\overrightarrow{C}$\n$\\overrightarrow{V_{CA}}=(\\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\\times\\left\\langle cos(\\phi),0,sin(\\phi) \\right\\rangle)\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle$\n$\\overrightarrow{V_{CA}}=\\left\\langle sin(\\theta)sin(\\phi),-cos(\\theta)sin(\\phi),-sin(\\theta)cos(\\phi) \\right\\rangle\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle$\nSince we're only interested in the $z$ component of the vector\n$\\overrightarrow{V_{CA}}=\\left\\langle V_{CA_{x}},V_{CA_{y}},sin^{2}(\\theta)sin(\\phi)+cos^{2}(\\theta)sin(\\phi) \\right\\rangle$\n$\\overrightarrow{V_{CA}}=\\left\\langle V_{CA_{x}},V_{CA_{y}},sin(\\phi) \\right\\rangle$\nVector $\\overrightarrow{V_{CB}}$:\n$\\overrightarrow{V_{CB}}=(\\overrightarrow{C}\\times\\overrightarrow{b})\\times\\overrightarrow{C}$\n$\\overrightarrow{V_{CB}}=(\\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\\times\\left\\langle -cos(\\phi),0,sin(\\phi) \\right\\rangle)\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle$\n$\\overrightarrow{V_{CB}}=\\left\\langle sin(\\theta)sin(\\phi),-cos(\\theta)sin(\\phi),sin(\\theta)cos(\\phi) \\right\\rangle\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle$\nSince we're only interested in the $z$ component of the vector\n$\\overrightarrow{V_{CB}}=\\left\\langle V_{CB_{x}},V_{CB_{y}},sin^{2}(\\theta)sin(\\phi)+cos^{2}(\\theta)sin(\\phi) \\right\\rangle$\n$\\overrightarrow{V_{CB}}=\\left\\langle V_{CB_{x}},V_{CB_{y}},sin(\\phi) \\right\\rangle$\nSince we're working with unit vectors, then we can use dot products on the vectors with their angles as follows:\n$cos(\\angle ACN) = \\overrightarrow{V_{CA}}\\cdot \\overrightarrow{V_{CN}}$\n$cos(\\angle ACN) = \\left\\langle V_{CA_{x}},V_{CA_{y}},sin(\\phi) \\right\\rangle \\cdot \\left\\langle 0,0,1 \\right\\rangle = 0*V_{CA_{x}}+0*V_{CA_{y}}+1*sin(\\phi)=sin(\\phi)$\nLikewise,\n$cos(\\angle BCN) = \\overrightarrow{V_{CB}}\\cdot \\overrightarrow{V_{CN}}$\n$cos(\\angle BCN) = \\left\\langle V_{CB_{x}},V_{CB_{y}},sin(\\phi) \\right\\rangle \\cdot \\left\\langle 0,0,1 \\right\\rangle = 0*V_{CB_{x}}+0*V_{CB_{y}}+1*sin(\\phi)=sin(\\phi)$\nTherefore,\n$cos(\\angle ACN) = cos(\\angle BCN)$ and thus $\\angle ACN = \\angle BCN$\nSince those angles are equal, it proves that the great circle through $C$ and $N$ bisects the $\\angle ACB$ in the spherical triangle $ABC$\n~Tomas Diaz.  orders@tomasdiaz.com\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "$a_1, a_2, \\ldots, a_n$ is an arbitrary sequence of positive integers. A member of the sequence is picked at \nrandom. Its value is $a$. Another member is picked at random, independently of the first. Its value is $b$. Then a third value, $c$. Show that the probability that $a + b +c$ is divisible by $3$ is at least $\\frac14$.",
        "Solution": "Let x equal the probability of picking an element of the sequence equivalent to 0 mod 3, y equal those 1 mod 3, and z equal those 2 mod 3. Then, considering that 0+0+0, 1+1+1, 2+2+2, and 0+1+2 are divisible by 3, we obtain the equivalent inequality in the First Hint. This simplifies to $x^3 + y^3 + z^3 + 6xyz \\ge x^2(y+z) + y^2(x+z) + z^2(x+y).$ Now, we attempt to prove that $x^3 + y^3 + z^3 + 3xyz \\ge x^2(y+z) + y^2(x+z) + z^2(x+y).$ for all nonnegative real numbers x, y, z; our inequality will follow from this. But this is just Schur's Inequality with r = 1! This completes the proof.\nNote: Schur's Inequality states that $x^r(x-y)(x-z) + y^r(y-z)(y-x) + z^r(z-x)(z-y) \\ge 0$. This can be proven by setting WLOG $x \\ge y \\ge z$ and bunching together terms that are nonnegative. Challenge: when does equality hold? When does equality hold in the problem?"
    },
    {
        "Problem": "$P$ lies between the rays $OA$ and $OB$. Find $Q$ on $OA$ and $R$ on $OB$ collinear with $P$ so that $\\frac{1}{PQ} + \\frac{1}{PR}$ is as large as possible.",
        "Solution_.28inversions.29": "Perform the inversion with center $P$ and radius $\\overline{PO}.$ Lines $OA,OB$ go to the circles $(O_1),(O_2)$ passing through $P,O$ and the line $QR$ cuts $(O_1),(O_2)$ again at the inverses $Q',R'$ of $Q,R.$ Hence\n$\\frac{1}{PQ}+\\frac{1}{PR}=\\frac{PQ'+PR'}{PO^2}=\\frac{Q'R'}{PO^2}$\nThus, it suffices to find the line through $P$ that maximizes the length of the  segment $\\overline{Q'R'}.$ If $M,N$ are the midpoints of $PQ',PR',$ i.e. the projections of $O_1,O_2$ onto $QR,$ then from the right trapezoid $O_1O_2NM,$ we deduce that $O_1O_2 \\ge MN = \\frac{_1}{^2}Q'R'.$ Consequently, $2 \\cdot O_1O_2$ is the greatest possible length of $Q'R',$ which obviously occurs when $O_1O_2NM$ is a rectangle. Hence, $Q,R$ are the intersections of $OA,OB$ with the perpendicular to $PO$ at $P.$",
        "Solution_.28trig_bash.29": "\nLet $r = OP, x = \\angle OPR, a = \\angle POR,$ and $b = \\angle POQ.$ Then $\\angle ORP = \\pi - x - a$ and $\\angle OQP = x - b.$ Using the Law of Sines on $\\triangle OPR$ gives\n\nand using the Law of Sines on $\\triangle OPQ$ gives\n\nNote that $r, a,$ and $b$ are given constants.\nHence,\nClearly, this quantity is maximized when $\\sin x = 1.$ Because $x$ must be less than $\\pi$, $\\frac{1}{PQ} + \\frac{1}{PR}$ is as large as possible when $x = \\frac{\\pi}{2},$ or when line $QR$ is perpendicular to line $PO$."
    },
    {
        "Problem": "Let $A_1,A_2,...,A_{n+1}$ be distinct subsets of $[n]$ with $|A_1|=|A_2|=\\cdots =|A_{n+1}|=3$.  Prove that $|A_i\\cap A_j|=1$ for some pair $\\{i,j\\}$.",
        "Solution": "Suppose the problem statement does not hold. It is clear that $n \\ge 4$. Choose the smallest $n$ such that $|A_i \\cap A_j| \\neq 1$ for each $\\{i, j\\}$.\nFirst, the $(n+1)$ subsets have $3(n+1)$ elements (some repeated) in conjunction. Because there are $n$ elements of $[n]$ total, by the Pigeonhole Principle one element of $[n]$, say $k$, is in at least four subsets. Let these subsets be $A_1, A_2, A_3, A_4$, without loss of generality, and let $A_1$ have elements $k, m, n$. Then without loss of generality let $A_2$ have elements $k, m, p$. If set $A_3$ has elements $k, n, p$, then simple casework shows that it is impossible to create $A_4$ without having $A_4$ intersect one of $A_1, A_2, A_3$ at exactly one element. Thus assume $A_3$ has elements $k, m, q$. Then $A_4$ has elements $k, m, r$. Consider each remaining set $A_i$. Then $A_i$ either contains both $k, m$ or none of them. Because there are $(n-2)$ distinct elements of $[n]$ apart from $k, m$, at most $(n-2)$ subsets can contain $k, m$. Then at least 3 subsets do not contain $k, m$, and it is easy to see that they are disjoint from those subsets that do contain $k, m$. Thus, we can partition $[n]$ into two subsets, one of which is the union of the $t$ subsets that do contain $k, m$, and the other is the union of the $(n+1)-t$ subsets that do not contain $k, m$. Because the latter subset has $(n-t-2) < (n+1)-t - 1$ elements, we may use infinite descent to contradict the minimality of $n$. The proof is complete."
    },
    {
        "Problem": "Find the maximum possible number of three term arithmetic progressions in a monotone sequence of $n$ distinct reals.",
        "Solution": "Consider the first few cases for $n$ with the entire $n$ numbers forming an arithmetic sequence \nIf $n = 3$, there will be one ascending triplet (123). Let's only consider the ascending order for now.\nIf $n = 4$, the first 3 numbers give 1 triplet, the addition of the 4 gives one more, for 2 in total.\nIf $n = 5$, the first 4 numbers give 2 triplets, and the 5th number gives 2 more triplets (135 and 345).\nRepeating a few more times, we can quickly see that if $n$ is even, the nth number will give  more triplets in addition to all the prior triplets from the first $n-1$ numbers. \nIf $n$ is odd, the $n$th number will give  more triplets. \nLet $f(n)$ denote the total number of triplets for $n$ numbers. The above two statements are summarized as follows: \nIf $n$ is even, \nIf $n$ is odd,\nLet's obtain the closed form for when $n$ is even:\nNow obtain the closed form when $n$ is odd by using the previous result for when $n$ is even:\nNote the ambiguous wording in the question! If the \"arithmetic progression\" is allowed to be a disordered subsequence, then every progression counts twice, both as an ascending progression and as a descending progression.\nDouble the expression to account for the descending versions of each triple, to obtain:\n~Lopkiloinm (corrected by integralarefun)"
    },
    {
        "Problem": "$A + B + C$ is an integral multiple of $\\pi$. $x, y,$ and $z$ are real numbers. If $x\\sin(A)+y\\sin(B)+z\\sin(C)=x^2\\sin(2A)+y^2\\sin(2B)+z^2\\sin(2C)=0$, show that $x^n\\sin(nA)+y^n \\sin(nB) +z^n \\sin(nC)=0$ for any positive integer $n$.",
        "Solution": "Let $a=xe^{iA}$, $b=ye^{iB}$, $c=ze^{iC}$ be numbers in the complex plane.\nNote that $A+B+C=k\\pi$ implies $abc=xyz(e^{ik\\pi})=\\pm xyz$ which is real. Also note that $x\\sin(A), y\\sin(B), z\\sin(C)$ are the imaginary parts of $a, b, c$ and that $x^2\\sin(2A), y^2\\sin(2B), z^2\\sin(2C)$ are the imaginary parts of $a^2, b^2, c^2$ by de Moivre's Theorem. Therefore, $a+b+c$ and $a^2+b^2+c^2$ are real because their imaginary parts sum to zero.\nFinally, note that $\\frac{1}{2}\\left((a+b+c)^2-(a^2+b^2+c^2)\\right)=ab+bc+ac$ is real as well.\nIt suffices to show that $P_n=a^n+b^n+c^n$ is real for all positive integer $n$, which can be shown by induction.\nNewton Sums gives the following relationship between sums of the form $P_k=a^k+b^k+c^k$\n\nWhere $S_1=a+b+c$, $S_2=ab+bc+ac$, and $S_3=abc$. It is given that $P_0, P_1, P_2$ are real. Note that if  $P_{k-1}, P_{k-2}, P_{k-3}$ are real, then clearly $P_k$ is real because all other parts of the above equation are real, completing the induction."
    },
    {
        "Problem": "The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular.",
        "Solution": "Let $ABCD$ be the tetrahedron, and let $P$ and $Q$ be the points at which the insphere touches faces $ABC$ and $ABD$ respectively (and therefore the centroids of those faces). Looking at the plane containing $P$, $A$, and $Q$, we see that the intersection of the sphere and the plane is a circle, and that $AP$ and $AQ$ are both tangent to said circle. $[AP]$ and $[AQ]$ are tangents to the circle from the same point and thus have the same length. The same goes for $[BP]$ and $[BQ]$. Thus, triangles $ABP$ and $ABQ$ are congruent, and $\\mathbf{\\angle PAB = \\angle QAB}$.\nLet the intersection of $AP$ and $BC$ be $M$, and let the intersection of $AQ$ and $BD$ be $N$. Then $AM$ and $AN$ are medians of $ABC$ and $ABD$, and thus $|AM| = \\frac{3}{2}|AP| = \\frac{3}{2}|AQ| = |AN|$. We already know from the previous congruence that $\\angle{MAB} = \\angle{PAB} = \\angle{QAB} = \\angle{NAB}$, and $|AB|$ is equal to itself. Thus, $MAB$ and $NAB$ are also congruent to each other. Finally, $|BC| = 2|BM| = 2|BN| = |BD|$ (Because $M$ and $N$ are midpoints of $BC$ and $BD$ respectively), and from the congruence of $MAB$ and $NAB$ we have $\\angle{CBA} = \\angle{MBA} = \\angle{NBA} = \\angle{DBA}$, and again $|AB|$ is equal to itself. Thus $ABC$ and $ABD$ are congruent, thus $\\mathbf{|AC| = |AD|}$ and $\\mathbf{|BC| = |BD|}$.\nBy applying the same logic to faces $BCD$ and $ACD$ we get $|AC| = |BC|$ and $|AD| = |BD|$. Finally, applying the same logic to faces $ACB$ and $ACD$ we get $|AB| = |AD|$ and $|CB| = |CD|$. Putting all these equalities together, we get that all edges of the tetrahedron are equal, and thus the tetrahedron is regular. $\\blacksquare$"
    },
    {
        "Problem": "If $x, y, z$ are reals such that $0\\le x, y, z \\le 1$, show that $\\frac{x}{y + z + 1} + \\frac{y}{z + x + 1} + \\frac{z}{x + y +  1} \\le 1 - (1 - x)(1 - y)(1 - z)$",
        "Solution": "Rewrite the given inequality so that $1$ is isolated on the right side. Set the left side to be $f(x, y, z)$. Now a routine computation shows\n$\\frac{\\partial^2 f}{\\partial x^2} = \\frac{2y}{(x + z + 1)^3} + \\frac{2z}{(x + y + 1)^3}\\geq 0$\nwhich shows that $f$ is convex (concave up) in all three variables. Thus the maxima can only occur at the endpoints, i.e. if and only if $x, y, z \\in \\{0,1\\}$. Checking all eight cases shows that the value of the expression cannot exceed 1."
    },
    {
        "Problem": "Prove that if $n$ is not a multiple of $3$, then the angle $\\frac{\\pi}{n}$ can be trisected with ruler and compasses.",
        "Solution": "Let $n=3k+1$.  Multiply throughout by $\\pi/3n$. We get\n$\\frac{\\pi}{3} = \\frac{\\pi \\times k}{n} + \\frac{\\pi}{3n}$\nRe-arranging, we get\n$\\frac{\\pi}{3} - \\frac{\\pi \\times k}{n} = \\frac{\\pi}{3n}$\nA way to interpret it is that if we know the value $k$, then the remainder angle of subtracting $k$ times the given angle from $\\frac{\\pi}{3}$ gives us $\\frac{\\pi}{3n}$, the desired trisected angle.\nThis can be extended to the case when $n=3k+2$ where now, the equation becomes\n$\\frac{\\pi}{3} - \\frac{\\pi \\times k}{n} = \\frac{2\\pi}{3n}$\nHence in this case, we will have to subtract $k$ times the original angle from $\\frac{\\pi}{3}$ to get twice the the trisected angle. We can bisect it after that to get the trisected angle."
    },
    {
        "Problem": "What is the largest number of towns that can meet the following criteria. Each pair is directly linked by just one of air, bus or train. At least one pair is linked by air, at least one pair by bus and at least one pair by train. No town has an air link, a bus link and a train link. No three towns, $A, B, C$ are such that the links between $AB, AC$ and $BC$ are all air, all bus or all train.",
        "Solution": "Assume $AB$, $AC$, and $AD$ are all rail.\nNone of $BC$, $CD$, or $CD$ can be rail, as those cities would form a rail triangle with $A$.\nIf $BC$ is bus, then $BD$ is bus as well, as otherwise $B$ has all three types.\nHowever, $CD$ cannot be rail (as $\\triangle ACD$ would be a rail triangle), bus (as $BCD$ would be a bus triangle), or ferry (as $C$ and $D$ would have all three types).\nTherefore, no city can have three connections of the same type.\nAssume there are 5 towns - $A$, $B$, $C$, $D$, and $E$.\nTwo connections from $A$ must be of one type, and two of another; otherwise there would be at least three connections of the same type from $A$, which has been shown to be impossible.\nLet $AB$ and $AC$ be rail connections, and $AD$ and $AE$ be bus.\nAssume $CD$ is air.\n$BC$ cannot be rail ($\\triangle ABC$ would be a rail triangle) or bus ($C$ would have all three types), so $BC$ must be air.\n$DE$ cannot be bus ($\\triangle ADE$ would be a bus triangle) or rail ($D$ would have all three types), so $DE$ must be air.\n$BE$ cannot be rail ($E$ would have all three types) or bus ($B$ would have all three types), so $BE$ must be air.\nHowever, $BD$ cannot be rail ($D$ would have all three types), bus ($B$ would have all three types), or air ($D$ would have three air connections).\nTherefore, the assumption that $CD$ is air is false.\n$CD$ can equally be rail or bus; assume it is bus.\n$BC$ cannot be rail ($\\triangle ABC$ would be a rail triangle) or air ($C$ would have all three types), so $BC$ must be bus.\n$BD$ cannot be air ($B$ would have all three types) or bus ($D$ would have three bus connections), so $BD$ must be rail.\n$DE$ cannot be air ($D$ would have all three types) or bus ($D$ would have three bus connections), so $DE$ must be rail.\n$CE$ cannot be air ($C$ would have all three types) or bus ($E$ would have three bus connections), so $CE$ must be rail.\nThe only connection remaining is $BE$, which cannot be orange as both $B$ and $D$ would have all three types, but this means there are no air connections.\nTherefore, it is impossible with five (or more) towns.\nA four-town mapping is possible:\n$AB$, $BC$, $CD$, and $DA$ are connected by bus.\n$AC$ is connected by rail.\n$BD$ is connected by air.\nTherefore, the maximum number of towns is $4$."
    },
    {
        "Problem": "Show that for any triangle, $\\frac{3\\sqrt{3}}{2}\\ge \\sin(3A) + \\sin(3B) + \\sin (3C) \\ge -2$.\nWhen does the equality hold?",
        "Solution": "Given three angles that add to 180°, one can construct a triangle from them.  This is true even if the angles are negative; however, the resulting triangle could then be recognized as having only positive angles, and the interpretation with negative angles (and, likely, negative side lengths) might be considered perverse.  In this case, the problem must be interpreted as ruling out such \"perverse\" triangles; otherwise, its angles could be -30°, -20°, and 230°, in which case sin(3x) becomes -1, -√3/2, and -1/2 respectively, which add to about -2.3, contradicting the problem statement.\nThe constraints on the angles of a non-perverse triangle are: $0\\degree\\leq A,B,C\\leq180\\degree$ (Error compiling LaTeX. Unknown error_msg), and $A+B+C=180\\degree$ (Error compiling LaTeX. Unknown error_msg).\nIn fact, at this point, we only care about 3A, 3B, and 3C.  Let us call them x, y, and z.  We have:\n0 ≤ x,y,z ≤ 540°; x+y+z = 540°.  Prove $\\frac{3\\sqrt{3}}{2}\\ge \\sin(x) + \\sin(y) + \\sin (z) \\ge -2$.\nNow.  Without loss of generality, assume x ≤ y ≤ z.  It follows that x ≤ 180° and z ≥ 180° [otherwise, x+y+z would be strictly greater than 180°, or strictly less than 180°, respectively].\nSince sin(u) is nonnegative over the interval [0, 180°], and -1 ≤ sin u ≤ 1 for all real u, we can immediately use the x ≤ 180° result to show that sin x + sin y + sin z ≤ 0 + -1 + -1 = -2.  This proves the lower bound on sin x + sin y + sin z; equality occurs when sin x = 0 and sin y = sin z = -1, and this is reachable only when y = z = 270° and x = 0°, which translates into A = 0° and B = C = 90°.  (This might be considered a degenerate triangle and ruled out; in that case, the lower bound could be stated as a strict one.)\nNow for the upper bound.  It is true that sin(u) is non-positive over the interval [180°, 360°].  Also, 3√3/2 is greater than 2 (it is roughly 2.6): if you square it, you get 27/4, while if you square 2, you get 16/4.  So... We know z ≥ 180°.  If z ≤ 360°, then sin z ≤ 0, and then sin x + sin y + sin z ≤ 1 + 1 + 0 = 2 < 3√3/2.  Therefore, we need merely handle the case where z ≥ 360°.  sin(u + 360°) = sin(u), so we may as well subtract 360° from z.  Then the problem effectively becomes:\n\"Given 0 ≤ x,y,z ≤ 180° and x+y+z = 180°, prove that sin x + sin y + sin z ≤ 3√3/2.  Also find when you get equality.\"\nThis is probably a well-known theorem, but I shall address it here anyway.\nLet us suppose that {x,y,z} is the selection of x,y,z given the above constraints with the largest value of sin x + sin y + sin z.  It suffices to show that, for this selection, sin x + sin y + sin z ≤ 3√3/2.  We shall see that, given these constraints, sin x + sin y + sin z increases when you move any two variables closer together, and it is maximized when x=y=z.\nLemma with some general expository use: A special case of a result called \"Jensen's inequality\".  If f(x) is well-behaved and f'(x) is strictly decreasing over an open interval (a, b), then, for any {x,y} selected from that interval with x ≠ y, $2f(\\frac{x + y}{2}) > f(x) + f(y).$\nSo.  sin(u) is well-behaved, and sin'(u) = cos(u), which is strictly decreasing over the open interval (0, 180°).  Suppose that {x,y,z} are not all the same value.  Then, without loss of generality, suppose x ≠ y.  Let a = (x+y)/2.  Then the selection {a,a,z} also satisfies the constraints 0 ≤ a,a,z ≤ 180° and a+a+z ≤ 180°.  Furthermore, by the above lemma, 2*sin a > sin x + sin y, so this new selection {a,a,z} has a larger corresponding value sin a + sin a + sin z than the selection {x,y,z}.  This contradicts our original assumption that {x,y,z} was chosen to have the largest value of sin x + sin y + sin z.  Therefore, x=y=z.\nThen, since x+y+z = 180°, we conclude x=y=z=60°.  Then sin x + sin y + sin z = 3 * sin 60° = 3 * √3/2.  So 3√3/2 is indeed the maximum value of sin x + sin y + sin z, which was to be proven.  Translating back into the previous problem (0 ≤ x,y,z ≤ 540°; x+y+z = 540°), the maximum occurs when x=y=60° and z=420°; translating into the original problem, we have proven the upper bound, and equality with the upper bound occurs when A=B=20° and C = 140°."
    },
    {
        "Problem": "A convex polygon has $n$ sides. Each vertex is joined to a point $P$ not in the same plane. If $A, B, C$ are adjacent vertices of the polygon take the angle between the planes $PAB$ and $PBC$. The sum of the $n$ such angles equals the sum of the $n$ angles in the polygon. Show that $n=3$.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "Show that for any positive real $x$, $[nx]\\ge \\sum_{1}^{n}\\left(\\frac{[kx]}{k}\\right)$",
        "Solution.28due_to_Pavel_Zatitskiy.29": "First of all we write $[kx]=kx-\\{kx\\}$. So, we need to prove that\n$\\sum_{1}^{n}\\left(\\frac{\\{kx\\}}{k}\\right)\\geq \\{nx\\}.$\nLet's denote $a_k=\\{kx\\}$. It is easy to see that $a_k+a_m \\geq a_{k+m}$. We need to prove\n$\\sum_{1}^{n}\\left(\\frac{a_k}{k}\\right)\\geq a_n.$\nWe will prove it by induction by $n$. The base is obvious, so we need to make a step.\nLet's take $m$ such that $\\frac{a_m}{m}$ is minimal. If $m=n$ then our inequality is obvious. So, $m<n$. Then, by induction, $\\sum_{1}^{n-m}\\left(\\frac{a_k}{k}\\right)\\geq a_{n-m}$ and $\\sum_{n-m+1}^{n}\\left(\\frac{a_k}{k}\\right)\\geq m\\cdot \\frac{a_m}{m}=a_m$. Now we can add these two inequalities and get"
    },
    {
        "Problem": "In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?",
        "Solution": "We induct on $n$ to prove that in a party with $n$ people, there must be at least $(n-3)$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people $A, B, C$, who do not know each other.)\nBase case: $n = 4$ is obvious.\nInductive step: Suppose in a party with $k$ people (with $k \\ge 4$), at least $(k-3)$ people know everyone else. Consider a party with $(k+1)$ people. Take $k$ of the people (leaving another person, $A$, out) and apply the inductive step to conclude that at least $(k-3)$ people know everyone else in the $k$-person group, $G$.\nNow suppose that everyone in the group $G$ knows each other. Then take $3$ of these people and $A$ to deduce that $A$ knows a person $B \\in G$, which means $B$ knows everyone else. Then apply the inductive step on the remaining $k$ people (excluding $B$) to find $(k-3)$ people out of them that know everyone else (including $B$, of course). Then these $(k-3)$ people and $B$, which enumerate $(k-2)$ people, know everyone else.\nSuppose that there exist two people $B, C \\in G$ who do not know each other. Because $k-3 \\ge 1$, there exist at least one person in $G$, person $D$, who knows everyone else in $G$. Now, take $A, B, C, D$ and observe that because $B, C$ do not know each other, either $A$ or $D$ knows everyone else of $A, B, C, D$ (by the problem condition), so in particular $A$ and $D$ know each other. Then apply the inductive step on the remaining $k$ people (excluding $D$) to find $(k-3)$ people out of them that know everyone else (including $D$, of course). Then these $(k-3)$ people and $D$, which enumerate $(k-2)$ people, know everyone else.\nThis completes the inductive step and thus the proof of this stronger result, which easily implies that at least $1982 - 3 = \\boxed{1979}$ people know everyone else."
    },
    {
        "Problem": "Let $S_r=x^r+y^r+z^r$ with $x,y,z$ real. It is known that if $S_1=0$,\n$(*)$ $\\frac{S_{m+n}}{m+n}=\\frac{S_m}{m}\\frac{S_n}{n}$\nfor $(m,n)=(2,3),(3,2),(2,5)$, or $(5,2)$. Determine all other pairs of integers $(m,n)$ if any, so that $(*)$ holds for all real numbers $x,y,z$ such that $x+y+z=0$.",
        "Solution_1": "Claim Both $m,n$ can not be even.\nProof\n$x+y+z=0$ ,$\\implies x=-(y+z)$.\nSince $\\frac{S_{m+n}}{m+n} = \\frac{S_m S_n}{mn}$,\nby equating cofficient of $y^{m+n}$ on LHS and RHS ,get\n$\\frac{2}{m+n}=\\frac{4}{mn}$.\n$\\implies  \\frac{m}{2} + \\frac {n}{2} = \\frac{m\\cdot n}{2\\cdot2}$.\nSo we have, $\\frac{m}{2} \\biggm{|} \\frac{n}{2}$ and $\\frac{n}{2} \\biggm{|} \\frac{m}{2}$.\n$\\implies m=n=4$.\nSo we have $S_8=2(S_4)^2$.\nNow since it will true for all real $x,y,z,x+y+z=0$.\nSo choose $x=1,y=-1,z=0$.\n$S_8=2$ and $S_4=2$ so $S_8 \\neq 2 S_4^2$.\nThis is contradiction. So, at least one of $m,n$ must be odd. WLOG assume $n$ is odd and m is even. The coefficient of $y^{m+n-1}$ in $\\frac{S_{m+n}}{m+n}$ is $\\frac{\\binom{m+n}{1} }{m+n} =1$\nThe coefficient of $y^{m+n-1}$ in $\\frac{S_m\\cdot S_n}{m\\cdot n}$ is $\\frac{2}{m}$.\nTherefore, $\\boxed{m=2}$.\nNow choose $x=y=\\frac1,z=(-2)$. (sic)\nSince $\\frac{S_{n+2}}{2+n}=\\frac{S_2}{2}\\frac{S_n}{n}$ holds for all real $x,y,z$ such that $x+y+z=0$.\nWe have $\\frac{2^{n+2}-2}{n+2} = 3\\cdot\\frac{2^n-2}{n}$. Therefore,\n\\begin{equation*}\n\\label{eq:l2}\n\\frac{2^{n+1}-1}{n+2} =3\\cdot\\frac{2^{n-1}-1}{n}\\ldots\n\\tag{**}\n\\end{equation*}\nClearly $(\\ref{eq:l2})$ holds for $n\\in\\{5,3\\}$.\nAnd one can say that for $n\\ge 6$,\n$\\text{RHS of (\\ref{eq:l2})}<\\text{LHS of (\\ref{eq:l2})}$.\nSo our answer is $(m,n)=(5,2),(2,5),(3,2),(2,3)$.\n-ftheftics (edited by integralarefun)"
    },
    {
        "Problem": "If a point $A_1$ is in the interior of an equilateral triangle $ABC$ and point $A_2$ is in the interior of $\\triangle{A_1BC}$, prove that\n$I.Q. (A_1BC) > I.Q.(A_2BC)$,\nwhere the isoperimetric quotient of a figure $F$ is defined by\n$I.Q.(F) = \\frac{\\text{Area (F)}}{\\text{[Perimeter (F)]}^2}$",
        "Solution": "First, an arbitrary triangle $ABC$ has isoperimetric quotient (using the notation $[ABC]$ for area and $s = \\frac{a + b + c}{2}$):\nLemma. $\\tan x \\tan (A - x)$ is increasing on $0 < x < \\frac{A}{2}$, where $0 < A < 90^\\circ$.\nProof. \n is increasing on the desired interval, because $\\cos (A - 2x)$ is increasing on $0 < x < \\frac{A}{2}.$\nLet $x_1, y_1, z_1$ and $x_2, y_2, z_2$ be half of the angles of triangles $A_1 BC$ and $A_2 BC$ in that order, respectively. Then it is immediate that $30^\\circ > y_1 > y_2$, $30^\\circ > z_1 > z_2$, and $x_1 + y_1 + z_1 = x_2 + y_2 + z_2 = 90^\\circ$. Hence, by Lemma it follows that\n\n\nMultiplying this inequality by $\\frac{1}{4}$ gives that $I.Q[A_1 BC] > I.Q[A_2 BC]$, as desired."
    },
    {
        "Problem": "Prove that there exists a positive integer $k$ such that $k\\cdot2^n+1$ is composite for every integer $n$.",
        "Solution": "Indeed, $\\boxed{k=2935363331541925531}$ has the requisite property.\nTo see why, consider the primes $3,\\ 5,\\ 17,\\ 257,\\ 65537,\\ 6700417,\\ 641$, and observe that\nMoreover,\nWe conclude that\nAnd $k>>3,5,17,257,65537,6700417,641$ so the relevant values will, in fact, always be composite. $\\blacksquare$"
    },
    {
        "Problem": "$A,B$, and $C$ are three interior points of a sphere $S$ such that $AB$ and $AC$ are perpendicular to the diameter of $S$ through $A$, and so that two spheres can be constructed through $A$, $B$, and $C$ which are both tangent to $S$. Prove that the sum of their radii is equal to the radius of $S$.",
        "Solution": "Let the two tangent spheres be $S_1$ and $S_2$, and let $O, O_1, O_2$ and $R, R_1, R_2$ be the origins and radii of $S, S_1, S_2$ respectively. Then $AO$ stands normal to the plane $P$ through $\\Delta ABC$. Because both spheres go through $A$, $B$, and $C$, the line $O_1 O_2$ also stands normal to $P$, meaning $AO$ and $O_1 O_2$ are both coplanar and parallel. Therefore the problem can be flattened to the plane $P'$ through $A$, $O$, $O_1$ and $O_2$.\nLet $X, Y, Z, M, N$ be points on $P'$ such that $X = S \\cap O O_1,  \\quad Y = S_1 \\cap S_2 \\neq A,  \\quad   Z = S \\cap O O_2,   \\quad  M = O_1 Y \\cap AO,    \\quad N = O_2 Y \\cap AO$\nLet $J$ be the three circles radical center, meaning $JX$ and $JZ$ are tangent segments to $S$ and $J \\in AY$.\nBecause  $Y \\in P \\iff \\angle NAY = \\angle YAM = 90 ^{\\circ},$ we have that $\\overline{YN}$ and $\\overline{YM}$ are diameters.\nThis means that $\\angle OZN = \\angle ZNY = \\angle JZY$ and $\\angle MXO = \\angle YMX= \\angle YXJ$.\nAnd because $\\angle ZNO = 180^{\\circ} - \\angle AYZ = \\angle ZYJ$ and $\\angle OMX = 180^{\\circ} -\\angle XYA = \\angle JYX,$ we have that $\\Delta JZY \\sim \\Delta OZN$ and $\\Delta JYX \\sim \\Delta OMX$.\nWe then conclude that $\\overline{OM} = \\overline{OX} \\enspace \\frac {\\overline{JY} }{ \\overline{JX}}  = \\overline{OZ} \\enspace \\frac {\\overline{JY} }{ \\overline{JZ}}  = \\overline{ON}$\nLet $a \\top b$ denote line $a$ bisecting line segment $b$.\nSince $O_1 O_2 || MN$ it follows that $OY \\, \\top \\, \\overline{MN} \\Rightarrow OY \\, \\top \\,  \\overline{O_1 O_2}$. \nSimilarly we have that $O_1 O_2 \\, \\top \\, \\overline{YM} \\Rightarrow O_1 O_2 \\, \\top \\,  \\overline{OY}$.\nAnd so $O_1 O O_2 Y$ is a parallelogram because $\\overline{OY}$ and $\\overline{O_1 O_2}$ bisect each other, meaning $R = \\overline{OZ} = \\overline{O O_2} + \\overline{O_2 Z} = \\overline{O_1 Y}+ R_2 = R_1 + R_2$\n$Quod \\enspace Erat \\enspace Demonstrandum$"
    },
    {
        "Problem": "On a given circle, six points $A$, $B$, $C$, $D$, $E$, and $F$ are chosen at random, independently and uniformly with respect to arc length. Determine the probability that the two triangles $ABC$ and $DEF$ are disjoint, i.e., have no common points.",
        "Solution": "First we give the circle an orientation (e.g., letting the circle be the unit circle in polar coordinates).  Then, for any set of six points chosen on the circle, there are exactly $6!$ ways to label them one through six. Also, this does not affect the probability we wish to calculate. This will, however, make calculations easier.\nNote that, for any unordered set of six points chosen from the circle boundary, the number of ways to number them such that they satisfy this disjoint-triangle property is constant: there are six ways to choose which triangle will be numbered with the numbers one through three, and there are $(3!)^2$ ways to arrange the numbers one through three and four through six on these two triangles. Therefore, for any given configuration of points, there are $6^3=216$ ways to label them to have this disjoint-triangle property. There are, however, $6!=720$ ways to label the points in all, so given any six unordered points, the probability that when we inflict an ordering we produce the disjoint-triangle property is $216/720=3/10$.\nSince this probability is constant for any configuration of six unordered points we choose, we must have that $3/10$ is the probability that we produce the disjoint-triangle property if we choose the points as detailed in the problem statement."
    },
    {
        "Problem": "Prove that the zeros of\n\ncannot all be real if $2a^2<5b$.",
        "Solution": "We prove the contrapositive: if the polynomial in question has the five real roots $x_1, x_2, x_3, x_4, x_5$, then $5b \\le 2a^2$.\nBecause $a = -(x_1 + x_2 + x_3 + x_4 + x_5)$ and $b = x_1x_2 + x_1x_3 + ... + x_4x_5$ by Vieta's Formulae, we have\n(by Cauchy-Schwarz)\nso $5b \\le 2a^2$, as desired.",
        "Solution_2": "Lemma:\nFor all real numbers $x_1,x_2,\\cdots x_5$,\n\n\nBy the trivial inequality,\n\nMaking such an inequality for all the variable pairs and summing them, we find the lemma is true.\nNow, let our roots be $x_1,x_2,\\cdots,x_5$. By Vieta's, $a=x_1+x_2+\\cdots+x_5$ and $b=x_1x_2+x_1x_3+\\cdots+x_4x_5$\nIf we show that for all real $x_1,x_2,\\cdots, x_5$ that $2a^2\\ge 5b$, then we have a contradiction and all of $x_1,x_2,\\cdots, x_5$ cannot be real. We start by rewriting $2a^2\\ge 5b$ as\n\nWe divide by $2$ and find\n\nExpanding the LHS, we have\n\nWe subtract the sum in brackets, and then multiply by $2$ to find\n\nwhich is true by our lemma."
    },
    {
        "Problem": "Each set of a finite family of subsets of a line is a union of two closed intervals. Moreover, any three of the sets of the family have a point in common. Prove that there is a point which is common to at least half the sets of the family.",
        "Solution": "Let us first see that this works for anything less than three sets.\nObviously, due to the second condition given, they all share at least one point in common.\nNow, let us see that it works for three or more sets.\nFirst, take any three of the subsets in the family of sets.\nLet them be A, B, and C. \nWe know that they share at least one point in common. Let us take the pair of segments that have the shortest length when you take the union of the pair. Let us denote this segment D. Evidently, if you take any other set of points, it must intersect D at least at one point. In order to minimize the number of intersections between the other varying segments, we must have at close to 1/2 of the remaining segments as possible intersecting each other. However, if we note, putting them that way forces at least 1/2 of the segments in each section."
    },
    {
        "Problem": "Six segments $S_1, S_2, S_3, S_4, S_5,$ and $S_6$ are given in a plane. These are congruent to the edges $AB, AC, AD, BC, BD,$ and $CD$, respectively, of a tetrahedron $ABCD$. Show how to construct a segment congruent to the altitude of the tetrahedron from vertex $A$ with straight-edge and compasses.",
        "Solution": "Throughout this solution, we denote the length of a segment $S$ by $|S|$.\nIn this solution, we employ several lemmas. Two we shall take for granted: given any point $A$ and a line $\\ell$ not passing through $A$, we can construct a line $\\ell'$ through $A$ parallel to $\\ell$; and given any point $A$ on a line $\\ell$, we can construct a line $\\ell'$ through $A$ perpendicular to $\\ell$.\nLemma 1: If we have two segments $S$ and $T$ on the plane with non-zero length, we may construct a circle at either endpoint of $S$ whose radius is $|T|$.\nProof: We can construct arbitrarily many copies of $T$ by drawing a circle about one of its endpoints through its other endpoint, and then connecting the center of this circle with any other point on the circle. We can construct copies of $T$ like this until we create a circle of radius $|T|$ and center $P_1$ that intersects segment $S$. We can then take this intersection point $P_2$ and draw a line $\\ell$ through it perpendicular to $S$, and draw a circle with center $P_2$ passing through $P_1$, and consider its intersection $P_3$ with $\\ell$. Note that $P_2P_3\\perp S$ and $|P_2P_3|=|T|$. Take an endpoint $P_4$ of $S$: then draw a line through $P_3$ parallel to $S$, and a line through $P_4$ parallel to $P_2P_3$. Let these two lines intersect at $P_5$. Then $P_2P_3P_5P_4$ is a rectangle, so $|P_4P_5|=|T|$. Our desired circle is then a circle centered at $P_4$ through $P_5$.\nLemma 2: Given three collinear points $A$, $B$, $C$ in this order, if $|AB|=a$ and $|BC|=b$ with $a>b$, then we can construct a segment of length $\\sqrt{a^2-b^2}$.\nProof: From Lemma 1, we can construct a circle through $C$ with radius $a$, and then construct a perpendicular through $B$ to $AC$: these two objects intersect at $D$ and $E$. Both $BD$ and $DE$ have length $\\sqrt{a^2-b^2}$, from the Pythagorean Theorem.\nProof of the original statement: Note that we can construct a triangle $B'C'D'$ congruent to triangle $BCD$ by applying Lemma 1 to segments $S_4$, $S_5$, and $S_6$. Similarly, we can construct $A_B$ and $A_C$ outside triangle $B'C'D'$ such that $A_BC'D'\\cong ACD$ and $A_CB'D'\\cong ABD$.\nLet $A'$ be a point outside of the plane containing $S_1$ through $S_6$ such that $A'B'C'D'\\cong ABCD$. Then the altitudes of triangles $A_BC'D'$ and $A'C'D'$ to segment $C'D'$ are congruent, as are the altitudes of triangles $A_CB'D'$ and $A'B'D'$ to segment $B'D'$. However, if we project the altitudes of $A'C'D'$ and $A'B'D'$ from $A'$ onto the plane, their intersection is the base of the altitude of tetrahedron $A'B'C'D'$ from $A'$. In addition, these altitude projections are collinear with the altitudes of triangles $A_BC'D'$ and $A_CB'D'$. Therefore, the altitudes of $A_BC'D'$ and $A_CB'D'$ from $A_B$ and $A_C$ intersect at the base $X'$ of the altitude of $A'B'C'D'$ from $A'$. In summary, we can construct $X'$ by constructing the perpendiculars from $A_B$ and $A_C$ to $C'D'$ and $B'D'$ respectively, and taking their intersection.\nLet $Y'$ be the intersection of $A_BX'$ with $C'D'$. Then the altitude length we seek to construct is, from the Pythagorean Theorem, $\\sqrt{|A_BY'|^2-|X'Y'|^2}$. We can directly apply Lemma 2 to segment $A_BY'X'$ to obtain this segment. This shows how to construct a segment of length $A'X'$."
    },
    {
        "Problem": "Consider an open interval of length $1/n$ on the real number line, where $n$ is a positive integer. Prove that the number of irreducible fractions $p/q$, with $1\\le q\\le n$, contained in the given interval is at most $(n+1)/2$.",
        "Solution": "Let $I$ be an open interval of length $1/n$ and $F_n$ the set of fractions $p/q\\in I$ with $p,q\\in\\mathbb{Z}$, $\\gcd(p,q)=1$ and $1\\leq q\\leq n$.\nAssume that $\\frac{p}{q}\\in F_n$. If $k\\in\\mathbb{Z}$ is such that $1\\leq kq\\leq n$, and $p'\\in\\mathbb{Z}$ is such that $\\gcd(p',kq)=1$, then\n\nTherefore $\\frac{p'}{kq}\\notin I\\supset F_n$. This means that $\\frac{p}{q}$ is the only fraction in $F_n$ with denominator $q$ or multiple of $q$.\nTherefore, from each of the pairs in $P=\\left\\{(k,2k):\\ 1\\leq k\\leq \\left\\lfloor\\frac{n+1}{2}\\right\\rfloor\\right\\}$ at most one element from each can be a denominator of a fraction in $F_n$.\nHence $|F_n|\\leq |P|\\leq\\frac{n+1}{2}$"
    },
    {
        "Problem": "In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$, the product of $2$ of its roots is $- 32$. Find $k$.",
        "Solution_1_.28ingenious.29": "Using Vieta's formulas, we have:\n\nFrom the last of these equations, we see that $cd = \\frac{abcd}{ab} = \\frac{-1984}{-32} = 62$. Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$, and so $ac+ad+bc+bd=k-30$. The key insight is now to factor the left-hand side as a product of two binomials: $(a+b)(c+d)=k-30$, so that we now only need to determine $a+b$ and $c+d$ rather than all four of $a,b,c,d$.\nLet $p=a+b$ and $q=c+d$. Plugging our known values for $ab$ and $cd$ into the third Vieta equation, $-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)$, we have $-200 = -32(c+d) + 62(a+b) = 62p-32q$. Moreover, the first Vieta equation, $a+b+c+d=18$, gives $p+q=18$. Thus we have two linear equations in $p$ and $q$, which we solve to obtain $p=4$ and $q=14$.\nTherefore, we have $(\\underbrace{a+b}_4)(\\underbrace{c+d}_{14}) = k-30$, yielding $k=4\\cdot 14+30 = \\boxed{86}$.",
        "Solution_2_.28cool.29": "We start as before: $ab=-32$ and $cd=62$. We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic $x^2+sx+62$.\nNow\n\nEquating the coefficients of $x^3$ and $x$ with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of $x^2$ and get $k=\\boxed{86}.$",
        "Solution_3": "Let the roots of the equation be $a,b,c,$ and $d$. By Vieta's,\n\nSince $abcd=-1984$ and $ab=-32$, then, $cd=62$. Notice thatcan be factored intoFrom the first equation, $c+d=18-a-b$. Substituting it back into the equation,Expanding,So, $a+b=4$ and $c+d=14$. Notice thatPlugging all our values in,\n~ kante314"
    },
    {
        "Problem": "The geometric mean of any set of $m$ non-negative numbers is the $m$-th root of their product.\n$\\quad (\\text{i})\\quad$ For which positive integers $n$ is there a finite set $S_n$ of $n$ distinct positive integers such that the geometric mean of any subset of $S_n$ is an integer?\n$\\quad (\\text{ii})\\quad$ Is there an infinite set $S$ of distinct positive integers such that the geometric mean of any finite subset of $S$ is an integer?",
        "Solution": "a) We claim that for any numbers $p_1$, $p_2$, ... $p_n$, $p_1^{n!}, p_2^{n!}, ... p_n^{n!}$ will satisfy the condition, which holds for any number $n$.\nSince $\\sqrt[n] ab = \\sqrt[n] a * \\sqrt[n] b$, we can separate each geometric mean into the product of parts, where each part is the $k$th root of each member of the subset and the subset has $k$ members.\nAssume our subset has $k$ members. Then, we know that the $k$th root of each of these members is an integer (namely $p^{n!/k}$), because $k \\leq n$ and thus $k  |  n!$. Since each root is an integer, the geometric mean will also be an integer.\nb) If we define $q$ as an arbitrarily large number, and $x$ and $y$ as numbers in set $S$, we know that ${\\sqrt[q]{\\frac{x}{y}}}$ is irrational for large enough $q$, meaning that it cannot be expressed as the fraction of two integers. However, both the geometric mean of the set of $x$ and $q-1$ other arbitrary numbers in $S$ and the set of $y$ and the same other $q-1$ numbers are integers, so since the other numbers cancel out, the geometric means divided, or ${\\sqrt[q]{\\frac{x}{y}}}$, must be rational. This is a contradiction, so no such infinite $S$ is possible.\n-aops111 (first solution dont bully me)",
        "Solution_2": "(i) The solution is the same as in the first solution\n(ii)\nLet $p$ and $q$ be some prime numbers, and let $v_p(n)$ be the largest number such that $p^{v_p(n)}$ divides $n$. $v_p(n)$ is also the exponent of $p$ in the prime factorization of $n$ (or $0$, if $p$ isn't a divisor of $n$). Then, because $S$ is infinite, we can choose $q-1$ numbers $a_1, a_2, ... a_{q-1}$ in $S$ such that $v_p(a_i)$ all give the same remainder when divided by $q$. (Otherwise, because $v_p(n) \\mod q$ has $q$ distinct values, $S$ could contain at most $q(p-2)$ numbers, which is finite.) Call that remainder $r$.\nNow take some integer $b$ in $S$, not part of the other $q-1$ values. Together, their product must be a $q$th power, meaning \n$q|v_p(a_1a_2...a_{q-1}b)$ Thus, $v_p(a_1a_2...a_{q-1}b) \\equiv 0 \\pmod q$. It can be seen easily that for two positive integers $x$ and $y$ we have $v_p(ab) = v_p(a) + v_p(b)$, and by definition $v_p(a_i) \\equiv r \\pmod q$, so we can write $v_p(b) + (q-1)r \\equiv 0 \\pmod q$. Adding $r$ to both sides we have $v_p(b) \\equiv v_p(b) + qr \\equiv r \\pmod q$.\nThe choice of $b$ was arbitrary, meaning for all numbers $n$ in $S$, $v_p(n)$ modulo $q$ is the constant $r$, whether part of $a_1, a_2... a_{q-1}$ or not. The choice of $q$ was also arbitrary, thus for two integers $m$ and $n$ in $S$, $v_p(m) = v_p(n)$. Otherwise, we could pick $q$ larger than both values, and get a contradiction. Thus for all  numbers $n$ in $S$, $v_p(n)$ is constant. The choice of $p$ was also arbitrary, meaning any two distinct numbers in $S$ have the same prime factorization. By the fundamental theorem of arithmetic, this means the two numbers are the same, contradicting with them being distinct. Thus, no such $S$ exists.\n-Circling"
    },
    {
        "Problem": "$P$, $A$, $B$, $C$, and $D$ are five distinct points in space such that $\\angle APB = \\angle BPC = \\angle CPD = \\angle DPA = \\theta$, where $\\theta$ is a given acute angle. Determine the greatest and least values of $\\angle APC + \\angle BPD$.",
        "Solution": "Greatest value is achieved when all the points are as close as possible to all being on a plane.\nSince $\\theta < \\frac{\\pi}{2}$, then $\\angle APC + \\angle BPD < \\pi$\nSmallest value is achieved when point P is above and the remaining points are as close as possible to colinear when $\\theta > 0$, then $\\angle APC + \\angle BPD > 0$\nand the inequality for this problem is:\n$0 < \\angle APC + \\angle BPD < \\pi$\n~Tomas Diaz. orders@tomasdiaz.com\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "A difficult mathematical competition consisted of a Part I and a Part II with a combined total of $28$ problems. Each contestant solved $7$ problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who, in Part I, solved either no problems or at least four problems.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "$P(x)$ is a polynomial of degree $3n$ such that\n\nDetermine $n$.",
        "Solution": "By Lagrange Interpolation Formula $f(x) = 2\\sum_{p=0}^{n}\\left ( \\prod_{0\\leq r\\neq3p\\leq 3n}^{{}}\\frac{x-r}{3p-r} \\right )+ \\sum_{p=1}^{n}\\left ( \\prod_{0\\leq r\\neq3p-2\\leq 3n}^{{}} \\frac{x-r}{3p-2-r}\\right )$\nand hence $f(3n+1) = 2\\sum_{p=0}^{n}\\left ( \\prod_{0\\leq r\\neq3p\\leq 3n}^{{}}\\frac{3n+1-r}{3p-r} \\right )+ \\sum_{p=1}^{n}\\left ( \\prod_{0\\leq r\\neq3p-2\\leq 3n}^{{}} \\frac{3n+1-r}{3p-2-r}\\right )$\nafter some calculations we get $f(3n+1) =\\left ( \\binom{3n+1}{0}- \\binom{3n+1}{3}+\\binom{3n+1}{6}- ... \\right )\\left ( 2.(-1)^{3n}-1 \\right )+1$\nGiven $f(3n+1)= 730$ so we have to find $n$ such that $\\left ( \\binom{3n+1}{0}- \\binom{3n+1}{3}+\\binom{3n+1}{6}- ... \\right )\\left ( 2.(-1)^{3n}-1 \\right )= 729$\nLemma: If $p$ is even  $\\binom{p}{0}- \\binom{p}{3}+ \\binom{p}{6}- \\cdots = \\frac{2^{p+1}sin^{p}\\left ( \\frac{\\pi}{3} \\right )(i)^{p}\\left ( cos\\left ( \\frac{p\\pi}{3} \\right ) \\right )}{3}$\nand if  $p$ is odd  $\\binom{p}{0}- \\binom{p}{3}+ \\binom{p}{6}- \\cdots = \\frac{-2^{p+1}sin^{p}\\left ( \\frac{\\pi}{3} \\right )(i)^{p+1}\\left ( sin\\left ( \\frac{p\\pi}{3} \\right ) \\right )}{3}$\n$i$ is $\\sqrt{-1}$\nUsing above lemmas we do not get any solution when $n$ is odd, but when $n$ is even $3n+1=13$ satisfies the required condition, hence $n=4$",
        "Solution_2": "The (3n+1)th differences of the polynomial are zero. Call it p(x), so we have p(3n+1) - (3n+1)C1 p(3n) + (3n+1)C2 p(3n-1) - ... + (-1)3n+1 p(0) = 0, where rCs is the binomial coefficient. Hence p(3n+1) = 2( (3n+1)C1 - (3n+1)C4 + ... ) + ( (3n+1)C3 - (3n+1)C6 + ... ). Putting n = 1, we get: p(4) = 2( 4C1 - 4C4) + 4C3 = 6 + 4 = 10. So n is not 1. Putting n = 2, we get: p(7) = 2( 7C1 - 7C4 + 7C7) + ( 7C3 - 7C6) = 2( 7 - 35 + 1) + (35 - 7) = -26. So n is not 2. Putting n = 3, we get: p(10) = 2( 10C1 - 10C4 + 10C7 - 10C10) + ( 10C3 - 10C6 + 10C9) = 2(10 - 210 + 120 - 1) + (120 - 210 + 10) = -162 -100 = -262. So n is not 3. Putting n = 4, we get: p(13) = 2( 13C1 - 13C4 + 13C7 - 13C10 + 13C13) + ( 13C3 - 13C6 + 13C9 - 13C12) = 2(13 - 715 + 1716 - 286 + 1) + (286 - 1716 + 715 - 13) = 1458 - 728 = 730. So n = 3 works."
    },
    {
        "Problem": "Determine whether or not there are any positive integral solutions of the simultaneous equations \n\nwith distinct integers $x_1,x_2,\\cdots,x_{1985}$.",
        "Solution": "Lemma: For a positive integer $n$, $1^3+2^3+\\cdots +n^3 = (1+2+\\cdots +n)^2$ (Also known as Nicomachus's theorem)\nProof by induction: The identity holds for $1$. Suppose the identity holds for a number $n$. It is well known that the sum of first $n$ positive integers is $\\frac{n(n+1)}{2} = \\frac{n^2+n}{2}$. Thus its square is $\\frac{n^4+2n^3+n^2}{4}$. Adding $(n+1)^3=n^3+3n^2+3n+1$ to this we get $\\frac{n^4+6n^3+13n^2+12n+4}{4}$, which can be rewritten as $\\frac{(n^4+4n^3+6n^2+4n+1)+2(n^3+3n^2+3n+1)+(n^2+2n+1)}{4}$ This simplifies to $\\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4} = ({\\frac{(n+1)^2+(n+1)}{2}})^2 = (1+2+\\cdots +n+(n+1))^2$. The induction is complete.\nLet $j$ be the sum $1+2+\\cdots 1985$, and let $k$ be the sum $1^2 + 2^2 + \\cdots + 1985^2$. Then assign $x_i$ the value $ik^4$ for each $i = 1, 2,\\cdots 1985$. Then:\nThus, a positive integral solution exists.\n-Circling"
    },
    {
        "Problem": "Determine each real root of\n$x^4-(2\\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0$\ncorrect to four decimal places.",
        "Solution": "The equation can be re-written as\nWe first prove that the equation has no negative roots.\nLet $x\\le 0.$ The equation above can be further re-arranged as\n\nThe right hand side of the equation is negative. Therefore  and we have\n$-1<(x+10^5)(x-10^5) <2.$ Then the left hand side of the equation is bounded by\n\nHowever, since $|(x+10^5)(x-10^5)|\\le 2$ and $x<0,$ it follows that $|x+10^5| <\\frac{2}{|x-10^5|}<2\\times 10^{-5}$ for negative $x.$ Then $x<2\\times 10^{-5}-10^5.$ The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9.\nNow let $x>0.$ When $x=10^5,$ the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of $10^5$, as its leading coefficient is positive. We will prove that $x=10^5$ is a good approximation of the roots (within $10^{-2}$). In fact, we can solve the \"quadratic\" equation (1) for $(x+10^5)(x-10^5)$:\n\nThen \nEasy to see that\n$|x-10^5| <1$ for positve $x.$ Therefore, $10^5-1<x<10^5+1.$ Then\nLet $x_1$ be a root of the equation with $x_1<10^5.$ Then $0<10^5-x_1<10^{-2}$ and \n\nAn aproximation of $x_1$ is defined as follows:\n\nWe check the error of the estimate:\nThe first absolute value\nThe second absolute value \n\nthrough a rationalized numerator.Therefore $|\\tilde{x}_1-x_1|\\le 10^{-6}.$\nFor a real root $x_2$ with $x_2>10^5,$ we choose \n\nWe can similarly prove it has the desired approximation."
    },
    {
        "Problem": "Let $A,B,C,D$ denote four points in space such that at most one of the distances $AB,AC,AD,BC,BD,CD$ is greater than $1$. Determine the maximum value of the sum of the six distances.",
        "Solution": "Suppose that $AB$ is the length that is more than $1$. Let spheres with radius $1$ around $A$ and $B$ be $S_A$ and $S_B$. $C$ and $D$ must be in the intersection of these spheres, and they must be on the circle created by the intersection to maximize the distance. We have $AC + BC + AD + BD = 4$.\nIn fact, $CD$ must be a diameter of the circle. This maximizes the five lengths $AC$, $BC$, $AD$, $BD$, and $CD$. Thus, quadrilateral $ACBD$ is a rhombus.\nSuppose that $\\angle CAD = 2\\theta$. Then, $AB + CD = 2\\sin{\\theta} + 2\\cos{\\theta}$. To maximize this, we must maximize $\\sin{\\theta} + \\cos{\\theta}$ on the range $0^{\\circ}$ to $90^{\\circ}$. However, note that we really only have to solve this problem on the range $0^{\\circ}$ to $45^{\\circ}$, since $\\theta > 45$ is just a symmetrical function.\nFor $\\theta < 45$, $\\sin{\\theta} \\leq \\cos{\\theta}$. We know that the derivative of $\\sin{\\theta}$ is $\\cos{\\theta}$, and the derivative of $\\cos{\\theta}$ is $-\\sin{\\theta}$. Thus, the derivative of $\\sin{\\theta} + \\cos{\\theta}$ is $\\cos{\\theta} - \\sin{\\theta}$, which is nonnegative between $0^{\\circ}$ and $45^{\\circ}$. Thus, we can conclude that this is an increasing function on this range.\nIt must be true that $2\\sin{\\theta} \\leq 1$, so $\\theta \\leq 30^{\\circ}$. But, because $\\sin{\\theta} + \\cos{\\theta}$ is increasing, it is maximized at $\\theta = 30^{\\circ}$. Thus, $AB = \\sqrt{3}$, $CD = 1$, and our sum is $5 + \\sqrt{3}$.\n~mathboy100"
    },
    {
        "Problem": "There are $n$ people at a party. Prove that there are two people such that, of the remaining $n-2$ people, there are at least $\\lfloor n/2\\rfloor -1$ of them, each of whom knows both or else  knows neither of the two. Assume that \"know\" is a symmetrical relation; $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "Let $a_1,a_2,a_3,\\cdots$ be a non-decreasing sequence of positive integers. For $m\\ge1$, define $b_m=\\min\\{n: a_n \\ge m\\}$, that is, $b_m$ is the minimum value of $n$ such that $a_n\\ge m$. If $a_{19}=85$, determine the maximum value of $a_1+a_2+\\cdots+a_{19}+b_1+b_2+\\cdots+b_{85}$.",
        "Solution": "We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the $i$th column we fill the first $a_i$ cells with one dot each. Then the $19$th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. In row $j$, we see that the number of blank cells is equal to $b_j-1$. Therefore the number of filled cells in the first 19 columns of row $j$ is equal to $20-b_j$.\nWe now count the number of cells in the first 19 columns of our array, but we do it in two different ways. First, we can sum the number of dots in each column: this is simply $a_1+\\cdots+a_{19}$. Alternatively, we can sum the number of dots in each row: this is $(20-b_1)+\\cdots +(20-b_{85})$. Since we have counted the same number in two different ways, these two sums must be equal. Therefore\n\nNote that this shows that the value of the desired sum is constant."
    },
    {
        "Problem": "$(\\text{a})$ Do there exist 14 consecutive positive integers each of which is divisible by one or more primes $p$ from the interval $2\\le p \\le 11$?\n$(\\text{b})$ Do there exist 21 consecutive positive integers each of which is divisible by one or more primes $p$ from the interval $2\\le p \\le 13$?",
        "Solution": "(a)     To solve part (a), we first note that for any 14 consecutive positive integers, exactly 7 are even (divisible by 2) and therefore satisfy the criteria. We can remove these from the problem, and simplify it to the following question, which is equivalent to part (a):\n\"Do there exist 7 consecutive positive odd integers each of which is divisible by one or more primes $p$ from the interval $3\\le p \\le 11$?\"\nAmong any 7 consecutive positive odd integers, the following holds:\n\n\n\n\nFor every one of these seven integers to be divisible by one of 3, 5, 7, or 11, there must be 3 multiples of 3, 2 multiples of 5, 1 multiple of 7, and 1 multiple of 11. Additionally, none of these integers may be a multiple of any two of the four aforementioned primes. Otherwise, the Pigeonhole Principle dictates that at least one integer is not divisible by any of the four, thus failing to meet the criteria. But this cannot be. Calling the consecutive odd integers $a_1, a_2, \\dots, a_7$, we note that $a_1$, $a_4$, and $a_7$ must be multiples of 3, and therefore cannot be multiples of 5, 7, or 11. But for there to be two multiples of 5, they must be one of the two pairs $(a_1,a_6)$ and $(a_2,a_7)$. But each of these pairs contains a multiple of 3, and so at least one of the 7 odd integers is divisible by none of the primes 3, 5, 7, or 11. Therefore the answer to part (a) is no.\n(b)    To solve part (b), we use a strategy similar to the one used in part (a), reducing part (b) to this equivalent question:\n\"Do there exist 10 consecutive positive odd integers each of which is divisible by one or more primes $p$ from the interval $3\\le p \\le 13$?\"\n(We will ignore the case where the first of the 21 integers is odd, resulting in 11 odd integers instead of 10, as it is only true if the weaker, 10-integer argument is also true.)\nAmong any 10 consecutive positive odd integers, the following holds:\n\n\n\n\n\nFor every one of these ten integers to be divisible by one of 3, 5, 7, 11, or 13, there must be 4 multiples of 3, 2 multiples of 5, 2 multiples of 7, 1 multiple of 11, and 1 multiple of 13. As before, none of these integers may be a multiple of any two of the five primes. Calling the consecutive odd integers $a_1, a_2, \\dots, a_{10}$, we note that $a_1$, $a_4$, $a_7$, and $a_{10}$ must be multiples of 3, and therefore cannot be multiples of 5, 7, 11, or 13. Unlike part (a), however, this stipulation is not a dead end. Let the multiples of 5 be $a_3$ and $a_8$ and let the multiples of 7 be $a_2$ and $a_9$. The multiples of 11 and 13 are $a_5$ and $a_6$, in some order (it doesn't really matter which). An example of a sequence of 21 consecutive positive integers satisfying part (b) is the integers from 9440 to 9460 (inclusive), which can be obtained by solving modular equations that result from these statements. So the answer to part (b) is yes."
    },
    {
        "Problem": "During a certain lecture, each of five mathematicians fell asleep exactly twice. For each pair of mathematicians, there was some moment when both were asleep simultaneously. Prove that, at some moment, three of them were sleeping simultaneously.",
        "Solution": "We shall assume to the contrary that there was never a time when three mathematicians were sleeping simultaneously, and derive a contradiction.\nAs a subtle but logically necessary note, we will assume without loss of generality that no two events (an event is one mathematician either falling asleep or waking up) happen at the same time.\nWe have (5 choose 2) = 10 sleeping-pairs of mathematicians to account for.  Also, since 5 mathematicians each fell asleep 2 times, we have a total of 10 occasions on which a mathematician fell asleep.  Now, if two mathematicians are ever asleep simultaneously, then one of them fell asleep while the other was already asleep.  However, because of the \"no three asleep at once\" rule, if a mathematician falls asleep, then at most one other mathematician could have been asleep already.  Therefore, each occasion when a mathematician falls asleep can account for at most one sleeping-pair.  It would appear that we have just enough to make it.  However, when the first mathematician falls asleep, no other mathematicians are asleep...\nHey, wait a minute.  Couldn't one mathematician, or even two, have been already asleep when the lecture started?  In fact, the wording of the problem does not forbid this, and this permits us to construct an easy counterexample:\nCall the mathematicians A,B,C,D,E.  A starts out asleep.  B sleeps; A wakes; C sleeps; B wakes; D sleeps; C wakes; E sleeps; D wakes; A sleeps; E wakes; then, C sleeps; A wakes; E sleeps; C wakes; B sleeps; E wakes; D sleeps; B wakes; A sleeps; D wakes.  This creates, in order, the sleeping-pairs AB, BC, CD, DE, EA, then AC, CE, EB, BD, DA; each mathematician falls asleep and wakes up exactly twice, and at no time are three mathematicians asleep.\nWell, this must be considered a hole in the problem as written.  However, if we add the (nontrivial) assumption that the mathematicians were all awake when the lecture began, then it follows that the first occasion when a mathematician falls asleep can only account for zero sleeping-pairs, and the remaining 9 occasions can only account for 9 of 10 sleeping-pairs, and so there must be some pair of mathematicians that are not simultaneously asleep during the lecture.  So our assumption to the contrary implies an impossible sequence of events.  Therefore, that assumption must be wrong, and there must be a moment when three mathematicians are asleep."
    },
    {
        "Problem": "What is the smallest integer $n$, greater than one, for which the root-mean-square of the first $n$ positive integers is an integer?\n$\\mathbf{Note.}$ The root-mean-square of $n$ numbers $a_1, a_2, \\cdots, a_n$ is defined to be",
        "Solution": "Let's first obtain an algebraic expression for the root mean square of the first $n$ integers, which we denote $I_n$. By repeatedly using the identity\n$(x+1)^3 = x^3 + 3x^2 + 3x + 1$, we can write\n\n \nand\n We can continue this pattern indefinitely, and thus for any\npositive integer $n$,\n\nSince $\\sum_{j=1}^n j  = n(n+1)/2$, we obtain\n\nTherefore,\n\nRequiring that $I_n$ be an integer, we find that\n where $k$ is an integer. Using the Euclidean algorithm, we see that\n$\\gcd(2n+1, n+1) = \\gcd(n+1,n) = 1$, and so $2n+1$ and $n+1$ share no\nfactors greater than 1. The equation above thus implies that $2n+1$\nand $n+1$ is each proportional to a perfect square. Since $2n+1$ is\nodd, there are only two possible cases:\nCase 1: $2n+1 = 3 a^2$ and $n+1 = 2b^2$, where $a$ and $b$ are integers.\nCase 2: $2n+1 = a^2$ and $n+1 = 6b^2$.\nIn Case 1,  $2n+1 = 4b^2 -1 = 3a^2$. This means that $(4b^2 -1)/3 = a^2$ for some integers $a$ and $b$. We proceed by checking whether\n$(4b^2-1)/3$ is a perfect square for $b=2, 3, 4, \\dots$. (The solution\n$b=1$ leads to $n=1$, and we are asked to find a value of $n$ \ngreater than 1.) The smallest positive integer $b$ greater than 1 for\nwhich $(4b^2-1)/3$ is a perfect square is $b=13$, which results in $n=337$.\nIn Case 2, $2n+1 = 12b^2 - 1 = a^2$. Note that $a^2 = 2n+1$ is an odd square, and hence is congruent to $1 \\pmod 4$. But $12b^2 -1 \\equiv 3 \\pmod 4$ for any $b$, so Case 2 has no solutions.\nAlternatively, one can proceed by checking whether $12b^2 -1$ is a perfect square for $b=1, 2 ,3 ,\\dots$. We find that $12b^2 -1$ is not a perfect square for $b = 1,2, 3, ..., 7, 8$, and $n= 383$ when $b=8$. Thus the smallest positive integers $a$ and $b$ for which $12b^2- 1 = a^2$ result in a value of $n$ exceeding the value found in Case 1, which was 337.\nIn summary, the smallest value of $n$ greater than 1 for which $I_n$ is an integer is $\\boxed{337}$."
    },
    {
        "Problem": "Two distinct circles $K_1$ and $K_2$ are drawn in the plane. They intersect at points $A$ and $B$, where $AB$ is the diameter of $K_1$. A point $P$ on $K_2$ and inside $K_1$ is also given.\nUsing only a \"T-square\" (i.e. an instrument which can produce a straight line joining two points and the perpendicular to a line through a point on or off the line), find a construction for two points $C$ and $D$ on $K_1$ such that $CD$ is perpendicular to $AB$ and $\\angle CPD$ is a right angle.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "By a partition $\\pi$ of an integer $n\\ge 1,$ we mean here a representation of $n$ as a sum of one or more positive integers where the summands must be put in nondecreasing order. (E.g., if $n=4,$ then the partitions $\\pi$ are $1+1+1+1,$ $1+1+2,$ $1+3, 2+2,$ and $4$).\nFor any partition $\\pi,$ define $A(\\pi)$ to be the number of $1$'s which appear in $\\pi,$ and define $B(\\pi)$ to be the number of distinct integers which appear in $\\pi$ (E.g., if $n=13$ and $\\pi$ is the partition $1+1+2+2+2+5,$ then $A(\\pi)=2$ and $B(\\pi) = 3$).\nProve that, for any fixed $n,$ the sum of $A(\\pi)$ over all partitions of $\\pi$ of $n$ is equal to the sum of $B(\\pi)$ over all partitions of $\\pi$ of $n.$",
        "Solution": "Let $S(n) = \\sum\\limits_{\\pi} A(\\pi)$ and let $T(n) = \\sum\\limits_{\\pi} B(\\pi).$ We will use generating functions to approach this problem -- specifically, we will show that the generating functions of $S(n)$ and $T(n)$ are equal.\nLet us start by finding the generating function of $S(n).$ This function counts the total number of 1's in all the partitions of $n.$ Another way to count this is by counting the number of partitions of $n$ that contain $x$ 1's and multiplying this by $x,$ then summing for $1\\leq x \\leq n.$ However, the number of partitions of $n$ that contain $x$ 1's is the same as the number of partitions of $n-x$ that contain no 1's, so\n\nThe number of partitions of $m$ with no 1's is the coefficient of $x^m$ in\n\nNote that there is no $(1+x+x^2+x^3+\\ldots)$ term in $F(x)$ because we cannot have any 1's in the partition.\nLet $c_m$ be the coefficient of $x^m$ in the expansion of $F(x),$ so we can rewrite it as $F(x)=c_0+c_1x+c_2x^2+c_3x^3+\\ldots.$ We wish to compute $S(n)=1\\cdot c_{n-1}+2\\cdot c_{n-2}+\\ldots+n\\cdot c_0.$\nConsider the power series $G(x)=(x+2x^2+3x^3+4x^4+\\ldots)F(x).$\n\nIf we expand the first line, we see that the coefficient of $x^n$ in $G(x)$ for any $n$ is $1\\cdot c_{n-1}+2\\cdot c_{n-2}+\\ldots+n\\cdot c_0,$ which is exactly $S(n)$! So by definition, $G(x)=\\frac{x}{1-x} \\prod\\limits_{i=1}^{\\infty}\\frac{1}{1-x^i}$ is the generating function of $S(n).$\nNow let's find the generating function of $T(n).$ Notice that counting the number of distinct elements in each partition and summing over all partitions is equivalent to counting how many partitions of $n$ contain i and then summing for all $1\\leq i \\leq n.$\nSo,\n\nHowever, the number of partitions of n that contain a $i$ is the same as the total number of partitions of $n-i,$ so\nThe generating function for the number of partitions is\nLet's write the expansion of $P(x)$ as $P(x)=d_0+d_1x+d_2x^2+d_3x^3+\\ldots,$ so we wish to find $T(n)=d_0+d_1+d_2+\\ldots+d_{n-1}.$\nConsider the power series $H(x)=(x+x^2+x^3+x^4+\\ldots)P(x).$\n\nIf we expand the first line, we see that the coefficient of $x^n$ in $H(x)$ for any $n$ is $d_{n-1}+d_{n-2}+\\ldots+d_0,$ which is precisely $T(n).$ This means $H(x)$ is the generating function of $T(n).$\nThus, the generating functions of $S(n)$ and $T(n)$ are the same, so $S(n)=T(n)$ for all $n$ and we are done.\n~Peggy"
    },
    {
        "Problem": "Find all solutions to $(m^2+n)(m + n^2)= (m - n)^3$, where m and n are non-zero integers.",
        "Solution": "Expanding both sides, \nNote that $m^3$ can be canceled and as $n \\neq 0$, $n$ can be factored out.\nWriting this as a quadratic equation in $n$: .\nThe discriminant $b^2-4ac$ equals \n, which we want to be a perfect square.\nMiraculously, this factors as $m(m-8)(m+1)^2$. This is square iff (if and only if) $m^2-8m$ is square or $m+1=0$. It can be checked that the only nonzero $m$ that work are $-1, 8, 9$. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs $(m, n)$ as ."
    },
    {
        "Problem": "The feet of the angle bisectors of $\\Delta ABC$ form a right-angled triangle. If the right-angle is at $X$, where $AX$ is the bisector of $\\angle A$, find all possible values for $\\angle A$.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "$X$ is the smallest set of polynomials $p(x)$ such that:\nShow that if $r(x)$ and $s(x)$ are distinct elements of $X$, then $r(x) \\neq s(x)$ for any $0 < x < 1$.",
        "Solution": "Let $s(x)$ be an arbitrary polynomial in $X.$ Then $0<s(x)<1$ when $0<x<1.$ Define $X_1=\\{s(x)\\in X:s(x)=x\\cdot s_1(x)$ for some $s_1(x)\\in X\\},$ and $X_2=\\{t(x)\\in X: t(x)=x+(1-x)t_1(x)$ for some $t_1(x)\\in X\\}.$\nIf $s(x) \\in X_1$ and $t(x)\\in X_2,$ we have $s(x) <x<t(x)$ for all $x$ with $0<x<1.$ Therefore $s(x)\\ne t(x)$ for any $0<x<1.$\nFor any $s(x), t(x) \\in X_1$, Let $s(x)=x\\cdot s_1(x)$ and $t(x)=x\\cdot t_1(x)$ for $s_1(x), t_1(x) \\in X.$ If $s_1(x) \\ne t_1(x)$ for $0<x<1,$ then\n$s(x)-t(x)=x(s_1(x)-t_1(x))\\ne 0$ for $0<x<1.$\nSimilarly, for any $s(x), t(x) \\in X_2$, Let $s(x)=x+(1-x) s_1(x)$ and $t(x)=x+(1-x) t_1(x)$ for $s_1(x), t_1(x) \\in X.$ If $s_1(x) \\ne t_1(x)$ for $0<x<1,$ then\n$s(x)-t(x)=(1-x)(s_1(x)-t_1(x))\\ne 0$ for $0<x<1.$\nThe proof is done by an induction.\nJ.Z."
    },
    {
        "Problem": "M is the midpoint of XY. The points P and Q lie on a line through Y on opposite sides of Y, such that $|XQ| = 2|MP|$ and $\\frac{|XY|}2 < |MP| < \\frac{3|XY|}2$. For what value of $\\frac{|PY|}{|QY|}$ is $|PQ|$ a minimum?",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "$a_1, a_2, \\cdots, a_n$ is a sequence of 0's and 1's. T is the number of triples $(a_i, a_j, a_k)$ with $i<j<k$ which are not equal to (0, 1, 0) or (1, 0, 1). For $1\\le i\\le n$, $f(i)$ is the number of $j<i$ with $a_j = a_i$ plus the number of $j>i$ with $a_j\\neq a_i$. Show that $T=\\sum_{i=1}^n f(i)\\cdot\\left(\\frac{f(i)-1}2\\right)$. If n is odd, what is the smallest value of T?",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "The repeating decimal $0.ab\\cdots k\\overline{pq\\cdots u}=\\frac mn$, where $m$ and $n$ are relatively prime integers, and there is at least one decimal before the repeating part. Show that $n$ is divisble by 2 or 5 (or both). (For example, $0.011\\overline{36}=0.01136363636\\cdots=\\frac 1{88}$, and 88 is divisible by 2.)",
        "Solution_1": "First, split up the nonrepeating parts and the repeating parts of the decimal, so that the nonrepeating parts equal to $\\frac{a}{b}$ and the repeating parts of the decimal is equal to $\\frac{c}{d}$.\nSuppose that the length of  $0.ab\\cdots k$ is $p$ digits. Then $\\frac{a}{b} = \\frac{0.ab\\cdots k}{10^{p+1}}$ Since $0.ab\\cdots k < 10^{p+1}$, after reducing the fraction, there MUST be either a factor of 2 or 5 remaining in the denominator. After adding the fractions $\\frac{a}{b}+\\frac{c}{d}$, the simplified denominator $n$ will be $\\\\lcm(b,d)$ and since $b$ has a factor of $2$ or $5$,  $n$ must also have a factor of 2 or 5.\n$\\blacksquare$",
        "Solution_2": "It is well-known that $0.ab...k \\overline{pq...u} = \\frac{ab...u - ab...k}{99...900...0}$, where there are a number of 9s equal to the count of digits in $pq...u$, and there are a number of 0s equal to the count of digits $c$ in $ab...k$. Obviously $ab...k$ is different from $pq...u$ (which is itself the repeating part), so the numerator cannot have $c$ consecutive terminating zeros, and hence the denominator still possesses a factor of 2 or 5 not canceled out. This completes the proof."
    },
    {
        "Problem": "The cubic polynomial $x^3+ax^2+bx+c$ has real coefficients and three real roots $r\\ge s\\ge t$. Show that $k=a^2-3b\\ge 0$ and that $\\sqrt k\\le r-t$.",
        "Solution_1": "By Vieta's Formulas, $a=-r-s-t$, $b=rs+st+rt$, and $c=-rst$.\nNow we know $k=a^2-3b$; in terms of r, s, and t, then, \n\n\nNow notice that we can multiply both sides by 2, and rearrange terms to get\n$2k=(r-s)^2+(s-t)^2+(r-t)^2$.\nBut since $r, s, t\\in \\mathbb{R}$, the three terms of the RHS are all non-negative (as the square of a real number is always non-negative), and therefore their sum is also non-negative -- that is, $2k\\ge 0 \\Rightarrow k\\ge 0$.\nNow, we will show that $\\sqrt k\\le r-t$.\nWe can square both sides, and the inequality will hold since they are both non-negative (it is given that $r\\ge t$, therefore $r-t\\ge 0$). This gives $k \\le r^2-2rt+t^2$.\nNow we already have $k=r^2+s^2+t^2-rs-st-rt$, so substituting this for k gives\n\n\n\nNote that this is a quadratic. Since its leading coefficient is positive, its value is less than 0 when s is between the two roots. Using the quadratic formula:\n\n\n\n\nThe quadratic is 0 when s is equal to r or t, and the inequality\nholds when its value is less than or equal to 0 -- that is, $r\\ge s\\ge t$.\n(Its value is less than or equal to 0 when s is between the roots, since the\ngraph of the quadratic opens upward.)\nIn fact, the problem tells us this is true. Q.E.D.",
        "Solution_2": "From Vieta's Formula (which tells us that $a = -(r+s+t)$ and $b = rs + st + rt$), we have that\n\nclearly non-negative. To prove $\\sqrt{k} \\le r - t$, it suffices to prove the square of this relation, or  This in turn simplifies to  or  which is clearly true as $r \\ge s \\ge t$. This completes the proof.",
        "Solution_3": "By Vieta's Formulas, $a = -(r+s+t)$ and $b = rs + st + rt$. $k = (r+s+t)^2 - 3(rs + st + rt) = r^2+s^2+t^2-rs-st-rt = \\frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2}$.\nTo show that $k \\ge 0$, simply note that by the trivial inequality, all three squares are greater than $0$ as they are the squares of real numbers.\nTo show that $\\sqrt{k} \\le r-t$, since both are positive, it is sufficient to show that $k \\le (r-t)^2$. $\\frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2} \\le (r-t)^2$ implies that $k \\le (r-t)^2$. $\\frac{(r-s)^2 + (s-t)^2 - (r-t)^2}{2} \\le 0$. Let $y = r-s$ and $z = s-t$. We then have $\\frac{y^2 + z^2 - (y+z)^2}{2} \\le 0 \\implies -2yz \\le 0$, which is clearly true as both $y$ and $z$ are positive."
    },
    {
        "Problem": "Let $X$ be the set $\\{ 1, 2, \\cdots , 20\\}$ and let $P$ be the set of all 9-element subsets of $X$. Show that for any map $f: P\\mapsto X$ we can find a 10-element subset $Y$ of $X$, such that $f(Y-\\{k\\})\\neq k$ for any $k$ in $Y$.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "$\\Delta ABC$ is a triangle with incenter $I$. Show that the circumcenters of $\\Delta IAB$, $\\Delta IBC$, and $\\Delta ICA$ lie on a circle whose center is the circumcenter of $\\Delta ABC$.",
        "Solution_1": "Let the circumcenters of $\\Delta IAB$, $\\Delta IBC$, and $\\Delta ICA$ be $O_c$, $O_a$, and $O_b$, respectively. It then suffices to show that $A$, $B$, $C$, $O_a$, $O_b$, and $O_c$ are concyclic.\nWe shall prove that quadrilateral $ABO_aC$ is cyclic first. Let $\\angle BAC=\\alpha$, $\\angle CBA=\\beta$, and $\\angle ACB=\\gamma$. Then $\\angle ICB=\\gamma/2$ and $\\angle IBC=\\beta/2$. Therefore minor arc $\\overarc{BIC}$ in the circumcircle of $IBC$ has a degree measure of $\\beta+\\gamma$. This shows that $\\angle CO_aB=\\beta+\\gamma$, implying that $\\angle BAC+\\angle BO_aC=\\alpha+\\beta+\\gamma=180^{\\circ}$. Therefore quadrilateral $ABO_aC$ is cyclic.\nThis shows that point $O_a$ is on the circumcircle of $\\Delta ABC$. Analagous proofs show that $O_b$ and $O_c$ are also on the circumcircle of $ABC$, which completes the proof. $\\blacksquare$",
        "Solution_2": "Let $M$ denote the midpoint of arc $AC$. It is well known that $M$ is equidistant from $A$, $C$, and $I$ (to check, prove $\\angle IAM = \\angle AIM = \\frac{\\angle BAC + \\angle ABC}{2}$), so that $M$ is the circumcenter of $AIC$. Similar results hold for $BIC$ and $CIA$, and hence $O_c$, $O_a$, and $O_b$ all lie on the circumcircle of $ABC$.",
        "Solution_3": "Extend $CI$ to point $L$ on $(ABC)$. By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle $IAB$. In other words, $L=O_c$, so $O_c$ is on $(ABC)$. Similarly, we can show that $O_a$ and $O_b$ are on $(ABC)$, and thus, $A,B,C,O_a,O_b,O_c$ are all concyclic. It follows that the circumcenters are equal.",
        "Solution_4": "Let the centers be $T, R, S$. We want to show that $ABC$ and $TRS$ have the same circumcircle. By Fact 5 we know that $CATB$ lie on a circle and similarly with the others. Thus the two triangles have the same circumcircle.\n~coolmath_2018"
    },
    {
        "Problem": "Let $p(x)$ be the polynomial $(1-x)^a(1-x^2)^b(1-x^3)^c\\cdots(1-x^{32})^k$, where $a, b, \\cdots, k$ are integers. When expanded in powers of $x$, the coefficient of $x^1$ is $-2$ and the coefficients of $x^2$, $x^3$, ..., $x^{32}$ are all zero. Find $k$.",
        "Solution_1": "First, note that if we reverse the order of the coefficients of each factor, then we will obtain a polynomial whose coefficients are exactly the coefficients of $p(x)$ in reverse order. Therefore, if\n\nwe define the polynomial $q(x)$ to be\n\nnoting that if the polynomial has degree $n$, then the coefficient of $x^{n-1}$ is $-2$, while the coefficients of $x^{n-k}$ for $k=2,3,\\dots, 32$ are all $0$.\nLet $P_n$ be the sum of the $n$th powers of the roots of $q(x)$. In particular, by Vieta's formulas, we know that $P_1=2$. Also, by Newton's Sums, as the coefficients of $x^{n-k}$ for $k=2,3,\\dots,32$ are all $0$, we find that\n\nThus $P_n=2^n$ for $n=1,2,\\dots, 32$. Now we compute $P_{32}$. Note that the roots of $(x^n-1)^{a_n}$ are all $n$th roots of unity. If $\\omega=e^{2\\pi i/n}$, then the sum of $32$nd powers of these roots will be\n\nIf $\\omega^{32}\\ne 1$, then we can multiply by $(\\omega^{32}-1)/(\\omega^{32}-1)$ to obtain\n\nBut as $\\omega^n=1$, this is just $0$. Therefore the sum of the $32$nd powers of the roots of $q(x)$ is the same as the sum of the $32$nd powers of the roots of\n\nThe $32$nd power of each of these roots is just $1$, hence the sum of the $32$nd powers of the roots is\n\nOn the other hand, we can use the same logic to show that\n\nSubtracting (2) from (1) and dividing by 32, we find\n\nTherefore, $a_{32}=2^{27}-2^{11}$.\n",
        "Solution_2": "By a limiting process, we can extend the problem to that of finding a sequence $b_1, b_2, \\ldots$ of integers such that\n\n(The notation comes from the Alcumus version of this problem.)\nIf we take logarithmic derivatives on both sides, we get\n\nand upon multiplying both sides by $-z$, this gives us the somewhat simple form\n\nExpanding all the fractions as geometric series, we get\n\nComparing coefficients, we get\n\nfor all positive integers $n$. In particular, as in Solution 1, we get\n\nfrom which the answer $b_{32} = 2^{27} - 2^{11}$ follows.\nRemark: To avoid the question of what an infinite product means in the context of formal power series, we could instead view the problem statement as saying that\n\nmodular arithmetic for polynomials can be defined in exactly the same way as modular arithmetic for integers. Uniqueness of the $b_n$'s comes from the fact that we have\n\nfor all $n\\leq 33$ by further reduction modulo $z^n$ (as $z^n\\mid z^{33}$ for $n\\leq 33$), so we could uniquely solve for the $b_n$'s one at a time. (This idea can be pushed further to explain why it's fine to pass to the infinite product version of the problem.)\nTo convert the above solution to one that works with polynomials modulo $z^{33}$, note that the derivative is not well-defined, as for instance, $1$ and $1 + z^{33}$ are equivalent modulo $z^{33}$, but their derivatives, $0$ and $33z^{32}$, are not. However, the operator $f(z)\\mapsto zf'(z)$ is well-defined. The other key idea is that for any $n$, modulo $z^n$, polynomials of the form $1 - zf(z)$ are invertible, with inverse\n\nTherefore, for the polynomial in the problem, call it $g(z)$, we can still form the expression $zg'(z)/g(z)$, which is what we originally got by taking the logarithmic derivative and multiplying by $z$, and expand it to eventually get\n\nwhich gets us the same relations (for $n\\leq 32$).",
        "Solution_3": "From the starting point of Solution 2,\n\ntaking reciprocals and expanding with geometric series gives us\n\nOn the right, we have the generating function for the number of monic polynomials of degree $n$ over the field $\\mathbb{F}_2$ of two elements, and on the left, we have the factorisation of this generating function that considers the breakdown of any given monic polynomial into monic irreducible factors. As such, we have the interpretation\n\nFrom here, to determine $b_n$, we analyse the elements of $\\mathbb{F}_{2^n}$, of which there are $2^{n}$ in total. Given $\\alpha\\in\\mathbb{F}_{2^n}$, if the minimal polynomial $f_{\\alpha}$ of $\\alpha$ has degree $d$, then $d\\mid n$ and all other roots of $f_{\\alpha}$ appear in $\\mathbb{F}_{2^n}$. Moreover, if $d\\mid n$ and $f$ is an irreducible polynomial of degree $d$, then all roots of $f$ appear in $\\mathbb{F}_{2^n}$. (These statements are all well-known in the theory of finite fields.) As such, for each $d\\mid n$, there are precisely $db_d$ elements of $\\mathbb{F}_{2^n}$ of degree $d$, and we obtain the same equation as in Solution 2,\n\nThe rest is as before.\n"
    },
    {
        "Problem": "For each positive integer $n$, let\n\nFind, with proof, integers $0 < a,\\ b,\\ c,\\ d < 1000000$ such that $T_{1988} = a S_{1989} - b$ and $U_{1988} = c S_{1989} - d$.",
        "Solution": "We note that for all integers $n \\ge 2$,\nIt then follows that\nIf we let $n=1989$, we see that $(a,b,c,d) = (1989,1989,1990, 2\\cdot 1989)$ is a suitable solution.  $\\blacksquare$\nNotice that it is also possible to use induction to prove the equations relating $T_n$ and $U_n$ with $S_n$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "The 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players.",
        "Solution_1": "Consider a graph with $20$ vertices and $14$ edges. The sum of the degrees of the vertices is $28$; by the Pigeonhole Principle at least $12$ vertices have degrees of $1$ and at most $8$ vertices have degrees greater than $1$. If we keep deleting edges of vertices with degree greater than $1$ (a maximum of $8$ such edges), then we are left with at least $6$ edges, and all of the vertices have degree either $0$ or $1$. These $6$ edges represent the $6$ games with $12$ distinct players.",
        "Solution_2": "Let a slot be a place we can put a member in a game, so there are two slots per game, and 28 slots total.  We begin by filling exactly 20 slots each with a distinct member since each member must play at least one game.  Let there be $m$ games with both slots filled and $n$ games with only one slot filled, so $2m+n=20$.  Since there are only 14 games, $m+n \\leq 14 \\Longrightarrow 2m+n \\leq 14+m \\Longleftrightarrow 20 \\leq 14+m \\Longrightarrow m \\geq 6$, so there must be at least 6 games with two distinct members each, and we must have our desired set of 6 games.",
        "Solution_3": "Assume the contrary.\nConsider the largest set of disjoint edges $E$. By assumption it has less than $6$ edges, i.e. maximum $10$ vertices. \nCall it a vertex set $V$.\n$10$ vertices remain outside $V$ and each has to be attached to at least one edge. \nNow, if any two vertices outside $V$ are connected by, say, edge $e$, we could have included $e$ in $E$ and gotten a larger disjoint set, so - a contradiction. \nTherefore the only option would be that all vertices outside $V$ are connected each by one edge to some vertices inside $V$. That would take $10$ edges, but $E$ already includes $5$ - again a contradiction.\nAll possibilities yield a contradiction, so our assumption can not be correct.\n(Cases when largest set $E$ is smaller than $6$ are equivalent and weaker)"
    },
    {
        "Problem": "Let $P(z)= z^n + c_1 z^{n-1} + c_2 z^{n-2} + \\cdots + c_n$ be a polynomial in the complex variable $z$, with real coefficients $c_k$. Suppose that $|P(i)| < 1$. Prove that there exist real numbers $a$ and $b$ such that $P(a + bi) = 0$ and $(a^2 + b^2 + 1)^2 < 4 b^2 + 1$.",
        "Solution": "Let $z_1, \\dotsc, z_n$ be the (not necessarily distinct) roots of $P$, so that\n\nSince all the coefficients of $P$ are real, it follows that if $w$ is a root of $P$, then $P( \\overline{w}) = \\overline{ P(w)} = 0$, so $\\overline{w}$, the complex conjugate of $w$, is also a root of $P$.\nSince\n\nit follows that for some (not necessarily distinct) conjugates $z_i$ and $z_j$,\n\nLet $z_i = a+bi$ and $z_j = a-bi$, for real $a,b$.  We note that\n\nThus\n\nSince $P(a+bi) = P(z_i) = 0$, these real numbers $a,b$ satisfy the problem's conditions.  $\\blacksquare$"
    },
    {
        "Problem": "Let $ABC$ be an acute-angled triangle whose side lengths satisfy the inequalities $AB < AC < BC$. If point $I$ is the center of the inscribed circle of triangle $ABC$ and point $O$ is the center of the circumscribed circle, prove that line $IO$ intersects segments $AB$ and $BC$.",
        "Solution": "Consider the lines that pass through the circumcenter $O$. Extend $AO$, $BO$, $CO$ to $D$,$E$,$F$ on $a$,$b$,$c$, respectively.\nWe notice that $IO$ passes through sides $a$ and $c$ if and only if $I$ belongs to either regions $AOF$ or $COD$.\n\nSince $AO = BO = CO = R$, we let $\\alpha = \\angle OAC = \\angle OCA$, $\\beta = \\angle BAO = \\angle ABO$, $\\gamma = \\angle BCO = \\angle CBO$.\nWe have $c < b < a\\implies C < B < A\\implies$ $\\gamma+\\alpha <\\beta+\\gamma <\\alpha+\\beta\\implies\\gamma <\\alpha <\\beta$\nSince $IA$ divides angle $A$ into two equal parts, it must be in the region marked by the $\\beta$ of angle $A$, so $I$ is in $ABD$.\nSimilarly, $I$ is in $ACF$ and $ABE$. Thus, $I$ is in their intersection, $AOF$. From above, we have $IO$ passes through $a$ and $c$. $\\blacksquare$"
    },
    {
        "Problem": "Let $u$ and $v$ be real numbers such that\n\nDetermine, with proof, which of the two numbers, $u$ or $v$, is larger.",
        "Solution": "The answer is $v$.\nWe define real functions $U$ and $V$ as follows:\n\nWe wish to show that if $U(u)=V(v)=8$, then $u <v$.\nWe first note that when $x \\le 0$, $x^{12}-x \\ge 0$, $x-1 < 0$, and $9x^9 \\le 0$, so\n\nSimilarly, $V(x) \\le 0 < 8$.\nWe also note that if $x \\ge 9/10$, then\n\nSimilarly $V(x) > 8$.  It then follows that $u, v \\in (0,9/10)$.\nNow, for all $x \\in (0,9/10)$,\n\nSince $V$ and $U$ are both strictly increasing functions over the nonnegative reals, it then follows that\n\nso $u<v$, as desired.  $\\blacksquare$"
    },
    {
        "Problem": "A certain state issues license plates consisting of six digits (from 0 through 9). The state requires that any two plates differ in at least two places. (Thus the plates $\\boxed{027592}$ and $\\boxed{020592}$ cannot both be used.) Determine, with proof, the maximum number of distinct license plates that the state can use.",
        "Solutions": "Consider license plates of $n$ digits, for some fixed $n$, issued with the same criteria.\nWe first note that by the pigeonhole principle, we may have at most $10^{n-1}$ distinct plates.  Indeed, if we have more, then there must be two plates which agree on the first $n-1$ digits; these plates thus differ only on one digit, the last one.\nWe now show that it is possible to issue $10^{n-1}$ distinct license plates which satisfy the problem's criteria.  Indeed, we issue plates with all $10^{n-1}$ possible combinations for the first $n-1$ digit, and for each plate, we let the last digit be the sum of the preceding digits taken mod 10.  This way, if two plates agree on the first $n-1$ digits, they agree on the last digit and are thus the same plate, and if two plates differ in only one of the first $n-1$ digits, they must differ as well in the last digit.\nIt then follows that $10^{n-1}$ is the greatest number of license plates the state can issue.  For $n=6$, as in the problem, this number is $10^5$.  $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "A sequence of functions $\\, \\{f_n(x) \\} \\,$ is defined recursively as follows:\n\n(Recall that $\\sqrt {\\makebox[5mm]{}}$ is understood to represent the positive square root.) For each positive integer $n$, find all real solutions of the equation $\\, f_n(x) = 2x \\,$.",
        "Solution": "We define $f_0(x) = 8$.  Then the recursive relation holds for $n=0$, as well.\nSince $f_n (x) \\ge 0$ for all nonnegative integers $n$, it suffices to consider nonnegative values of $x$.\nWe claim that the following set of relations hold true for all natural numbers $n$ and nonnegative reals $x$:\n\nTo prove this claim, we induct on $n$.  The statement evidently holds for our base case, $n=0$.\nNow, suppose the claim holds for $n$.  Then\n\nThe claim therefore holds by induction.  It then follows that for all nonnegative integers $n$, $x=4$ is the unique solution to the equation $f_n(x) = 2x$.  $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Suppose that necklace $\\, A \\,$ has 14 beads and necklace $\\, B \\,$ has \n19. Prove that for any odd integer $n \\geq 1$, there is a way to number \neach of the 33 beads with an integer from the sequence\n\nso that each integer is used once, and adjacent beads correspond to \nrelatively prime integers. (Here a \"necklace\" is viewed as a circle in \nwhich each bead is adjacent to two other beads.)",
        "Solution": "Lemma.  For every positive odd integer $n$, there exists a nonnegative integer $k \\le 17$ such that $n+k$ is relatively prime to $n+k+15$, $n+k+1$ is relatively prime to $n+k+14$.\nProof.  Consider the positive integers $n, n+1, n+2$.  Note that at most one of these is divisible by 3, and at most one is divisible by 5.  Therefore one of these, say $n+a$, is divisible by neither, and is therefore relatively prime to 15.  Furthermore, $n+a+1$ and $n+(a+15)+1$ have different residues mod 13, so one of them, say $n+b+1$, is relatively prime to 13.  Since $n+b \\equiv n+a \\pmod{15}$, $n+b$ is relatively prime to 15.  But\n\nso $n+b$ is relatively prime to $n+b+15$.  Also,\n\nso $n+b+1$ is relatively prime to $n+b+14$.  Finally, $b \\le a+15 \\le 17$.  It follows that setting $k=b$ satisfies the lemma.  $\\blacksquare$\nLet $k$ be an integer as described in the lemma.  We place the integers $n, \\dotsc, n+k, n+k+15, \\dotsc, n+32$ on the necklace with 19 beads, and the integers $n+k+1, \\dotsc, n+k+14$ on the necklace with 14 beads, in those orders.  Since $n$ is odd, $n$ is relatively prime to $n+32 = n+2^5$.  By definition, $n+k$ and $n+k+15$ are relatively prime, as are $n+k+1$ and $n+k+14$; finally, $a$ and $a+1$ are relatively prime for all integers $a$.  It follows that each bead in this arrangement is relatively prime to its neighbors.  $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Find, with proof, the number of positive integers whose base-$n$ representation consists of distinct digits with the property that, except for the leftmost digit, every digit differs by $\\pm 1$ from some digit further to the left.  (Your answer should be an explicit function of $n$ in simplest form.)",
        "Solution": "Let a $k$-good sequence be a sequence of distinct integers $\\{ a_i \\}_{i=1}^k$ such that for all integers $2\\le i \\le k$, $a_i$ differs from some preceding term by $\\pm 1$.\nLemma.  Let $a$ be an integer.  Then there are $2^{k-1}$ $k$-good sequences starting on $a$, and furthermore, the terms of each of these sequences constitute a permutation of $k$ consecutive integers.\nProof. We induct on $k$.  For $k=1$, the lemma is trivially true.  Now, suppose the lemma holds for $k$.  If $\\{ a_i \\}_{i=1}^{k+1}$ is a $(k+1)$-good sequence, then $\\{ a_i \\}_{i=1}^k$ is a $k$-good sequence which starts on $a$, so it is a permutation of $k$ consecutive integers, say $m, \\dotsc, M$.  Then the only possibilities for $a_{k+1}$ are $m-1$ and $M+1$; either way, $\\{ a_i \\}_{i=1}^{k+1}$ constitutes a permutation of $k+1$ consecutive integers.  Since there are $2^k$ possible sequences $\\{a_i\\}_{i=1}^k$, and 2 choices of $a_{k+1}$ for each of these sequences, it also follows that there are $2^k \\cdot 2 = 2^{k+1}$ $(k+1)$-good sequences which start on $a$.  Thus the lemma holds by induction.  $\\blacksquare$\nWe now consider the number of desired positive integers with $k$ digits.  Evidently, $k$ must be less than or equal to $n$.  We also note that the digits of such an integer must constitute a $k$-good sequence.  Since the minimum of this sequence can be any of the digits $0, \\dotsc, n-k$, unless the minimum is 0 and is the first digit (in which case the only possible sequence is an increasing arithmetic sequence), and there are $2^{k-1}$ $k$-good sequences up to translation, it follows that there are $(n-k+1) 2^{k-1}-1$ desired positive integers with $k$ digits.  Thus the total number of desired positive integers is\n\nwhich is equal to $2^{n+1} - 2(n+1)$, our answer.  $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $\\, AB \\,$ intersects altitude $\\, CC' \\,$ and its extension at points $\\, M \\,$ and $\\, N \\,$, and the circle with diameter $\\, AC \\,$ intersects altitude $\\, BB' \\,$ and its extensions at $\\, P \\,$ and $\\, Q \\,$. Prove that the points $\\, M, N, P, Q \\,$ lie on a common circle.",
        "Solution_1": "Let $A'$ be the intersection of the two circles (other than $A$). $AA'$ is perpendicular to both $BA'$, $CA'$ implying $B$, $C$, $A'$ are collinear. Since $A'$ is the foot of the altitude from $A$: $A$, $H$, $A'$ are concurrent, where $H$ is the orthocentre.\nNow, $H$ is also the intersection of $BB'$, $CC'$ which means that $AA'$, $MN$, $PQ$ are concurrent. Since $A$, $M$, $N$, $A'$ and $A$, $P$, $Q$, $A'$ are cyclic, $M$, $N$, $P$, $Q$ are cyclic by the radical axis theorem.",
        "Solution_2": "Define $A'$ as the foot of the altitude from $A$ to $BC$. Then, $AA' \\cap BB' \\cap CC'$ is the orthocenter. We will denote this point as $H$.\nSince $\\angle AA'C$ and $\\angle AA'B$ are both $90^{\\circ}$, $A'$ lies on the circles with diameters $AC$ and $AB$.\nNow we use the Power of a Point theorem with respect to point $H$. From the circle with diameter $AB$ we get $AH \\cdot A'H = MH \\cdot NH$. From the circle with diameter $AC$ we get $AH \\cdot A'H = PH \\cdot QH$. Thus, we conclude that $PH \\cdot QH = MH \\cdot NH$, which implies that $P$, $Q$, $M$, and $N$ all lie on a circle.",
        "Solution_3_.28Radical_Lemma.29": "Let $\\omega_1$ be the circumcircle with diameter $AB$ and $\\omega_2$ be the circumcircle with diameter $AC$.  We claim that the second intersection of $\\omega_1$ and $\\omega_2$ other than $A$ is $A'$, where $A'$ is the feet of the perpendicular from $A$ to segment $BC$.  Note that  so $A'$ lies on $\\omega_1.$  Similarly, $A'$ lies on $\\omega_2$.  Hence, $AA'$ is the radical axis of $\\omega_1$ and $\\omega_2$.  By the Radical Lemma, it suffices to prove that the intersection of lines $MN$ and $PQ$ lie on $AA'$.  But, $MN$ is the same line as $CC'$ and $PQ$ is the same line as $BB'$.  Since $AA', BB'$, and $CC'$ intersect at the orthocenter $H$, $H$ lies on the radical axis $AA'$ and we are done. $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "In triangle $ABC$, angle $A$ is twice angle $B$, angle $C$ is obtuse, and the three side lengths $a, b, c$ are integers.  Determine, with proof, the minimum possible perimeter.",
        "Solution_1": "(diagram by integralarefun)\nAfter drawing the triangle, also draw the angle bisector of $\\angle A$, and let it intersect $\\overline{BC}$ at $D$. Notice that $\\triangle ADC\\sim \\triangle BAC$, and let $AD=x$. Now from similarity,\n\nHowever, from the angle bisector theorem, we have \n\nbut $\\triangle ABD$ is isosceles, so\n\nso all sets of side lengths which satisfy the conditions also meet the boxed condition.\nNotice that $\\text{gcd}(a, b, c)=1$ or else we can form a triangle by dividing $a, b, c$ by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since $a$ is squared, $b$ must also be a square because if it isn't, then $b$ must share a common factor with $b+c$, meaning it also shares a common factor with $c$, which means $a, b, c$ share a common factor—a contradiction. Thus we let $b = x^2, b+c = y^2$, so $a = xy$, and we want the minimal pair $(x,y)$.\nBy the Law of Cosines,\nSubstituting $a^2 = b^2 + bc$ yields $\\cos B = \\frac{b+c}{2a} = \\frac{y}{2x}$. Since $\\angle C > 90^{\\circ}$, $0^{\\circ} < \\angle B < 30^{\\circ} \\Longrightarrow \\sqrt{3} < \\frac{y}{x} < 2$. For $x \\le 3$ there are no integer solutions. For $x = 4$, we have $y = 7$ that works, so the side lengths are $(a, b, c)=(28, 16, 33)$ and the minimal perimeter is $\\boxed{77}$."
    },
    {
        "Problem": "For any nonempty set $\\,S\\,$ of numbers, let $\\,\\sigma(S)\\,$ and $\\,\\pi(S)\\,$ denote the sum and product, respectively, of the elements of $\\,S\\,$.  Prove that\n\nwhere \"$\\Sigma$\" denotes a sum involving all nonempty subsets $S$ of $\\{1,2,3, \\ldots,n\\}$.",
        "Solution": "Let $N(m)$ denote the set $\\{1, \\dotsc, m\\}$.  Since $\\sigma(\\varnothing)$, the empty sum, is equal to zero, and $\\pi(\\varnothing)$, the empty product, is equal to 1, the equation\n\nis equivalent to the desired equation when $n >0$.  We will prove this equation, but first, we prove a lemma.\nLemma.  For all nonnegative integers $n$, $\\sum_{S \\subseteq N(n)} \\frac{1}{\\pi(S)} = n+1$.\nProof. Evidently,\n\nBut the terms of this sum are the coefficients of the polynomial $P(x) = \\prod_{k=1}^n (x + 1/k)$, and the sum of the coefficients of this polynomial is\n\nas desired.  $\\blacksquare$\nWe now prove our equation by induction on $n$.  For $n=0$, we have the simple equation 0=0.\nSuppose the equation holds when $n=a$.  Then\n\nBy inductive hypothesis,\n\nand by the lemma,\n\nSince $(a+2)(s+1) = (a+3)(a+1) = (a+2)^2 -1$, our equation thus holds by induction.  Thus the problem statement is proven.  $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Show that, for any fixed integer $\\,n \\geq 1,\\,$ the sequence\n\nis eventually constant.\n[The tower of exponents is defined by $a_1 = 2, \\; a_{i+1} = 2^{a_i}$. Also $a_i \\pmod{n}$ means the remainder which results from dividing $\\,a_i\\,$ by $\\,n$.]",
        "Solution_1": "Suppose that the problem statement is false for some integer $n \\ge 1$.  Then there is a least $n$, which we call $b$, for which the statement is false.\nSince all integers are equivalent mod 1, $b\\neq 1$.\nNote that for all integers $b$, the sequence $2^0, 2^1, 2^2, \\dotsc$ eventually becomes cyclic mod $b$.  Let $k$ be the period of this cycle.  Since there are $k-1$ nonzero residues mod $b$. $1 \\le k\\le b-1 < b$.  Since\n\ndoes not become constant mod $b$, it follows the sequence of exponents of these terms, i.e., the sequence\n\ndoes not become constant mod $k$.  Then the problem statement is false for $n=k$.  Since $k<b$, this is a contradiction.  Therefore the problem statement is true.  $\\blacksquare$\nNote that we may replace 2 with any other positive integer, and both the problem and this solution remain valid.",
        "Solution_2": "We’ll prove by strong induction that for every natural number $n$, the sequence $a_1, a_2, \\ldots$ is eventually constant. Since every term of the sequence is $0 \\mathrm{\\ mod\\ } 1$, the claim is true when $n = 1$. Assuming that it’s true for $1, \\ldots, n$, we’ll now show that it’s true for $n + 1$ as well.\nSuppose first that $n + 1$ is odd. Since $\\varphi(n + 1) < n + 1$, by our inductive hypothesis there exists an $m$ such that\n\nSince $n + 1$ is coprime to powers of $2$, it follows by Euler’s theorem that\n\nor equivalently\n\nwhich is what we wanted to show.\nNow suppose that $n + 1$ is even. Write $n + 1 = 2^{k} \\cdot s$, where $1 \\leq s < n + 1$ is odd. The series must eventually be constant $\\textrm{mod\\ } 2^k$, since $a_m = 0 \\textrm{\\ mod\\ } {2^k}$ for large enough $m$. And by our inductive hypothesis, the series must also eventually be constant $\\textrm{mod\\ } s$. So for large enough $m$,\n\nSince $2^k$ and $s$ are coprime, these equations are also true modulo $2^k \\cdot s = n + 1$. So\n\nwhich completes the proof."
    },
    {
        "Problem": "Let $\\, a =(m^{m+1} + n^{n+1})/(m^m + n^n), \\,$ where $\\,m\\,$ and $\\,n\\,$ are positive integers.  Prove that $\\,a^m + a^n \\geq m^m + n^n$.\n[You may wish to analyze the ratio $\\,(a^N - N^N)/(a-N),$ for real $\\, a \\geq 0 \\,$ and integer $\\, N \\geq 1$.]",
        "Solution": "Let us assume without loss of generality that $m\\ge n$.  We then note that\n\nSimilarly,\nWe note that equations $(*)$ and $(**)$ imply that $n \\le a \\le m$.  Then $a/m \\le 1 \\le a/n$, so\n\nMultiplying this inequality by $m^m n^n(m-n)/(m^m+n^n)$, we have\n\nIt then follows that\n\nRearranging this inequality, we find that $a^m + a^n \\ge m^m + n^n$, as desired.  $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Let $\\, D \\,$ be an arbitrary point on side $\\, AB \\,$ of a given triangle $\\, ABC, \\,$ and let $\\, E \\,$ be the interior point where $\\, CD \\,$ intersects the external common tangent to the incircles of triangles $\\, ACD \\,$ and $\\, BCD$. As $\\, D \\,$ assumes all positions between $\\, A \\,$ and $\\, B \\,$, prove that the point $\\, E \\,$ traces the arc of a circle.",
        "Solution_1": "Let the incircle of $ACD$ and the incircle of $BCD$ touch line $AB$ at points $D_a,D_b$, respectively; let these circles touch $CD$ at $C_a$, $C_b$, respectively; and let them touch their common external tangent containing $E$ at $T_a,T_b$, respectively, as shown in the diagram below.\nWe note that\n\nOn the other hand, since $EC_a$ and $ET_a$ are tangents from the same point to a common circle, $EC_a = T_aE$, and similarly $EC_b = ET_b$, so\n\nOn the other hand, the segments $T_a T_b$ and $D_a D_b$ evidently have the same length, and $D_a D_b = D_aD + DD_b$, so $EC_a + EC_b = D_aD + DD_b$. Thus\n\nIf we let $s_a$ be the semiperimeter of triangle $ACD$, then $CC_a = s_a - AD$, and $D_aD = s_a - AC$, so\n\nSimilarly,\n\nso that\n\nThus $E$ lies on the arc of the circle with center $C$ and radius $(AB+BC-AB)/2$ intercepted by segments $CA$ and $CB$.  If we choose an arbitrary point $X$ on this arc and let $D$ be the intersection of lines $CX$ and $AB$, then $X$ becomes point $E$ in the diagram, so every point on this arc is in the locus of $E$. $\\blacksquare$",
        "Solution_2": "Define the same points as in the first solution.  First extend $T_aT_b$ to intersect $AB$ at a point $P$; without loss of generality let $A$ lie in between $B$ and $P$.  Then the incircle of $\\triangle ACD$ is also the incircle of $\\triangle PED$, while the incircle of $\\triangle BCD$ is the $P$-excircle of $\\triangle PED$.  It follows that $EC_a = C_bD$; denote this equality by $(*)$.\nNow remark that\n\nHence\n\nis a constant equal to $r := \\tfrac{AC+BC-AB}2$, and so $C$ lies on the circle with center $C$ and radius $r$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Find, as a function of $\\, n, \\,$ the sum of the digits of \n \nwhere each factor has twice as many digits as the previous one.",
        "Solution": "The answer is $9 \\cdot 2^n$.\nLet us denote the quantity $\\prod_{k=0}^n \\bigl( 10^{2^k}-1 \\bigr)$ as $P_n$.  We wish to find the sum of the digits of $P_n$.\nWe first note that\n\nso $P_{n-1}$ is a number of at most $2^n$ digits.  We also note that the units digit is not equal to zero.  We may thus represent $P_{n-1}$ as\n\nwhere the $d_k$ are digits and $d_0 \\neq 0$.  Then\n\nThus the digits of $P_n$ are\n\nand the sum of these is evidently $9 \\cdot 2^n$, as desired.  $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Prove",
        "Solution_1": "Consider the points $M_k = (1, \\tan k^\\circ)$ in the coordinate plane with origin $O=(0,0)$, for integers $0 \\le k \\le 89$.\nEvidently, the angle between segments $OM_a$ and $OM_b$ is $(b-a)^\\circ$, and the length of segment $OM_a$ is $1/\\cos a^\\circ$.  It then follows that the area of triangle $M_aOM_b$ is $\\tfrac{1}{2} \\sin(b-a)^\\circ \\cdot OM_a \\cdot OM_b = \\tfrac{1}{2} \\sin(b-a)^\\circ / (\\cos a^\\circ \\cdot \\cos b^\\circ)$.  Therefore\n\nso\n\nas desired.  $\\blacksquare$",
        "Solution_2": "First multiply both sides of the equation by $\\sin 1$, so the right hand side is $\\frac{\\cos 1}{\\sin 1}$.  Now by rewriting $\\sin 1=\\sin((k+1)-k)=\\sin(k+1)\\cos(k)+\\sin(k)\\cos(k+1)$, we can derive the identity $\\tan(n+1)-\\tan(n)=\\frac{\\sin 1}{\\cos(n)\\cos(n+1)}$.  Then the left hand side of the equation simplifies to $\\tan 89-\\tan 0=\\tan 89=\\frac{\\sin 89}{\\cos 89}=\\frac{\\cos 1}{\\sin 1}$ as desired.",
        "Solution_3": "Multiply by $\\sin{1}$. We get:\n$\\frac {\\sin{1}}{\\cos{0}\\cos{1}} + \\frac {\\sin{1}}{\\cos{1}\\cos{2}} + ... + \\frac {\\sin{1}}{\\cos{88}\\cos{89}} = \\frac {\\cos{1}}{\\sin{1}}$\nwe can write this as:\n$\\frac {\\sin{1 - 0}}{\\cos{0}\\cos{1}} + \\frac {\\sin{2 - 1}}{\\cos{1}\\cos{2}} + ... + \\frac {\\sin{89 - 88}}{\\cos{88}\\cos{89}} = \\frac {\\cos{1}}{\\sin{1}}$\nThis is an identity $\\tan{a} - \\tan{b} = \\frac {\\sin{(a - b)}}{\\cos{a}\\cos{b}}$\nTherefore;\n$\\sum_{i = 1}^{89}[\\tan{k} - \\tan{(k - 1)}] = \\tan{89} - \\tan{0} = \\cot{1}$, because of telescoping.\nbut since we multiplied $\\sin{1}$ in the beginning, we need to divide by $\\sin{1}$. So we get that:\n$\\frac {1}{\\cos{0}\\cos{1}} + \\frac {1}{\\cos{1}\\cos{2}} + \\frac {1}{\\cos{2}\\cos{3}} + .... + \\frac {1}{\\cos{88}\\cos{89}} = \\frac {\\cos{1}}{\\sin^2{1}}$\nas desired. QED",
        "Solution_4": "Let $S = \\frac{1}{\\cos 0^\\circ\\cos 1^\\circ} + \\frac{1}{\\cos 1^\\circ\\cos 2^\\circ} + ... + \\frac{1}{\\cos 88^\\circ\\cos 89^\\circ}$.\nMultiplying by $\\sin 1^\\circ$ gives\nNotice that $\\frac{\\sin((x+1^\\circ)-x)}{\\cos 0^\\circ\\cos 1^\\circ} = \\tan (x+1^\\circ) - \\tan x$ after expanding the sine, and so\n so"
    },
    {
        "Problem": "For a nonempty set $S$ of integers, let $\\sigma(S)$ be the sum of the elements of $S$. Suppose that $A = \\{a_1, a_2, \\ldots, a_{11}\\}$ is a set of positive integers with $a_1 < a_2 < \\cdots < a_{11}$ and that, for each positive integer $n \\le 1500$, there is a subset $S$ of $A$ for which $\\sigma(S) = n$. What is the smallest possible value of $a_{10}$?",
        "Solution": "Let's a $n$-$p$ set be a set $Z$ such that $Z=\\{a_1,a_2,\\cdots,a_n\\}$, where $\\forall i<n$, $i\\in \\mathbb{Z}^+$, $a_i<a_{i+1}$, and for each $x\\le p$, $x\\in \\mathbb{Z}^+$, $\\exists Y\\subseteq Z$, $\\sigma(Y)=x$, $\\nexists \\sigma(Y)=p+1$.\n(For Example $\\{1,2\\}$ is a $2$-$3$ set and $\\{1,2,4,10\\}$ is a $4$-$8$ set)\nFurthermore, let call a $n$-$p$ set a $n$-$p$ good set if $a_n\\le p$, and a $n$-$p$ bad set if $a_n\\ge p+1$ (note that $\\nexists \\sigma(Y)=p+1$ for any $n$-$p$ set. Thus, we can ignore the case where $a_n=p+1$).\nFurthermore, if you add any amount of elements to the end of a $n$-$p$ bad set to form another $n$-$p$ set (with a different $n$), it will stay as a $n$-$p$ bad set because $a_{n+x}>a_{n}>p+1$ for any positive integer $x$ and $\\nexists \\sigma(Y)=p+1$.\nLemma) If $Z$ is a $n$-$p$ set, $p\\leq 2^n-1$.\nFor $n=1$, $p=0$ or $1$ because $a_1=1 \\rightarrow p=1$ and $a_1\\ne1\\rightarrow p=0$.\nAssume that the lemma is true for some $n$, then $2^n$ is not expressible with the $n$-$p$ set. Thus, when we add an element to the end to from a $n+1$-$r$ set, $a_{n+1}$ must be $\\le p+1$ if we want $r>p$ because we need a way to express $p+1$. Since $p+1$ is not expressible by the first $n$ elements, $p+1+a_{n+1}$ is not expressible by these $n+1$ elements. Thus, the new set is a $n+1$-$r$ set, where $r\\leq p+1+a_{n+1} \\leq 2^{n+1}-1$\nLemma Proven\nThe answer to this question is $\\max{(a_{10})}=248$.\nThe following set is a $11$-$1500$ set:\n$\\{1,2,4,8,16,32,64,128,247,248,750\\}$\nNote that the first 8 numbers are power of $2$ from $0$ to $7$, and realize that any $8$ or less digit binary number is basically sum of a combination of the first $8$ elements in the set. Thus, $\\exists Y\\subseteq\\{1,2,4,8,16,32,64,128\\}$, $\\sigma(Y)=x \\forall 1\\le x\\leq 255$.\n$248\\le\\sigma(y)+a_9\\le502$ which implies that $\\exists A\\subseteq\\{1,2,4,8,16,32,64,128,247\\}$, $\\sigma(A)=x \\forall 1\\le x\\leq 502$.\nSimilarly $\\exists B\\subseteq\\{1,2,4,8,16,32,64,128,247,248\\}$, $\\sigma(A)=x \\forall 1\\le x\\le750$ and $\\exists C\\subseteq\\{1,2,4,8,16,32,64,128,247,248,750\\}$, $\\sigma(A)=x \\forall 1\\leq x\\leq 1500$.\nThus, $\\{1,2,4,8,16,32,64,128,247,248,750\\}$ is a $11$-$1500$ set.\nNow, let's assume for contradiction that $\\exists a_{10}\\leq 247$ such that ${a_1, a_2, \\dots, a_{11}}$ is a $11$-$q$ set where $q\\geq 1500$\n${a_1, a_2, \\dots a_8}$ is a $8$-$a$ set where $a\\leq 255$ (lemma).\n$max{(a_9)}=a_{10}-1\\leq 246$\nLet ${a_1, a_2, \\dots, a_{10}}$ be a $10$-$b$ set where the first $8$ elements are the same as the previous set. Then, $256+a_9+a_{10}$ is not expressible as $\\sigma(Y)$. Thus, $b\\leq 255+a_9+a_{10}\\leq 748$.\nIn order to create a $11$-$d$ set with $d>748$ and the first $10$ elements being the ones on the previous set, $a_{11}\\leq 749$ because we need to make $749$ expressible as $\\sigma(Y)$. Note that $b+1+a_{11}$ is not expressible, thus $d<b+1+a_{11}\\leq 1498$.\nDone but not elegant..."
    },
    {
        "Problem": "Chords $AA'$, $BB'$, and $CC'$ of a sphere meet at an interior point $P$ but are not contained in the same plane.  The sphere through $A$, $B$, $C$, and $P$ is tangent to the sphere through $A'$, $B'$, $C'$, and $P$.  Prove that $AA'=BB'=CC'$.",
        "Solution_1": "Consider the plane through $A,A',B,B'$.  This plane, of course, also contains $P$. We can easily find the $\\triangle APB$ is isosceles because the base angles are equal. Thus, $AP=BP$. Similarly, $A'P=B'P$. Thus, $AA'=BB'$. By symmetry, $BB'=CC'$ and $CC'=AA'$, and hence $AA'=BB'=CC'$ as desired.\n$\\mathbb{QED.}$\nBy another person ^v^\nThe person that came up with the solution did not prove that  $\\triangle APB$ is isosceles nor the base angles are congruent. I will add on to the solution.\nThere is a common tangent plane that pass through $P$ for the $2$ spheres that are tangent to each other.\nSince any cross section of sphere is a circle. It implies that $A$, $A'$, $B$, $B'$ be on the same circle ($\\omega_1$), $A$, $B$, $P$ be on the same circle ($\\omega_2$), and $A'$, $B'$, $P$ be on the same circle ($\\omega_3$).\n$m\\angle APB= m\\angle A'PB'$ because they are vertical angles. By power of point, $(AP)(A'P)=(BP)(B'P)\\rightarrow\\frac{AP}{B'P}=\\frac{BP}{A'P}$\nBy the SAS triangle simlarity theory, $\\triangle APB \\sim\\triangle B'PA'$. That implies that $\\angle ABP\\cong\\angle B'PA'$.\n\nLet's call the interception of the common tangent plane and the plane containing $A$, $A'$, $B$, $B'$, $P$, line $l$.\n$l$ must be the common tangent of $\\omega_2$ and $\\omega_3$.\nThe acute angles form by $l$ and $\\overline{AA'}$ are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord $\\overline{AP}$ and $\\overline{A'P}$ are equal.\nSimilarly the central angle of chord $\\overline{BP}$ and $\\overline{B'P}$ are equal.\nThe length of any chord with central angle $2\\theta$ and radius $r$ is $2r\\sin\\left({\\theta}\\right)$, which can easily been seen if we drop the perpendicular from the center to the chord.\nThus, $\\frac{AP}{A'P}=\\frac{BP}{B'P}$.\nBy the SAS triangle simlarity theory, $\\triangle APB \\sim\\triangle A'PB'$. That implies that $\\angle ABP\\cong\\angle B'PA'$.\nThat implies that $\\angle ABP\\cong\\angle A'PB'\\cong\\angle B'PA'$. Thus, $\\triangle A'PB'$ is an isosceles triangle and since $\\triangle APB \\sim\\triangle A'PB'$, then$\\triangle APB$ must be an isosceles triangle too.",
        "Solution_2": "Call the large sphere $O_1$, the one containing $A$ $O_2$, and the one containing $A'$ $O_3$.  The centers are $c_1$, $c_2,$ and $c_3$.\nSince two spheres always intersect in a circle , we know that A,B, and C must lie on a circle ($w_1$)completely contained in $O_1$ and $O_2$\nSimilarly, A', B', and C' must lie on a circle ($w_2$) completely contained in $O_1$ and $O_3$.\nSo, we know that 3 lines connecting a point on $w_1$ and P hit a point on $w_2$.  This implies that $w_1$ projects through P to $w_2$, which in turn means that $w_1$ is in a plane parallel to that of $w_2$.  Then, since $w_1$ and $w_2$ lie on the same sphere, we know that they must have the same central axis, which also must contain P (since the center projects through P to the other center).\nSo, all line from a point on $w_1$ to P are of the same length, as are all lines from a point on $w_2$ to P.  Since AA', BB', and CC' are all composed of one of each type of line, they must all be equal."
    },
    {
        "Problem_5": "Let $\\, P(z) \\,$ be a polynomial with complex coefficients which is of degree $\\, 1992 \\,$ and has distinct zeros. Prove that there exist complex numbers $\\, a_1, a_2, \\ldots, a_{1992} \\,$ such that $\\, P(z) \\,$ divides the polynomial $\\left( \\cdots \\left( (z-a_1)^2 - a_2 \\right)^2 \\cdots - a_{1991} \\right)^2 - a_{1992}$.",
        "Solution": "Since the zeros $z_{1}, z_{2}, \\ldots, z_{1992}$ of the polynomial $P(z)$ are distinct, the polynomial $P(z)$ divides the polynomial \n\nif and only if $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;\\ldots;\\;1992\\right\\}$. So it is enough to show that for any $1992$ complex numbers $z_{1}, z_{2}, \\dots, z_{1992}$, there exist $1992$ complex numbers $a_{1}, a_{2}, \\ldots, a_{1992}$, such that the polynomial \n\nsatisfies $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;\\ldots;\\;1992\\right\\}$. This is a special case of the following lemma.\nLemma: Let $n$ be a positive integer, and $z_{1}, z_{2}, \\ldots, z_{n}$ be $n$ complex numbers. Then, there exist $n$ complex numbers $a_{1}, a_{2},\\ldots, a_{n}$ such that the polynomial $Q\\left(z\\right) = \\left( \\ldots \\left( \\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}\\ldots-a_{n-1}\\right)^{2}-a_{n}$ satisfies $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;\\ldots;\\;n\\right\\}$.\nProof: We use induction over $n$.\nFor $n = 1$, the lemma is trivial, since $Q\\left(z\\right)=z-a_{1}$, so we can take $a_{1}=z_{1}$, and then the polynomial $Q\\left(z\\right)=z-z_{1}$ clearly satisfies $Q\\left(z_{1}\\right)=0$.\nLet $k$ be a positive integer. Assume that for $n = k$, the lemma is true. Therefore, there exist $k$ complex numbers $a_{1}, a_{2}, \\ldots, a_{k}$ such that the polynomial $Q\\left(z\\right)=\\left(\\ldots\\left( \\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}\\ldots-a_{k-1}\\right)^{2}-a_{k}$ satisfies $Q\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;\\ldots;\\;k\\right\\}$. Then we wish to prove that for any $k + 1$ complex numbers $z_{1}, z_{2}, \\ldots, z_{k+1}$, there exist $k + 1$ complex numbers $b_{1}, b_{2}, \\ldots, b_{k+1}$ such that the polynomial $R\\left(z\\right)=\\left(\\ldots\\left( \\left(z-b_{1}\\right)^{2}-b_{2}\\right)^{2}\\ldots-b_{k}\\right)^{2}-b_{k+1}$ satisfies $R\\left(z_{i}\\right)=0$ for every $i\\in\\left\\{1;\\;2;\\;\\ldots;\\;k+1\\right\\}$. We will construct this polynomial $R$ from the polynomial $Q$.\nLet $b_{i}=a_{i}$ for every $i\\le k-1$, and let $b_{k}=a_{k}+\\frac{1}{2}Q\\left(z_{k+1}\\right)$ and $b_{k+1}=\\frac{1}{4}\\left(Q\\left(z_{k+1}\\right)\\right)^{2}$. Then,\n\nThen as $Q\\left(z_{i}\\right)=0$ for $i\\in\\left\\{1;\\;2;\\;\\ldots;\\;k\\right\\}$, we have \n\nfor all $i\\in\\left\\{1;\\;2;\\;\\ldots;\\;k\\right\\}$. Also, \n\nThus, $R\\left(z_{i}\\right)=0$ holds for every $i\\in\\left\\{1;\\;2;\\;\\ldots;\\;k+1\\right\\}$, and this proves the lemma for $n = k + 1$. Hence, the induction step is complete, and the lemma is proven.\nAs the problem is a special case of the lemma ($n=1992$), we are done."
    },
    {
        "Problem": "For each integer $n\\ge 2$, determine, with proof, which of the two positive real numbers $a$ and $b$ satisfying\n\nis larger.",
        "Solution_1": "Square and rearrange the first equation and also rearrange the second.\n\nIt is trivial that\n\nsince $a-1$ clearly cannot equal $0$ (Otherwise $a^n=1\\neq 1+1$).  Thus\n\nwhere we substituted in equations (1) and (2) to achieve (5).  Notice that from $a^{n}=a+1$ we have $a>1$.  Thus, if $b>a$, then $b^{2n-1}-1>a^{2n-1}-1$.  Since $a>1\\Rightarrow a^{2n-1}-1>0$, multiplying the two inequalities yields $b^{2n}-b>a^{2n}-a$, a contradiction, so $a> b$.  However, when $n$ equals $0$ or $1$, the first equation becomes meaningless, so we conclude that for each integer $n\\ge 2$, we always have $a>b$.",
        "Solution_2": "Define $f(x)=x^n-x-1$ and $g(x)=x^{2n}-x-3a$. By Descarte's Rule of Signs, both polynomials' only positive roots are $a$ and $b$, respectively. With the Intermediate Value Theorem and the fact that $f(1)=-1$ and $f(2)=2^n-3>0$, we have $a\\in(1,2)$.\nThus, $-3a\\in(-6,-3)$, which means that $g(1)=-3a<0$. Also, we find that $g(a)=a^{2n}-4a$. All that remains to prove is that $g(a)>0$, or $a^{2n}-4a>0$. We can then conclude that $b$ is between $1$ and $a$ from the Intermediate Value Theorem. From the first equation given, $a^{2n}=(a+1)^2=a^2+2a+1$. Subtracting $4a$ gives us $a^2-2a+1>0$, which is clearly true, as $a\\neq1$. Therefore, we conclude that $1<b<a<2$.\n"
    },
    {
        "Problem_3": "Consider functions $f : [0, 1] \\rightarrow \\mathbb{R}$ which satisfy\nFind, with proof, the smallest constant $c$ such that\nfor every function $f$ satisfying (i)-(iii) and every $x$ in $[0, 1]$.",
        "Solution": "My claim: $c\\ge2$\nLemma 1) $f\\left(\\left(\\frac{1}{2}\\right)^n\\right)\\le\\left(\\frac{1}{2}\\right)^n$ for $n\\in \\mathbb{Z}, n\\ge0$\nFor $n=0$, $f(1)=1$ (ii)\nAssume that it is true for $n-1$, then $f\\left(\\left(\\frac{1}{2}\\right)^{n}\\right)+f\\left(\\left(\\frac{1}{2}\\right)^{n}\\right)\\le f\\left(\\left(\\frac{1}{2}\\right)^{n-1}\\right)\\le \\left(\\frac{1}{2}\\right)^{n-1}$\n$f\\left(\\left(\\frac{1}{2}\\right)^{n}\\right)\\le \\left(\\frac{1}{2}\\right)^{n}$\nBy principle of induction, lemma 1 is proven.\nLemma 2) For any $x$, $\\left(\\frac{1}{2}\\right)^{n+1}<x\\le\\left(\\frac{1}{2}\\right)^n\\le1$ and $n\\in \\mathbb{Z}$, $f(x)\\le\\left(\\frac{1}{2}\\right)^n$.\n$f(x)+f\\left(\\left(\\frac{1}{2}\\right)^n-x\\right)\\le f\\left(\\left(\\frac{1}{2}\\right)^{n}\\right)\\le \\left(\\frac{1}{2}\\right)^{n}$ (lemma 1 and (iii) )\n$f(x)\\le\\left(\\frac{1}{2}\\right)^n$ (because $f\\left(\\left(\\frac{1}{2}\\right)^n-x\\right)\\ge0$ (i) )\n\n$\\forall 0\\le x\\le 1$, $\\left(\\frac{1}{2}\\right)^{n-1}\\ge2x\\ge \\left(\\frac{1}{2}\\right)^n\\ge f(x)$. Thus, $c=2$ works.\n\nLet's look at a function $g(x)=\\left\\{\\begin{array}{ll}0&0\\le x\\le \\frac{1}{2};\\\\1&\\frac{1}{2}<x\\le1;\\\\\\end{array}\\right\\}$\nIt clearly have property (i) and (ii). For $0\\le x\\le\\frac{1}{2}$ and WLOG let $x\\le y$, $f(x)+f(y)=0+f(y)\\le f(y)$\nFor $\\frac{1}{2}< x\\le 1$, $x+y>1$. Thus, property (iii) holds too. Thus $g(x)$ is one of the legit function.\n\n$\\lim_{x\\rightarrow\\frac{1}{2}^+} cx \\ge \\lim_{x\\rightarrow\\frac{1}{2}^+} g(x)=1$\n$\\frac{1}{2}c>1$\n$c>2$ but approach to $2$ when $x$ is extremely close to $\\frac{1}{2}$ from the right side.\n$\\mathbb{Q.E.D}$"
    },
    {
        "Problem_4": "Let $a$, $b$ be odd positive integers. Define the sequence $(f_n)$ by putting $f_1 = a$,\n$f_2 = b$, and by letting $f_n$ for $n\\ge3$ be the greatest odd divisor of $f_{n-1} + f_{n-2}$.\nShow that $f_n$ is constant for $n$ sufficiently large and determine the eventual\nvalue as a function of $a$ and $b$.",
        "Solution": "Part 1) Prove that $f_n$ is constant for sufficiently large $n$.\nNote that if there is some $f_n=f_{n-1}$ for any $n$, then $\\frac{f_{n}+f_{n-1}}{2}=f_n$, which is odd. Thus, $f_{n+1}=f_n=f_{n-1}$ and by induction, all $f_p$ is constant for $p\\ge n$.\nAlso note that $f_n>0$ since average of $2$ positive number is always positive.\nThus, assume for contradiction, $\\nexists n$, $f_n=f_{n-1}$.\nThen, $f_{n+1}\\le\\frac{f_{n}+f_{n-1}}{2}< \\max (f_{n},f_{n-1})$, $f_{n+2}\\le\\frac{f_{n+1}+f_{n}}{2}< \\max (f_{n},f_{n+1})\\le\\max (f_{n},f_{n-1})$\nThus, $\\max (f_{n},f_{n-1})> \\max (f_{n+1},f_{n+2})$ and that means that $\\max (f_{2n},f_{2n+1})$ is a strictly decreasing function and it must reach $0$ as $n\\rightarrow\\infty$, which contradict with the fact that $f_n>0$.\nPart 1 proven.\n\n\nPart 2) Show that the constant is $\\gcd(a,b)$.\nFor any $f_n$ where $\\gcd(a,b)=d\\ne1$. $f_n=dg_n$ for $g_n$ with the same property except with $g_1=\\frac{a}{d}$ and $g_2=\\frac{b}{d}$.\nTherefore, if I prove that the constant for any $f_n$ with relatively prime $a$, $b$ is $1$, then I have shown that part 2 is true.\nLemma) If $\\gcd(f_n,f_{n-1})=1$, then $\\gcd(f_n,f_{n+1})=1$.\nAssume for contradiction that $\\gcd(f_n,f_{n+1})=d\\ne1$, since both $f_n$ and $f_{n+1}$ are odd, $d$ is not divisible by $2$.\n$f_{n+1}=\\frac{f_n+f_{n-1}}{2^n}$ for some $n\\in \\mathbb{Z}^+$ such that $f_{n+1}$ is odd.\n$(2^n)f_{n+1}-f_n=f_{n-1}$\n$d(p+q)=f_{n-1}$, where $p$ and $q$ is another integer.\nThus, $f_{n-1}$ is divisible by $d$ which contradicts with the assumption that $\\gcd(f_n,f_{n-1})=1$.\nLemma proven\nBy induction, $\\gcd(f_n,f_{n-1})=1$ since $\\gcd(a,b)=\\gcd(f_1,f_2)=1$.\nSince there must exist some $n$ where $f_n=f_{n+1}$ (part 1), $\\gcd(f_n,f_{n+1})=f_n=1$.\n$\\mathbb{Q.E.D}$"
    },
    {
        "Problem_5": "Let $a_0, a_1, a_2,\\cdots$ be a sequence of positive real numbers satisfying $a_{i-1}a_{i+1}\\le a^2_i$\nfor $i = 1, 2, 3,\\cdots$ . (Such a sequence is said to be log concave.) Show that for\neach $n > 1$,\n",
        "Solution": "Notice that because\n\n\n\nwe may subtract $\\dfrac{(a_1 + a_2 + \\dots + a_{n-1})(a_0 + a_1 + \\dots + a_n)}{n^2}$ from both sides of the inequality and observe that it is sufficient to prove that\n\nor\nFortunately, this is an easy inequality. Indeed, from AM-GM applied on each group of terms we have\n\nand so it suffices to prove\n\nor, after taking both sides to the $(n^2 - 1)$ power, simplifying, and taking the $n$-th root of both sides, to prove\n\nThis easily follows from the Fact that $a_0 a_n \\le a_i a_{n-i}$ for $1 \\le i \\le n-1$. Indeed, we are given that\n\n\n\n\n\n\nMultiply all inequalities together and cancel $a_1, a_2^2, a_3^2, \\dots, a_{n-2}^2, a_{n-1}$ to give $a_0 a_n \\le a_1 a_{n-1}$. Similarly, by multiplying all inequalities except the first and the last, we deduce that $a_2 a_{n-2} \\ge a_1 a_{n-1} \\ge a_0 a_n$, and a simple induction argument proves the verity of the Fact for $1 \\le i \\le \\frac{n}{2}$, and so by the Commutative Property the Fact is true for all $1 \\le i \\le n-1$, as desired. Now multiply each inequality of the Fact for $i = 1, 2, 3, \\dots, n-1$ to give the desired result.\nNote: The Fact can be generalized into a Lemma: $a_x a_w \\le a_y a_z$ whenever $x \\le y \\le z \\le w$ and $x + w = y + z$. The proof is similar to that of the Fact and is left as an exercise to the reader.\n--User suli, April 15, 2015."
    },
    {
        "Problem": "Let $\\, k_1 < k_2 < k_3 <\\cdots\\,$, be positive integers, no two consecutive, and let $\\, s_m = k_1+k_2+\\cdots+k_m\\,$, for $\\, m = 1,2,3,\\ldots\\;\\;$. Prove that, for each positive integer $n$, the interval $\\, [s_n, s_{n+1})\\,$, contains at least one perfect square.",
        "Solution": "We want to show that the distance between $s_n$ and $s_{n+1}$ is greater than the distance between $s_n$ and the next perfect square following $s_n$.\nGiven $s_n=\\sum_{i=1}^{n}k_i$, where no $k_i$ are consecutive, we can put a lower bound on $k_n$. This occurs when all $k_{i+1}=k_i+2$:\n\nRearranging, $k_{n,min}=\\frac{s_n}{n}+n-1$. So, $k_n\\geq\\frac{s_n}{n}+n-1$, and the distance between $s_n$ and $s_{n+1}$ is $k_{n+1}\\geq k_n+2\\geq\\frac{s_n}{n}+n+1$.\nAlso, let $d(s_n)$ be the distance between $s_n$ and the next perfect square following $s_n$. Let's look at the function $d(x)$ for all positive integers $x$.\nWhen $x$ is a perfect square, it is easy to see that $d(x)=2\\sqrt{x}+1$.\nProof: Choose $x=m^2$. $d(m^2)=(m+1)^2-m^2=2m+1=2\\sqrt{m^2}+1$.\nWhen $x$ is not a perfect square, $d(x)<2\\sqrt{x}+1$.\nProof: Choose $x=m^2+p$ with $0<p<2m+1$. $d(m^2+p)=(m+1)^2-m^2-p=2m+1-p<2m+1=2\\sqrt{m^2}+1<2\\sqrt{m^2+p}+1$.\nSo, $d(x)\\leq 2\\sqrt{x}+1$ for all $x$ and $d(s_n)\\leq 2\\sqrt{s_n}+1$ for all $s_n$.\nNow, it suffices to show that $k_{n+1}\\geq d(s_n)$ for all $n$.\n\nSo, $k_{n+1}\\geq d(s_n)$ and all intervals between $s_n$ and $s_{n+1}$ will contain at least one perfect square.",
        "Solution_2": "We see that by increasing $n$ by some amount, we simply shift our interval by a finite amount. It suffices to consider the case $n=1$ (since this can be inducted across all positive integers). Let $k_1=x$. We want the smallest interval, so we have $[x, 2x+2]$. Simple induction reveals that the ration of consecutive squares grows slower than our linear bound. We now consider sufficiently small $x$ (where $\\frac{(n+1)^2}{n^2}<2$). This first happens at $n=3$. By simple casework, our answer is as desired $\\blacksquare$.\n",
        "Solution_3": "We will first prove by Induction on $n\\in\\mathbb{N}$ that $(k_{n}+1)^2\\geq4(k_1+k_2+\\cdots +k_{n}).$ Denote this statement by $P(n).$\nFor the Base Case let $n=1,$ we know that;\nHence, the Base Case holds.\nFor the Inductive Step, suppose that $P(m)$ holds for some $k\\in\\mathbb{N},$ we will prove that $P(m+1)$ holds as well.\nAssume for contradiction that $P(m+1)$ doesn't hold, then we know that;\nWe know that since $k_m$ and $k_{m-1}$ are not consecutive, we have;\nBut this contradicts the Inductive Hypothesis that  thus our assumption was false and the Inductive Step is complete.\nHence, we have proved that $P(n)$ holds, and we will use $P(n)$ to solve the original problem.\nNow suppose that for some positive integer $n$, the interval $\\, [s_n, s_{n+1})\\,$ does not contain any perfect square, then we know that there must exist two perfect squares of consecutive integers, such that the smaller one is lesser than $s_n$ and the larger one greater than or equal to $s_{n+1}.$\nWe thus know that there exists some $x\\in\\mathbb{N}$ such that;\nBy Inequality 1;\nHence, we know that;\nCombining this with Inequality 2 gives;\nWe know that since $k_{n+1}$ are not consecutive, $k_n+1\\leq k_{n+1}-1,$ hence, we have;\nBut this contradicts the statement $P(n)$ which was proved earlier.\n$\\square$"
    },
    {
        "Problem": "The sides of a $99$-gon are initially colored so that consecutive sides are red, blue, red, blue,..., red, blue, yellow. We make a sequence of modifications in the coloring, changing the color of one side at a time to one of the three given colors (red, blue, yellow), under the constraint that no two adjacent sides may be the same color. By making a sequence of such modifications, is it possible to arrive at the coloring in which consecutive sides \nare red, blue, red, blue, red, blue,..., red, yellow, blue?",
        "Solution": "We proceed by representing the colors as numbers, i.e. Red = 0, Blue = 1, Yellow = 2. Thus, we start with some sequence 0101...012 and are trying to end up with the sequence 1010...0102. Generate a second sequence of terms by subtracting each term by the following term and taking it modulus 3, i.e. (1-0, 0-1, 1-0, 0-1, ... 0-1, 1-0, 0-2, 2-1) = 1212...2111 is the sequence generated using the start sequence. The sequence generated using the end sequence is (0-1, 1-0, 0-1, 1-0, ... 1-0, 0-1, 1-2, 2-0) = 2121...2122. Observe that the sum of terms generated by the start sequence is different from the sum of terms generated by the end sequence.\nWe must now prove that changing the color of a side does not change the sum of the generated sequence. We do this by noting that we can only change the color of a side if its adjacent sides have the same color as each other. Thus, changing a side does not change the sum of terms in the generated sequence, as the term generated by the changing side and the side to its left is simply the \"negative\" of the term generated by the changing side and the side to its right (i.e. 212 to 202 changes 12 to 21).\nSince it is impossible to change the sum of terms in the generated sequence, and the sequences generated by the start and end sequences have two different sums, it is impossible to perform a series of such modifications that change the start sequence to the end sequence."
    },
    {
        "Problem": "A convex hexagon $ABCDEF$ is inscribed in a circle such that $AB=CD=EF$ and diagonals $AD,BE$, and $CF$ are concurrent. Let $P$ be the intersection of $AD$ and $CE$. Prove that $\\frac{CP}{PE}=(\\frac{AC}{CE})^2$.",
        "Solution": "Let the diagonals $AD$, $BE$, $CF$ meet at $Q$.\nFirst, let's show that the triangles $\\triangle AEC$ and $\\triangle QED$ are similar.\n$\\angle ACE=\\angle ADE$ because $A$,$C$,$D$ and $E$ all lie on the circle, and $\\angle ADE=\\angle QDE$. $\\angle AEB=\\angle CED$ because $AB=CD$, and $A$,$B$,$C$,$D$ and $E$ all lie on the circle. Then,\n$\\angle AEB=\\angle CED \\rightarrow \\angle AEB+\\angle BEC=\\angle CED+\\angle BEC \\rightarrow \\angle AEC=\\angle QED$\nTherefore, $\\triangle AEC$ and $\\triangle QED$ are similar, so $AC/CE=QD/DE$.\nNext, let's show that $\\triangle AEC$ and $\\triangle CDQ$ are similar.\n$\\angle AEC=\\angle ADC$ because $A$,$C$,$D$ and $E$ all lie on the circle, and $\\angle ADC=\\angle CDQ$. $\\angle EAD=\\angle ECD$ because $A$,$C$,$D$ and $E$ all lie on the circle. $\\angle DAC=\\angle ECF$ because $CD=EF$, and $A$,$C$,$D$,$E$ and $F$ all lie on the circle. Then,\n$\\angle EAC=\\angle EAD+\\angle DAC=\\angle ECD+\\angle ECF=\\angle DCQ$\nTherefore, $\\triangle AEC$ and $\\triangle CDQ$ are similar, so $AC/CE=CQ/QD$.\nLastly, let's show that $\\triangle CPQ$ and $\\triangle EPD$ are similar.\nBecause $CD=EF$ and $C$,$D$,$E$ and $F$ all lie on the circle, $CF$ is parallel to $DE$. So, $\\triangle CPQ$ and $\\triangle EPD$ are similar, and $CQ/DE=CP/PE$.\nPutting it all together, $\\frac{CP}{PE}=\\frac{CQ}{DE}=\\frac{AC}{CE}\\cdot \\frac{QD}{DE}=(\\frac{AC}{CE})^2$.\nBorrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html"
    },
    {
        "Problem": "Let $\\, a_1, a_2, a_3, \\ldots \\,$ be a sequence of positive real numbers satisfying $\\, \\sum_{j = 1}^n a_j \\geq \\sqrt {n} \\,$ for all $\\, n \\geq 1$. Prove that, for all $\\, n \\geq 1, \\,$\n",
        "Solution": "Note that if we try to minimize $(a_j)^2$, we would try to make the $a_j$ as equal as possible. However, by the condition given in the problem, this isn't possible, the $a_j$'s have to be an increasing sequence. Thinking of minimizing sequences, we realize that the optimal equation is $a_n = \\sqrt{n} - \\sqrt{n-1} = 1/(\\sqrt{n} + \\sqrt{n-1})$. Note that this is strictly greater than $1/(2\\sqrt{n})$. So it is greater than the sum of $(1/\\sqrt{4n})^2$ over all n from 1 to $\\infty$. So the sum we are looking to minimize is strictly greater than the sum of $1/4n$ over all $n$ from 1 to $\\infty$, which is what we wanted to do."
    },
    {
        "Problem": "Let $\\, |U|, \\, \\sigma(U) \\,$ and $\\, \\pi(U) \\,$ denote the number of elements, the sum, and the product, respectively, of a finite set $\\, U \\,$ of positive integers.  (If $\\, U \\,$ is the empty set, $\\, |U| = 0, \\, \\sigma(U) = 0, \\, \\pi(U) = 1$.) Let $\\, S \\,$ be a finite set of positive integers. As usual, let $\\, \\binom{n}{k} \\,$ denote $\\, n! \\over k! \\, (n-k)!$. Prove that  for all integers $\\, m \\geq \\sigma(S)$.",
        "Solution": "Let, for $|S|\\geq k \\geq 1$, $A_k$ be a collection of sets such that $\\left|\\bigcap_{k \\in U}A_k\\right| = \\binom{m- \\sigma(U)}{|S|}$. We'll try to find such a collection to prove $\\left|\\bigcup_{k \\in U}A_k\\right| = \\binom{m}{|S|}-\\pi(S)$, since we can use PIE to show that $\\left|\\bigcup_{k \\in U}A_k\\right| = \\sum_{\\emptyset \\neq U \\subseteq S} (-1)^{|U|+1}\\left|\\bigcap_{k \\in U}A_k\\right|$.\nLet $Z$ be the set of binary sequences with size $m$, which we'll divide into blocks, in particular, with $S=\\{x_1, x_2, ..., x_n\\}$, the first block has size $x_1$, the next has size $x_2$ and so on, until we reach $\\sigma(S)$, the last few we shall consider as an extra block with size $m-\\sigma(S)=R$.\nNow we perceive that the $\\sigma(U)$ function has the property, that, for $U \\subseteq S$, $\\sigma(S)-\\sigma(U)= \\sigma(S\\setminus U)$, so, $m- \\sigma(U)= \\sigma(S\\setminus U)+R$.\nLet $A_k$ be the number of binary sequences with exactly $|S|$ times $1$, but no appearances of $1$ in the $k$-th block. It is easy to see that $\\bigcap_{k \\in U}A_k$ is all of the binary sequences with exactly $|S|$ times $1$, but no appearances of $1$ in any of the $k$-th blocks, for $k \\in U$. So, we can simply eliminate those blocks, with is equivalent to eliminating $U$ from $S$ , so we are left with $\\sigma(S\\setminus U)+R$ places to put the $1$'s $|S|$ times. This is $\\binom{\\sigma(S\\setminus U)+R}{|S|}$, and we saw that is is equal to $\\binom{m- \\sigma(U)}{|S|}$. So, we've found a collection that fits the description.\nThe number of ways to pick $1$'s $|S|$ times from $m$ places is $\\binom{m}{|S|}$. But we don't want for the $1$'s to fill every block, since this is not in the union. We have $\\pi(S)$ ways to put the $1$'s $|S|$ times, where each $1$ is in exacty one block ($x_1$ places $\\times$ $x_2$ places and so on). So, we find that $| \\bigcup_{k \\in U}A_k | = \\binom{m}{|S|}-\\pi(S)$. Which implies, by the Principle of Inclusion Exclusion, \n$\\binom{m}{|S|}-\\pi(S) = \\sum_{\\emptyset \\neq U \\subseteq S} (-1)^{|U|+1}\\binom{m- \\sigma(U)}{|S|} \\iff \\pi(S) = \\sum_{U \\subseteq S} (-1)^{|U|}\\binom{m- \\sigma(U)}{|S|}$"
    },
    {
        "Problem": "Let $\\, p \\,$ be an odd prime.  The sequence $(a_n)_{n \\geq 0}$ is defined as follows: $\\, a_0 = 0,$ $a_1 = 1, \\, \\ldots, \\, a_{p-2} = p-2 \\,$ and, for all $\\, n \\geq p-1, \\,$ $\\, a_n \\,$ is the least positive integer that does not form an arithmetic sequence of length $\\, p \\,$ with any of the preceding terms. Prove that, for all $\\, n, \\,$ $\\, a_n \\,$ is the number obtained by writing $\\, n \\,$ in base $\\, p-1 \\,$ and reading the result in base $\\, p$.",
        "Solution": "I have to make the assumption that $a_{n+1}>a_{n}$ (without this assumption, I can have the sequence\n${1,\\cdots,p-2,1,1,\\cdots,1,2,1,\\cdots}$)\nAll of the following work are in base $p$ otherwise stated\n\nLemma 1: for an arithmetic sequence of length $\\, p \\,$ to exist, there must be a number in the sequence with $(p-1)$ as a digit.\nA arithmetic sequence of length $\\, p \\,$ can be represented as\n$a,a+d,a+2d,\\cdots,a+(p-1)d$\nSince no number repeats, $d\\neq0$. Thus, d must have a rightmost non-zero digits, and every term in the sequence $0,d,2d,\\cdots,(p-1)d$ have the same number of tailing zeros, let's say there are $x$ tailing zeros.\nand let $\\dfrac{{0,d,2d,\\cdots,(p-1)d}}{p^x}=S$\n$S\\equiv{1,2,...,(p-1)}(mod)p$ since p is an odd prime and the operation divide by $p^x$ has remove all factors of p in S\nThus, there must be a number with $(p-1)$ as a digit if a length-p sequence exist.\n\nNow, I'm going to prove the statement by strong induction.\nI'm going to assume that for all $\\, k \\,$  less than $\\, n \\,$\n$\\, a_k \\,$ is the number obtained by writing $\\, k \\,$ in base $\\, p-1 \\,$ and reading the result in base $\\, p$.\n(which is true for $n=p-2$ already)\nOr in another word, the terms precede $a_n$ contains all number less than $n$ written in base $\\, p-1 \\,$ and reading the result in base $\\, p$.\n\nLemma 2: Any term containing the digit $p-1$ will form an arithmetic sequence of length-p with preceding terms\nlet $p-1$ be the $x^{th}$ digit from the right (There may be more than 1 $(p-1)$)\n$a=$ the number with all $(p-1)$ replaced by 0 (e.g: $p=7$, the number$=1361264$, then $a=1301204$)\n$d=\\sum p^{x-1}$ (for all $x$'s)\n$a,a+d,a+2d,\\cdots,a+(p-2)d$ all precede $a_n$, thus, $a+(p-1)d$ can't be in the sequence $(a_n)_{n \\geq 0}$\nTherefore, any term containing the digit $p-1$ can't be a term in the sequence ${(a_n)}_{n\\ge0}$\n\nSince the digit $p-1$ won't appear (lemma 2) and as long as it doesn't appear, the arithmetic sequence of length $\\, p \\,$ won't be formed (lemma 1), and $a_n$ must be as small as possible,\nTherefore, for all $\\, n, \\,$ $\\, a_n \\,$ is the number obtained by writing $\\, n \\,$ in base $\\, p-1 \\,$ and reading the result in base $\\, p$."
    },
    {
        "Problem": "A calculator is broken so that the only keys that still work are the $\\, \\sin, \\; \\cos,$ $\\tan, \\; \\sin^{-1}, \\; \\cos^{-1}, \\,$ and $\\, \\tan^{-1} \\,$ buttons.  The display initially shows 0. Given any positive rational number $\\, q, \\,$ show that pressing some finite sequence of buttons will yield $\\, q$.   Assume that the calculator does real number calculations with infinite precision.  All functions are in terms of radians.",
        "Solution": "We will prove the following, stronger statement :  If $m$ and $n$ are relatively prime nonnegative integers such that $n>0$, then the some finite sequence of buttons will yield $\\sqrt{m/n}$.\nTo prove this statement, we induct strongly on $m+n$.  For our base case, $m+n=1$, we have $n=1$ and $m=0$, and $\\sqrt{m/n} = 0$, which is initially shown on the screen.  For the inductive step, we consider separately the cases $m=0$, $0<m\\le n$, and $n<m$.\nIf $m=0$, then $n=1$, and we have the base case.\nIf $0< m \\le n$, then by inductive hypothesis, $\\sqrt{(n-m)/m}$ can be obtained in finitely many steps; then so can\nIf $n<m$, then by the previous case, $\\sqrt{n/m}$ can be obtained in finitely many steps.  Since $\\cos \\tan^{-1} \\sqrt{n/m} = \\sin \\tan^{-1} \\sqrt{m/n}$, it follows that\n\ncan be obtained in finitely many steps.  Thus the induction is complete.  $\\blacksquare$\n"
    },
    {
        "Problem": "Given a nonisosceles, nonright triangle $ABC,$ let $O$ denote the center of its circumscribed circle, and let $A_1, \\, B_1,$ and $C_1$ be the midpoints of sides $BC, \\, CA,$ and $AB,$ respectively.  Point $A_2$ is located on the ray $OA_1$ so that $\\triangle OAA_1$ is similar to $\\triangle OA_2A$.  Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1,$ respectively, are defined similarly.  Prove that lines $AA_2, \\, BB_2,$ and $CC_2$ are concurrent, i.e. these three lines intersect at a point.\n",
        "Solution": "LEMMA 1:  In $\\triangle ABC$ with circumcenter $O$, $\\angle OAC = 90 - \\angle B$.\nPROOF of Lemma 1: The arc $AC$ equals $2\\angle B$ which equals $\\angle AOC$.  Since $\\triangle AOC$ is isosceles we have that $\\angle OAC = \\angle OCA = 90 - \\angle B$. \nQED\nDefine $H \\in BC$ s.t. $AH \\perp BC$.  Since $OA_1 \\perp BC$, $AH \\parallel OA_1$.  Let $\\angle AA_2O = \\angle A_1AO = x$ and $\\angle AA_1O = \\angle A_2AO = y$.  Since we have $AH \\parallel OA_1$, we have that $\\angle HAA_2 = x$.  Also, we have that $\\angle A_2AA_1 = y-x$.  Furthermore, $\\angle BAH = 90 - \\angle B = \\angle OAC$, by lemma 1.  Therefore, $\\angle A_1AC = 90 - \\angle B + x = \\angle BAA_2$.  Since $A_1$ is the midpoint of $BC$, $AA_1$ is the median.  However $\\angle A_1AC =  \\angle BAA_2$ tells us that $AA_2$ is just $AA_1$ reflected across the internal angle bisector of $A$.  By definition, $AA_2$ is the $A$-symmedian.  Likewise, $BB_2$ is the $B$-symmedian and $CC_2$ is the $C$-symmedian.  Since the symmedians concur at the symmedian point, we are done.\nQED"
    },
    {
        "Problem": "Suppose $q_1,q_2,...$ is an infinite sequence of integers satisfying the following two conditions:\n(a) $m - n$ divides $q_m - q_n$ for $m>n \\geq 0$\n(b) There is a polynomial $P$ such that $|q_n|<P(n)$ for all $n$.\nProve that there is a polynomial $Q$ such that $q_n = Q(n)$ for each $n$.",
        "Solution": "Step 1: Suppose $P$ has degree $d$. Let $Q$ be the polynomial of degree at most $d$ with $Q(x)=q_x$ for $0\\leq x\\leq d$. Since the $q_x$ are all integers, $Q$ has rational coefficients, and there exists $k$ so that $kQ$ has integer coefficients. Then $m-n|kQ(m)-kQ(n)$ for all $m,n\\in \\mathbb N_0$.\nStep 2: We show that $Q$ is the desired polynomial.\nLet $x>n$ be given. Now\n\nSince $kQ(x)$ satisfies these relations as well, and $kq_m=kQ(m)$,\n\nand hence\n\nNow\n\nso by induction $\\text{lcm}(x,x-1,\\ldots, x-d)\\geq \\frac{x(x-1)\\cdots (x-d)}{d!(d-1)!\\cdots 1!}$. Since $P(x), Q(x)$ have degree $d$, for large enough $x$ (say $x>L$) we have $\\left|Q(x)\\pm\\frac{x(x-1)\\cdots (x-d)}{kd!(d-1)!\\cdots 1!}\\right|>P(x)$. By (1) $kq_x$ must differ by a multiple of $\\text{lcm}(x,x-1,\\ldots, x-d)$ from $kQ(x)$; hence $q_x$ must differ by a multiple of $\\frac{x(x-1)\\cdots (x-d)}{kd!(d-1)!\\cdots 1!}$ from $Q(x)$, and for $x>L$ we must have $q_x=Q(x)$.\nNow for any $y$ we have $kQ(y)\\equiv kQ(x)\\equiv kq_x \\equiv kq_y\\pmod{x-y}$ for any $x>L$. Since $x-y$ can be arbitrarily large, we must have $Q(y)=q_y$, as needed."
    },
    {
        "Problem": "Suppose that in a certain society, each pair of persons can be classified as either amicable or hostile. We shall say that each member of an amicable pair is a friend of the other, and each member of a hostile pair is a  foe of the other.  Suppose that the society has $\\, n \\,$ persons and $\\, q \\,$ amicable pairs, and that for every set of three persons, at least one pair is hostile.  Prove that there is at least one member of the society whose foes include $\\, q(1 - 4q/n^2) \\,$ or fewer amicable pairs.",
        "Solution": "Consider the graph with two people joined if they are friends.\nFor each person $X$, let $A(X)$ be the set of its friends and $B(X)$ the set of its foes. Note that any edge goes either: from $X$ to $A(X)$ (type 1), from $A(X)$ to $B(X)$ (type 2) or from a point of $B(X)$ to another (type 3), but there's no edge joining two points of $A(X)$ (since they would form a triangle with $X$). Let the number of type 1, type 2, type 3 edges of $X$ be $x_1, x_2, x_3$ respectively, so that $x_1$ is the degree of $X$ and we want to show that for some $X$, we have $x_3 \\leq q(1-4q/n^2)$.\nSince each edge is of one of those types, we have $x_1+x_2+x_3=q$. Thus\n\nThat is, what we want is equivalent to proving that for some vertex $X$, the set of edges touching either $X$ or a vertex joined to $X$ is at least $4q^2/n^2$. Obviously now we'll sum $x_1+x_2$ over all vertices $X$. In the resulting sum, an edge joining $X, Y$ is counted once for each edge that $X, Y$ have, that is, it is counted $D(X)+D(Y)$ times, where $D(X)=x_1$ is the degree of $X$. Thus each vertex $X$ contributes to the overall sum with $D(X)$ for each edge it has, and since it has $D(X)$ edges, it contributes with $D(X)^2$. Thus the considered sum is equal to\n\n(That's Cauchy.) Since we are summing over $n$ vertices, one of the summands is at least $4q^2/n^2$ by pigeonhole, which is what we wanted to prove.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$.",
        "Solution_1": "First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. \nThis simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired.\n$\\Box$",
        "Solution_2": "Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is  We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired.\n$\\Box$",
        "Solution_3_-hashtagmath": "We know that the average of a list of numbers is the sum of all the terms, divided by the number of terms. So we will set up an average. This average will average to $\\cot 1^\\circ$. So we can set it equal to $\\frac{\\cos 1^\\circ}{\\sin 1^\\circ}$. Doing so, gives us\nWe should try to simplify the numerator as it looks pretty messy.\nNow we know that $\\sin x = \\sin (180-x)$, so maybe we can use this to clear things up a bit. Applying this to $(1)$ gives us\nNow, we know that $\\sin 2^\\circ = \\sin 178^\\circ$, and $\\sin 4^\\circ = \\sin 176^\\circ$ and so on. We also know that $\\sin 0^\\circ = 0$, so we can omit the last term. Also, we know that $\\sin 90 = 1$. Thus $90 \\sin 90 = 90$. So we will use all these when we further simplify. Another thing that we can simplify is to notice that when we reach the term $92\\sin 88^\\circ$, we can rewrite it as $92\\sin 92^\\circ$. Thus we can rewrite everything in the form $x\\sin y^\\circ$ as $x \\sin x^\\circ$, starting with the term $92\\sin 88^\\circ$, then $94\\sin 86^\\circ$, and so on.\nThus we can further simplify our equation. Doing so, gives us\nThen, to make things easier, we can rewrite all the numbers with degrees less than $90^\\circ$ and combine like terms. Doing so gives us\nNow we can factor out $180$ to get\nThen we can simplify it by getting rid of the denominator to get to\nNow we notice that we can multiply both sides by $\\sin 1^\\circ$. Doing so gives us\nNow, simplifying $\\sin 2^\\circ+ \\sin 4^\\circ + \\sin 6^\\circ + \\sin 8^\\circ + \\sin 10^\\circ ... + \\sin 88^\\circ$ doesn't look too promising. So maybe if we expand again, we can maybe somehow use our product to sum formulas.\nDoing so, gives us\nNow we can use our product of sines to sum formula and see if we can find a pattern.\nWe will start with expanding $(\\sin 1^\\circ)(2 \\sin 2^\\circ)$. Because the format of our formula for the product of sines is $\\sin \\alpha + \\sin \\beta$, we can factor out the $2$ and find the product, then multiply by $2$. So, we are at $2(\\sin 1^\\circ)(\\sin 2^\\circ)$. Our formula for the product of sines is $\\sin \\alpha + \\sin \\beta = \\frac{1}{2}[\\cos(\\alpha - \\beta) - \\cos(\\alpha + \\beta)]$. Plugging in our values into the formula and simplifying gives us $2\\left(\\frac{1}{2}(\\cos 1^\\circ - \\cos 3^\\circ)\\right)$. We know that $2\\left(\\frac{1}{2}\\right)$ will just cancel out to $1$. So we are left with $\\cos 1^\\circ - \\cos 3^\\circ$. $(9)$\nNext we expand $(\\sin 1^\\circ)(2\\sin 4^\\circ)$. We use the same steps or strategy as we did above and get $\\cos 3^\\circ - \\cos 5^\\circ$. $(10)$\nWe may be noticing a pattern. Just to make sure it is true, we will expand $(\\sin 1^\\circ)(2\\sin 6^\\circ)$. When we expand this using the same strategy above, we get $\\cos 5^\\circ - \\cos 7^\\circ$. $(11)$ So our pattern is that $(\\sin 1^\\circ)(2\\sin \\mu^\\circ) = \\cos (\\mu-1)^\\circ - \\cos (\\mu+1)^\\circ. (12)$\nNow we notice that when we add $(9)$ and $(10)$, $-\\cos 3^\\circ$ and $+\\cos 3^\\circ$ cancel themselves out to just $\\cos 1^\\circ - \\cos 5^\\circ$. Then, when we add $(11)$ to it, $+ \\cos 5^\\circ$ and $- \\cos 5^\\circ$ also cancel themselves out.\nNow, we need to figure out when we should stop canceling out. We can use formula $(12)$ and replace $88$ with $\\mu$ since $88$ is the last one in $(8)$. Applying this formula and simplifying gives us $\\cos 87^\\circ - \\cos 89$, so the last term would be $- \\cos 89$.\nThus we know that everything we cancel themselves out except the first and the last term. So, we are left with\nNow we notice that we have $\\cos 1^\\circ$ on both sides. Thus we can subtract $\\cos 1^\\circ$ from both sides to get $- \\cos 89^\\circ + \\sin 1^\\circ = 0$. Now we can add $\\cos 89^\\circ$ to both sides and get $\\sin 1^\\circ = \\cos 89^\\circ$. Now we also know that $\\sin (\\omega)^\\circ = \\cos(90-\\omega)$. Thus we know that $\\sin 1^\\circ = \\cos 89^\\circ$. Hence, our proof is complete. $\\square$"
    },
    {
        "Problem": "For any nonempty set $S$ of real numbers, let $\\sigma(S)$ denote the sum of the elements of $S$. Given a set $A$ of $n$ positive integers, consider the collection of all distinct sums $\\sigma(S)$ as $S$ ranges over the nonempty subsets of $A$. Prove that this collection of sums can be partitioned into $n$ classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.",
        "Solution_1": "Let set $A$ consist of the integers $a_1\\le a_2\\le a_3\\le\\dots\\le a_n$. For $k\\ge 2$, call $a_k$ greedy if $\\sum_{i=1}^{k-1}a_i < a_k$. Also call $a_1$ greedy. Now put all elements of $A$ into groups of consecutive terms in such a way that each group $G$ begins with a greedy term, call it $a_p$, and ends on the term $a_{q-1}$ just before the next greedy term after $a_p$. (If $a_p$ is the last greedy term, let $q-1=n$.) We introduce some more terminology. A sum $\\sigma(S)$ is said to \"belong to\" a group $G$ if $\\max(S)\\in G$. Denote by $\\mathcal{S}(G)$ the set of all sums belonging to $G$.\nWe now show that we can divide $\\mathcal{S}(G)$ into $|G|$ (the cardinality of $G$) classes in such a way that the maximum and minimum sum belonging to each class differ by no more than a factor of 2. Using the previous notation, we first prove that $\\frac{\\max{\\mathcal{S}(G)}}{\\min{\\mathcal{S}(G)}}< 2^{|G|}=2^{q-p}$. Note that  and  Taking note of that fact that $a_{p+1},a_{p+2},\\dots,a_{q-1}$ are not greedy numbers, we write:\n\nwhere the inequalities after the ellipses result from the fact that $a_p$ is a greedy number (which implies by definition that $\\sum_{i=1}^{p-1}a_i<a_p$). This proves the desired inequality. Now we can prove the result in bold above. Divide $\\mathcal{S}(G)$ into $|G|=q-p$ classes by taking those terms in $[a_p,2a_p)$ and placing them in the first class, taking those terms in $[2a_p,2^2a_p)$ and placing them in another, and so on, until we reach $[2^{q-p-1}a_p,2^{q-p}a_p)$. The inequality we proved above shows that all of the sums in $\\mathcal{S}(G)$ will fall in one of these classes, as the intervals into which the classes fall form a continuous range bounded by $\\min\\mathcal{S}(G)=a_p$ on the bottom and $2^{q-p}a_p>\\max\\mathcal{S}(G)$ on the top. This proves the result in bold.\nHowever, that clearly implies the desired conclusion, as every sum belongs to a group, and every sum belonging to a group is a member of a class. Moreover, there will be precisely $n$ classes, because every term $a_i$ belongs to a group and for each group $G$, there are as many classes for $\\mathcal{S}(G)$ as there are terms in $G$.",
        "Solution_2": "We proceed to solve this problem by induction. Our base case is when $A$ has 1 integer - there is 1 class: itself.\nLet $A$ now have $n$ elements, and assume it works for this case. We add $k$, another element, to the set. WLOG, let $k$ be the maximum element. Note that $\\sigma(A)$ has to be the maximum element of a class.\nWe always get a new class that works: $\\sigma(A+\\{k\\})$ being the maximum element. This way all elements including $k$ will be in that class if $k>\\sigma(A)$. When $k<\\sigma(A)$ it also works since $k$ must be the maximum element from our WLOG condition above, so our induction step is done."
    },
    {
        "Problem": "Let $ABC$ be a triangle. Prove that there is a line $l$ (in the plane of triangle $ABC$) such that the intersection of the interior of triangle $ABC$ and the interior of its reflection $A'B'C'$ in $l$ has area more than $\\frac{2}{3}$ the area of triangle $ABC$.",
        "Solution": "Let the triangle be $ABC$. Assume $A$ is the largest angle. Let $AD$ be the altitude. Assume $AB \\le AC$, so that $BD \\le BC/2$. If $BD > \\frac{BC}{3}$, then reflect in $AD$. If $B'$ is the reflection of $B$, then $B'D = BD$ and the intersection of the two triangles is just $ABB'$. But $BB' = 2BD > \\frac{2}{3} BC$, so $ABB'$ has more than $\\frac{2}{3}$ the area of $ABC$.\nIf $BD < BC/3$, then reflect in the angle bisector of $C$. The reflection of $A'$ is a point on the segment $BD$ and not $D$. (It lies on the line $BC$ because we are reflecting in the angle bisector. $A'C > DC$ because $\\angle{CAD} < \\angle{CDA} = 90^{\\circ}$. Finally, $A'C \\le BC$ because we assumed $\\angle B$ does not exceed $\\angle A$). The intersection is at least $AA'C$. But  $\\frac{[AA'C]}{[ABC]} = \\frac{CA'}{CB} > CD/CB \\ge 2/3$."
    },
    {
        "Problem": "An $n$-term sequence $(x_1, x_2, \\ldots, x_n)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_n$ be the number of binary sequences of length n containing no three consecutive terms equal to 0, 1, 0 in that order. Let $b_n$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.\n(proposed by Kiran Kedlaya)",
        "Solution": "Given any binary sequence $B=(b_1,b_2,b_3,\\dots,b_k)$, define $f(B)=(|b_2-b_1|,|b_3-b_2|,\\dots,|b_k-b_{k-1}|)$. The operator $f$ basically takes pairs of consecutive terms and returns 0 if the terms are the same and 1 otherwise. Note that for every sequence $S$ of length $n$ there exist exactly two binary sequences $B$ of length $n+1$ such that $f(B)=S$.\nIf $f(B)$ does not contain the string 0, 1, 0, $B$ cannot contain either of the strings 0, 0, 1, 1 or 1, 1, 0, 0. Conversely, if $B$ does not contain the sequences 0, 0, 1, 1 or 1, 1, 0, 0, $f(B)$ cannot contain 0, 1, 0. There are $a_n$ such $f(B)$ and $b_{n+1}$ such $B$. Since each $S$ corresponds with two $B$, there are twice as many such $B$ as such $S$; thus, $b_{n+1}=2a_n$."
    },
    {
        "Problem": "Let $ABC$ be a triangle, and $M$ an interior point such that $\\angle MAB=10^\\circ$, $\\angle MBA=20^\\circ$ , $\\angle MAC= 40^\\circ$ and $\\angle MCA=30^\\circ$. Prove that the triangle is isosceles.",
        "Solution_1": "Clearly, $\\angle AMB = 150^\\circ$ and $\\angle AMC = 110^\\circ$. Now by the Law of Sines on triangles $ABM$ and $ACM$, we have  and  Combining these equations gives us  Without loss of generality, let $AB = \\sin 150^\\circ \\sin 30^\\circ = \\frac{1}{4}$ and $AC = \\sin 20^\\circ \\sin 110^\\circ$. Then by the Law of Cosines, we have\n\nThus, $AB = BC$, our desired conclusion.",
        "Solution_2": "By the law of sines, $\\frac{BM}{sin(10^\\circ)}=\\frac{AM}{sin(20^\\circ)}$ and $\\frac{CM}{sin(40^\\circ)}=\\frac{AM}{sin(30^\\circ)}$, so $\\frac{BM}{CM}=\\frac{sin(10^\\circ)sin(30^\\circ)}{sin(20^\\circ)sin(40^\\circ)}$.\nLet $\\angle MBC=x$. Then, $\\angle MCB=80^\\circ-x$. By the law of sines, $\\frac{BM}{CM}=\\frac{sin(80^\\circ-x)}{sin(x)}$.\nCombining, we have $\\frac{sin(80^\\circ-x)}{sin(x)}=\\frac{sin(10^\\circ)sin(30^\\circ)}{sin(20^\\circ)sin(40^\\circ)}$. From here, we can use the given trigonometric identities at each step:\nThe only acute angle satisfying this equality is $x=60^\\circ$. Therefore, $\\angle ACB=80^\\circ-x+30^\\circ=50^\\circ$ and $\\angle BAC=10^\\circ+40^\\circ=50^\\circ$. Thus, $\\triangle ABC$ is isosceles.",
        "Solution_3": "If $\\angle{MBC} = x$ then by Angle Sum in a Triangle we have $\\angle{MCB} = 80^\\circ - x$. By Trig Ceva we have\n\nBecause $\\dfrac{\\sin x}{\\sin (80^\\circ - x)}$ is monotonic increasing over $(0, \\dfrac{\\pi}{2})$, there is only one solution $0 \\le x \\le \\dfrac{\\pi}{2}$ to the equation. We claim it is $x = 60^\\circ$, which will make $ABC$ isosceles with $\\angle{A} = \\angle{C}$.\nNotice that\n\n\n\n\n\nas desired."
    },
    {
        "Problem": "Determine (with proof) whether there is a subset $X$ of the integers with the following property: for any integer $n$ there is exactly one solution of $a + 2b = n$ with $a,b \\in X$.",
        "Solution": "Start with an incomplete subset $S = (S_1, S_2, S_3, ... S_m)$, such that for any integer n, there is exactly zero or one solutions to $a + 2b = n$ with $a,b \\in S$. Let $N$ be the smallest integer such that for any $S_i$, $|S_i| < N$. Note that $|S_i+2S_j| < 3N$ for any $S_i$ and $S_j$\nSuppose $M$ is the smallest non-negative integer without a solution in $S$ yet. Clearly, $0 \\le M \\le 3N$. Generate $S_{m+1}$ and $S_{m+2}$ such that $S_{m+1} = -10N - M$, and $S_{m+2} = 5N + M$. Thus, we now have the solution $S_{m+1}+2S_{m+2} = M$.\nNote: The values 10 and 5 can be replaced by any sufficiently large values such that the first is twice the second.\nNow, we must prove that the addition of these two terms to $S$ does not result in an integer n that has two solutions. Of course, $S_{m+1} + 2S_{m+2} = M$ which previously had no solutions. Furthermore, $S_{m+1} + 2S_{m+2} = -15N - M$.\nSince all of these sums (other than $M$) are either greater than $3N$ or less than $-3N$, they are all sums that previously had no solutions. Furthermore, none of these sums are duplicated, as sums of different forms are contained in disjoint ranges of integers.\nThus, we have proved that we can generate a subset $S$ such that all non-negative integers n have a unique solution $a + 2b = n$.\nFor negative integers M that have no solutions in $S$ a similar proof holds, but instead generating the terms $S_{m+1} = 10N - M$ and $S_{m+2} = -5N + M$.\nFor any integer M that currently has no solution in S, we can always add two terms $S_{m+1}$ and $S_{m+2}$ such that $S_{m+1} + 2S_{m+2} = M$ that do not result in duplicated sums.\nThus, there does exist a subset $X$ of the integers such that for any integer $n$ there is exactly one solution to $a + 2b = n$ with $a, b \\in X$."
    },
    {
        "Problem": "Let $p_1,p_2,p_3,...$ be the prime numbers listed in increasing order, and let $x_0$ be a real number between $0$ and $1$. For positive integer $k$, define\n$x_{k}=\\begin{cases}0&\\text{ if }x_{k-1}=0\\\\ \\left\\{\\frac{p_{k}}{x_{k-1}}\\right\\}&\\text{ if }x_{k-1}\\ne0\\end{cases}$\nwhere $\\{x\\}$ denotes the fractional part of $x$. (The fractional part of $x$ is given by $x-\\lfloor{x}\\rfloor$ where $\\lfloor{x}\\rfloor$ is the greatest integer less than or equal to $x$.) Find, with proof, all $x_0$ satisfying $0<x_0<1$ for which the sequence $x_0,x_1,x_2,...$ eventually becomes $0$.",
        "Solution": "All rational numbers between 0 and 1 inclusive will eventually yield some $x_k = 0$. To begin, note that by definition, all rational numbers can be written as a quotient of coprime integers. Let $x_0 = \\frac{m}{n}$, where $m,n$ are coprime positive integers. Since $0<x_0<1$, $0<m<n$. Now  From this, we can see that applying the iterative process will decrease the value of the denominator, since $m<n$. Moreover, the numerator is always smaller than the denominator, thanks to the fractional part operator. So we have a strictly decreasing denominator that bounds the numerator. Thus, the numerator will eventually become 0.\nOn the other hand, if $x_0$ is irrational, then a simple induction will show that $x_k$ will always be irrational. Indeed, the base case has been established, and, if $x_k$ is irrational, then $\\dfrac{p_{k+1}}{x_k}$ must be too, and likewise for its fractional portion, which differs from it by an integer. Hence $x_{k+1}$ is irrational, completing the proof."
    },
    {
        "Problem": "$\\triangle ABC$ is a triangle. Take points $D, E, F$ on the perpendicular bisectors of $BC, CA, AB$ respectively. Show that the lines through $A, B, C$ perpendicular to $EF, FD, DE$ respectively are concurrent.",
        "Solution_1": "Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Theorem, The three lines are concurrent if\nBut this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC.\nQED",
        "Solution_2": "We split this into two cases:\nCase 1: $D,E,F$ are non-collinear\nObserve that since $D,E,F$ lie on perpendicular bisectors, then we get that $DC=DB$, $EC=EA$, and $FA=FB$. This motivates us to construct a circle $\\omega_1$ centered at $D$ with radius $DC$, and similarly construct $\\omega_2$ and $\\omega_3$ respectively for $E$ and $F$.\nNow, clearly $\\omega_1$ and $\\omega_2$ intersect at $C$ and some other point. Now, we know that $DE$ is the line containing the two centers. So, the line perpendicular to $DE$ and passing through $C$ must be the radical axis of $\\omega_1$ and $\\omega_2$, which is exactly the line that the problem describes! We do this similarly for the others pairs of circles.\nNow by the radical lemma, the pairwise radical axes of $\\omega_1,\\omega_2,\\omega_3$ are concurrent, as desired, and they intersect at the radical center.\nCase 2: $D,E,F$ are collinear\nNow, we are drawing perpendicular lines from $A$, $B$, and $C$ onto the single line $DEF$. Clearly, these lines are parallel and concur at the point of infinity."
    },
    {
        "Problem": "Prove that for any integer $n$, there exists a unique polynomial $Q(x)$ with coefficients in $\\{0,1,...,9\\}$ such that $Q(-2)=Q(-5)=n$.",
        "Solution": "The problem implies that $Q(x) = (x+2)(x+5) \\cdot H_y(x) + n$. We define $H_y(x) = a_0 + a_1x + ... + a_nx^n$. Then, after expanding, the coefficient of $x^k$ is $10a_k + 7a_{k-1} + a_{k-2}$ where $2 \\le k \\le n$. The constant term is $10a_0 + n$. Since this has to be within the set ${0,1,...,9}$, $a_0$ must be unique according to $n$. The linear term is determined by $10a_1 + 7a_0$, which also must be within the set ${0,1,...,9}$. Since $a_0$ is determined uniquely by $n$, in order to keep the linear coefficient in the set ${0,1,...,9}$, $a_1$ is also determined uniquely. With $a_0$ and $a_1$ determined uniquely, we move on to the general coefficient.\nNote that $10a_k + 7a_{k-1} + a_{k-2}$ is in the set ${0,1,...,9}$. We define $J_h(x)$ to be a function determining the remainder when $x$ is divided by $10$. Thus, for all $x$, $0 \\le J_h(x) \\le 9$. Note also that $J_h(10a_k + 7a_{k-1} + a_{k-2}) = 10a_k + 7a_{k-1} + a_{k-2}$, since $10a_k + 7a_{k-1} + a_{k-2}$ is in the range of $J_h(x)$. Therefore, $J_h(7a_{k-1} + a_{k-2}) = 10a_k + 7a_{k-1} + a_{k-2}$, which results in $a_k = \\lfloor \\frac{7a_{k-1} + a_{k-2}}{10} \\rfloor$ for all $2 \\le k \\le n$. Thus, $a_k$ is determined uniquely, since all coefficients $a_i$ where $0 \\le i < k$ are uniquely determined, which in turn determines a unique polynomial $Q(x)$."
    },
    {
        "Problem": "To clip a convex $n$-gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by three segments $AM, MN,$ and $NC,$ where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$. In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon. A regular hexagon $P_6$ of area $1$ is clipped to obtain a heptagon $P_7$. Then $P_7$ is clipped (in one of the seven possible ways) to obtain an octagon $P_8$, and so on. Prove that no matter how the clippings are done, the area of $P_n$ is greater than $\\frac{1}{3}$, for all $n\\ge6$.",
        "Solution": "\n$\\textbf{Claim:}$  It is impossible to choose two non-adjacent sides and clip a whole part of it off.\n$\\textbf{Proof:}$  If you clip adjacent sides, you can cut off at most up to the blue lines; Clipping more is impossible due to the degrees getting larger and larger and more and more circular.\nThus, after infinitely many clips, the hexagon bounded by the blue lines is left, so after finitely many clips, the area left is more than that hexagon.\nThe side length of that hexagon is $\\frac{\\sqrt{3}}{3}$ of the large one, because of 30-30-120 triangles. Thus, the area is $\\frac13$ of the larger one, so we are done."
    },
    {
        "Problem": "Prove that, for all positive real numbers $a, b, c,$\n$\\frac{1}{a^3+b^3+abc}+\\frac{1}{b^3+c^3+abc}+\\frac{1}{a^3+c^3+abc} \\le \\frac{1}{abc}$.",
        "Solution_1": "Because the inequality is homogenous (i.e. $(a, b, c)$ can be replaced with $(ka, kb, kc)$ without changing the inequality other than by a factor of $k^n$ for some $n$), without loss of generality, let $abc = 1$.\nLemma:\n\nProof: Rearranging gives $(a^3 + b^3) c + c \\ge a + b + c$, which is a simple consequence of $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and\nThus, by $abc = 1$:",
        "Solution_2": "Rearranging the AM-HM inequality, we get $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\le \\frac{9}{x+y+z}$. Letting $x=a^{3}+b^{3}+abc$, $y=b^{3}+c^{3}+abc$, and $z=c^{3}+a^{3}+abc$, we get  By AM-GM on $a^{3}$, $b^{3}$, and $c^{3}$, we have  So, $\\frac{1}{a^{3}+b^{3}+abc}+\\frac{1}{b^{3}+c^{3}+abc}+\\frac{1}{c^{3}+a^{3}+abc} \\le \\frac{1}{abc}$.\n-Tigerzhang\n$\\textbf{WARNING:}$\nThis solution doesn’t work because $(x+y+z)\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right)\\geq 9$, so $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\geq \\frac{9}{x+y+z}$",
        "Solution_3": "If we multiply each side by $abc$, we get that we must just prove that  If we divide our LHS equation, we get that   Make the astute observation that by Titu's Lemma,   Therefore:  If we expand it out, we get that  Since our original equation is less than this, we get that\n-KEVIN_LIU"
    },
    {
        "Problem": "Suppose the sequence of nonnegative integers $a_1,a_2,...,a_{1997}$ satisfies\n$a_i+a_j \\le a_{i+j} \\le a_i+a_j+1$\nfor all $i, j \\ge 1$ with $i+j \\le 1997$. Show that there exists a real number $x$ such that $a_n=\\lfloor{nx}\\rfloor$ (the greatest integer $\\le nx$) for all $1 \\le n \\le 1997$.",
        "Solution": "Given nonnegative integers $a_1,a_2,\\dots, a_{1997}$ satisfying the given inequalities, let $I_n$ be the set of all $x$ such that $a_n=\\lfloor nx\\rfloor$. Therefore,\n\nWe can rewrite this as\n\nSo we know that each $I_n$ is an interval. We start by proving that if $n$ and $k$ are positive integers, then $I_n\\cap I_{k}\\ne \\emptyset$. To prove this, we need the following lemma.\nLemma 1: If $n$ and $k$ are positive integers, then\nProof: We prove this by induction. If $k=1$, then we wish to show that $na_1< a_n+1$. By the given lower bound, we know that\n\nTherefore, $na_1\\le a_n<a_n+1$, so the hypothesis is true for $k=1$. If $n=1$, then we wish to prove that $a_k < k(a_1+1)$. By the given upper bound, we know that\n\nTherefore, $a_k<ka_1+k=k(a_1+1)$, so the hypothesis is true for $n=1$.\nNow suppose that (2) holds for all $k,n<j$. If $k=n=j$, then (2) is equivalent to $0<j$, which is obviously true. If $n=j>k$, then by the division algorithm, we can write $n=qk+r$ for nonnegative integers $q$ and $r$ with $0\\le r<k$. Thus by applying the lower bound repeatedly (with one of the indices equal to $1$), we find\n\nBut then as $k,r<j$, we can apply the inductive hypothesis with $n=r$ and $k=k$ to find $k(a_r+1)> ra_k$. Substituting this into (3), we find\n\nThis proves the hypothesis for $n=j>k$.\nSuppose that $k=j>n$. Then by the division algorithm, we can write $k=qn+r$ for nonnegative integers $q$ and $r$ with $0\\le r<n$. Then by applying the upper bound repeatedly (with one of the indices equal to $1$), we find\n\nBut as $n,r<j$, we can apply the inductive hypothesis with $n=n$ and $k=r$, finding that $na_r< r(a_n+1)$. Subsituting this into (4), we find\n\nThis proves the hypothesis for $k=j>n$.\nTherefore, if the hypothesis is true for all $k,n<j$, then it must be true for all $k,n\\le j$. Hence by induction, it must be true for all positive integers $k$ and $n$. $\\hfill\\ensuremath{\\square}$\nNow suppose for the sake of contradiction that $I_n$ and $I_k$ are disjoint intervals. Without loss of generality, we may assume that $I_n$ precedes $I_k$ on the number line. Hence the upper bound of $I_n$ is less than or equal to the minimum value of $I_k$, or rather\n\nThis simplifies to $na_k\\ge k(a_n+1)$. But this contradicts the statement of Lemma 1. Therefore, $I_k\\cap I_n\\ne \\emptyset$.\nNow we claim that $I_1\\cap I_2\\cap \\cdots \\cap I_{1997}\\ne \\emptyset$. We prove this by induction. By the above, we know that $I_1\\cap I_2\\ne \\emptyset$. Now suppose that $I_1\\cap I_2\\cap\\cdots\\cap I_k\\ne \\emptyset$. The intersection of two overlapping intervals of the form $[a,b)$ and $[c,d)$ is an interval of the form $[e,f)$, where $e=\\max\\{a,c\\}$ and $f=\\min\\{b,d\\}$. Therefore, by induction, we know that if the intersection of $k$ overlapping intervals is nonempty, then it must also be an interval, say\n\nIf $I_{k+1}$ does not intersect $[x_k,y_k)$, then as an interval, it must appear either completely before $x_k$ or completely after $y_k$.  If $I_{k+1}$ appears completely before $x_k$, then it has a nonempty intersection with each of $I_1, I_2, \\dots, I_k$. But we also know that $x_k$ is a lower bound of one of the intervals, hence $I_{k+1}$ cannot intersect that interval, a contradiction. A similar contradiction arises if $I_{k+1}$ appears completely after $y_k$. Therefore,\n\nThus by induction, $I_1\\cap I_2\\cap\\cdots\\cap I_{1997}\\ne \\emptyset$. So let $x\\in I_1\\cap I_2\\cap\\cdots\\cap I_{1997}$. Then by the definition of $I_n$, we know that $a_n=\\lfloor nx\\rfloor$ for all $1\\le n\\le 1997$, and we are done."
    },
    {
        "Problem": "Suppose that the set $\\{1,2,\\cdots, 1998\\}$ has been partitioned into disjoint pairs $\\{a_i,b_i\\}$ ($1\\leq i\\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum  ends in the digit $9$.",
        "Solution": "Notice that $|a_i - b_i| \\equiv 1 \\pmod 5$, so $S=|a_1-b_1|+|a_2-b_2|+\\cdots +|a_{999}-b_{999}| \\equiv 1+1+\\cdots + 1 \\equiv 999 \\equiv 4 \\bmod{5}$.\nAlso, for integers $M, N$ we have $|M-N| \\equiv M-N \\equiv M+N \\bmod{2}$.\nThus, we also have $S \\equiv a_1+b_1+a_2+b_2+\\cdots +a_{999}+b_{999} \\equiv 1+2+ \\cdots +1998 \\equiv 999*1999 \\equiv 1 \\bmod{2}$ also, so by the Chinese Remainder Theorem $S \\equiv 9\\bmod{10}$. Thus, $S$ ends in the digit 9, as desired."
    },
    {
        "Problem": "Let ${\\cal C}_1$ and ${\\cal C}_2$ be  concentric circles, with ${\\cal C}_2$ in the interior of  ${\\cal C}_1$. From a point $A$ on ${\\cal C}_1$ one draws the tangent $AB$ to ${\\cal C}_2$ ($B\\in {\\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular  bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof,  the ratio $AM/MC$.",
        "Solution": "First, $AD=\\frac{AB}{2}=\\frac{AC}{4}$. Because $E$,$F$ and $B$ all lie on a circle, $AE \\cdot AF=AB \\cdot AB=\\frac{AB}{2} \\cdot 2AB=AD \\cdot AC$. Therefore, $\\triangle ACF \\sim \\triangle AED$, so $\\angle ACF = \\angle AED$. Thus, quadrilateral $CFED$ is cyclic, and $M$ must be the center of the circumcircle of $CFED$, which implies that $MC=\\frac{CD}{2}$. Putting it all together,\n$\\frac{AM}{MC}=\\frac{AC-MC}{MC}=\\frac{AC-\\frac{CD}{2}}{\\frac{CD}{2}}=\\frac{AC-\\frac{AC-AD}{2}}{\\frac{AC-AD}{2}}=\\frac{AC-\\frac{3AC}{8}}{\\frac{3AC}{8}}=\\frac{\\frac{5AC}{8}}{\\frac{3AC}{8}}=\\frac{5}{3}$\nBorrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html",
        "Solution_2": "We will use signed lengths. WLOG let $AC = 4$, $AM = x$, and $AE = y > 0$. Then $AB = 2$ and $AD = 1$. Therefore, $DM = x - 1$ and $MC = 4 - x$. By Power of a Point, $AE \\cdot AF = AB^2 = 4$, so $AF = \\frac{4}{AE} = \\frac{4}{y}$, and $EF = \\frac{4}{y} - y = \\frac{4 - y^2}{y}$. Since $M$ is on the perpendicular bisectors of $DE$ and $CF$, we have $|DM| = |EM| = |x - 1|$ and $|CM| = |FM| = |4 - x|$.\nBy Stewart's Theorem on $\\triangle MFA$ and cevian $DM$,\nSo either $8x - 20 = 0$ or $y^2 - 1 = 0$. If $y^2 - 1 = 0$, then $AE = y = 1 = AD$ and $AF = AC = 4$, so the perpendicular bisectors of $CF$ and $DE$ are the same line, and they do not intersect at a point. Therefore, $AM = x = \\frac{5}{2}$ and $MC = \\frac{3}{2}$, so $\\frac{AM}{MC} = \\boxed{\\frac{5}{3}}$."
    },
    {
        "Problem": "Let $a_0,\\cdots a_n$ be real numbers in the interval $\\left(0,\\frac {\\pi}{2}\\right)$ such that\n\nProve that $\\tan{\\left(a_0\\right)}\\tan{\\left(a_1\\right)}\\cdots \\tan{\\left(a_n\\right)}\\ge n^{n + 1}$.",
        "Solution": "Let $y_i = \\tan{(a_i - \\frac {\\pi}{4})}$, where $0\\le i\\le n$. Then we have\nBy AM-GM,\n\nNote that by the addition formula for tangents, .\nSo $\\prod_{i = 0}^n{\\frac {1 + y_i}{1 - y_i}} = \\tan{(a_0)}\\tan{(a_1)}\\cdots \\tan{(a_n)}\\ge n^{n + 1}$, as desired. $\\blacksquare$"
    },
    {
        "Problem": "A computer screen shows a $98 \\times 98$ chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result,  the colors in the selected rectangle switch (black becomes white, white becomes black). Find, with proof, the minimum number of mouse clicks needed to make the chessboard all one color.",
        "Solution": "Answer: $98$.\nThere are $4\\cdot97$ adjacent pairs of squares in the border and each pair has one black and one white square. Each move can fix at most $4$ pairs, so we need at least $97$ moves. However, we start with two corners one color and two another, so at least one rectangle must include a corner square. But such a rectangle can only fix two pairs, so at least $98$ moves are needed.\nIt is easy to see that 98 moves suffice: take 49 $1\\times98$ rectangles (alternate rows), and 49 $98\\times1$ rectangles (alternate columns).\ncredit: https://mks.mff.cuni.cz/kalva/usa/usoln/usol984.html\neditor: Brian Joseph\nsecond editor: integralarefun"
    },
    {
        "Problem": "Prove that for each $n\\geq 2$, there is a set $S$ of $n$ integers such that $(a-b)^2$ divides $ab$ for every distinct $a,b\\in S$.",
        "Solution": "Proof by induction.  For n=2, the proof is trivial, since $S = (1,2)$ satisfies the condition.  Assume now that there is such a set S of n elements, $a_1, a_2,...a_n$ which satisfy the condition.  The key is to note that if $m=a_1a_2...a_n$, then if we define $b_i=a_i + km$ for all $i\\le n$, where k is a positive integer, then $a_i \\mid b_i$ and $b_i - b_j = a_i - a_j$, and so $(b_i - b_j)^2 = (a_i - a_j)^2 \\mid a_ia_j \\mid b_ib_j$.\nLet $b_{n+1}=m +km$.   Consider the set $T = (b_1,b_2,...,b_n,b_{n+1})$. To finish the proof, we simply need to choose a k such that $(b_{n+1}-b_i)^2 \\mid b_{n+1}b_i$ for all $i\\le n$.  Since $(b_{n+1}-b_i)^2 = (m-a_i)^2$, simply choose k so that $k+1 = (m-a_1)^2(m-a_2)^2...(m-a_n)^2$."
    },
    {
        "Problem": "Let $n \\geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$) such that there exists a convex $n$-gon $A_{1}A_{2}\\dots A_{n}$ for which exactly $k$ of the quadrilaterals $A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle. (Here $A_{n+j} = A_{j}$.)",
        "Solution": "Lemma: If quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ in an equiangular $n$-gon are tangential, and $A_iA_{i+3}$ is the longest side quadrilateral $A_iA_{i+1}A_{i+2}A_{i+3}$ for all $i$, then quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not tangential.\nProof:\n\nIf quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ are tangential, then $A_iA_{i+3}$ must have side length of $j+y-u$, and $A_{i+2}A_{i+5}$ must have side length of $h + y - n$ (One can see this from what is known as walk-around). Suppose quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is tangential. Then, again, we see that $A_{i+1}A_{i+4}$ must have side length $u + n - y$. We assumed by lemma that $A_iA_{i+3} > A_{i}A_{i+1}$ for all $i$, so we have $A_iA_{i+3} > j$, $A_{i+1}A_{i+4} > y$, and $A_{i+2}A_{i+5} > h$. If we add up the side lengths $A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5}$, we get:\nHowever, by the lemma, we assumed that $A_iA_{i+3} > j$, $A_{i+1}A_{i+4} > y$, and $A_{i+2}A_{i+5} > h$. Adding these up, we get:\nwhich is a contradiction. Thus, quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not tangential, proving the lemma.\n\nBy lemma, the maximum number of quadrilaterals in a $n$-gon occurs when the tangential quadrilaterals alternate, giving us $k = \\lfloor \\frac{n}{2} \\rfloor$.\nNote that one can find the ratio of side of an equiangular $n$-gon in order for alternating quadrilaterals to be tangential.\n\nSince exterior angles of a equiangular $n$-gon have degree measure $\\frac{2pi}{n}$, one can write the equation:\n\nThus, we can find the ratio of sides of an equiangular $n$-gon which fits the maximum to be $1 : 1- \\cos \\frac{2\\pi}{n}$. Note that if $n$ is even, we can easily alternate them, but if $n$ is odd, we must have two adjacent sides be the same length, and that length must be the larger side in the ratio  of adjacent sides. The proof is left as an exercise for the reader."
    },
    {
        "Problem": "Some checkers placed on an $n\\times n$ checkerboard satisfy the following conditions:\n(a) every square that does not contain a checker shares a side with one that does;\n(b) given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares, such that every two consecutive squares of the sequence share a side.\nProve that at least $(n^{2}-2)/3$ checkers have been placed on the board.",
        "Solution_1": "For the proof let's look at the checkers board as a graph $G$, where the checkers are the vertices and the edges are every pair of checkers sharing a side.\nDefine $R$ as the circuit rank of $G$(the minimum number of edges to remove from a graph to remove all its cycles).\nDefine $x$ as the number of checkers placed on the board.\n1. From (a) for every square containing a checker there can be up to 4 distinct squares without checkers; thus, $4\\times x + x = 5\\times x \\ge n^2$ (this is the most naive upper bound).\n2. From (b) this graph has to be connected. which means that no square which contains a checker can share 4 sides with squares that doesn't contain a checker (because it has to share at least one side with a square that  contains a checker).\n3. Now it is possible to improve the upper bound. Since every edge in G represents a square with a checker that shares a side with a square with another checker, the new upper bound is $5\\times x - \\text{[sum of degrees in G]} \\ge n^2.$\n4. Thus, $5x - (2x - 2 + 2R) \\ge n^2$ (see lemma below).\n5. Thus,  $3x + 2 - 2R \\ge n^2.$\n6. Thus,  $x \\ge \\frac{n^2 - 2 + 2R}3,$ where $R\\ge0$ and is equal to 0 if $G$ is a tree.\n$\\textit{Clarification/Lemma:}$ The sum of degrees of a connected graph $G = (V,E)$ is $2V -2 + 2R = 2E,$ where $R$ is the circuit rank of $G$.\n$\\textit{Proof.}$ $2 \\times V -2$ is the sum of ranks of the spanning tree created by decircuiting the graph $G$. Since the circuit rank of $G$ is $R$, $2R$ is the sum of ranks removed from $G$. Thus,  $2V -2 + 2R = 2 E.$ $\\blacksquare$",
        "Solution_2": "Call a square of the checkerboard “good” if it either has a checker on it or\nshares a side with a square with a checker on it.\nSay there are $k$ checkers on the board, sitting on squares $s_1, \\ldots, s_k$. We can assume WLOG that $s_1, \\ldots, s_k$ is ordered in such a way that if $1 \\leq i \\leq k$, then $s_i$ shares a side with at least one of $s_1, \\ldots, s_{i - 1}$. (If such an ordering were impossible, then there would be an $s_i$ which isn’t connected to $s_1$ by a sequence of squares with checkers on them, violating condition (b).)\nNow imagine removing all the checkers and then putting them back onto the squares $s_1, \\ldots, s_k$ in that order, counting the number of good squares after each\nstep. When we put a checker on $s_1$, we increase the number of good squares\nby at most 5: the square $s_1$ becomes good, and each square it\nshares a side with (at most 4) becomes good. When we put a checker onto $s_i$,\nwhere $i > 1$, we increase the number of good squares by at most $3$, since $s_i$ is\nalready good, and at least $1$ of its at most $4$ neighbors already has a checker on it.\nSo the total number of good squares when we’ve put back every checker is at most $5 + 3(k - 1) = 3k + 2$. In order to satisfy condition (a), we have to make each of the $n^2$ squares good. In other words, we need $3k + 2 \\geq n^2$, or"
    },
    {
        "Problem": "Let $ABCD$ be a cyclic quadrilateral. Prove that",
        "Solution": "Let arc $AB$ of the circumscribed circle (which we assume WLOG has radius 0.5) have value $2x$, $BC$ have $2y$, $CD$ have $2z$, and $DA$ have $2w$. \nThen our inequality reduces to, for $x+y+z+w = 180^\\circ$:\nThis is equivalent to by sum-to-product and use of $\\cos x = \\sin (90^\\circ - x)$:\n\nClearly $90^\\circ \\ge \\frac{x+z}{2} > \\frac{x-z}{2} \\ge 0^\\circ$. As sine is increasing over $[0, \\pi/2]$, $|\\sin \\frac{x+z}{2}| > |\\sin \\frac{x-z}{2}|$.\nSimilarly, $|\\sin \\frac{y+w}{2}| > |\\sin \\frac{y-w}{2}|$. The result now follows after multiplying the first inequality by $|\\sin \\frac{x-z}{2}|$, the second by $|\\sin \\frac{y-w}{2}|$, and adding. (Equality holds if and only if $x=z$ and $y=w$, ie. $ABCD$ is a parallelogram.)\n--Suli 11:23, 5 October 2014 (EDT)"
    },
    {
        "Problem": "Let $p > 2$ be a prime and let $a,b,c,d$ be integers not divisible by $p$, such that\n\nfor any integer $r$ not divisible by $p$. Prove that at least two of the numbers $a+b$, $a+c$, $a+d$, $b+c$, $b+d$, $c+d$ are divisible by $p$.\n(Note: $\\{x\\} = x - \\lfloor x \\rfloor$ denotes the fractional part of $x$.)",
        "Solution": "We see that $\\biggl\\{\\frac{ra+rb+rc+rd}{p}\\biggr\\}=0$ means that $p|r(a+b+c+d)$. Now, since $p$ does not divide $r$ and $p$ is prime, their GCD is 1 so $p\\mathrel{|}a+b+c+d$.\nSince $\\biggl\\{\\frac{ra}{p}\\biggr\\}+\\biggl\\{\\frac{rb}{p}\\biggr\\}+\\biggl\\{\\frac{rc}{p}\\biggr\\}+\\biggl\\{\\frac{rd}{p}\\biggr\\}=2$, then we see that they have to represent mods $\\bmod\\medspace p$, and thus, our possible values of $p$ are all such that $k^4 \\equiv 1\\pmod{p}$  for all $k$ that are relatively prime to $p$. This happens when $p=3$ or $5$.\nWhen $p=3$ then $r$ is not divisible by 3, thus two are $1$, and the other two are $2$. Thus, four pairwise sums sum to 3.\nWhen $p=5$ then $r$ is not divisible by 5 so $a, b, c, d$ are $1, 2, 3,$ and $4$, so two pairwise sums sum to 5.\nAll three possible cases work so we are done.\n(This solution makes absolutely no sense. Why is $k^4\\equiv 1$? And how do we know that only $3$ and $5$ work!?)"
    },
    {
        "Problem": "Let $a_{1}, a_{2}, \\dots, a_{n}$ ($n > 3$) be real numbers such that\n\nProve that $\\max(a_{1}, a_{2}, \\dots, a_{n}) \\geq 2$.",
        "Solution_1": "First, suppose all the $a_i$ are positive.  Then\n\nSuppose, on the other hand, that without loss of generality,\n\nwith $1\\le k <n$.  If $a_1 >2$ we are done, so suppose that $a_1 \\le2$.  Then\n$\\sum_{i=1}^k a_i \\le 2k$, so \n\nSince $-a_i$ is a positive real for all $k+1 \\le i \\le n$, it follows that\n\nThen\n\nSince $k<n$, $4(n-k) > 4$.  It follows that $\\max(a_1, \\dotsc, a_n) \\ge \\sqrt{4} = 2$, as desired.  $\\blacksquare$",
        "Solution_2": "Assume the contrary and suppose each $a_i$ is less than 2.\nWithout loss of generality let $a_1 \\le a_2 \\le a_3 \\le ... \\le a_n < 2$, and let $k$ be the largest integer such that $a_k \\le 0$ and $0 \\le a_{k+1}$ if it exists, or 0 if all the $a_i$ are non-negative. If $k = 0$, then (as $n \\ge 4$) $\\sum_{i=1}^n a_i^2 < \\sum_{i=1}^n 4 = 4n \\le n^2$, a contradiction. Hence, assume $n \\ge k \\ge 1$. Then\n\nBecause $a_i \\le 0$ for $i \\le k$, both sides of the inequality are non-positive, so squaring flips the sign. But we also know that $a_i a_j \\ge 0$ for $i, j \\le k$, so\n\nwhich results in\n\na contradiction to our given condition. The proof is complete.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "The Y2K Game is played on a $1 \\times 2000$ grid as follows. Two players in turn write either an S or an O in an empty square. The first player who produces three consecutive boxes that spell SOS wins. If all boxes are filled without producing SOS then the game is a draw. Prove that the second player has a winning strategy.",
        "Solution": "Label the squares $1$ through $2000$. The first player writes a letter in any square $k$. The second player now selects any empty square $l$ such that $\\text{min}\\,(|l-1|,|l-2000|,|l-k|) > 3$, and fills in a S. The first player may again play anywhere, say $m$. If the second player can now complete SOS, he does so; otherwise, if $l > m$ he fills an S in $l + 3$ and if $l < m$ he fills an S in $l-3$.\nWe thus have two Ss separated by two empty squares. Note that if any player plays any letter between the two Ss, the other player can always complete a SOS.\nNow the second player repeats the following algorithm:\nEventually, assuming the second player has not won yet (and since the first player does not have a winning opportunity yet), this algorithm must stop. It follows that all of the empty squares are now in pairs. Since there are an even number of empty squares, it is the first player's turn. If the first player plays on one of any pair of empty squares that does not have Ss on both its sides, then the second player plays any letter in the other square of the pair. Then, we will only have pairs of empty squares with Ss on both sides remaining (since from the first moves we have constructed at least one, and no one has played between them otherwise the game will have ended), and the first player will have to play between them, so that the second player will win."
    },
    {
        "Problem": "Let $ABCD$ be an isosceles trapezoid with $AB \\parallel CD$. The inscribed circle $\\omega$ of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a point on the (internal) angle bisector of $\\angle DAC$ such that $EF \\perp CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$. Prove that the triangle $AFG$ is isosceles.",
        "Solution": "Quadrilateral $ABCD$ is cyclic since it is an isosceles trapezoid. $AD=BC$. Triangle $ADC$ and triangle $BCD$ are reflections of each other with respect to diameter which is perpendicular to $AB$. Let the incircle of triangle $ADC$ touch $DC$ at $K$. The reflection implies that $DK=CE$, which then implies that the excircle of triangle $ADC$ is tangent to $DC$ at $E$. Since $EF$ is perpendicular to $DC$ which is tangent to the excircle, this implies that $EF$ passes through center of excircle of triangle $ADC$.\nWe know that the center of the excircle lies on the angular bisector of $DAC$ and the perpendicular line from $DC$ to $E$. This implies that $F$ is the center of the excircle.\nNow $\\angle GFA=\\angle GCA=\\angle DCA$. \n$\\angle ACF=90+\\frac{\\angle DCA}{2}$.\nThis means that $\\angle AGF=90-\\frac{\\angle ACD}{2}$. (due to cyclic quadilateral $ACFG$ as given).\nNow $\\angle FAG - (\\angle AFG + \\angle FGA)=90-\\frac{\\angle ACD}{2}=\\angle AGF$.\nTherefore $\\angle FAG=\\angle AGF$. \nQED."
    },
    {
        "Problem": "Call a real-valued function $f$ very convex if\n\nholds for all real numbers $x$ and $y$. Prove that no very convex function exists.",
        "Solution_1": "Let $y \\ge x$, and substitute $a = x, 2b = y-x$. Then a function is very convex if $\\frac{f(a) + f(a+2b)}{2} \\ge f(a + b) + 2b$, or rearranging,\n\nLet $g(a) = \\frac{f(a+b) - f(a)}{b}$, which is the slope of the secant between $(a,f(a))(a+b,f(a+b))$. Let $b$ be arbitrarily small; then it follows that $g(a+b) - g(a) > 4$, $g(a+2b) - g(a+b) > 4,\\, \\cdots, g(a+kb) - g(a+ [k-1]b) > 4$. Summing these inequalities yields $g(a+kb)-g(a) > 4k$. As $k \\rightarrow \\infty$ (but $b << \\frac{1}{k}$, so $bk < \\epsilon$ is still arbitrarily small), we have $\\lim_{k \\rightarrow \\infty} g(a+kb) - g(a) = g(a + \\epsilon) - g(a) > \\lim_{k \\rightarrow \\infty} 4k = \\infty$. This implies that in the vicinity of any $a$, the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.\n",
        "Solution_2": "Suppose, for the sake of contradiction, that there exists a very convex function $f.$ Notice that $f(x)$ is convex if and only if $f(x) + C$ is convex, where $C$ is a constant. Thus, we may set $f(0) = 0$ for convenience.\nSuppose that $f(1) = A$ and $f(-1) = B.$ By the very convex condition,\n\nA straightforward induction shows that:\n\nfor all nonnegative integers $n.$\nUsing a similar line of reasoning as above,\n\nTherefore, for every nonnegative integer $n,$\n\nNow, we choose $n$ large enough such that $n > \\frac{A+B}{4} - 1.$ This is always possible because $A$ and $B$ are fixed for any particular function $f.$ It follows that:\n\nHowever, by the very convex condition,\n\nThis is a contradiction. It follows that no very convex function exists."
    },
    {
        "Problem": "Let $S$ be the set of all triangles $ABC$ for which\n\nwhere $r$ is the inradius and $P, Q, R$ are the points of tangency of the incircle with sides $AB, BC, CA,$ respectively. Prove that all triangles in $S$ are isosceles  and similar to one another.",
        "Solution": "We let $x = AP = s - a, y = BQ = s-b, z = CR = s-c$, and without loss of generality let $x \\le y \\le z$. Then $x + y + z = 3s - (a+b+c) = s$, so $r = \\frac {A}{s} = \\frac{\\sqrt{(x+y+z)xyz}}{x+y+z} = \\sqrt{\\frac{xyz}{x+y+z}}$. Thus,\n\nSquaring and simplifying yields (after much grueling work)\n\nWe claim that the inequality\n\nholds true, with equality iff $4x = y = z$.\nNote that $(*)$ is homogeneous in $x,y,z$, so without loss of generality, scale so that $x=1$. Then\n\nwhich is a quadratic in $y$. As $4z^2 - 16z + 25 = 4(z-4)^2 + 9 \\ge 0$, it suffices to show that the quadratic cannot have more than one root, or the discriminant $\\Delta \\le 0$. Then,\n\nas desired. Equality comes when $z = 4$; since $(*)$ is symmetric in $y$ and $z$, it follows that $y = 4$ is also necessary for equality. Reversing our scaling, it follows that $x:y:z = 1:4:4$.\nThen $s = x+y+z = 9x$, and $s-a = x, s-b = 4x, s-c = 4x$ yields $a:b:c = 8:5:5$. Thus, we have proved that all possible ABC are isosceles (as b = c), and that they are similar to a 5-5-8 triangle.\n$\\blacksquare$",
        "Solution_2": "Let $A$, $B$, $C$ be the three angles of the triangle, then $r/AP = \\tan{(A/2)}$, $r/BQ=\\tan{(B/2)}$, and $r/CR=\\tan{(C/2)}$. Without loss of generality, assume $A \\leq B \\leq C$, then $CR=min{(AP,BQ,CR)}$. Denote $x=\\tan{(A/2)}$, $x=\\tan{(B/2)}$, $z=\\tan{(C/2)}$, then we have,\n\nMeanwhile,\n\nSubstitute $z$ in $(1)$, we get\n\nTreating (2) as a quadratic equation for $x$, the discriminant is:\n\nFor $x$, $y$ to be both real numbers, $\\Delta$ must not be negative, so $\\Delta = 0$, which yields $y=1/3$. Solving for x, we get $x=\\frac{-(8y-6)}{2*5}=1/3$. Using the formula $\\tan(A)=\\frac{2\\tan(A/2)}{1-\\tan^2(A/2)}$, we have $\\tan(A)=\\tan(B)=3/4$. Therefore $A=B=\\arctan{(3/4)}$. This proves that all possible triangle $ABC$ are isosceles with the same set of angles, i.e., they are similar to each other.\nIt's inconsequential to the proof, but it can be easily found that, since $tan(A) = 3/4$, the altitude to the base of the isosceles triangle divides the triangle into two 3-4-5 right triangles with $4$ on the base; therefore, the entire triangle is 5-5-8. $\\square$.\nSolution by $Mathdummy$."
    },
    {
        "Problem": "A game of solitaire is played with $R$ red cards, $W$ white cards, and $B$ blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of $R, W,$ and $B,$ the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.",
        "Solution": "We claim (inductively) that the minimum is just going to be $\\min(BW,2WR,3RB)$. We'll start our induction with the case where one of the three quantities is zero, in which case we verify that we can indeed get away without any penalty by, for example, discarding blue if we are out of white.\nNow, for the inductive step, let $f(B,W,R)$ be the minimum we seek. Note that\n\nBy our inductive hypothesis, $f(B-1,W,R) = \\min((B-1)W,2WR,3R(B-1))$. In order for this to cause our inductive step not to hold, we would require that $W+\\min((B-1)W,2WR,3R(B-1)) < \\min(BW,2WR,3RB)$. It is evident that the first two entries in the $min$ expression cannot cause this to happen, so that we need only consider $W+3R(B-1) < \\min(BW,2WR,3RB)$. So $W+3R(B-1) < BW$, whence $3R < W$. But $W+3R(B-1) < 3RB$, so that $W < 3R$, a contradiction.\nFor the other two cases, we can get similar contradictions, so that our inductive step must hold, and so $f(B,W,R)$ is indeed $\\min(BW,2WR,3RB)$.\nWe now need only establish how many ways to do this. If one of these quantities is smaller, our induction and the fact that it is eventually zero guarantees that it will continue to be the smallest quantity as cards are discarded. (For example, if it is currently optimal to discard a blue card, it will continue to be so until we run out of blue cards.) Therefore, assuming that there is currently only one best choice of card to discard, this will continue to be so in the future, whence if $BW \\neq 2WR \\neq 3RB$, there is only $1$ optimal strategy.\nSuppose, now, that $BW = 2WR$. It is thus optimal to discard either a $B$ or $W$ card. If we ever discard a blue card, then we will cause $BW < 2WR$, whence there is only one possible strategy from then on. However, if we discard a white card, then we will still have $BW = 2WR$, meaning that we continue to have the choice of discarding a white or blue card. Since we can discard a white card at most $W$ times, there are $W+1$ choices for how many $W$ cards to discard ($0$ to $W$), meaning that there are $W+1$ optimal strategies.\nBy similar logic, we get $R+1$ optimal strategies if $2WR = 3RB$, and $B+1$ optimal strategies if $3RB = BW$.\nThe final case, then, is if $BW = 2WR = 3RB$. In this case, if we first discard a white card, we are left with the $BW = 2WR$ case, and similarly for a blue and red card. The total number of optimal strategies in this case is then the sum of the optimal strategies in those cases, or, in other words, $B+W+R$.\nTo summarize:\nThe minimum penalty is $\\min(BW,2WR,3RB)$.\nIf $BW \\neq 2WR \\neq 3RB$, there is $1$ optimal strategy.\nIf $BW = 2WR < 3RB$, there are $W+1$ strategies.\nIf $2WR = 3RB < BW$, there are $R+1$ strategies.\nIf $3RB = BW < 2WR$, there are $B+1$ strategies.\nIf $BW = 2WR = 3RB$, there are $R+B+W$ strategies.\nBy J Steinhardt, from AoPS Community"
    },
    {
        "Problem": "Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \\times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.",
        "Solution": "We claim that $n = 1999$ is the smallest such number. For $n \\le 1998$, we can simply color any of the $1998$ squares forming the top row and the left column, but excluding the top left corner square.\nWe now show that no configuration with no colored right triangles exists for $n = 1999$. We call a row or column filled if all $1000$ of its squares are colored. Then any of the remaining $999$ colored squares must share a column or row, respectively, with one of the colored squares in a filled row or column. These two squares, and any other square in the filled row or column, form a colored right triangle, giving us a contradiction. Hence, no filled row or column may exist.\nLet $m$ be the number of columns with $1$ colored square. Then there are $1999-m$ colored squares in the remaining columns, and in each of these $< 1999-m$ columns that have a colored square must have at least two colored squares in them. These two colored squares will form a triangle with any other colored square in either of the rows containing the colored squares. Hence, each of the $1999-m$ colored squares must be placed in different rows, but as there are only $1000$ rows, the inequality $1999 - m \\le 1000 \\Longrightarrow m \\ge 999$ holds. If $m = 1000$, then each column only has $1$ colored square, leaving no place for the remaining $999$, contradiction. If $m = 999$, then each of the $1000$ rows has $1$ black square, leaving no place for the other $999$, contradiction. Hence $n = \\boxed{1999}$ is the minimal value."
    },
    {
        "Problem": "Let $A_1A_2A_3$ be a triangle and let $\\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\\omega_2, \\omega_3, \\dots, \\omega_7$ such that for $k = 2, 3, \\dots, 7,$ $\\omega_k$ is externally tangent to $\\omega_{k - 1}$ and passes through $A_k$ and $A_{k + 1},$  where $A_{n + 3} = A_{n}$ for all $n \\ge 1$. Prove that $\\omega_7 = \\omega_1.$",
        "Solution_1": "Let the circumcenter of $\\triangle ABC$ be $O$, and let the center of $\\omega_k$ be $O_k$. $\\omega_k$ and $\\omega_{k-1}$ are externally tangent at the point $A_k$, so $O_k, A_k, O_{k-1}$ are collinear.\n$O$ is the intersection of the perpendicular bisectors of $\\overline{A_1A_2}, \\overline{A_2A_3}, \\overline{A_3A_1}$, and each of the centers $O_k$ lie on the perpendicular bisector of the side of the triangle that determines $\\omega_k$. It follows from $OA_k = OA_{k+1}, O_kA_k = O_kA_{k+1}, OO_k = OO_k \\Longrightarrow \\triangle OA_kO_k \\cong \\triangle OA_{k+1}O_k$ that $\\angle OA_kO_k = \\angle OA_{k+1}O_k$.\nSince $O, A_k$, and the perpendicular bisector of $\\overline{A_kA_{k+1}}$ are fixed, the angle $OA_kO_k$ determines the position of $O_k$ (since $O_k$ lies on the perpendicular bisector). Let $\\theta_k = m\\angle OA_kO_k$; then, $\\theta_i = \\theta_j$ and $i \\equiv j \\pmod{3}$ together imply that $O_i \\equiv O_j$.\nNow $\\theta_1 = \\angle OA_1O_1 = \\angle OA_2O_1 = 180 - \\angle OA_2O_2 = 180 - \\theta_2$ (due to collinearility). Hence, we have the recursion $\\theta_k = 180 - \\theta_{k-1}$, and so $\\theta_k = \\theta_{k-2}$. Thus, $\\theta_{1} = \\theta_{7}$.\n$\\theta_{1} = \\theta_{7}$ implies that $O_1 \\equiv O_7$, and circles $\\omega_1$ and $\\omega_7$ are the same circle since they have the same center and go through the same two points.",
        "Solution_2": "Using the collinearity of certain points and the fact that $A_k A_{k+1} O_k$ is isosceles, we quickly deduce that\n\nFrom ASA Congruence we deduce that $A_1 A_2 O_1$ and $A_7 A_8 O_7$ are congruent triangles, and so $O_1 A_1 = O_7 A_7$, that is $\\omega_1 = \\omega_7$."
    },
    {
        "Problem": "Let $a_1, b_1, a_2, b_2, \\dots , a_n, b_n$ be nonnegative real numbers. Prove that\n",
        "Solution": "Credit for this solution goes to Ravi Boppana.\nLemma 1: If $r_1, r_2, \\ldots , r_n$ are non-negative reals and $x_1, x_2, \\ldots x_n$ are reals, then\n\nProof: Without loss of generality assume that the sequence $\\{r_i\\}$ is increasing. For convenience, define $r_0=0$. The LHS of our inequality becomes\n\nThis expression is equivalent to the sum\n\nEach term in the summation is non-negative, so the sum itself is non-negative. $\\blacksquare$\nWe now define $r_i=\\frac{\\max(a_i,b_i)}{\\min(a_i,b_i)}-1$. If $\\min(a_i,b_i)=0$, then let $r_i$ be any non-negative number. Define $x_i=\\text{sgn}(a_i-b_i)\\min(a_i,b_i)$.\nLemma 2: $\\min(a_{i}b_{j}, a_{j}b_{i})-\\min(a_{i}a_{j}, b_{i}b_{j}) =\\min(r_{i}, r_{j}) x_{i}x_{j}$\nProof: Switching the signs of $a_i$ and $b_i$ preserves inequality, so we may assume that $a_i>b_i$. Similarly, we can assume that $a_j>b_j$. If $b_ib_j=0$, then both sides are zero, so we may assume that $b_i$ and $b_j$ are positive. We then have from the definitions of $r_i$ and $x_i$ that\n\nThis means that\n\nThis concludes the proof of Lemma 2. $\\blacksquare$\nWe can then apply Lemma 2 and Lemma 1 in order to get that\n\nThis implies the desired inequality."
    },
    {
        "Problem": "Each of eight boxes contains six balls. Each ball has been colored with one of $n$ colors, such that no two balls in the same box are the same color, and no two colors occur together in more than one box. Determine, with justification, the smallest integer $n$ for which this is possible.",
        "Solution_1": "We claim that $n=23$ is the minimum. Consider the following construction (replacing colors with numbers) which fulfills this:\n\nSuppose a configuration exists with $n \\le 22$.\nSuppose a ball appears $5$ or more times. Then the remaining balls of the $5$ boxes must be distinct, so that there are at least $n \\ge 5 \\cdot 5 + 1 = 26$ balls, contradiction. If a ball appears $4$ or more times, the remaining balls of the $4$ boxes must be distinct, leading to $5 \\cdot 4 + 1 = 21$ balls. The fifth box can contain at most four balls from the previous boxes, and then the remaining two balls must be distinct, leading to $n \\ge 2 + 21 = 23$, contradiction.\nHowever, by the Pigeonhole Principle, at least one ball must appear $3$ times. Without loss of generality suppose that $1$ appears three times, and let the boxes that contain these have balls with colors $\\{1,2,3,4,5,6\\},\\{1,7,8,9,10,11\\},\\{1,12,13,14,15,16\\}$. Each of the remaining five boxes can have at most $3$ balls from each of these boxes. Thus, each of the remaining five boxes must have $3$ additional balls $> 16$. Thus, it is necessary that we use $\\le 22 - 16 = 6$ balls to fill a $3 \\times 5$ grid by the same rules.\nAgain, no balls may appear $\\ge 4$ times, but by Pigeonhole, one ball must appear $3$ times. Without loss of generality, let this ball have color $17$; then the three boxes containing $17$ must have at least $2 \\cdot 3 + 1 = 7$ balls, contradiction.\nTherefore, $n = 23$ is the minimum.",
        "Solution_2": "Similar to the above solution, no color can appear 4 times or more.\nWe can use PIE. Let $S_1,S_2,\\cdots ,S_8$ be the colors in each of the $8$ boxes. Then $n=|\\cup_{i=1}^8S_i|$. By PIE we know thatNote however that no color can appear 4 times or more, so all the items after $\\sum_{1\\leq i<j<k\\leq 8}|S_i\\cap S_j\\cap S_k|$ must be equal to 0, soNow note that for each $i$, $|S_i|=6,$ so $\\sum_{i=1}^8 |S_i|=48.$ Also no pair of colors appears twice, so for any $i,j, |S_i\\cap S_j|\\leq 1,$ and so $\\sum_{1\\leq i<j\\leq 8}|S_i\\cap S_j|\\leq \\binom82=28.$ Thus, $n\\geq 48-28+\\sum_{1\\leq i<j<k\\leq 8}|S_i\\cap S_j\\cap S_k|.$ Given that there are $n$ colors, to minimize the number of colors in 3 boxes means each color is in either 2 or 3 boxes, and since there are 48 total spots, this means there are $48-2n$ colors in 3 boxes and $3n-48$ colors in 2. Thus $n\\geq 20+48-2n,$ so $3n\\geq 68$, and $n\\geq23$ and we are done.",
        "Solution_3": "Let $m_{i,j}$ be the number of balls which are the same color as the $j^{\\text{th}}$ ball in box $i$ (including that ball). For a fixed box $i$, $1\\leq i\\leq 8$, consider the sums\n\nFor each fixed $i$, since no pair of colors is repeated, each of the remaining seven boxes can contribute at most one ball to $S_i$. Thus, $S_i\\leq 13$. It follows by the convexity of $f(x) = 1/x$ that $s_i$ is minimized when one of the $m_{i,j}$ is equal to 3 and the other five equal to 2. Hence $s_i\\geq 17/6$. Note that\n\nHence there must be 23 colors. The construction for $n = 23$ is above."
    },
    {
        "Problem": "Let $ABC$ be a triangle and let $\\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ = D_2P$.",
        "Solution_1": "It is well known that the excircle opposite $A$ is tangent to $\\overline{BC}$ at the point $D_2$. (Proof: let the points of tangency of the excircle with the lines $BC, AB, AC$ be $D_3, F,G$ respectively. Then $AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3$. It follows that $2CD_3 = AB + BC - AC$, and $CD_3 = s-b = BD_1 = CD_2$, so $D_3 \\equiv D_2$.)\nNow consider the homothety that carries the incircle of $\\triangle ABC$ to its excircle. The homothety also carries $Q$ to $D_2$ (since $A,Q,D_2$ are collinear), and carries the tangency points $E_1$ to $G$. It follows that $\\frac{AQ}{QD_2} = \\frac{AE_1}{E_1G} = \\frac{s-a}{E_1C + CD_2} = \\frac{s-a}{CD_1 + BD_1} = \\frac{s-a}{a}$.\nBy Menelaus' Theorem on $\\triangle ACD_2$ with segment $\\overline{BE_2}$, it follows that $\\frac{CE_2}{E_2A} \\cdot \\frac{AP}{PD_2} \\cdot \\frac{BD_2}{BC} = 1 \\Longrightarrow \\frac{AP}{PD_2} = \\frac{(c - (s-a)) \\cdot a}{(a-(s-c)) \\cdot AE_1} = \\frac{a}{s-a}$. It easily follows that $AQ = D_2P$. $\\blacksquare$",
        "Solution_2": "The key observation is the following lemma.\nLemma: Segment $D_1Q$ is a diameter of circle $\\omega$.\nProof: Let $I$ be the center of circle $\\omega$, i.e., $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\\omega$ again at $Q'$, and extend segment $AQ'$ through $Q'$ to intersect segment $BC$ at $D'$. We show that $D_2 = D'$, which in turn implies that $Q = Q'$, that is, $D_1Q$ is a diameter of $\\omega$.\nLet $l$ be the line tangent to circle $\\omega$ at $Q'$, and let $l$ intersect the segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\\omega$ is an excircle of triangle $AB_1C_1$. Let $\\mathbf{H}_1$ denote the dilation with its center at $A$ and ratio $AD'/AQ'$. Since $l\\perp D_1Q'$ and $BC\\perp D_1Q'$, $l\\parallel BC$. Hence $AB/AB_1 = AC/AC_1 = AD'/AQ'$. Thus $\\mathbf{H}_1(Q') = D'$, $\\mathbf{H}_1(B_1) = B$, and $\\mathbf{H}_1(C_1) = C$. It also follows that an excircle $\\Omega$ of triangle $ABC$ is tangent to the side $BC$ at $D'$.\nIt is well known that  We compute $BD'$. Let $X$ and $Y$ denote the points of tangency of circle $\\Omega$ with rays $AB$ and $AC$, respectively. Then by equal tangents, $AX = AY$, $BD' = BX$, and $D'C = YC$. Hence  It follows that  Combining these two equations yields $BD' = CD_1$. Thus  that is, $D' = D_2$, as desired. $\\blacksquare$\nNow we prove our main result. Let $M_1$ and $M_2$ be the respective midpoints of segments $BC$ and $CA$. Then $M_1$ is also the midpoint of segment $D_1D_2$, from which it follows that $IM_1$ is the midline of triangle $D_1QD_2$. Hence  and $AD_2\\parallel M_1I$. Similarly, we can prove that $BE_2\\parallel M_2I$.\n2001usamo2-2.png\nLet $G$ be the centroid of triangle $ABC$. Thus segments $AM_1$ and $BM_2$ intersect at $G$. Define transformation $\\mathbf{H}_2$ as the dilation with its center at $G$ and ratio $-1/2$. Then $\\mathbf{H}_2(A) = M_1$ and $\\mathbf{H}_2(B) = M_2$. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since $AD_2\\parallel M_1I$ and $BE_2\\parallel M_2I$, $\\mathbf{H}_2$ maps lines $AD_2$ and $BE_2$ to lines $M_1I$ and $M_2I$, respectively. It also follows that $\\mathbf{H}_2(I) = P$ and  or  This yields  as desired.\nNote: We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle $ABC$.",
        "Solution_3": "Here is a rather nice solution using barycentric coordinates:\nLet $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Let the side lengths of the triangle be $a,b,c$ and the semi-perimeter $s$.\nNow,  Thus,\nTherefore, $D_2=(0:s-b:s-c)$ and $E_2=(s-a:0:s-c).$ Clearly then, the non-normalized coordinates of $P=(s-a:s-b:s-c).$\nNormalizing, we have that\nNow, we find the point $Q'$ inside the triangle on the line $AD_2$ such that $AQ'=D_2P$. It is then sufficient to show that this point lies on the incircle.\n$P$ is the fraction $\\frac{s-a}{s}$ of the way \"up\" the line segment from $D_2$ to $A$. Thus, we are looking for the point that is $\\frac{s-a}{s}$ of the way \"down\" the line segment from $A$ to $D_2$, or, the fraction $1-\\frac{s-a}{s}$ of the way \"up\".\nThus, $Q'$ has normalized $x$-coordinate $1-\\frac{s-a}{s}=\\frac{a}{s}$.\nAs the line $AD_2$ has equation $(s-c)y=(s-b)z$, it can easily be found that\nRecalling that the equation of the incircle is  We must show that this equation is true for $Q'$'s values of $x,y,z$.\nPlugging in our values, this means showing that\n\nDividing by $a(s-a)$, this is just\nPlugging in the value of $s:$\n\n\n\nThe first bracket is just\n\n\nand the second bracket is \nDividing everything by $2a$ gives\n which is $0$, as desired.\nAs $Q'$ lies on the incircle and $AD_2$, $Q'=Q$, and our proof is complete.\n",
        "Solution_4": "We again use Barycentric coordinates. As before, let $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Also\n\nNow, consider a point $Q'$ for which $AQ'=PD_{2}$. Then working component-vise, we get $Q'+P=A+D_{2}$ from which we can easily get the coordinates of $Q'$ as;\n\nIt suffices to show that $Q'=Q$.\nLet $I=\\left(\\frac{a}{2s},\\frac{b}{2s},\\frac{c}{2s}\\right)$  be the incenter of triangle $ABC$. We claim that $I$ is the midpoint of $Q'D_{1}$. Indeed,\n\n\n\nHence the claim has been proved.\nSince $I$ is the center of $\\omega$ and the midpoint of  $Q'D_{1}$, thus $Q'$ is the point diametrically opposite to $D_{1}$, and hence it lies on $\\omega$ and closer to $A$ off the 2 points. Thus $Q=Q'$ as desired."
    },
    {
        "Problem": "Let $a, b, c \\geq 0$ and satisfy\nShow that",
        "Solution_1": "First we prove the lower bound.\nNote that we cannot have $a, b, c$ all greater than 1.\nTherefore, suppose $a \\le 1$.\nThen\n\nNote that, by the Pigeonhole Principle, at least two of $a,b,c$ are either both greater than or less than $1$. Without loss of generality, let them be $b$ and $c$. Therefore, $(b-1)(c-1)\\ge 0$. From the given equation, we can express $a$ in terms of $b$ and $c$ as\n\nThus,\nFrom the Cauchy-Schwarz Inequality,\nThis completes the proof.",
        "Solution_2": "The proof for the lower bound is the same as in the first solution.\nNow we prove the upper bound. Let us note that at least two of the three numbers $a$, $b$, and $c$ are both greater than or equal to 1 or less than or equal to 1. Without loss of generality, we assume that the numbers with this property are $b$ and $c$. Then we have\n\nThe given equality $a^2 + b^2 + c^2 + abc = 4$ and the inequality $b^2 + c^2\\geq 2bc$ imply that\n\nor\n\nDividing both sides of the last inequality by $2 + a$ yields\n\nThus,\n\nas desired.\nThe last equality holds if and only if $b = c$ and $a(1 - b)(1 - c) = 0$. Hence equality for the upper bound holds if and only if $(a,b,c)$ is one of the triples $(1,1,1)$, $(0,\\sqrt{2},\\sqrt{2})$, $(\\sqrt{2},0,\\sqrt{2})$, and $(\\sqrt{2},\\sqrt{2},0)$. Equality for the lower bound holds if and only if $(a,b,c)$ is one of the triples $(2,0,0)$, $(0,2,0)$ and $(0,0,2)$.",
        "Solution_3": "The proof for the lower bound is the same as in the first solution.\nNow we prove the upper bound. It is clear that $a,b,c\\leq 2$. If $abc = 0$, then the result is trivial. Suppose that $a,b,c > 0$. Solving for $a$ yields\n\nThis asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then\n\nand if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have\n\nwhere the inequality step follows from AM-GM. Likewise,\n\nTherefore\n\nas desired."
    },
    {
        "Problem": "Let $P$ be a point in the plane of triangle $ABC$ such that the segments $PA$, $PB$, and $PC$ are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to $PA$. Prove that $\\angle BAC$ is acute.",
        "Solution_1": "We know that $PB^2+PC^2 < PA^2$ and we wish to prove that $AB^2 + AC^2 > BC^2$.\nIt would be sufficient to prove that\n\nSet $A(0,0)$, $B(1,0)$, $C(x,y)$, $P(p,q)$.\nThen, we wish to show\n\nwhich is true by the trivial inequality.",
        "Solution_2": "Let $A$ be the origin. For a point $Q$, denote by $q$ the vector $\\overrightarrow{AQ}$, and denote by $|q|$ the length of $q$. The given conditions may be written as\n\nor\n\nAdding $2b\\cdot c$ on both sides of the last inequality gives\n\nSince the left-hand side of the last inequality is nonnegative, the right-hand side is positive. Hence\n\nthat is, $\\angle BAC$ is acute.",
        "Solution_3": "For the sake of contradiction, let's assume to the contrary that $\\angle BAC$. Let $AB = c$, $BC = a$, and $CA = b$. Then $a^2\\geq b^2 + c^2$. We claim that the quadrilateral $ABPC$ is convex. Now applying the generalized Ptolemy's Theorem to the convex quadrilateral $ABPC$ yields\n\nwhere the second inequality is by Cauchy-Schwarz. This implies $PA^2\\leq PB^2 + PC^2$, in contradiction with the facts that $PA$, $PB$, and $PC$ are the sides of an obtuse triangle and $PA > \\max\\{PB, PC\\}$.\nWe present two arguments to prove our claim.\nFirst argument: Without loss of generality, we may assume that $A$, $B$, and $C$ are in counterclockwise order. Let lines $l_1$ and $l_2$ be the perpendicular bisectors of segments $AB$ and $AC$, respectively. Then $l_1$ and $l_2$ meet at $O$, the circumcenter of triangle $ABC$. Lines $l_1$ and $l_2$ cut the plane into four regions and $A$ is in the interior of one of these regions. Since $PA > PB$ and $PA > PC$, $P$ must be in the interior of the region that opposes $A$. Since $\\angle BAC$ is not acute, ray $AC$ does not meet $l_1$ and ray $AB$ does not meet $l_2$. Hence $B$ and $C$ must lie in the interiors of the regions adjacent to $A$. Let $\\mathcal{R}_X$ denote the region containing $X$. Then $\\mathcal{R}_A$, $\\mathcal{R}_B$, $\\mathcal{R}_P$, and $\\mathcal{R}_C$ are the four regions in counterclockwise order. Since $\\angle BAC\\geq 90^\\circ$, either $O$ is on side $BC$ or $O$ and $A$ are on opposite sides of line $BC$. In either case $P$ and $A$ are on opposite sides of line $BC$. Also, since ray $AB$ does not meet $l_2$ and ray $AC$ does not meet $l_1$, it follows that $\\mathcal{R}_P$ is entirely in the interior of $\\angle BAC$. Hence $B$ and $C$ are on opposite sides of $AP$. Therefore $ABPC$ is convex.\nSecond argument: Since $PA > PB$ and $PA > PC$, $A$ cannot be inside or on the sides of triangle $PBC$. Since $PA > PB$, we have $\\angle ABP > \\angle BAP$ and hence $\\angle BAC\\geq 90^\\circ > \\angle BAP$. Hence $C$ cannot be inside or on the sides of triangle $BAP$. Symmetrically, $B$ cannot be inside or on the sides of triangle $CAP$. Finally, since $\\angle ABP > \\angle BAP$ and $\\angle ACP > \\angle CAP$, we have\n\nTherefore $P$ cannot be inside or on the sides of triangle $ABC$. Since this covers all four cases, $ABPC$ is convex.",
        "Solution_4": "Let $P$ be the origin in vector space, and let $a, b, c$ denote the position vectors of $A, B, C$ respectively. Then the obtuse triangle condition, $PA^2 > PB^2 + PC^2$, becomes $a^2 > b^2 + c^2$ using the fact that the square of a vector (the dot product of itself and itself) is the square of its magnitude. Now, notice that to prove $\\angle{BAC}$ is acute, it suffices to show that $(a - b)(a - c) > 0$, or $a^2 - ab - ac + bc > 0$. But this follows from the observation that\n\nwhich leads to\n\nand therefore our desired conclusion.",
        "Solution_5": "Let $M, N$ be midpoints of $AP$ and $BC$, respectively. For the points $A, B, P, C$; let's apply Euler's quadrilateral formula,\n\nGiven that $AP^2 > BP^2 + PC^2$. Thus,\n \nand we get $\\angle BAC$ is acute.\n(Lokman GÖKÇE)",
        "Solution_6": "Without loss of generality, assume that in a Cartesian coordinate system, $A$ is at the point $(0, 0)$ and $C$ is at the point $(1,0)$. Let $B$ be at the point $(b_x,b_y)$ and $P$ be at the point $(p_x,p_y)$. Without loss of generality, also assume that $b_y>0$.\nNow, assume for contradiction that $\\angle BAC$ is not acute. Since $PA$, $PB$, and $PC$ are the sides of an obtuse triangle, with $PA$ the longest side, it follows that $PA^2>PB^2+PC^2$, implying that $p_x^2+p_y^2>(p_x-b_x)^2+(p_y-b_y)^2+(p_x-1)^2+p_y^2$. This inequality simplifies to $b_x^2-2p_x b_x+b_y^2-2p_y b_y+p_x^2-2p_x+1+p_y^2<0$. Note that since $p_x^2-2p_x+1$ and $b_y^2-2p_y b_y+p_y^2$ are both perfect squares, all terms of this inequality except for $-2p_x b_x$ are already guaranteed to be nonnegative.\nIf $p_x<0$, then $P$ would be closer to $A$ than to $C$, but since $PA^2=PB^2+PC^2$, this is not possible. Therefore, $p_x \\geq 0$. Since $\\angle BAC$ not being acute implies that $b_x \\leq 0$, it follows that $-2p_x b_x \\geq 0$. But now since all terms of $b_x^2-2p_x b_x+b_y^2-2p_y b_y+p_x^2-2p_x+1+p_y^2<0$ are guaranteed to be nonnegative, this entire expression cannot be negative, leading to a contradiction. Therefore, $\\angle BAC$ is acute."
    },
    {
        "Problem": "Let $S$ be a set of integers (not necessarily positive) such that\n(a) there exist $a,b \\in S$ with $\\gcd(a,b) = \\gcd(a - 2,b - 2) = 1$;\n(b) if $x$ and $y$ are elements of $S$ (possibly equal), then $x^2 - y$ also belongs to $S$.\nProve that $S$ is the set of all integers.",
        "Solution": "In the solution below we use the expression $S$ is stable under $x\\mapsto f(x)$ to mean that if $x$ belongs to $S$ then $f(x)$ also belongs to $S$. If $c,d\\in S$, then by (B), $S$ is stable under $x\\mapsto c^2 - x$ and $x\\mapsto d^2 - x$, hence stable under $x\\mapsto c^2 - (d^2 - x) = x + (c^2 - d^2)$. Similarly $S$ is stable under $x\\mapsto x + (d^2 - c^2)$. Hence $S$ is stable under $x\\mapsto x + n$ and $x\\mapsto x - n$ whenever $n$ is an integer linear combination of numbers of the form $c^2 - d^2$ with $c,d\\in S$. In particular, this holds for $n = m$, where $m = \\gcd\\{c^2 - d^2 : c,d\\in S\\}$.\nSince $S\\neq \\emptyset$ by (A), it suffices to prove that $m = 1$. For the sake of contradiction, assume that $m\\neq 1$. Let $p$ be a prime dividing $m$. Then $c^2 - d^2\\equiv 0\\pmod{p}$ for all $c,d\\in S$. In other words, for each $c,d\\in S$, either $d\\equiv c\\pmod{p}$ or $d\\equiv -c\\pmod{p}$. Given $c\\in S$, $c^2 - c\\in S$ by (B), so $c^2 - c\\equiv c\\pmod{p}$ or $c^2 - c\\equiv -c\\pmod{p}$. Hence\n\nBy (A), there exist some $a$ and $b$ in $S$ such that $\\gcd(a,b) = 1$, that is, at least one of $a$ or $b$ cannot be divisible by $p$. Denote such an element of $S$ by $\\alpha$: thus, $\\alpha\\not\\equiv 0\\pmod{p}$. Similarly, by (A), $\\gcd(a - 2, b - 2) = 1$, so $p$ cannot divide both $a - 2$ and $b - 2$. Thus, there is an element of $S$, call it $\\beta$, such that $\\beta\\not\\equiv 2\\pmod{p}$. By $(*)$, $\\alpha\\equiv 2\\pmod{p}$ and $\\beta\\equiv 0\\pmod{p}$. By (B), $\\beta^2 - \\alpha\\in S$. Taking $c = \\beta^2 - \\alpha$ in $(*)$ yields either $-2\\equiv 0\\pmod{p}$ or $-2\\equiv 2\\pmod{p}$, so $p = 2$. Now $(*)$ says that all elements of $S$ are even, contradicting (A). Hence our assumption is false and $S$ is the set of all integers."
    },
    {
        "Problem": "Each point in the plane is assigned a real number such that, for any triangle, the number at the center of its inscribed circle is equal to the arithmetic mean of the three numbers at its vertices. Prove that all points in the plane are assigned the same number.",
        "Solution": "We label each upper case point with the corresponding lower case letter as its assigned number. The key step is the following lemma.\nLemma: If $ABCD$ is an isosceles trapezoid, then $a + c = b + d$.\nProof: Assume without loss of generality that $BC\\parallel AD$, and that rays $AB$ and $DC$ meet at $P$. Let $I$ be the incenter of triangle $PAC$, and let line $l$ bisect $\\angle APD$. Then $I$ is on $l$, so reflecting everything across line $l$ shows that $I$ is also the incenter of triangle $PDB$. Therefore,\n\nHence $a + c = b + d$, as desired.$\\blacksquare$\nFor any two distinct points $A_1$ and $A_2$ in the plane, we construct a regular pentagon $A_1A_2A_3A_4A_5$. Applying the lemma to isosceles trapezoids $A_1A_3A_4A_5$ and $A_2A_3A_4A_5$ yields\n\nHence $a_1 = a_2$. Since $A_1$ and $A_2$ were arbitrary, all points in the plane are assigned the same number."
    },
    {
        "Problem": "Let $S$ be a set with 2002 elements, and let $N$ be an integer with $0 \\le N \\le 2^{2002}$.  Prove that it is possible to color every subset of $S$ either blue or red so that the following conditions hold:\n(a) the union of any two red subsets is red;\n(b) the union of any two blue subsets is blue;\n(c) there are exactly $N$ red subsets.",
        "Solution_1": "Let a set colored in such a manner be properly colored.  We prove that any set with $n$ elements can be properly colored for any $0 \\le N \\le 2^n$.  We proceed by induction.\nThe base case, $n = 0$, is trivial.\nSuppose that our claim holds for $n = k$.  Let $s \\in S$, $|S| = k + 1$, and let $S'$ denote the set of all elements of $S$ other than $s$.\nIf $N \\le 2^k$, then we may color all subsets of $S$ which contain $s$ blue, and we may properly color $S'$.  This is a proper coloring because the union of any two red sets must be a subset of $S'$, which is properly colored, and any the union of any two blue sets either must be in $S'$, which is properly colored, or must contain $s$ and therefore be blue.\nIf $N > 2^k$, then we color all subsets containing $s$ red, and we color $N - 2^k$ elements of $S'$ red in such a way that $S'$ is colored properly.  Then $S$ is properly colored, using similar reasoning as before.  Thus the induction is complete.",
        "Solution_2": "If $N = 0$, color every subset blue.  If $N = 2^{2002}$, color every subset red.  Otherwise, let $S$ be $\\{ 0, 1, \\ldots , 2001 \\}$.  Write $N$ in binary, i.e., let\nwhere each of the $a_i$ is an element of $S$.  We color each of the $a_i$ red and all the other elements of $S$ blue.  We color the empty set blue, and we color any other set the color of its largest element.  This satisfies the problem's first two conditions, as the largest element of the union of two red (or blue) sets will have a red (or blue) number as its largest element.  In addition, for each integer $n \\in S$, there are $2^{n}$ subsets of $S$ with $n$ as a maximal element, so $\\sum_{i=1}^{k} 2^{a_i} = N$ subsets of $S$ are colored red, as desired.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Let $ABC$ be a triangle such that\nwhere $s$ and $r$ denote its semiperimeter and inradius, respectively.  Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisor and determine those integers.",
        "Solution": "Let $a,b,c$ denote $BC, CA, AB$, respectively.  Since the line from a triangle's incenter to one of its vertices bisects the angle at the triangle's vertex, the condition of the problem is equivalent to the\nor\nBut by the Cauchy-Schwarz Inequality, we know\nwith equality only when $\\frac{(s-a)^2}{36}, \\frac{(s-b)^2}{9}, \\frac{(s-c)^2}{4}$ are directly proportional to 36, 9, 4, respectively.  Therefore (clearing denominators and taking square roots) our problem requires that $(s-a), (s-b), (s-c)$ be directly proportional to 36, 9, 4, and since $a = (s-b) + (s-c)$ etc., this is equivalent to the condition that $a,b,c$ be in proportion with 13, 40, 45, Q.E.D.\nSidenote: A 13, 40, 45 obtuse triangle has an integer area of 252 and an inradius of $\\frac{36}{7}$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree $n$ with real coefficients is the average of two monic polynomials of degree $n$ with $n$ real roots.",
        "Solution_1": "Lemma.  For any set of ordered pairs of reals $\\{ (x_i , y_i ) \\}_{i=1}^{n}$ with $x_i \\neq x_j$ for all $i \\neq j$, there exists a unique monic real polynomial  $P(x)$ of degree $n$ such that $P(x_i) = y_i$ for all integers $i \\in [1,n]$.\nProof 1.  By the Lagrange Interpolation Formula, there exists a unique real polynomial $Q(x)$ of degree less than $n$ such that $Q(x_i) = y_i - x_i^n$.  Then it is necessary and sufficient that $P(x) = x^n + Q(x)$.\nProof 2.  Again by the Lagrange Interpolation Formula, there exists a unique real polynomial $Q(x)$ of degree less than $n$ such that $Q(x_i) = y_i$.  Then it is necessary and sufficient that $P(x) = Q(x) + \\prod_{i=1}^{n}(x-x_i)$.\nLet $F(x)$ be the monic polynomial of degree $n$ given in the problem.  Let $x_1, \\ldots, x_n$ be a sequence of strictly increasing reals, and choose $y_1, \\ldots, y_n$ such that for odd $i$, $y_i > \\max (0, 2F(x_i))$, but for even $i$, $y_i < \\min (0, 2F(x_i))$.  Let $P(x)$ be the unique monic polynomial such that $P(x_i) = y_i$.  We shall prove that $P(x)$ and $Q(x) = 2F(x) - P(x)$ satisfy the problem's conditions.\nNote that for $i < n$, $P(x_i)$ and $P(x_{i+1})$ have different signs, so $P(x)$ has a real root between $x_i$ and $x_{i+1}$.  It follows that $P(x)$ has at least $n-1$ real roots.  But a polynomial with real coefficients must have an even number of non-real roots, so $P(x)$ must have $n$ real roots.  Similarly, $Q(x)$ must have $n$ real roots.  Q.E.D.",
        "Solution_2": "Let $F(x)$ be our polynomial.  If $n = 1$, then we may let $F = x + a$, which is the average of the polynomials $x$ and $x + 2a$, each of which has a real root.  Otherwise, let\nand let\nand\nWe will prove that for sufficiently large $k$, $P(x)$ and $Q(x)$ satisfy the problem's conditions.\nWe note that for the values $1, 3, \\ldots, 2n-1$ of $x$, $G(x)$ alternates in sign, and always has magnitude at least 1 (since it is the product of several factors of magnitude at least one.  On the other hand, there exists some $c$ such that for all $x \\in [1, 2n-1]$, $|x^n| < c$ and $|F(n) - x^n | < c$.  Let $k$ be greater than $c$.  Then both $P(x)$ and $Q(x)$ alternate in sign for $x = 1, 3, \\ldots, 2n-1$.  It follows that each has at least $n-1$ real roots.  But since a polynomial with real coefficients can only have an even number of complex roots, this means that both $P$ and $Q$ must have $n$ roots.  Q.E.D.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Let $\\mathbb{R}$ be the set of real numbers.  Determine all functions $f : \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\nfor all pairs of real numbers $x$ and $y$.",
        "Solution_1": "We first prove that $f$ is  odd.\nNote that $f(0) = f(x^2 - x^2) = xf(x) - xf(x) = 0$, and for nonzero $y$, $xf(x) + yf(-y) = f(x^2 - y^2) = xf(x) - yf(y)$, or $yf(-y) = -yf(y)$, which implies $f(-y) = -f(y)$.  Therefore $f$ is odd.  Henceforth, we shall assume that all variables are non-negative.\nIf we let $y = 0$, then we obtain $f(x^2) = xf(x)$.  Therefore the problem's condition becomes\nBut for any $a,b$, we may set $x = \\sqrt{a}$, $y = \\sqrt{b}$ to obtain\n(It is well known that the only continuous solutions to this functional equation are of the form $f(x) = kx$, but there do exist other solutions to this which are not solutions to the equation of this problem.)\nWe may let $a = 2t$, $b = t$ to obtain $2f(t) = f(2t)$.\nLetting $x = t+1$ and $y = t$ in the original condition yields\nBut we know $f(2t + 1) = f(2t) + f(1) = 2f(t) + f(1)$, so we have $2f(t) + f(1) = f(t) + tf(1) + f(1)$, or\nHence all solutions to our equation are of the form $f(x) = kx$.  It is easy to see that real value of $k$ will suffice.",
        "Solution_2": "As in the first solution, we obtain the result that $f$ satisfies the condition\nWe note that\nSince $f(2t) = 2f(t)$, this is equal to\nIt follows that $f$ must be of the form $f(x) = kx$.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Let $a, b$ be integers greater than 2.  Prove that there exists a positive integer $k$ and a finite sequence $n_1, n_2, \\ldots, n_k$ of positive integers such that $n_1 = a$, $n_k = b$, and $n_in_{i+1}$ is divisible by $n_i + n_{i+1}$ for each $i$ ($1 \\le i \\le k$).",
        "Solutions": "We may say that two integers $a$ and $b$ are connected (and write $a \\leftrightarrow b$) if there exists such a sequence of integers as described in the problem.  For reference, we note that $\\leftrightarrow$ is an equivalence relation: it is reflexive ($a \\leftrightarrow a$), symmetric ($a \\leftrightarrow b$ implies $b \\leftrightarrow a$), and transitive ($a \\leftrightarrow b \\leftrightarrow c$ implies $a \\leftrightarrow c$ ).",
        "Solution_1": "Note that for any divisor $d$ of some $n$, $n(d-1) + n = nd \\mid n^2(d-1)$, so $n \\leftrightarrow n(d-1)$.  It follows that in fact $n \\leftrightarrow n(d-1)^k$, for any nonnegative integer $k$, and\nfor any positive integer $c < d$.\nFor all integers $a > b > 2$, there exists some integer $\\lambda$ such that $(b-1)^{\\lambda} > a$.  Let\nWe have\nBut we also have\nHence $a \\leftrightarrow X \\leftrightarrow b$, so $a \\leftrightarrow b$, as desired.",
        "Solution_2": "We note that for any integer $n \\ge 3$, $n \\leftrightarrow 2n$, for\nIt follows that for $n \\ge 4$,\nThus all integers greater than 2 are connected.",
        "Solution_3": "We note that $n_i + n_{i+1} \\mid n_in_{i+1}$ if and only if\nTherefore for any divisor $d > n_i$ of $n_i^2$, $d - n_i \\leftrightarrow n_i$.\nNow, for $a \\ge 2$,\nAlso,\nThis means that all positive multiples of 4 are connected.\nFurthermore, for $a \\ge 2$,\nwhich implies that every even number greater than 2 is connected to a multiple of 4.\nFinally, for $a \\ge 3$,\nSince $a(a-1)$ is even, this means that all integers greater than or equal to 2 are connected to some even number.\nTogether, these imply that all integers greater than 2 are connected.",
        "Solution_4": "We note that if $d > 2$ is a divisor of $n$, then $n \\leftrightarrow n(d-1)$.\nWe say a positive integer $k$ is safe if for all integers $n \\ge 3$, $n \\leftrightarrow kn$.  Note that the product of two safe numbers is also a safe number.  Define $f(n)$ ($n>2$) to be the smallest divisor of $n$ that is greater than 2.  We will prove that 2 is safe, by strong induction on $f(n)$.  For the case $f(n) = 3$, we have $n \\leftrightarrow (3-1)n = 2n$.  If $f(n) > 3$, we note that $f[n(f(n) - 1)] < f(n)$, since $f(n) - 1$ must be less than all the divisors of $n$.  Thus the inductive hypothesis gives us $[f(n) - 1]n \\leftrightarrow 2[f(n) - 1]n$.  Furthermore, we have $n \\leftrightarrow [f(n)-1]n$ and $2n \\leftrightarrow 2[f(n)-1]n$, both from our initial note.  Thus $n \\leftrightarrow 2n$.\nWe will now prove that every prime $p$ is safe, by strong induction.  We have already proven the base case $p = 2$.  Now, for odd $p$, $p+1$ is the product of odd primes less than $p$, so $p+1$ is safe.  Then we have\nThus the induction is complete, and all primes are safe.\nWe have now shown that all integers greater than 1 are safe.  Specifically, $a$ and $b$ are safe, and $a \\leftrightarrow ab \\leftrightarrow b$.",
        "Solution_5": "Similarly to the first solution, we observe that for any integer $k < a$ and any positive integers $e_1, \\ldots, e_k$,\nWe also note that if $a \\leftrightarrow b$, then for any positive integer $k$, $ka \\leftrightarrow kb$, for $(a+b) \\mid (ab)$ implies $k(a+b) \\mid k^2(ab)$.\nIt is sufficient to prove that for any $a > 3$, $a \\leftrightarrow (a-1)$.  If $a$ is composite, then there exist $m \\ge n > 0$ such that $(a-1-m)(a-1-n) = a$, and\nIf, on the other hand, $a$ is prime, then we use strong induction.  For the base case, $a = 5$, we note $4 \\leftrightarrow 3 \\leftrightarrow 3(3-1) = 6 \\leftrightarrow 5$.  Now, assuming that $a>5$ and that this holds for all primes less than $a$ (we know it holds for all composites), it is sufficient to show that $(a+1) \\leftrightarrow (a-1)$, since $(a+1)$ is composite and therefore $(a+1) \\leftrightarrow a$.  But since $(a+1)$ and $(a-1)$ are both even, it suffices to show that $\\frac{a+1}{2} \\leftrightarrow \\frac{a-1}{2} = \\frac{a+1}{2} - 1$.  But this is true, since $\\frac{a+1}{2}$ either composite, or a prime less than $a$, and it is greater than 3, since $a > 5$.  Thus the induction is complete."
    },
    {
        "Problem": "I have an $n \\times n$ sheet of stamps, from which I've been asked to tear out blocks of three adjacent stamps in a single row or column. (I can only tear along the perforations separating adjacent stamps, and each block must come out of the sheet in one piece.) Let $b(n)$ be the smallest number of blocks I can tear out and make it impossible to tear out any more blocks. Prove that there are real constants $c$ and $d$ such that\nfor all $n > 0$.",
        "Solution": "The upper bound requires an example of a set of $\\frac{1}{5}n^2 + dn$ blocks whose removal makes it impossible to remove any further blocks. It suffices to show that we can tile the plane by tiles containing one block for every five stamps such that no more blocks can be chosen. Two such tilings are shown below with one tile outlined in heavy lines. Given an $n\\times n$ section of the tiling take all blocks lying entirely within that section and add as many additional blocks as possible. If the basic tile is contained in an $(m + 1)\\times (m + 1)$ square, then the $n\\times n$ section is covered by tiles contained in a concentric $(n + 2m)\\times (n + 2m)$ square. Hence there are at most $\\frac{1}{5}(n + 2m)^2$ blocks entirely within the section. For an $n\\times n$ section of the tiling, there are at most $4n$ blocks which lie partially in and partially out of that section (hence these blocks contain at most $8n$ stamps in the $n\\times n$ square) and each of the additional blocks must contain one of these stamps. Thus there are at most $8n$ additional blocks. Thus there are at most\n\nblocks total.\nThe lower bound requires an argument. Suppose that we have a set of $b(n)$ blocks whose removal makes removing any further blocks impossible.",
        "Solution_1": "There are $2n(n - 2)$ potential blocks of three consecutive stamps in a row or column. Each of these must meet at least one of the $b(n)$ blocks removed. Conversely, each of the $b(n)$ blocks removed meets at most 14 of these potential blocks (5 oriented the same way, including itself, and 9 oriented the orthogonal way). Therefore $14b(n)\\geq 2n(n - 2)$ or",
        "Solution_2": "Call a stamp used if it belongs to one of the $b(n)$ removed blocks. Consider the $(n - 2)^2$ five-stamp crosses centered at each stamp not on an edge of the sheet. Each cross must contain two used stamps. (One stamp not in the center is not enough to prevent another block from being torn out, and it is impossible to use one stamp in the center and use no other stamps in the cross.) In addition, each block not lying along an edge of the sheet lies entirely inside one cross, which thus contains three used stamps. There are at most $4n/3$ of the $b(n)$ blocks lying along the edges, hence there are at least $b(n) - 4n/3$ crosses containing three used stamps.\nNow count the number of pairs of a used stamp and a cross containing that stamp, in two ways. First counting block by block, we get $3b(n)$ used stamps, and each used stamp is contained in at most five crosses (exactly five if it is not on an edge), for a total of at most $15b(n)$ pairs. Next, counting cross by cross, each of the $(n - 2)^2$ crosses contains at least two used stamps and we have at least $b(n) - 4n/3$ crosses containing three used stamps, for a total of at least $2(n - 2)^2 + b(n) - 4n/3$ pairs. Therefore\n\nor",
        "Solution_3": "Call a stamp used if it belongs to one of the $b(n)$ removed blocks. Count the number of pairs consisting of a used stamp and an adjacent unused stamp, in two ways.\nThere are at least $(n - 2)^2 - 3b(n)$ unused stamps which are not on an edge. Since no more blocks can be torn out, either the stamp to the left or right and either the stamp or below such an unused stamp must be used. Thus we have at least $2n^2 - 8n - 6b(n)$ such pairs.\nEach block removed is adjacent to at most eight other stamps. However these eight stamps contain two blocks of three consecutive stamps. Hence at most six of these eight stamps can be unused. Thus each of the $b(n)$ blocks removed is involved in at most six pairs. Thus there are at most $6b(n)$ pairs.\nCombining these we have\n\nor\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Titu Andreescu)\nProve that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.",
        "Solution_1": "We proceed by induction.  For our base case, $n=1$, we have the number 5.  Now, suppose that there exists some number $a \\cdot 5^{n-1}$ with $n-1$ digits, all of which are odd.  It is then sufficient to prove that there exists an odd digit $k$ such that $k\\cdot 10^{n-1} + a \\cdot 5^{n-1} = 5^{n-1}(k \\cdot 2^{n-1} + a)$ is divisible by $5^n$.  This is equivalent to proving that there exists an odd digit $k$ such that $k \\cdot 2^{n-1} + a$ is divisible by 5, which is true when $k \\equiv -3^{n-1}a \\pmod{5}$.  Since there is an odd digit in each of the residue classes mod 5, $k$ exists and the induction is complete.",
        "Solution_2": "First, we note that there are $5^n$ $n$ digit numbers with only odd digits.  Now, we will prove that none of these numbers have the same residue mod $5^n$, and therefore one of them must be 0 mod $5^n$.\nProof by contradiction:  \nAssume we have two distinct numbers $A_1A_2A_3...A_n$ and $B_1B_2B_3...B_n$ with only odd digits that leave the same residue mod $5^n$.  Then, subtracting the larger from the smaller would yield a new number that is a multiple of $5^n$ and has only even digits.  We could then halve all of the digits in that number to get a second multiple of $5^n$ with at most n digits that only uses the digits 0 through 4.\nLemma: Every multiple of $5^n$ with n digits or less has a 5 as one of its digits.\nAll numbers of this type can be written as $k5^n$.  Then, let $k$ have $x$ factors of $2$ in it.  ($x<n$, or else our number would have more than n digits).  So, we have $k5^n=a2^x5^n=a*10^x*5^{n-x}$ for some odd a.  Now $a*10^x*5^{n-x}$ is an odd multiple of 5  ($a5^{n-x}$)  with x zeroes after it, and all multiples of 5 end in 5.  Therefore, $a*10^x*5^{n-x}$ always contains a 5 as its $(x+1)^{st}$ digit, and we have proven our lemma.\nBy our lemma, our number with only the digits 0 through 4 cannot be a multiple of $5^n$, and so we have reached a contradiction.  QED\nNote:  Not only does this prove the desired claim that there exists such a number, but it also proves that there is exactly one such number.",
        "Solution_3_.28Construct_digits_of_exemplar..29": "$5^n \\pmod {10} \\equiv 5$.\n$2^k 5^n \\pmod {10^{k +1} } \\equiv  10^k 5^{n-k} \\equiv 5 \\cdot 10^k$.\nFor any $n$, we can construct a $a = 5^n \\cdot \\sum_{0 \\leq i < n}{2^{x_i}}$:\n$x_0 = 1$\n$a_i = 5^n \\sum_{0 \\leq j \\leq i}{2^{x_j}}$\n$x_{i} = \\frac{a_{i-1}}{10^{i}} + 1 \\pmod 2$. (This makes the $10^{i}$ place of $a_{i}$ odd, without changing the the smaller place digits).\n$a = a_{n-1} = 5^n \\cdot  \\sum_{0 \\leq j < n}{2^{x_j}}$ ,  with digits in place $10^i$ odd, for $0 \\leq i<n$.\nSince $\\sum_{0 \\leq j < n}{2^{x_j}}  < 2^n$,  $a < 5^n 2^n = 10^n$.\nBy construction, the digits in all the places up through  $10^{n-1}$ are odd, and since $a<10^n$, there are no other digits.\nIn fact, if $A_m = a$ that solves the case $n=m$,  then $A_{m+1} \\mod 10^ m \\equiv A_m$.\nNote that in some cases (like $n=5$) , both of $x_{n-1} \\in \\{0,1\\}$ yield distinct numbers $a-5\\cdot10^{n-1}$ and $a$ with all odd digits. \n$a-5\\cdot10^{n-1}$ has $n-1<n$ digits, and so $x_{n-1}=1$ is needed for padding.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "A convex polygon $\\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.",
        "Solution": "When $\\mathcal{P}$ is a triangle, the problem is trivial.  Otherwise, it is sufficient to prove that any two diagonals of the polygon cut each other into rational lengths.  Let two diagonals which intersect at a point $P$ within the polygon be $AC$ and $BD$.  Since $ABCD$ is a convex quadrilateral with sides and diagonals of rational length, we consider it in isolation.\nBy the Law of Cosines, $\\cos BAC = \\frac{AB^2 + CA^2 - BC^2}{2AB \\cdot CA}$, which is rational.  Similarly, $\\cos CAD$ is rational, as well as $\\cos BAD = \\cos (BAC + CAD) = \\cos BAC \\cos CAD - \\sin BAC \\sin CAD$.  It follows that $\\sin BAC \\sin CAD$ is rational.  Since $\\sin^2 CAD = 1 - \\cos^2 CAD$ is rational, this means that $\\frac{\\sin BAC \\sin CAD}{\\sin^2 CAD} = \\frac{\\sin BAC}{\\sin CAD} = \\frac{\\sin BAP}{\\sin PAD}$ is rational.  This implies that $\\frac{ AB \\sin BAP}{ AD \\sin PAD} = \\frac{\\frac{1}{2} AB \\cdot AP \\sin BAP}{\\frac{1}{2} AP \\cdot AD \\sin PAD} = \\frac{[BAP]}{[PAD]} = \\frac{BP}{PD}$ is rational.  Define $r$ to be equal to $\\frac{BP}{PD}$. We know that $\\frac{BP}{PD}$ is rational; hence $r$ is rational. We also have $(1+r)PD = (1+\\frac{BP}{PD} )PD = PD + BP = BD$, which is, of course, rational.  It follows that $BP$ and $PD$ both have rational length, as desired.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Let $n \\neq 0$. For every sequence of integers\nsatisfying $0 \\le a_i \\le i$, for $i=0,\\dots,n$, define another sequence\nby setting $t(a_i)$ to be the number of  terms in the sequence $A$ that precede the term $a_i$ and are different from $a_i$. Show that, starting from any sequence $A$ as above, fewer than $n$ applications of the transformation $t$ lead to a sequence $B$ such that $t(B) = B$.",
        "Solution": "Consider some sequence $C = c_0, \\ldots, c_n$ as the image of $A$ after $t$ has been applied some finite number of times.\nLemma 1.  If $t(c_k) = c_k = j$, then ${} c_j = \\cdots = c_{j+i} = \\cdots = c_k = j$ ($0 \\le i \\le k-j$).\nProof. Since the $j$ terms $c_0, \\ldots, c_{j-1}$ are all less than $j$, no other terms that precede $c_{k}$ can be unequal to $c_k$.  The lemma follows.\nLemma 2. If $t(c_k) = c_k = j$, then $t^2(c_k) = t(c_k)$ (where $f^2(x) = f(f(x))$).\nProof.  Since only $i$ terms precede the term $c_i$, we will have $t^m(c_i) \\le i$, for any integer $m$.  This means that we will always have $t^m (c_0), \\ldots, t^m (c_{j-1}) < j$.  This means that $c_j = t(c_j) = \\cdots = c_k = t(c_k)$, and the lemma follows by iteration.\nThus we may regard a term $c_k$ as stable if $t(c_k) = c_k$.  We will call a sequence stable if all of its terms are stable.\nLemma 3.  If $c_k =j$ is not stable, then $t(c_k) > c_k$.\nProof. We have $c_0, \\ldots, c_{j-1} < j$, so we always have $t(c_k) \\ge c_k$.  Equality implies that $c_k$ is stable.\nWe will now prove the problem by induction.  For the base case, $n=1$, we have either 0,0 or 0,1, both of which are stable.  Now, suppose that $t^{n-2}(a_0), \\ldots, t^{n-2}(a_{n-1})$ are stable.  We must then have $t^{n-2}(a_n) \\in \\{ n-2, n-1, n\\}$.  If $t^{n-2}(a_n) = n$, then it is already stable.  If $t^{n-2}(a_n) = n-1$, then either it is already stable or $t^{n-1}(a_n) = n$, which is stable.  If $t^{n-2}(a_n) = t^{n-2}(a_{n-1})$, then $t^{n-2}(a_n)$ must already be stable.  The only possibilities remaining are $t^{n-2}(a_n) = n-2, t^{n-2}(a_{n-1}) = n-1$ and $t^{n-2}(a_n) = n-2, t^{n-2}(a_{n-1}) < n-2$.  In the first case, we must have $t^{n-1}(a_n)$ equal to $n-1$ or $n$, both of which will make it stable.  In the second case, we must have $t^{n-2}(a_{n-2}) = t^{n-2}(a_{n-1}) < n-2$, giving us $t^{n-1}(a_n) = n$, which will make it stable.  Thus the induction is complete.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\n"
    },
    {
        "Problem": "Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\\cdot MD = MC^2$.",
        "Solution_1": "Extend segment $DM$ through $M$ to $G$ such that $FG\\parallel CD$.\n\nThen $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD\\parallel CG$. Hence $MC = MF$ if and only if $\\angle GCD = \\angle FDA$, that is, $\\angle FDA + \\angle CGF = 180^\\circ$.\nBecause quadrilateral $ABED$ is cyclic, $\\angle FDA = \\angle ABE$. It follows that $MC = MF$ if and only if\n\nthat is, quadrilateral $CBFG$ is cyclic, which is equivalent to\n\nBecause $\\angle DMC = \\angle CMB$, $\\angle CBM = \\angle DCM$ if and only if triangles $BCM$ and $CDM$ are similar, that is\n\nor $MB\\cdot MD = MC^2$.",
        "Solution_2": "We first assume that $MB\\cdot MD = MC^2$. Because $\\frac{MC}{MD} = \\frac{MB}{MC}$ and $\\angle CMD = \\angle BMC$, triangles $CMD$ and $BMC$ are similar. Consequently, $\\angle MCD = \\angle MBC$. This exact condition can also be visualized through power of M with respect to the circumcircle of triangle $BDC$ and getting the angle condition from the alternate segment theorem.\n\nBecause quadrilateral $ABED$ is cyclic, $\\angle DAE = \\angle DBE$. Hence\n\nimplying that $AE\\parallel CF$, so $\\angle AEF = \\angle CFE$. Because quadrilateral $ABED$ is cyclic, $\\angle ABD = \\angle AED$. Hence\n\nBecause $\\angle FBM = \\angle DFM$ and $\\angle FMB = \\angle DMF$, triangles $BFM$ and $FDM$ are similar. Consequently, $\\frac{FM}{DM} = \\frac{BM}{FM}$, or $FM^2 = BM\\cdot DM = CM^2$. Therefore $MC^2 = MB\\cdot MD$ implies $MC = MF$.\nNow we assume that $MC = MF$. Applying Ceva's Theorem to triangle $BCF$ and cevians $BM, CA, FE$ gives\n\nimplying that $\\frac{BA}{AF} = \\frac{BE}{EC}$, so $AE\\parallel CF$.\nConsequently, $\\angle DCM = \\angle DAE$. Because quadrilateral $ABED$ is cyclic, $\\angle DAE = \\angle DBE$. Hence\n\nBecause $\\angle CBM = \\angle DCM$ and $\\angle CMB = \\angle DMC$, triangles $BCM$ and $CDM$ are similar. Consequently, $\\frac{CM}{DM} = \\frac{BM}{CM}$, or $CM^2 = BM\\cdot DM$.\nCombining the above, we conclude that $MF = MC$ if and only if $MB\\cdot MD = MC^2$.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Let $a$, $b$, $c$ be positive real numbers. Prove that",
        "Solution_1": "Since all terms are homogeneous, we may assume WLOG that $a + b + c = 3$.\nThen the LHS becomes $\\sum \\frac {(a + 3)^2}{2a^2 + (3 - a)^2} = \\sum \\frac {a^2 + 6a + 9}{3a^2 - 6a + 9} = \\sum \\left(\\frac {1}{3} + \\frac {8a + 6}{3a^2 - 6a + 9}\\right)$.\nNotice $3a^2 - 6a + 9 = 3(a - 1)^2 + 6 \\ge 6$, so $\\frac {8a + 6}{3a^2 - 6a + 9} \\le \\frac {8a + 6}{6}$.\nSo $\\sum \\frac {(a + 3)^2}{2a^2 + (3 - a)^2} \\le \\sum \\left(\\frac {1}{3} + \\frac {8a + 6}{6}\\right) = 1 + \\frac {8(a + b + c) + 18}{6} = 8$, as desired.",
        "Solution_2": "Note that\n\nSetting $x = a$ and $y = b + c$ yields\n\nThus, we have\n\nand its analogous forms. Thus, the desired inequality is equivalent to\n\nBecause $(b + c)^2\\leq 2(b^2 + c^2)$, we have $2a^2 + (b + c)^2\\leq 2(a^2 + b^2 + c^2)$ and its analogous forms. It suffices to show that\n\nor,\n\nMultiplying this out the left-hand side of the last inequality gives $3(a^2 + b^2 + c^2) - 2(ab + bc + ca)$. Therefore the inequality $(*)$ is equivalent to $2[a^2 + b^2 + c^2 - (ab + bc + ca)]\\geq 0$, which is evident because\n\nEquality holds when $a = b = c$.",
        "Solution_3": "Given a function $f$ of three variables, define the cyclic sum\n\nWe first convert the inequality into\n\nSplitting the 5 among the three terms yields the equivalent form\n\nThe numerator of the term shown factors as $(2a - x)(2a - 5x)$, where $x = b + c$. We will show that\n\nIndeed, $(3)$ is equivalent to\n\nwhich reduces to\n\nevident. We proved that\n\nhence $(2)$ follows. Equality holds if and only if $2a = b + c, 2b = c + a, 2c = a + b$, i.e., when $a = b = c$.",
        "Solution_4": "Given a function $f$ of $n$ variables, we define the symmetric sum\n\nwhere $\\sigma$ runs over all permutations of $1, \\ldots, n$ (for a total of $n!$ terms).\nWe combine the terms in the desired inequality over a common denominator and use symmetric sum notation to simplify the algebra. The numerator of the difference between the two sides is\n\nRecalling Schur's Inequality, we have\n\nor\n\nHence,\n\nand by repeated AM-GM inequality,\n\nand\n\nAdding these three inequalities yields the desired result.",
        "Solution_5": "Since, the inequality is homogeneous we may assume that $a+b+c=1$ and $0<a,b, c<1$.\nThe first time on the LHS is the inequality will be: \n\nNote that equality holds when $a=b=c=1/3$. \nA simple sketch of $f(x)$ on $[0,1]$ shows that the curve lies below the tangent line at $1/3$.\nWhich has the equation of the form $y=\\frac{12x+4}{3}$.\nSo we claim that\nUpon clearing the denominators, it is equivalent to:\nNote that since the curve and the line intersect at $1/3, 3a-1$ would be a factor.\n\nAdding the similar inequalities for $b$ and $c$ gives:\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "At the vertices of a regular hexagon are written six nonnegative integers whose sum is 2003. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices.",
        "Solution": "Assume the original numbers are $a,b,c,d,e,f$. Since $a+b+c+d+e+f$ is odd, either $a+c+e$ or $b+d+f$ must be odd. WLOG let $a+c+e$ be odd and $a\\ge c\\ge e \\ge 0$."
    },
    {
        "Problem": "(Titu Andreescu) Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that\n\nWhen does equality hold?",
        "Solution_1": "It suffices to show that  because the other side of the inequality follows by symmetry. By Pitot's Theorem, we have $AB-AD=BC-CD$. WLOG, let $AB \\geq AD$. Then we have that $BC \\geq CD$, so we must show $BC^3-CD^3 \\leq 3(AB^3-AD^3)$. If $AB=AD$, we are done. Therefore, we will assume for the rest of the proof that $AB>AD$, $BC>CD$.\nThen we would like to show that \nBy the Law of Cosines, we have that  where $-1 \\leq k_1, k_2 \\leq 1$ (this is because $60^{\\circ} \\leq \\angle BAD,\\angle BCD \\leq 120^{\\circ}$). Therefore,  Then we wish to show that  and we are done. $\\blacksquare$",
        "Solution_2": "By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence $a + c = b + d \\Rightarrow a - b = d - c \\Rightarrow |a - b| = |d - c|$ Now we factor the desired expression into $\\frac {|d - c|(c^2 + d^2 + cd)}{3} \\le|a - b|(a^2 + b^2 + ab)\\le 3|d - c|(c^2 + d^2 + cd)$. Temporarily discarding the case where $a = b$ and $c = d$, we can divide through by the $|a - b| = |d - c|$ to get the simplified expression $(c^2 + d^2 + cd)/3\\le a^2 + b^2 + ab\\le 3(c^2 + d^2 + cd)$.\nNow, draw diagonal $BD$. By the law of cosines, $c^2 + d^2 + 2cd\\cos A = BD$. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that $A\\in [60^{\\circ},120^{\\circ}]$. Cosine is monotonically decreasing on this interval, so by setting $A$ at the extreme values, we see that $c^2 + d^2 - cd\\le BD^2 \\le c^2 + d^2 + cd$. Applying the law of cosines analogously to $a$ and $b$, we see that $a^2 + b^2 - ab\\le BD^2 \\le a^2 + b^2 + ab$; we hence have $c^2 + d^2 - cd\\le BD^2 \\le a^2 + b^2 + ab$ and $a^2 + b^2 - ab\\le BD^2 \\le c^2 + d^2 + cd$.\nWe wrap up first by considering the second inequality. Because $c^2 + d^2 - cd\\le BD^2 \\le a^2 + b^2 + ab$, $\\text{RHS}\\ge 3(a^2 + b^2 - ab)$. This latter expression is of course greater than or equal to $a^2 + b^2 + ab$ because the inequality can be rearranged to $2(a - b)^2\\ge 0$, which is always true. Multiply the first inequality by $3$ and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.\nEquality occurs when $a = b$ and $c = d$, or when $ABCD$ is a kite."
    },
    {
        "Problem": "(Kiran Kedlaya) Suppose $a_1, \\dots, a_n$ are integers whose greatest common divisor is 1. Let $S$ be a set of integers with the following properties:\n(a) For $i = 1, \\dots, n$, $a_i \\in S$.\n(b) For $i,j = 1, \\dots, n$ (not necessarily distinct), $a_i - a_j \\in S$.\n(c) For any integers $x,y \\in S$, if $x + y \\in S$, then $x - y \\in S$.\nProve that $S$ must be equal to the set of all integers.",
        "Solution": "Suppose $a_i$ has only one element; then for the greatest common divisor to be 1, $1$ has to be the sole element. Then $1$ is in $S$ by (a), $0$ is in $S$ by (b), $0 + 1 = 1\\in S\\Rightarrow 0 - 1 = - 1\\in S$ by (c), and we can apply (c) analogously to get that $n\\cdot 1 \\in S$ for integers $n$ and hence $S$ is the set of all integers, as desired.\nLemma: If $x,y\\in a_i$, then $ax + by\\in S$ for integers $a,b$.\nProof: Assume $a_i$ has at least two elements; $x$ and $y$. By (b), $x - y$ is in $S$, and by the application of (c) above, we get that $n(x - y)$ for integers $n$ is in $S$. Then apply (c) to $n(x - y)$ and $ny$ or $nx$ to get that $ax + by\\in S$ for all $a,b\\in \\mathbb{Z}$.\nNow let the terms be $a_1,a_2,\\ldots,a_{n}$. By applying our lemma many times, all numbers in  the form $\\sum c_1a_1$ for a sequence of integers $c_i$ are attainable if the sequence is of a length which is a power of 2. If not, we \"pad\" the sequence with many copies of an existing element of the sequence until it does have a length which is a power of 2 - it is apparent that this will not change $S$.\nBy Schur's theorem (a generalization of the more well-known Chicken McNugget theorem), every integer greater than some integer $n$ is attainable, and hence there are two members of $S$ in the form $\\sum c_1a_1$ which are consecutive integers. Furthermore, because such numbers are closed under addition, their sum is in $S$, and hence so is their difference; $1$. Thus, by the argument at the beginning at this proof, $S$ is the set of all integers, as desired."
    },
    {
        "Problem": "(Ricky Liu) For what values of $k > 0$ is it possible to dissect a $1 \\times k$ rectangle into two similar, but incongruent, polygons?",
        "Solution": "We will show that a dissection satisfying the requirements of the problem is possible if and only if $k\\neq 1$.\nWe first show by contradiction that such a dissection is not possible when $k = 1$. Assume that we have such a dissection. The common boundary of the two dissecting polygons must be a single broken line connecting two points on the boundary of the square (otherwise either the square is subdivided in more than two pieces or one of the polygons is inside the other). The two dissecting polygons must have the same number of vertices. They share all the vertices on the common boundary, so they have to use the same number of corners of the square as their own vertices. Therefore, the common boundary must connect two opposite sides of the square (otherwise one of the polygons will contain at least three corners of the square, while the other at most two). However, this means that each of the dissecting polygons must use an entire side of the square as one of its sides, and thus each polygon has a side of length 1. A side of longest length in one of the polygons is either a side on the common boundary or, if all those sides have length less than 1, it is a side of the square. But this is also true of the polygon, which means that the longest side length in the two polygons is the same. This is impossible since they are similar but not congruent, so we have a contradiction.\nWe now construct a dissection satisfying the requirements of the problem when $k\\neq 1$. Notice that we may assume that $k > 1$, because a $1\\times k$ rectangle is similar to a $1\\times\\frac{1}{k}$ rectangle.\nWe first construct a dissection of an appropriate chosen rectangle (denoted by $ABCD$ below) into two similar noncongruent polygons. The construct depends on two parameters ($n$ and $r$ below). By appropriate choice of these parameters we show that the constructed rectangle can be made similar to a $1\\times k$ rectangle, for any $k > 1$. The construction follows.\nLet $r > 1$ be a real number. For any positive integer $n$, consider the following sequence of $2n + 2$ points:\n\nand so on, until\n\nDefine a rectangle $ABCD$ by\n\nThe sides of the $(2n+2)$-gon $A_1A_2\\ldots A_{2n+1}B$ have lengths\n\nand the sides of the $(2n+2)$-gon $A_0A_1A_2\\ldots A_{2n}D$ have lengths\n\nrespectively. These two polygons dissect the rectangle $ABCD$ and, apart from orientation, it is clear that they are similar but noncongruent, with coefficient of similarity $r > 1$. The rectangle $ABCD$ and its dissection are thus constructed.\nThe rectangle $ABCD$ is similar to a rectangle of size $1\\times f_n(r)$, where\n\nIt remains to show that $f_n(r)$ can have any value $k > 1$ for appropriate choices of $n$ and $r$. Choose $n$ sufficiently large so that $1 + \\frac{1}{n} < k$. Since\n\nand $f_n(r)$ is a continuous function for positive $r$, there exists an $r$ such that $1 < r < k$ and $f_n(r) = k$, so we are done.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Melanie Wood)\nAlice and Bob play a game on a 6 by 6 grid.  On his or her turn, a player chooses a rational number not yet appearing on the grid and writes it in an empty square of the grid.  Alice goes first and then the players alternate.  When all squares have numbers written in them, in each row, the square with the greatest number is colored black.  Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't.  (If two squares share a vertex, Alice can draw a line from on to the other that stays in those two squares.)  Find, with proof, a winning strategy for one of the players.",
        "Solution_1": "Before the game, Bob may select three useless squares per row.  He may then move according to the following rules:\nBob can always do this because there are an even number of squares in each row.\nIn this way, Bob can prevent any useless square from being colored black at the end.  Then it will be sufficient for Bob to make the squares (1,1), (2,1), (2,2), (3,2), (3,3), (4,3), (4,4), (5,4), (5,5), (6,5), (6,6) useless, where $(m,n)$ is the square in the $m$th column and $n$th row.  Thus Bob can always win.\nNote that it is not necessary for Bob to move first; even if he moves second, he will always be able to ensure that at least two squares per row of his choosing are not colored black.",
        "Solution_2": "Bob may move according to the following rules:\nIn this way, there will be 2 columns in between the black squares in rows 1 and 2, so they cannot possibly touch, which means that Bob will win.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem_5": "(Titu Andreescu)\nLet $a$, $b$, and $c$ be positive real numbers.  Prove that",
        "Solutions": "We first note that for positive $x$, $x^5 + 1 \\ge x^3 + x^2$.  We may prove this in the following ways:\nIt thus becomes sufficient to prove that\nWe present two proofs of this inequality:\nWe get the desired inequality by taking $m_{1,1} = a^3$, $m_{2,2} = b^3$, $m_{3,3} = c^3$, and $m_{x,y} = 1$ when $x \\neq y$. We have equality if and only if $a = b = c = 1$.\nby Chebyshev's Inequality. By the Cauchy-Schwarz Inequality we have\nApplying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get\nand\nMultiplying the above four inequalities together yields\nas desired, with equality if and only if $x = y = z = 1$.\nBy the AM-GM inequality, it is true that $a^3 + a^3b^3 + 1 \\ge 3a^2b$, and so it is clear that\nAdditionally, again by AM-GM, it is true that $a^3b^3c^3 + a^3 + b^3 + c^3 + 1 + 1 \\ge 6abc$, and so\nas desired.\n\nIt is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that $x^5 - x^2 + 3 \\ge x^3 + 2$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Zuming Feng) A circle $\\omega$ is inscribed in a quadrilateral $ABCD$.  Let $I$ be the center of $\\omega$.  Suppose that\nProve that $ABCD$ is an isosceles trapezoid.",
        "Solution": "Our proof is based on the following key Lemma.\nLemma: If a circle $\\omega$, centered at $I$, is inscribed in a quadrilateral $ABCD$, then\nProof: Since circle $\\omega$ is inscribed in $ABCD$, we get $m\\angle DAI = m\\angle IAB = a$, $m\\angle ABI = m\\angle IBC = b$, $m\\angle BCI = m\\angle ICD = c$, $m\\angle CDI = m\\angle IDA = d$, and $a + b + c + d = 180^\\circ$. Construct a point $P$ outside of the quadrilateral such that $\\triangle ABP$ is similar to $\\triangle DCI$. We obtain\n\nimplying that the quadrilateral $PAIB$ is cyclic. By Ptolemy's Theorem, we have $AI\\cdot BP + BI\\cdot AP = AB\\cdot IP$, or\n\nBecause $PAIB$ is cyclic, it is not difficult to see that, as indicated in the figure, $m\\angle IPB = m\\angle IAB = a$, $m\\angle API = m\\angle ABI = b$, $m\\angle AIP = m\\angle ABP = c$, and $m\\angle PIB = m\\angle PAB = d$. Note that $\\triangle AIP$ and $\\triangle ICB$ are similar, implying that\n\nSubstituting the above equalities into the identity $(2)$, we arrive at\n\nor\n\nNote also that $\\triangle BIP$ and $\\triangle IDA$ are similar, implying that $\\frac{BP}{BI} = \\frac{IA}{ID}$, or\n\nSubstituting the above identity back into $(3)$ gives the desired relation $(1)$, establishing the Lemma. $\\blacksquare$\nNow we prove our main result. By the Lemma and symmetry, we have\n\nAdding the two identities $(1)$ and $(4)$ gives\n\nBy the AM-GM Inequality, we have $\\frac{AI}{DI} + \\frac{DI}{AI}\\geq 2$. Thus,\n\nwhere the equality holds if and only if $AI = DI$. Likewise, we have\n\nwhere the equality holds if and only if $BI = CI$. Adding the last two identities gives\n\nbecause $AD + BC = AB + CD$. (The latter equality is true because the circle $\\omega$ is inscribed in the quadrilateral $ABCD$.)\nBy the given condition in the problem, all the equalities in the above discussion must hold, that is, $AI = DI$ and $BI = CI$. Consequently, we have $a = d$, $b = c$, and so $\\angle DAB + \\angle ABC = 2a + 2b = 180^\\circ$, implying that $AD\\parallel BC$. It is not difficult to see that $\\triangle AIB$ and $\\triangle DIC$ are congruent, implying that $AB = CD$. Thus, $ABCD$ is an isosceles trapezoid.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Zuming Feng) Determine all composite positive integers $n$ for which it is possible to arrange all divisors of $n$ that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.",
        "Solution_1_.28official_solution.29": "No such circular arrangement exists for $n=pq$, where $p$ and $q$ are distinct primes. In that case, the numbers to be arranged are $p$; $q$ and $pq$, and in any circular arrangement, $p$ and $q$ will be adjacent. We claim that the desired circular arrangement exists in all other cases. If $n=p^e$ where $e\\ge2$, an arbitrary circular arrangement works. Henceforth we assume that $n$ has prime factorization $p^{e_1}_{1}p^{e_2}_{2}\\cdots p^{e_k}_k$, where $p_1<p_2<\\cdots<p_k$ and either $k>2$ or else $\\max(e1,e2)>1$. To construct the desired circular arrangement of $D_n:=\\lbrace d:d|n\\ \\text{and}\\ d>1\\rbrace$, start with the circular arrangement of $n,p_{1}p_{2},p_{2}p_{3}\\ldots,p_{k-1}p_{k}$ as shown.\n\nThen between $n$ and $p_{1}p_{2}$, place (in arbitrary order) all other members of $D_n$ that have $p_1$ as their smallest prime factor. Between $p_{1}p_{2}$ and $p_{2}p_{3}$, place all members of $D_n$ other than $p_{2}p_{3}$ that have $p_2$ as their smallest prime factor. Continue in this way, ending by placing $p_k,p^{2}_{k},\\ldots,p^{e_k}_{k}$ between $p_{k-1}p_k$ and $n$. It is easy to see that each element of $D_n$ is placed exactly one time, and any two adjacent elements have a common prime factor. Hence this arrangement has the desired property.\nNote. In graph theory terms, this construction yields a Hamiltonian cycle in the graph with vertex set $D_n$ in which two vertices form an edge if the two corresponding numbers have a common prime factor. The graphs below illustrate the construction for the special cases $n=p^{2}q$ and $n=pqr$.\n",
        "Solution_2": "The proof that no arrangement exists for $n=pq$, where $p,q$ are distinct primes follows from above. Apply induction to prove all other cases are possible\nBase case:\nInductive step: Suppose the desired arrangement exists for a composite $n$, show the arrangement exists for $np^r$, where $p$ is a prime relatively prime to $n$ and $r$ is a positive integer\nLet $a_1,a_2,\\cdots,a_m$ be the arrangement of divisors of $n$, then $(a_i,a_{i+1})>1$ for $i=1,2,\\cdots,m$, where $a_{m+1}=a_1$. The divisors of $np^r$ greater than 1 are of the form\n\nThe following sequence works\n\nsince all other divisors are divisible by $p$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Răzvan Gelca) Prove that the system\n\nhas no solutions in integers $x$, $y$, and $z$.",
        "Solution_1": "It suffices to show that there are no solutions to this system in the integers mod 19.  We note that $152 = 8 \\cdot 19$, so $157 \\equiv -147 \\equiv 5 \\pmod{19}$.  For reference, we construct a table of powers of five:\n\nEvidently, the order of 5 is 9.  Hence 5 is the square of a multiplicative generator of the nonzero integers mod 19, so this table shows all nonzero squares mod 19, as well.\nIt follows that $147^{157} \\equiv (-5)^{13} \\equiv -5^4 \\equiv 2$, and $157^{147} \\equiv 5^3 \\equiv -8$.  Thus we rewrite our system thus:\n\nAdding these, we have\n\nor\n\nBy Fermat's Little Theorem, the only possible values of $z^9$ are $\\pm 1$ and 0, so the only possible values of $(x^3+y+1)^2$ are $-4,-5$, and $-6$.  But none of these are squares mod 19, a contradiction.  Therefore the system has no solutions in the integers mod 19.  Therefore the solution has no equation in the integers.  $\\blacksquare$",
        "Solution_2": "Note that the given can be rewritten as\n(1) $(x^3+y)(x^3+1) = (x^3+y)(x+1)(x^2-x+1) = 147^{157} = 7^{314}3^{152}$,\n(2) $(x^3+y)(y+1)+z^9 = 157^{147} \\rightarrow (x^3+y)(y+1) = (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)$.\nWe can also see that\n$x^2-x+1 = (x+1)^2-3(x+1)+3 \\rightarrow \\gcd(x+1, x^2-x+1) \\le 3$.\nNow we notice\n$x^3+1 = 3^a7^b$\nfor some pair of non-negative integers $(a,b)$. We also note that\n$x^2 \\ge 2x$ when $|x| \\ge 2 \\rightarrow x^2-x+1 \\ge x+1$\nwhen $|x| \\ge 2$. If $x = 1$ or $x = -1$ then examining (1) would yield $147^{157} \\equiv 0 \\pmod{2}$ which is a contradiction. If $x = 0$ then from (1) we can see that $y+1 = 147^{157}$, plugging this into 2 yields\n(3) $(147^{157}-1)(147^{157}) = (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)$, $146 | (147^{157}-1)$, $146 = 2*73$.\nNoting that 73 is a prime number we see that it must divide at least 1 of the 2 factors on the right hand side of 3. Let us consider both cases.\n$73 | (157^{49}-z^3) \\rightarrow z^3 \\equiv 11^{49} \\pmod{73} \\rightarrow (11^{49})^{24} \\equiv 1 \\pmod{73}$.\nHowever\n$(11^{49})^{24} \\equiv 8 \\pmod{73}$\nThus we can see that 73 cannot divide the first factor in the right hand side of (3). Let us consider the next case.\n$73 | (157^{98}+157^{49}z^3+z^6) \\rightarrow 11^{98}+11^{49}z^3+z^6 \\equiv 0 \\pmod{73}$.\nHowever\n$11^{98}+11^{49}z^3+z^6 \\equiv z^6+47z^3+19 \\equiv z^6-26z^3+19 \\equiv (z^3-13)^2-150 \\equiv (z^3-13)^2-4 \\equiv (z^3-11)(z^3-15) \\pmod{73}$.\nIt can be seen that 11 and 15 are not perfect cubes $\\pmod{73}$ from the following.\n$15^{24} \\equiv 11^{24} \\equiv 8 \\not\\equiv 1 \\pmod{73}$\nWe can now see that $|x| \\ge 2$. Furthermore, notice that if\n$x+1 = \\pm3^k7^j$\nfor a pair of positive integers $(j,k)$ means that\n$|x^2-x+1| \\le 3$\nwhich cannot be true. We now know that\n$x+1 = \\pm3^k, \\pm7^k \\rightarrow x^2-x+1 = 3*7^m, 3^m$.\nSuppose that\n$x+1 = \\pm7^k \\rightarrow (x+1)^2-3(x+1)+3 = 3^m \\rightarrow 7^{2k}\\pm3*7^k+3  = 3^m \\rightarrow 7^{2k} \\equiv 0\\pmod{3}$\nwhich is a contradiction. Now suppose that\n$x+1 = \\pm3^k \\rightarrow (x+1)^2-3(x+1)+3 = 3*7^m \\rightarrow 3^{2k}\\pm3^{k+1}+3 = 3*7^m \\rightarrow 3^{2k-1}\\pm3^{k}+1 = 7^m \\rightarrow 3^k(3^{k-1}\\pm1) = 7^m-1$.\nWe now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation when $m > 0$ to obtain\n$k \\le 2+v_3(m) \\le 2+log_3(m) \\rightarrow 7^m-1 = 3^k(3^{k-1}\\pm1)\\le 3^{2+log_3(m)}(3^{1+log_3(m)}\\pm1) = 9m(3m\\pm1)$.\nFor $m > 2$ we can see that\n$7^m-1 > 9m(3m\\pm1)$\nwhich is a contradiction. Therefore there only possible solution is when\n$m = 0 \\rightarrow k = 1 \\rightarrow x = 2, m = 1 \\rightarrow k = 1 \\rightarrow x = -4, m = 2 \\rightarrow$ no integer solutions for k.\nPlugging this back into (1) and (2) yields\n(4) $3^{155} | (x^3+y) \\rightarrow 3^{155}| (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)$.\nIn order for (4) to be true we must have 9 dividing at least 1 of the factors on the right hand side of the equation. Let us consider both cases.\n$9 | 157^{49}-z^3 \\rightarrow 157^{49}-z^3 \\equiv 0 \\pmod{9}$.\nHowever,\n$z^3 \\equiv 0, 1, 8\\pmod{9}, 157^{49}-z^3 \\equiv 4-z^3 \\not\\equiv 0 \\pmod{9}$.\nWe now consider the second case.\n$9 | (z^6+157^{49}z^3+157^{98}) \\rightarrow 157^{98}+157^{49}z^3+z^6\\equiv 0\\pmod{9}$.\nHowever\n$z^6+157^{49}z^3+157^{98} \\equiv z^6+4z^3+7 \\equiv (z^3+2)^2+3 \\not\\equiv 0\\pmod{9}$\nTherefore there are no solutions to the given system of diophantine equations. $\\blacksquare$",
        "Solution_3": "We will show there is no solution to the system modulo 13. Add the two equations and add 1 to obtain\n\nBy Fermat's Theorem, $a^{12}\\equiv 1\\pmod{13}$ when $a$ is not a multiple of 13. Hence we compute $147^{157}\\equiv 4^1\\equiv 4\\pmod{13}$ and $157^{147}\\equiv 1^3\\equiv 1\\pmod{13}$. Thus\n\nThe cubes mod 13 are $0$, $\\pm 1$, and $\\pm 5$. Writing the first equation as\n\nwe see that there is no solution in case $x^3\\equiv -1\\pmod{13}$ and for $x^3$ congruent to $0, 1, 5, -5\\pmod{13}$, correspondingly $x^3 + y$ must be congruent to $4, 2, 5, -1$. Hence\n\nAlso $z^9$ is a cube, hence $z^9$ must be $\\text{0, 1, 5, 8, or 12}\\pmod{13}$. It is easy to check that $6\\pmod{13}$ is not obtained by adding one of $0, 9, 10, 12$ to one of $0, 1, 5, 8, 12$. Hence the system has no solutions in integers.\nNote: This argument shows there is no solution even if $z^9$ is replaced by $z^3$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Zuming Feng) Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $BC$. Construct point $C_1$ in such a way that convex quadrilateral $APBC_1$ is cyclic, $QC_1 \\parallel CA$, and $C_1$ and $Q$ lie on opposite sides of line $AB$. Construct point $B_1$ in such a way that convex quadrilateral $APCB_1$ is cyclic, $QB_1 \\parallel BA$, and $B_1$ and $Q$  lie on opposite sides of line $AC$.  Prove that points $B_1, C_1,P$, and $Q$ lie on a circle.",
        "Solution_1": "Let $B_1'$ be the second intersection of the line $C_1A$ with the circumcircle of $APC$, and let $Q'$ be the second intersection of the circumcircle of $B_1' C_1P$ and line $BC$.  It is enough to show that $B_1'=B_1$ and $Q' =Q$.  All our angles will be directed, and measured mod $\\pi$.\n\nSince points $C_1, P, Q', B_1'$ are concyclic and points $C_1, A,B_1'$ are collinear, it follows that\n\nBut since points $A, B_1', P, C$ are concyclic,\n\nIt follows that lines $AC$ and $C_1 Q'$ are parallel.  If we exchange $C$ with $B$ and $C_1$ with $B_1'$ in this argument, we see that lines $AB$ and $B_1' Q'$ are likewise parallel.\nIt follows that $Q'$ is the intersection of $BC$ and the line parallel to $AC$ and passing through $C_1$.  Hence $Q' = Q$.  Then $B_1'$ is the second intersection of the circumcircle of $APC$ and the line parallel to $AB$ passing through $Q$.  Hence $B_1' = B_1$, as desired.  $\\blacksquare$\nMotivation: One can notice that if you take such a $B_1'$ then $PQB_1'C_1$ is cyclic, and that similarly $PQB_1C_1'$ is also cyclic. One gets the intuition that only one such circle should exist where the other chord passes through A, and so sets up a ghost point for $Q$, which works. ~cocohearts",
        "Solution_2": "Lemma. $B_1, A, C_1$ are collinear.\nSuppose they are not collinear. Let line $B_1 A$ intersect circle $ABP$ (i.e. the circumcircle of $ABP$) again at $C_2$ distinct from $C_1$. Because $\\angle C_2 B_1 Q = \\angle C_2 AB = \\angle C_2 PB = 180^\\circ - \\angle C_2 PQ$, we have that $B_1 C_2 PQ$ is cyclic. Hence $\\angle C_2 QP = \\angle C_2 B_1 P = \\angle P B_1 A = \\angle C$, so $C_2 Q // AC$. Then $C_2$ must be the other intersection of the parallel to $AC$ through $Q$ with circle $ABP$. Then $C_2$ is on segment $C_1 Q$, so $C_2$ is contained in triangle $ABQ$. However, line $AB_1$ intersects this triangle only at point $A$ because $B_1$ lies on arc $AC$ not containing $P$ of circle $APC$, a contradiction. Hence, $B_1, A, C_1$ are collinear, as desired.\nAs a result, we have $\\angle C_1 B_1 Q = \\angle C_1 AB = \\angle C_1 PB = 180^\\circ - \\angle C_1 PQ$, so $B_1 C_1 PQ$ is cyclic, as desired.",
        "Solution_3": "Due to the parallel lines, $m\\angle C_1QB_1=m\\angle A.$ Therefore, it suffices to prove that  Note that $m\\angle BC_1A+m\\angle AB_1C=180^{\\circ}$ by the cyclic quadrilaterals. Now, the condition simplifies to proving $C_1,A,B_1$ collinear.\nUse barycentric coordinates. Let  By the parallel lines,  for some $y',z'.$ As $A$ is a vertex of the reference triangle, we must prove that $q(1-q)=y'z'.$\nNow we find the circumcircle of $\\triangle APC.$ Let its equation be  Substituting in $A,C$ gives $u=w=0.$ Substituting in $P$ yields  Substituting in $B_1$ yields\nSimilarly, the circumcircle of $\\triangle APB$ is  Now, we substitute $C_1$ into the equation\n\nLet $k=a^2p-a^2q-b^2(1-q)+c^2q,$ and note that both $y',z'$ are negative. By the quadratic formula, we have\n\nMultiplying these two, we have  as desired. $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Elgin Johnston) Legs $L_1, L_2, L_3, L_4$ of a square table each have  length $n$, where $n$ is a positive integer.  For how many ordered  4-tuples $(k_1, k_2, k_3, k_4)$ of nonnegative integers can we cut  a piece of length $k_i$ from the end of leg $L_i \\; (i = 1,2,3,4)$ and still have a stable table?\n(The table is stable if it can be placed so that  all four of the leg ends touch the floor. Note that a cut leg of length 0 is permitted.)",
        "Solution_1": "First flip the table upside down and observe the plane that the ends of the legs form. Let this plane be $ax+by+c=z$. Assume WLOG that the table is the square with endpoints $(0,0), (0,1), (1,0), (1,1)$ on the xy plane. Thus the lengths of the legs will be $c, c+b, c+a, c+a+b$ respectively. In other words $k_1+k_3=k_2+k_4=x$ for some $0 \\le x \\le 2n$.\nFor $0 \\le x \\le n$, it is easy to see that the number of stable tables is $(x+1)^2$.\nFor $n+1 \\le x \\le 2n$, we can biject the stable tables of $x$ to the stable of $2n-x$ by replacing all of the legs with the cut off portions.\nSo the total number of stable tables is $1^2+2^2+...+n^2+(n+1)^2+n^2+...+2^2+1^2 =$ $(1^2+2^2+...+(n+1)^2)+(1^2+2^2+...+n^2) =$ $\\dfrac{(n+1)(n+2)(2n+3)}{6}+\\dfrac{n(n+1)(2n+1)}{6}=\\boxed{\\dfrac{2n^3+6n^2+7n+3}{3}}$.\n",
        "Solution_2": "Let $C_i$ be the number of sets of cuts $\\{k_1,k_2,k_3,k_4\\}$ that result in the longest leg being of length $i$.  Then $C_i=C_{i-1}+x$, and we will evaluate $x$ to develop a recursion.  Note that if a set of cuts has non-zero cuts, then if that set is counted in $C_i$, subtract $1$ from each of the cuts to obtain a set of cuts that is counted in $C_{i-1}$.  For example, if $\\{2,3,4,5\\}$ was counted in $C_5$, then $\\{1,2,3,4\\}$ would have been counted in $C_4$ because subtracting the same length from each leg preserves the quality of the legs being coplanar, and there is a bijection.\nNow we must evaluate those sets of cuts that don't fall in this bijection, which are clearly those where one leg is completely cut off.  After some simple geometric reasoning that I won't include here, we see that one corner has a leg of length of zero, and the opposite leg, called the initial leg, will have length $i$  If the remaining legs have length $y$ and $z$, then $(y,z)=(0,i),(1,i-1),...,(i-1,1),(i,0)$, so $i+1$ options.  The initial corner can be any of the four legs, but each of the four permutations of the set of cuts ${0,0,i,i}$ is repeated twice, so we have $4(i+1)-4=4i$ total additional sets of cuts.  We conclude that $C_i=C_{i-1}+4i$, and note that $C_0=1$.  Now we can create generating functions.  $F(x)=\\sum_{i=0}^\\infty C_ix^i$.  Also, $G(x)=\\sum_{i=1}^\\infty 4ix^i$. From the recursion, we have\n\nThis final equation is easy to analyze.  Simply use $G(x)$ as the first term of a geometric series with constant multiplier of $x$.  This gives $C_i=2i^2+2i+1$.  The total number of sets of cuts for a table of legs of length $n$ is the sum of all the $C_i$, $0\\le i\\le n$, from which we deduce the answer $\\frac{2n^3 + 6n^2 +7n + 3}{3}$, or $\\frac{(n+1)(2n^2 + 4n +3)}{3}$.\n",
        "Solution_3": "The problem's condition is that the ends of the legs must be coplanar. So the plane formed by legs $L_1$, $L_2$, and $L_3$ is the same as the plane formed by the legs $L_2$, $L_3$, and $L_4$ iff the problem's condition is met.\nWe now assign coordinates to the ends of the legs. Let the table be on the xy-plane such that $L_1$ is at $(0,0,k_1)$, $L_2$ is at $(s,0,k_2)$, $L_3$ is at $(0,s,k_3)$, and $L_4$ is at $(s,s,k_4)$, where $s$ is the side length of the table. (Note that we have turned the table over so that the top lies on the xy plane while the legs point in the +z direction - this does not affect the result.) A plane in 3-space has the equation form $Ax+By+Cz=1$. Plugging the coordinates of $L_1$, $L_2$, and $L_3$ into $Ax+By+Cz=1$, we eventually get $A_1=\\dfrac{k_1-k_2}{sk_1}$, $B_1=\\dfrac{k_1-k_3}{sk_1}$, $C_1=\\dfrac{1}{k_1}$. Plugging the coordinates of $L_2$, $L_3$, and $L_4$ into $Ax+By+Cz=1$, we get $A_2=\\dfrac{k_3-k_4}{s(k_2+k_3-k_4)}$, $B_2=\\dfrac{k_2-k_4}{s(k_2+k_3-k_4)}$, $C_2=\\dfrac{1}{k_2+k_3-k_4}$.\nSince we need these to be coplanar, we set $A_1=A_2$, $B_1=B_2$, and $C_1=C_2$. (Notice that the $s$'s cancel.) From $A_1=A_2$, we get $(k_1-k_2)(k_2+k_3-k_4)=k_1(k_3-k_4)$ [1]. From $B_1=B_2$, we get $(k_1-k_3)(k_2+k_3-k_4)=k_1(k_2-k_4)$ [2]. From $C_1=C_2$, we get $k_1=k_2+k_3-k_4$ [3]. Notice that [1] and [2] can both be obtained directly from [3] and are therefore redundant. The only condition we have left to ensure that the ends of the legs are coplanar is $k_1+k_4=k_2+k_3$. Let $k_1+k_4=k_2+k_3=m$.\nWe proceed using casework by $m$. $m$ is an integer between $0$ and $2n$, inclusive. Furthermore, we note that $k_i$ range from $0$ to $n$, inclusive. For $m=0$, each of $(k_1,k_4)$ and $(k_2,k_3)$ have one choice, $(0,0)$. For $m=1$, they each have two choices, $(0,1)$ and $(1,0)$. And so on, until $m=n$, where they each have $n+1$ choices: $(0,n),(1,n-1),\\cdots,(n-1,1),(n,0)$. When $m=n+1$, the restriction that $0\\le k_i\\le n$ takes effect: there are only $n$ choices for each ordered pair, which are $(1,n),(2,n-1),\\cdots,(n-1,2),(n,1)$. And so on, until $m=2n$, where there is one choice for each ordered pair, $(n,n)$.\nThe number of possible 4-tuples $(k_1,k_2,k_3,k_4)$ is thus $1^2+2^2+...+n^2+(n+1)^2+n^2+...+2^2+1^2$, which is equivalent to $(1^2+2^2+...+(n+1)^2)+(1^2+2^2+...+n^2)$. Our final answer is $\\dfrac{(n+1)(n+2)(2n+3)}{6}+\\dfrac{n(n+1)(2n+1)}{6}=\\boxed{\\dfrac{2n^3+6n^2+7n+3}{3}}$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Kiran Kedlaya) Let $n$ be an integer greater than 1. Suppose $2n$ points are given in the plane, no three of which are collinear. Suppose $n$ of the given $2n$ points are colored blue and the other $n$ colored red. A line in the plane is called a balancing line if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side.\nProve that there exist at least two balancing lines.",
        "Solution": "Consider the convex hull of the the $2n$ points, or the points that would form the largest convex polygon. If the points in the convex hull contain both red and blue points, then there must be at least 2 edges of the graph of the convex hull such that the edge connects a blue and a red point. Drawing a line through those points would give a balancing line, as we have n-1 blue points and n-1 red points on one side, and 0 points on the other.\nTherefore it suffices to show that there exist at least 2 balancing lines when the convex hull is colored all the same color.\nPick a random point on the convex hull, and without loss of generality we can say it is blue (if there are no red we can change all the colors, and end up with an equivalent setup). Consider a line going through it and not any other points. As we rotate the line clockwise, we encounter the red points in some order. Let the ith point encountered be $R_i$. Let $b_i$ and $r_i$ be the number of points encountered before $R_i$. Then $r_i=i-1$.\nDefine a sequence $f(i)=b_i-r_i$. Then $f(1)=b_i-(1-1)=b_i\\geq0$\n$f(n)=b_n-(n-1)=b_n-n+1\\leq0$, because we can only encounter up to n-1 blue points.\nThus, $f(i)$ goes from negative to positive as $i$ goes from 1 to $n$. We can also see that $f(i)$ can only increase by one for each change in i, so we know $f(i)$ must be 0 for some value of $i$. Take the first $f(i)$ where $f(i) = 0$. Since $f(i-1)$ must have been positive, the ith point must have been red. Thus there is a balancing line for every point on the convex hull. Since a polygon must have at least 3 vertices, there must be at least $3>2$ balancing lines for the set of points when the convex hull is all the same color, and the statement is true as we desired.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\n\nCase 1 One can notice That if any blue and red point have not points on one of it's sides this mean the line between them is a balancing line (as on the other side we will have $n-1$ blue points and $n-1$ red).\nCase 2: Now we consider the case in which all the points are inside a polygon with all vertices being one color (assume WLOG it's blue\nNow consider a line that is pivoting at any vertex and rotating till reaching another vertex of the polygon (convex hull) passing through every point in side the polygon. and let $x$ be the number of red points.\nminus the number of blue points on one side of the line.\nIn other words every time the line passes through a blue point $x$ will increase by $1$, and decrease by $1$ if it passes through a red point.\nFirstly we have that $x=1$ and when the line reaches the other vertex of the polygon  it will be equal to $-1$.\nSo when $x$ will be equal to zero, we have a balancing line.\nNow repeating this process on all the other vertices of the polygon (convex hull) then we will have balancing lines equal to the number of the vertices."
    },
    {
        "Problem": "(Titu Andreescu, Gabriel Dospinescu) For $m$ a positive integer, let $s(m)$ be the sum of the digits of $m$. For $n\\ge 2$, let $f(n)$ be the minimal $k$ for which there  exists a set $S$ of $n$ positive integers such that  $s\\left(\\sum_{x\\in X} x\\right) = k$ for any nonempty subset $X\\subset S$.  Prove that there are constants $0 < C_1 < C_2$ with",
        "Solution": "For the upper bound, let $p$ be the smallest integer such that $10^p\\geq n(n+1)/2$ and let\n\nThe sum of any nonempty set of elements of $S$ will have the form $k(10^p - 1)$ for some $1\\leq k\\leq n(n+1)/2$. Write $k(10^p - 1) = [(k - 1)10^p] + [(10^p - 1) - (k - 1)]$. The second term gives the bottom $p$ digits of the sum and the first term gives at most $p$ top digits. Since the sum of a digit of the second term and the corresponding digit of $k - 1$ is always 9, the sum of the digits will be $9p$. Since $10^{p-1} < n(n+1)/2$, this example shows that\n\nSince $n\\geq 2$, $5(n + 1) < n^4$, and hence\n\nFor the lower bound, let $S$ be a set of $n\\geq 2$ positive integers such that any nonempty $X\\subset S$ has $s\\left(\\sum_{x\\in X} x\\right) = f(n)$. Since $s(m)$ is always congruent to $m$ modulo 9, $\\sum_{x\\in X} x\\equiv f(n)\\pmod{9}$ for all nonempty $X\\subset S$. Hence every element of $S$ must be a multiple of 9 and $f(n)\\geq 9$. Let $q$ be the largest positive integer such that $10^q - 1\\leq n$. Lemma 1 below shows that there is a nonempty subset $X$ of $S$ with $\\sum_{x\\in X} x$ a multiple of $10^q - 1$, and hence Lemma 2 shows that $f(n)\\geq 9q$.\nLemma 1. Any set of $m$ positive integers contains a nonempty subset whose sum is a multiple of $m$.\nProof. Suppose a set $T$ has no nonempty subset with sum divisible by $m$. Look at the possible sums mod $m$ of nonempty subsets of $T$. Adding a new element $a$ to $T$ will give at least one new sum mod $m$, namely the least multiple of $a$ which does not already occur. Therefore the set $T$ has at least $|T|$ distinct sums mod $m$ of nonempty subsets and $|T| < m$. $\\blacksquare$\nLemma 2. Any positive multiple $M$ of $10^q - 1$ has $s(M)\\geq 9q$.\nProof. Suppose on the contrary that $M$ is the smallest positive multiple of $10^q - 1$ with $s(M) < 9q$. Then $M\\neq 10^q - 1$, hence $M > 10^q$. Suppose the most significant digit of $M$ is the $10^m$ digit, $m\\geq q$. Then $N = M - 10^{m-q}(10^q - 1)$ is a smaller positive multiple of $10^q - 1$ and has $s(N)\\leq s(M) < 9q$, a contradiction. $\\blacksquare$\nFinally, since $10^{q+1} > n$, we have $q + 1 > \\log_{10} n$. Since $f(n)\\geq 9q$ and $f(n)\\geq 9$, we have\n\nWeaker versions of Lemmas 1 and 2 are still sufficient to prove the desired type of lower bound.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Kiran Kedlaya) Let $p$ be a prime number and let $s$ be an integer with $0 < s < p$. Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and\nif and only if $s$ is not a divisor of $p-1$.\nNote: For $x$ a real number, let $\\lfloor x \\rfloor$ denote the greatest integer less than or equal to $x$, and let $\\{x\\} = x - \\lfloor x \\rfloor$ denote the fractional part of $x$.",
        "Solution_1": "We proceed by contradiction.  Assume that $s|(p-1)$.  Then for some positive integer $k$, $sk=p-1$.  The conditions given are equivalent to stating that $sm \\bmod p < sn \\bmod p< s\\bmod p$.\nNow consider the following array modulo p:\n\nObviously, there are $s$ rows and $k$ columns. The first entry of each row is simply $((r-1)k+1)s\\mod p$.  Since we wish for $sm\\mod p$ and $sn\\mod p$ to both be in the first column, while also satisfying the given conditions, we can easily see that $sm$ must be in a row $m_r$ with $m_r>n_r$, where $n_r$ denotes the row $sn \\mod p$ is in.  However, since the values of each entry decreases while $((r-1)k+1)s$ keeps increasing, we can see that the condition can never be satisfied and thus, $s\\not|(p-1).$\nTo prove the other direction, let $sj+r=p-1$ for positive integers $j$ and $r$ with $j$ being the largest integer such that $sj<p-1$ and $r<s$. Since $s\\not|(p-1)$, $1<s<p-1$.\nNote that for $0<l<j$, $ls\\ge s\\mod p$.  Thus, the first integer multiplied by s modulo p that will be less than s is $j+1.$ $(j+1)s\\equiv s-r-1\\mod p$.  Since $\\lbrace s,2s,\\hdots,js,(j+1)s,\\hdots,xs,\\hdots,(p-1)s \\rbrace$ is a complete residue system mod p with the exception of the 0 term, we can find an $x>j+1$ such that $xs \\equiv s-1 \\mod p$.\nThus, choose $m=j+1$ and $n=x$ to complete the proof.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Dick Gibbs) For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k+1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $N/2$.",
        "Solution_1": "Let one optimal set of integers be $\\{a_1,\\dots,a_{2k+1}\\}$ with $a_1 > a_2 > \\cdots > a_{2k+1} > 0$.\nThe two conditions can now be rewritten as $a_1+\\cdots + a_k \\leq N/2$ and $a_1+\\cdots +a_{2k+1} > N$.\nSubtracting, we get that $a_{k+1}+\\cdots + a_{2k+1} > N/2$, and hence $a_{k+1}+\\cdots + a_{2k+1} > a_1+\\cdots + a_k$.\nIn words, the sum of the $k+1$ smallest numbers must exceed the sum of the $k$ largest ones.\nLet $a_{k+1}=C$. As all the numbers are distinct integers, we must have $\\forall i \\in\\{1,\\dots,k\\}:~ a_{k+1-i} \\geq C+i$, and also $\\forall i \\in\\{1,\\dots,k\\}:~ a_{k+1+i} \\leq C-i$.\nThus we get that $a_1+\\cdots + a_k \\geq kC + \\dfrac{k(k+1)}2$, and $a_{k+1}+\\cdots + a_{2k+1} \\leq (k+1)C - \\dfrac{k(k+1)}2$.\nAs we want the second sum to be larger, clearly we must have $(k+1)C - \\dfrac{k(k+1)}2 > kC + \\dfrac{k(k+1)}2$.\nThis simplifies to $C > k(k+1)$.\nHence we get that:\n\nOn the other hand, for the set $\\{ k^2+k+1+i ~|~ i\\in\\{-k,\\dots,k\\} \\, \\}$ the sum of the largest $k$ elements is exactly $k^3 + k^2 + k + \\dfrac{k(k+1)}2$, and the sum of the entire set is $(k^2+k+1)(2k+1) = 2k^3 + 3k^2 + 3k + 1$, which is more than twice the sum of the largest set.\nHence the smallest possible $N$ is $\\boxed{ N = 2k^3 + 3k^2 + 3k }$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Titu Andreescu, Gabriel Dospinescu) For integral $m$, let $p(m)$ be the greatest prime divisor of $m$. By convention, we set $p(\\pm 1)=1$ and $p(0)=\\infty$. Find all polynomials $f$ with integer coefficients such that the sequence $\\{ p(f(n^2))-2n) \\}_{n \\in \\mathbb{Z} \\ge 0}$ is bounded above. (In particular, this requires $f(n^2)\\neq 0$ for $n\\ge 0$.)",
        "Solution_1": "Let $f(x)$ be a non-constant polynomial in $x$ of degree $d$ with\ninteger coefficients, suppose further that no prime divides all the\ncoefficients of $f$ (otherwise consider the polynomial obtained by\ndividing $f$ by the gcd of its coefficients). We further normalize\n$f$ by multiplying by $-1$, if necessary, to ensure that the\nleading coefficient (of $x^d$) is positive.\nLet $g(n) = f(n^2)$, then $g(n)$ is a polynomial of degree $2$ or\nmore and $g(n) = g(-n)$. Let $g_1, \\ldots, g_k$ be the factorization\nof $g$ into irreducible factors with positive leading coefficients.\nSuch a factorization is unique. Let $d(g_i)$ denote the degree of\n$g_i$. Since $g(-n) = g(n)$ the factors $g_i$ are either even\nfunctions of $n$ or come in pairs $(g_i, h_i)$ with $g_i(-n) = (-1)^{d(g_i)} h_i(n)$.\nLet $P(0) = \\infty$, $P(\\pm 1) = 1$. For any other integer $m$ let\n$P(m)$ be the largest prime factor of $m$.\nSuppose that for some finite constant $C$ and all $n \\ge 0$ we have\n$P(g(n)) - 2n < C$. Since the polynomials $g_i$ divide $g$, the\nsame must be true for each of the irreducible polynomials $g_i$.\nA theorem of T. Nagell implies that if $d(g_i) \\ge 2$ the ratio\n$P(g_i(n))/n$ is unbounded for large values of $n$. Since in our case the $P(g_i(n))/n$ is asymptotically bounded above\nby $2$ for large $n$, we conclude that all the irreducible\nfactors $g_i$ are linear. Since linear polynomials are not even\nfunctions of $n$, they must occur in pairs $g_i(n) = a_in + b_i$,\n$h_i(n) = a_in - b_i$. Without loss of generality, $b_i \\ge 0$.\nSince the coefficients of $f$ are relatively prime, so are $a_i$\nand $b_i$, and since $P(0) = \\infty$, neither polynomial can have\nany non-negative integer roots, so $a_i > 1$ and thus $b_i > 0$.\nOn the other hand, by Dirichlet's theorem, $a_i \\le 2$, since\notherwise the sequence $a_in + b_i$ would yield infinitely many\nprime values with $P(g_i(n)) = a_in + b_i \\ge 3n.$ So $a_i = 2$ and\ntherefore $b_i$ is a positive odd integer. Setting $b_i = 2c_i + 1$, clearly $P(g_i(n)) - 2n < 2c_i + 2$.  Since this holds for each\nfactor $g_i$, it is true for the product $g$ of all the factors\nwith the bound determined by the factor with the largest value of\n$c_i$.\nTherefore, for suitable non-negative integers $c_i$, $g(n)$ is a\nproduct of polynomials of the form $4n^2 - (2c_i + 1)^2$.  Now,\nsince $g(n) = f(n^2)$, we conclude that $f(n)$ is a product of\nlinear factors of the form $4n - (2c_i + 1)^2$.\nSince we restricted ourselves to non-constant polynomials with \nrelatively prime coefficients, we can now relax this condition and\nadmit a possibly empty list of linear factors as well as an arbitrary\nnon-zero integer multiple $M$. Thus for a suitable non-zero integer\n$M$ and $k \\ge 0$ non-negative integers $c_i$, we have:",
        "Solution_2": "The polynomial $f$ has the required properties if and only if\n\nwhere $a_1, a_2, \\ldots, a_k$ are odd positive integers and $c$ is a nonzero integer. It is straightforward to verify that polynomials given by $(*)$ have the required property. If $p$ is a prime divisor of $f(n^2)$ but not of $c$, then $p|(2n - a_j)$ or $p|(2n + a_j)$ for some $j\\leq k$. Hence $p - 2n\\leq \\max\\{a_1, a_2, \\ldots, a_k\\}$. The prime divisors of $c$ form a finite set and do not affect whether or not the given sequence is bounded above. The rest of the proof is devoted to showing that any $f$ for which $\\{p(f(n^2)) - 2n\\}_{n\\geq 0}$ is bounded above is given by $(*)$.\nLet $\\mathbb{Z}[x]$ denote the set of all polynomials with integer coefficients. Given $f\\in\\mathbb{Z}[x]$, let $\\mathcal{P}(f)$ denote the set of those primes that divide at least one of the numbers in the sequence $\\{f(n)\\}_{n\\geq 0}$. The solution is based on the following lemma.\nLemma. If $f\\in\\mathbb{Z}[x]$ is a nonconstant polynomial then $\\mathcal{P}(f)$ is infinite.\nProof. Repeated use will be made of the following basic fact: if $a$ and $b$ are distinct integers and $f\\in\\mathbb{Z}[x]$, then $a - b$ divides $f(a) - f(b)$. If $f(0) = 0$, then $p$ divides $f(p)$ for every prime $p$, so $\\mathcal{P}(f)$ is infinite. If $f(0) = 1$, then every prime divisor $p$ of $f(n!)$ satisfies $p > n$. Otherwise $p$ divides $n!$, which in turn divides $f(n!) - f(0) = f(n!) - 1$. This yields $p|1$, which is false. Hence $f(0) = 1$ implies that $\\mathcal{P}(f)$ is infinite. To complete the proof, set $g(x) = f(f(0)x)/f(0)$ and observe that $g\\in\\mathcal{Z}[x]$ and $g(0) = 1$. The preceding argument shows that $\\mathcal{P}(g)$ is infinite, and it follows that $\\mathcal{P}(f)$ is infinite. $\\blacksquare$\nSuppose $f\\in\\mathbb{Z}[x]$ is nonconstant and there exists a number $M$ such that $p(f(n^2)) - 2n\\leq M$ for all $n\\geq 0$. Application of the lemma to $f(x^2)$ shows that there is an infinite sequence of distinct primes $\\{p_j\\}$ and a corresponding infinite sequence of nonnegative integers $\\{k_j\\}$ such that $p_j|f(k_j)^2$ for all $j\\geq 1$. Consider the sequence $\\{r_j\\}$ where $r_j = \\min\\{k_j\\pmod{p_j}, p_j - k_j\\pmod{p_j}\\}$. Then $0\\leq r_j\\leq (p_j - 1)/2$ and $p_j|f(r_j)^2$. Hence $2r_j + 1\\leq p_j\\leq p(f(r_j^2))\\leq M + 2r_j$, so $1\\leq p_j - 2r_\\leq M$ for all $j\\geq 1$. It follows that there is an integer $a_1$ such that $1\\leq a_1\\leq M$ and $a_1 = p_j - 2r_j$ for infinitely many $j$. Let $m = \\deg f$. Then $p_j|4^mf(((p_j - a_1)/2)^2)$ and $4^mf(((x - a_1)/2)^2)\\in\\mathbb{Z}[x]$. Consequently, $p_j|f((a_1/2)^2)$ for infinitely many $j$, which shows that $(a_1/2)^2$ is a zero of $f$. Since $f(n^2)\\leq 0$ for $n\\geq 0$, $a_1$ must be odd. Then $f(x) = (4x - a_1)^2g(x)$, where $g\\in\\mathbb{Z}[x]$. (See the note below.) Observe that $\\{p(g(n^2)) - 2n\\}_{n\\geq 0}$ must be bounded above. If $g$ is constant, we are done. If $g$ is nonconstant, the argument can be repeated to show that $f$ is given by $(*)$.\nNote. The step that gives $f(x) = (4x - a_1^2)g(x)$ where $g\\in\\mathbb{Z}[x]$ follows immediately using a lemma of Gauss. The use of such an advanced result can be avoided by first writing $f(x) = r(4x - a_1^2)g(x)$ where $r$ is rational and $g\\in\\mathbb{Z}[x]$. Then continuation gives $f(x) = c(4x - a_1^2)\\cdots (4x - a_k^2)$ where $c$ is rational and the $a_i$ are odd. Consideration of the leading coefficient shows that the denominator of $c$ is $2^s$ for some $s\\geq 0$ and consideration of the constant term shows that the denominator is odd. Hence $c$ is an integer.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Ricky Liu) Find all positive integers $n$ such that there are $k\\ge 2$ positive rational numbers $a_1, a_2, \\ldots, a_k$ satisfying $a_1 + a_2 + \\cdots + a_k = a_1\\cdot a_2\\cdots a_k = n$.",
        "Solution_1": "First, consider composite numbers.  We can then factor $n$ into $p_1p_2.$  It is easy to see that $p_1+p_2\\le n$, and thus, we can add $(n-p_1-p_2)$ 1s in order to achieve a sum and product of $n$.  For $p_1+p_2=n$, which is only possible in one case, $n=4$, we consider $p_1=p_2=2$.\nSecondly, let $n$ be a prime.  Then we can find the following procedure: Let $a_1=\\frac{n}{2}, a_2=4, a_3=\\frac{1}{2}$ and let the rest of the $a_k$ be 1.  The only numbers we now need to check are those such that $\\frac{n}{2}+4+\\frac{1}{2}>n\\Longrightarrow n<9$.  Thus, we need to check for $n=1,2,3,5,7$.  One is included because it is neither prime nor composite.\nFor $n=1$, consider $a_1a_2\\hdots a_k=1$.  Then by AM-GM, $a_1+a_2+\\hdots+a_k\\ge k\\sqrt[k]{1}>1$ for $k\\ge 2$.  Thus, $n=1$ is impossible.\nIf $n=2$, once again consider $a_1a_2\\hdots a_k=2$.  Similar to the above, $a_1+a_2+\\hdots\\ge k\\sqrt[k]{2}>2$ for $k\\ge 2$ since $\\sqrt[k]{2}>1$ and $k>2$. Obviously, $n=2$ is then impossible.\nIf $n=3$, let $a_1a_2\\hdots a_k=3$.  Again, $a_1+a_2+\\hdots\\ge k\\sqrt[k]{3}>3$.  This is obvious for $k\\ge 3$.  Now consider $k=2$.  Then $2\\sqrt{3}\\approx 3.4$ is obviously greater than $3$.  Thus, $n=3$ is impossible.\nIf $n=5$, proceed as above and consider $k=2$.  Then $a_1+a_2=5$ and $a_1a_2=5$.  However, we then come to the quadratic $a_1^2-5a_1+5=0 \\Longrightarrow a_1=\\frac{5\\pm\\sqrt{5}}{2}$, which is not rational.  For $k=3$ and $k=4$ we note that $\\sqrt[3]{5}>\\frac{5}{3}$ and $\\sqrt[4]{5}>\\frac{5}{4}$.  This is trivial to prove.  If $k\\ge 5$, it is obviously impossible, and thus $n=5$ does not work.\nThe last case, where $n=7$, is possible using the following three numbers.  $a_1=\\frac{9}{2}, a_2=\\frac{4}{3}, a_3=\\frac{7}{6}$ shows that $n=7$ is possible.\nHence, $n$ can be any positive integer greater than $3$ with the exclusion of $5$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Zoran Sunik) A mathematical frog jumps along the number line. The frog starts at 1, and jumps according to the following rule: if the frog is at integer $n$, then it can jump either to $n+1$ or to $n+2^{m_n+1}$ where $2^{m_n}$ is the largest power of 2 that is a factor of $n$. Show that if $k\\ge 2$ is a positive integer and $i$ is a nonnegative integer, then the minimum number of jumps needed to reach $2^i k$ is greater than the minimum number of jumps needed to reach $2^i$.",
        "Solution_1": "For $i\\geq 0$ and $k\\geq 1$, let $x_{i,k}$ denote the minimum number of jumps needed to reach the integer $n_{i,k} = 2^i k$. We must prove that\n\nfor all $i\\geq 0$ and $k\\geq 2$. We prove this using the method of descent.\nFirst note that $(1)$ holds for $i = 0$ and all $k\\geq 2$, because it takes 0 jumps to reach the starting value $n_{0,1} = 1$, and at least one jump to reach $n_{0,k} = k\\geq 2$. Now assume that $(1)$ is not true for all choices of $i$ and $k$. Let $i_0$ be the minimal value of $i$ for which $(1)$ fails for some $k$, let $k_0$ be the minimal value of $k > 1$ for which $x_{i_0,k}\\leq x_{i_0,1}$. Then it must be the case that $i_0\\geq 1$ and $k_0\\geq 2$.\nLet $J_{i_0,k_0}$ be a shortest sequence of $x_{i_0,k_0} + 1$ integers that the frog occupies in jumping from 1 to $2^{i_0} k_0$. The length of each jump, that is, the difference between consecutive integers in $J_{i_0,k_0}$, is either 1 or a positive integer power of 2. The sequence $J_{i_0,k_0}$ cannot contain $2^{t_0}$ because it takes more jumps to reach $2^{t_0} k_0$ than it does to reach $2^{t_0}$. Let $2^{M+1}$, $M\\geq 0$ be the length of the longest jump made in generating $J_{i_0,k_0}$. Such a jump can only be made from a number that is divisible by $2^M$ (and by no higher power of 2). Thus we must have $M < i_0$, since otherwise a number divisible by $2^{i_0}$ is visited before $2^{i_0} k_0$ is reached, contradicting the definition of $k_0$.\nLet $2^{m+1}$ be the length of the jump when the frog jumps over $2^{i_0}$. If this jump starts at $2^m(2t - 1)$ for some positive integer $t$, then it will end at $2^m(2t - 1) + 2^{m+1} = 2^m(2t + 1)$. Since it goes over $2^{i_0}$ we see $2^m(2t - 1) < 2^{i_0} < 2^m(2t + 1)$ or $(2^{i_0-m} - 1)/2 < t < (2^{i_0-m} + 1)/2$. Thus $t = 2^{i_0-m-1}$ and the jump over $2^{i_0}$ is from $2^m(2^{i_0-m} - 1) = 2^{i_0} - 2^m$ to $2^m(2^{i_0-m}+1) = 2^{i_0} + 2^m$.\nConsidering the jumps that generate $J_{i_0,k_0}$, let $N_1$ be the number of jumps from 1 to $2^{i_0} + 2^m$, and let $N_2$ be the number of jumps from $2^{i_0} + 2^m$ to $2^{i_0}k$. By definition of $i_0$, it follows that $2^m$ can be reached from 1 in less than $N_1$ jumps. On the other hand, because $m < i_0$, the number $2^{i_0}(k_0 - 1)$ can be reached from $2^m$ in exactly $N_2$ jumps by using the same jump length sequence as in jumping from $2^m + 2^{i_0}$ to $2^{i_0} k_0 = 2^{i_0}(k_0 - 1) + 2^{i_0}$. The key point here is that the shift by $2^{i_0}$ does not affect any of divisibility conditions needed to make jumps of the same length. In particular, with the exception of the last entry, $2^{i_0} k_0$, all of the elements of $J_{i_0,k_0}$ are of the form $2^p(2t + 1)$ with $p < i_0$, again because of the definition of $k_0$. Because $2^p(2t + 1) - 2^{i_0} = 2^p(2t - 2^{i_0-p} + 1)$ and the number $2t + 2^{i_0-p} + 1$ is odd, a jump of size $2^{p+1}$ can be made from $2^p(2t + 1) - 2^{i_0}$ just as it can be made from $2^p(2t + 1)$.\nThus the frog can reach $2^m$ from 1 in less than $N_1$ jumps, and can then reach $2^{i_0}(k_0 - 1)$ from $2^m$ in $N_2$ jumps. Hence the frog can reach $2^{i_0}(k_0 - 1)$ from 1 in less than $N_1 + N_2$ jumps, that is, in fewer jumps than needed to get to $2^{i_0} k_0$ and hence in fewer jumps than required to get to $2^{i_0}$. This contradicts the definition of $k_0$.",
        "Solution_2": "Suppose $x_0 = 1, x_1, \\ldots, x_t = 2^i k$ are the integers visited by the frog on his trip from 1 to $2^i k$, $k\\geq 2$. Let $s_j = x_j - x_{j-1}$ be the jump sizes. Define a reduced path $y_j$ inductively by\n\nSay a jump $s_j$ is deleted in the second case. We will show that the distinct integers among the $y_j$ give a shorter path from 1 to $2^i$. Clearly $y_j\\leq 2^i$ for all $j$. Suppose $2^i - 2^{r+1} < y_j\\leq 2^i - 2^r$ for some $0\\leq r\\leq i - 1$. Then every deleted jump before $y_j$ must have length greater than $2^r$, hence must be a multiple of $2^{r+1}$. Thus $y_j\\equiv x_j\\pmod{2^{r+1}}$. If $y_{j+1} > y_j$, then either $s_{j+1} = 1$ (in which case this is a valid jump) or $s_{j+1}/2 = 2^m$ is the exact power of 2 dividing $x_j$. In the second case, since $2^r\\geq s_{j+1} > 2^m$, the congruence says $2^m$ is also the exact power of 2 dividing $y_j$, thus again this is a valid jump. Thus the distinct $y_j$ form a valid path for the frog. If $j = t$ the congruence gives $y_t\\equiv x_t\\equiv 0\\pmod{2^{r+1}}$, but this is impossible for $2^i - 2^{r+1} < y_t\\leq 2^i - 2^r$. Hence we see $y_t = 2^i$, that is, the reduced path ends at $2^i$. Finally since the reduced path ends at $2^i < 2^i k$ at least one jump must have been deleted and it is strictly shorter than the original path.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Zuming Feng, Zhonghao Ye) Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC$, respectively, such that $AE/ED = BF/FC$.  Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$ respectively. Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$ pass through a common point.",
        "Solution_1": "Let the intersection of the circumcircles of $SAE$ and $SBF$ be $X$, and let the intersection of the circumcircles of $TCF$ and $TDE$ be $Y$.\n$BXF=BSF=AXE$ because $BSF$ tends both arcs $AE$ and $BF$.\n$BFX=XSB=XEA$ because $XSB$ tends both arcs $XA$ and $XB$.\nThus, $XAE\\sim XBF$ by AA similarity, and $X$ is the center of spiral similarity for $A,E,B,$ and $F$.\n$FYC=FTC=EYD$ because $FTC$ tends both arcs $ED$ and $FC$.\n$FCY=FTY=EDY$ because $FTY$ tends both arcs $YF$ and $YE$.\nThus, $YED\\sim YFC$ by AA similarity, and $Y$ is the center of spiral similarity for $E,D,F,$ and $C$.\nFrom the similarity, we have that $XE/XF=AE/BF$. But we are given $ED/AE=CF/BF$, so multiplying the 2 equations together gets us $ED/FC=XE/XF$. $DEX,CFX$ are the supplements of $AEX, BFX$, which are congruent, so $DEX=CFX$, and so $XED\\sim XFC$ by SAS similarity, and so $X$ is also the center of spiral similarity for $E,D,F,$ and $C$. Thus, $X$ and $Y$ are the same point, which all the circumcircles pass through, and so the statement is true.",
        "Solution_2": "We will give a solution using complex coordinates. The first step is the following lemma.\nLemma. Suppose $s$ and $t$ are real numbers and $x$, $y$ and $z$ are complex. The circle in the complex plane passing through $x$, $x + ty$ and $x + (s + t)z$ also passes through the point $x + syz/(y - z)$, independent of $t$.\nProof. Four points $z_1$, $z_2$, $z_3$ and $z_4$ in the complex plane lie on a circle if and only if the cross-ratio\n\nis real. Since we compute\n\nthe given points are on a circle. $\\blacksquare$\nLay down complex coordinates with $S = 0$ and $E$ and $F$ on the positive real axis. Then there are real $r_1$, $r_2$ and $R$ with $B = r_1A$, $F = r_2E$ and $D = E + R(A - E)$ and hence $AE/ED = BF/FC$ gives\n\nThe line $CD$ consists of all points of the form $sC + (1 - s)D$ for real $s$. Since $T$ lies on this line and has zero imaginary part, we see from $\\text{Im}(sC + (1 - s)D) = (sr_1R + (1 - s)R)\\text{Im}(A)$ that it corresponds to $s = -1/(r_1 - 1)$. Thus\n\nApply the lemma with $x = E$, $y = A - E$, $z = (r_2 - r_1)E/(r_1 - 1)$, and $s = (r_2 - 1)(r_1 - r_2)$. Setting $t = 1$ gives\n\nand setting $t = R$ gives\n\nTherefore the circumcircles to $SAE$ and $TDE$ meet at\n\nThis last expression is invariant under simultaneously interchanging $A$ and $B$ and interchanging $E$ and $F$. Therefore it is also the intersection of the circumcircles of $SBF$ and $TCF$.",
        "Solution_3": "Let $M$ be the Miquel point of $ABCD$; then $M$ is the center of the spiral similarity that takes $AD$ to $BC$. Because $\\frac{AE}{ED} = \\frac{BF}{FC}$, the same spiral similarity also takes $E$ to $F$, so $M$ is the center of the spiral similarity that maps $AE$ to $BF$ and $ED$ to $FC$. Then it is obvious that the circumcircles of $SAE$, $SBF$, $TCF$, and $TDE$ pass through $M$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Sam Vandervelde) Let $n$ be a positive integer.  Define a sequence by setting $a_1 = n$ and, for each $k>1$, letting $a_k$ be the unique integer in the range $0 \\le a_k \\le k-1$ for which $a_1 + a_2 + \\cdots + a_k$ is divisible by $k$.  For instance, when $n=9$ the obtained sequence is $9, 1, 2, 0, 3, 3, 3, \\ldots$.  Prove that for any $n$ the sequence $a_1, a_2, a_3, \\ldots$ eventually becomes constant.",
        "Solution_1": "Let $S_k = a_1 + a_2 + \\cdots + a_k$ and $b_k = \\frac{S_k}{k}$. Thus, because $S_{k+1} = S_k + a_{k+1}$,\n\n$\\frac{k}{k+1} < 1$, and by definition, $\\frac{a_{k+1}}{k+1} < 1$. Thus, $b_{k+1} < b_k + 1$. Also, both $b_k$ and $b_{k+1}$ are integers, so $b_{k+1} \\le b_k$. As the $b_k$'s form a non-increasing sequence of positive integers, they must eventually become constant.\nTherefore, $b_k = b_{k+1}$ for some sufficiently large value of $k$. Then $a_{k+1} = S_{k+1} - S_k = b_k(k + 1) - b_k(k) = b_k$, so eventually the sequence $a_k$ becomes constant.",
        "Solution_2": "Let $a_1 = n$. Since $a_k\\leq k - 1$, we have that\n\nSince $a_1 + a_2 + \\cdots + a_n = nk$ for some integer $k$, we can keep adding $k$ to satisfy the conditions, provided that $k\\leq n$. This is true since $k\\leq\\frac{n + 1}{2}\\leq n$, so the sequence must eventually become constant.",
        "Solution_3": "Define $S_k = a_1 + a_2 + \\cdots + a_k$, and $b_k = \\frac{S_k}{k}$. By the problem hypothesis, $b_k$ is an integer valued sequence.\nLemma: There exists a $k$ such that $b_k < k$.\nProof: Choose any $k$ such that $k^2 + 3k - 2 > 2n$. Then\n\nas desired.\nEnd Lemma\nLet $k$ be the smallest $k$ such that $b_k < k$. Then $b_k = m < k$, and $S_k = km$. To make $b_{k+1}$ an integer, $S_{k+1} = S_k + a_{k+1}$ must be divisible by $k+1$. Thus, because $km + m$ is divisible by $k+1$, $a_{k+1} \\equiv m \\pmod{k+1}$, and, because $0 \\le a_{k+1} < k$, $a_{k+1} = m$. Then $b_{k+1} = \\frac{(k+1)m}{k+1} = m$ as well. Repeating the same process using $k+1$ instead of $k$ gives $a_{k+2} = m$, and an easy induction can prove that for all $N > k+1$, $a_N = m$. Thus, $a_k$ becomes a constant function for arbitrarily large values of $k$.",
        "Solution_4": "For $k\\geq 1$, let\n\nWe claim that for some $m$ we have $s_m = m(m - 1)$. To this end, consider the sequence which computes the differences between $s_k$ and $k(k - 1)$, i.e., whose $k$-th term is $s_k - k(k - 1)$. Note that the first term of this sequence is positive (it is equal to $n$) and that its terms are strictly decreasing since\n\nFurther, a negative term cannot immediately follow a positive term. Suppose otherwise, namely that $s_k > k(k - 1)$ and $s_{k+1} < (k + 1)k$. Since $s_k$ and $s_{k+1}$ are divisible by $k$ and $k + 1$, respectively, we can tighten the above inequalities to $s_k\\geq k^2$ and $s_{k+1}\\leq (k + 1)(k - 1) = k^2 - 1$. But this would imply that $s_k > s_{k+1}$, a contradiction. We conclude that the sequence of differences must eventually include a term equal to zero.\nLet $m$ be a positive integer such that $s_m = m(m - 1)$. We claim that\n\nThis follows from the fact that the sequence $a_1, a_2, a_3, \\ldots$ is uniquely determined and choosing $a_{m+i} = m - 1$, for $i\\geq 1$, satisfies the range condition\n\nand yields",
        "Solution_5.28like_solution_2.29": "First, we may make an observation and say that for $n \\equiv p (mod k)$, $\\sum_{m=2}^{k} a_m \\equiv k-p (mod k)$ must occur for the whole sum to be divisible by $k$. Thus, the following is apparent:\n\nThen, we may make another observation that when $n=k$, the sum also has to be divisible by n. We may then explore when n=k:\n\nand\n\nThen,\n\nAlso,\n for $s \\ge 1$. This is because:\n\nThis must be true since $r(n+1)$ will be divisible by $n+1$ and $k=n+1$, we may then generalize this to all $r(n+s), s \\in \\mathbb{Z}, s \\ge 1$\n\nThus, we may say that the sequence $a_1,a_2,...a_k$ must converge to some integer value $r \\le n$ when $k \\ge n + 1$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Gregory Galperin) A square grid on the Euclidean plane consists of all points $(m,n)$, where $m$ and $n$ are integers.  Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least 5?",
        "Solution_1": "Lemma. Among 3 tangent circles with radius greater than or equal to 5, one can always fit a circle with radius greater than $\\frac{1}{\\sqrt{2}}$ between those 3 circles.\nProof. Descartes' Circle Theorem states that if $a$ is the curvature of a circle ($a=\\frac 1{r}$, positive for externally tangent, negative for internally tangent), then we have that \n\nSolving for $a$, we get \n\nTake the positive root, as the negative root corresponds to internally tangent circle.\nNow clearly, we have $b+c+d \\le \\frac 35$, and $bc+cd+db\\le \\frac 3{25}$.\nSumming/square root/multiplying appropriately shows that $a \\le \\frac{3 + 2 \\sqrt{3}}5$. Incidently, $\\frac{3 + 2\\sqrt{3}}5 < \\sqrt{2}$, so $a< \\sqrt{2}$, $r > \\frac 1{\\sqrt{2}}$, as desired. $\\blacksquare$\nFor sake of contradiction, assume that we have a satisfactory placement of circles. Consider 3 circles, $p,\\ q,\\ r$ where there are no circles in between. By Appolonius' problem, there exists a circle $t$ tangent to $p,\\ q,\\ r$ externally that is between those 3 circles. Clearly, if we move $p,\\ q,\\ r$ together, $t$ must decrease in radius. Hence it is sufficient to consider 3 tangent circles. By lemma 1, there is always a circle of radius greater than $\\frac{1}{\\sqrt{2}}$ that lies between $p,\\ q,\\ r$. However, any circle with $r>\\frac 1{\\sqrt{2}}$ must contain a lattice point. (Consider placing an unit square parallel to the gridlines in the circle.) That is a contradiction. Hence no such tiling exists.",
        "Solution_2": "It is not possible. The proof is by contradiction. Suppose that such a covering family $\\mathcal{F}$ exists. Let $D(P,\\rho)$ denote the disc with center $P$ and radius $\\rho$. Start with an arbitrary disc $D(O,r)$ that does not overlap any member of $\\mathcal{F}$. Then $D(O,r)$ covers no grid point. Take the disc $D(O,r)$ to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then $D(O,r)$ is tangent to at least three discs in $\\mathcal{F}$. Observe that there must be two of the three tangent discs, say $D(A,a)$ and $D(B,b)$ such that $\\angle AOB\\leq 120^\\circ$. By the Law of Cosines applied to triangle $ABO$,\n\nwhich yields\n\nand thus\n\nNote that $r < 1/\\sqrt{2}$ because $D(O,r)$ covers no grid point, and $(a - 3r)(b - 3r)\\geq (5 - 3r)^2$ because each disc in $\\mathcal{F}$ has radius at least 5. Hence $2\\sqrt{3}r\\geq 5 - 3r$, which gives $5\\leq (3 + 2\\sqrt{3})r < (3 + 2\\sqrt{3})/\\sqrt{2}$ and thus $5\\sqrt{2} < 3 + 2\\sqrt{3}$. Squaring both sides of this inequality yields $50 < 21 + 12\\sqrt{3} < 21 + 12\\cdot 2 = 45$. This contradiction completes the proof.\nRemark: The above argument shows that no covering family exists where each disc has radius greater than $(3 + 2\\sqrt{3})/\\sqrt{2}\\approx 4.571$. In the other direction, there exists a covering family in which each disc has radius $\\sqrt{13}/2\\approx 1.802$. Take discs with this radius centered at points of the form $\\left(2m + 4n + \\frac{1}{2}, 3m + \\frac{1}{2}\\right)$, where $m$ and $n$ are integers. Then any grid point is with $\\sqrt{13}/2$ of one of the centers and the distance between any two centers is at least $\\sqrt{13}$. The extremal radius of a covering family is unknown.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(András Gyárfás) Let $S$ be a set containing $n^2+n-1$ elements, for some positive integer $n$.  Suppose that the $n$-element subsets of $S$ are partitioned into two classes.  Prove that there are at least $n$ pairwise disjoint sets in the same class.",
        "Solution_1": "Claim: If we have instead $k(n+1)-1$ elements, then we must have k disjoint subsets in a class.\nWe proceed by Induction\nBase Case: $k=1$ is trivial.\nCase 1: If there exists an $n+1$ element subset $X$ s.t. $X$ has 2 n-element subsets in different classes.\nConsider the $(k-1)(n+1)-1$ elements that are not in $X$, from the Induction hypothesis, we have $k-1$ disjoint subsets in one of the classes. Those $k-1$ subsets paired with a subset of $X$ in the same class forms $k$ disjoint subsets of size n.\nCase 2: All $n+1$ element subsets have all of their n-element subsets in one class.\nLook at any 2 n-element subsets $a_1$ and $a_p$. Then we can create a path $a_1,a_2,...,a_p$ s.t. any 2 consecutive terms belong to the same n+1-element subset.(i.e. we change 1 element each time.) Thus, $a_1$ and $a_p$ are in the same class and we see that all subsets are in one class. Obviously we can find $k$ disjoint subsets of size n.\nThe question is asking for the case when $k=n$ which we have proved.\n",
        "Solution_2": "Call an $n+1$-element subset of $S$ separable if it has a subset in each class of the partition. We recursively build a set $Q$ of disjoint separable subsets of $S$: begin with $Q$ empty and at each step if there is a separable subset which is disjoint from all sets in $Q$ add that set to $Q$. The process terminates when every separable subset intersects a set in $Q$. Let $T$ be the set of elements in $S$ which are not in any set in $Q$. We claim that one class contains every $n$-element subset of $T$.\nSuppose that $a_1, a_2, \\ldots a_k$ are elements of $T$. Denote by $A_i$ the set $\\left\\{a_i, a_{i+1}, \\ldots a_{i+n-1}\\right\\}$. Note that for each $i$, $A_i \\cup A_{i+1}$ is not separable, so that $A_i$ and $A_{i+1}$ are in the same class. But then $A_i$ is in the same class for each $1 \\leq i \\leq k - n + 1$ — in particular, $A_1$ and $A_{k-n+1}$ are in the same class. But for any two sets we may construct such a sequence with $A_1$ equal to one and $A_{k-n+1}$ equal to the other.\nWe are now ready to construct our $n$ disjoint sets. Suppose that $|Q| = q$. Then $|T| = (n+1)(n-q) - 1 \\geq n(n-q)$, so we may select $n - q$ disjoint $n$-element subsets of $T$. Then for each of the $q$ sets in $Q$, we may select a subset which is in the same class as all the subsets of $T$, for a total of $n$ disjoint sets.",
        "Solution_3": "In order to apply induction, we generalize the result to be proved so that it reads as follows:\nProposition. If the $n$-element subsets of a set $S$ with $(n + 1)m - 1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.\nProof. Fix $n$ and proceed by induction on $m$. The case of $m = 1$ is trivial. Assume $m > 1$ and that the proposition is true for $m - 1$. Let $\\mathcal{P}$ be the partition of the $n$-element subsets into two classes. If all the $n$-element subsets belong to the same class, the result is obvious. Otherwise select two $n$-element subsets $A$ and $B$ from different classes so that their intersection has maximal size. It is easy to see that $|A\\cap B| = n - 1$. (If $|A\\cap B| = k < n - 1$, then build $C$ from $B$ by replacing some element not in $A\\cap B$ with an element of $A$ not already in $B$. Then $|A\\cap C| = k + 1$ and $|B\\cap C| = n - 1$ and either $A$ and $C$ or $B$ and $C$ are in different classes.) Removing $A\\cup B$ from $S$, there are $(n + 1)(m - 1) - 1$ elements left. On this set the partition induced by $\\mathcal{P}$ has, by the inductive hypothesis, $m - 1$ pairwise disjoint sets in the same class. Adding either $A$ or $B$ as appropriate gives $m$ pairwise disjoint sets in the same class. $\\blacksquare$\nRemark: The value $n^2 + n - 1$ is sharp. A set $S$ with $n^2 + n - 2$ elements can be split into a set $A$ with $n^2 - 1$ elements and a set $B$ of $n - 1$ elements. Let one class consist of all $n$-element subsets of $A$ and the other consist of all $n$-element subsets that intersect $B$. Then neither class contains $n$ pairwise disjoint sets.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Reid Barton) An animal with $n$ cells is a connected figure consisting of $n$ equal-sized square cells.${}^1$ The figure below shows an 8-cell animal.\nA dinosaur is an animal with at least 2007 cells.  It is said to be primitive if its cells cannot be partitioned into two or more dinosaurs.  Find with proof the maximum number of cells in a primitive dinosaur.\n${}^1$Animals are also called polyominoes.  They can be defined inductively.  Two cells are adjacent if they share a complete edge.  A single cell is an animal, and given an animal with $n$ cells, one with $n+1$ cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.",
        "Solution_1": "Let a $n$-dino denote an animal with $n$ or more cells.\nWe show by induction that an $n$-dino with $4n-2$ or more animal cells is not primitive. (Note: if it had more, we could just take off enough until it had $4n-2$, which would have a partition, and then add the cells back on.)\nBase Case: If $n=1$, we have two cells, which are clearly not primitive.\nInductive Step: Assume any $4n-2$ cell animal can be partitioned into two or more $n$-dinos.\nFor a given $(4n+2)$-dino, take off any four cells (call them $w,\\ x,\\ y,\\ z$) to get an animal with $4n-2$ cells.\nThis can be partitioned into two or more $n$-dinos, let's call them $A$ and $B$. This means that $A$ and $B$ are connected.\nIf both $A$ and $B$ are $(n+1)$-dinos or if $w,\\ x,\\ y,\\ z$ don't all attach to one of them, then we're done.\nSo assume $A$ has $n$ cells and thus $B$ has at least $3n-2$ cells, and that $w,\\ x,\\ y,\\ z$ are added to $B$. So $B$ has $3n+2$ cells total.\nLet $C$ denote the cell of $B$ attached to $A$. There are $3n+1$ cells on $B$ besides $C$. Thus, of the three (or less) sides of $C$ not attached to $A$, one of them must have $n+1$ cells by the pigeonhole principle. It then follows that we can add $A$, $C$, and the other two sides together to get an $(n+1)$ dino, and the side of $C$ that has $n+1$ cells is also an $n+1$-dino, so we can partition the animal with $4n+2$ cells into two $(n+1)$-dinos and we're done.\nThus, our answer is $4(2007) - 3 = 8025$ cells.",
        "Solution_2": "For simplicity, let $k=2007$ and let $n$ be the number of squares. Let the centers of the squares be vertices, and connect any centers of adjacent squares with edges. Suppose we have some loops. Just remove an edge in the loop. We are still connected since you can go around the other way in the loop. Now we have no loops. Each vertex can have at most 4 edges coming out of it. For each point, assign it the quadruple: $(a,b,c,d)$ where $a$, $b$, $c$, $d$ are the numbers of vertices on each branch, WLOG $a\\ge b\\ge c\\ge d$. Note $a+b+c+d=n-1$.\nClaim: If $n=4k-2$, then we must be able to divide the animal into two dinosaurs. \nChose a vertex, $v$, for which $a$ is minimal (i.e. out of all maximal elements in a quadruple, choose the one with the least maximal element).  We have that $4a \\ge a+b+c+d=4k-3$, so $a\\ge k$. Hence we can just cut off that branch, that forms a dinosaur.\nBut suppose the remaining vertices do not make a dinosaur. Then we have $b+c+d+1\\le k-1 \\iff n-a\\le k-1\\iff a\\ge 3k-1$. Now move to the first point on the branch at $a$. We have a new quadruple $p,\\ q,\\ r,\\ b+c+d+1$) where $p+q+r=a-1\\ge 3k-2$.\nNow consider the maximal element of that quadruple. We already have $b+c+d+1\\le k-1$. WLOG $p\\ge q\\ge r\\ge 0$, then $3p\\ge p+q+r=a-1\\ge 3k-2\\implies p\\ge k$ so $p>k-1=b+c+d+1$, so $p$ is the maximal element of that quadruple.\nAlso $a-1=p+q+r\\ge p+0+0$, so $p<a$. But that is a contradiction to the minimality of $a$. Therefore, we must have that $b+c+d+1\\ge k$, so we have a partition of two dinosaurs.\nMaximum: $n=4k-3$.\nConsider a cross with each branch having $k-1$ verticies. Clearly if we take partition $k$ vertices, we remove the center, and we are not connected.\nSo $k=2007$: $4\\cdot 2007-3=8025$.",
        "Solution_3_.28Generalization.29": "Turn the dinosaur into a graph (cells are vertices, adjacent cells connected by an edge) and prove this result about graphs. A connected graph with $V$ vertices, where each vertex has degree less than or equal to $D$, can be partitioned into connected components of sizes at least $\\frac{V-1}{D}$. So then in this special case, we have $D = 4$, and so $V = 2006 \\times 4+1$ (a possible configuration of this size that works consists of a center and 4 lines of cells each of size 2006 connected to the center). We next throw out all the geometry of this situation, so that we have a completely unconstrained graph. If we prove the above-mentioned result, we can put the geometry back in later by taking the connected components that our partition gives us, then filling back all edges that have to be there due to adjacent cells. This won't change any of the problem constraints, so we can legitimately do this.\nGoing, now, to the case of arbitrary graphs, we WOP on the number of edges. If we can remove any edge and still have a connected graph, then we have found a smaller graph that does not obey our theorem, a contradiction due to the minimality imposed by WOP. Therefore, the only case we have to worry about is when the graph is a tree. If it's a tree, we can root the tree and consider the size of subtrees. Pick the root such that the size of the largest subtree is minimized. This minimum must be at least $\\frac{V-1}{D}$, otherwise the sum of the size of the subtrees is smaller than the size of the graph, which is a contradiction. Also, it must be at most $\\frac{V}{2}$, or else pick the subtree of size greater than $\\frac{V}{2}$ and you have decreased the size of the largest subtree if you root from that vertex instead, so you have some subtree with size between $\\frac{V-1}{D}$ and $\\frac V2$. Cut the edge connecting the root to that subtree, and use that as your partition.\nIt is easy to see that these partitions satisfy the contention of our theorem, so we are done.",
        "Solution_4": "Let $s$ denote the minimum number of cells in a dinosaur; the number this year is $s = 2007$.\nClaim: The maximum number of cells in a primitive dinosaur is $4(s - 1) + 1$.\nFirst, a primitive dinosaur can contain up to $4(s - 1) + 1$ cells. To see this, consider a dinosaur in the form of a cross consisting of a central cell and four arms with $s - 1$ cells apiece. No connected figure with at least $s$ cells can be removed without disconnecting the dinosaur.\nThe proof that no dinosaur with at least $4(s - 1) + 2$ cells is primitive relies on the following result.\nLemma. Let $D$ be a dinosaur having at least $4(s - 1) + 2$ cells, and let $R$ (red) and $B$ (black) be two complementary animals in $D$, i.e. $R\\cap B = \\emptyset$ and $R\\cup B = D$. Suppose $|R|\\leq s - 1$. Then $R$ can be augmented to produce animals $\\~{R}\\subset R$ (Error compiling LaTeX. Unknown error_msg) and $\\{B} = D\\backslash\\{R}$ (Error compiling LaTeX. Unknown error_msg) such that at least one of the following holds:\n(i) $|\\{R}|\\geq s$ (Error compiling LaTeX. Unknown error_msg) and $|\\~{B}|\\geq s$ (Error compiling LaTeX. Unknown error_msg),\n(ii) $|\\{R}| = |R| + 1$ (Error compiling LaTeX. Unknown error_msg),\n(iii) $|R| < |\\{R}|\\leq s - 1$ (Error compiling LaTeX. Unknown error_msg).\nProof. If there is a black cell adjacent to $R$ that can be made red without disconnecting $B$, then (ii) holds. Otherwise, there is a black cell $c$ adjacent to $R$ whose removal disconnects $B$. Of the squares adjacent to $c$, at least one is red, and at least one is black, otherwise $B$ would be disconnected. Then there are at most three resulting components $\\mathcal{C}_1, \\mathcal{C}_2, \\mathcal{C}_3$ of $B$ after the removal of $c$. Without loss of generality, $\\mathcal{C}_3$ is the largest of the remaining components. (Note that $\\mathcal{C}_1$ or $\\mathcal{C}_2$ may be empty.) Now $\\mathcal{C}_3$ has at least $\\lceil (3s - 2)/3\\rceil = s$ cells. Let $\\{B} = \\mathcal{C}_3$ (Error compiling LaTeX. Unknown error_msg). Then $|\\{R}| = |R| + |\\mathcal{C}_1| + |\\mathcal{C}_2| + 1$ (Error compiling LaTeX. Unknown error_msg). If $|\\{B}|\\leq 3s - 2$ (Error compiling LaTeX. Unknown error_msg), then $|\\{R}|\\geq s$ (Error compiling LaTeX. Unknown error_msg) and (i) holds. If $|\\{B}|\\geq 3s - 1$ (Error compiling LaTeX. Unknown error_msg) then either (ii) or (iii) holds, depending on whether $|\\{R}|\\geq s$ (Error compiling LaTeX. Unknown error_msg) or not. $\\blacksquare$\nStarting with $|R| = 1$, repeatedly apply the Lemma. Because in alternatives (ii) and (iii) $|R|$ increases but remains less than $s$, alternative (i) eventually must occur. This shows that no dinosaur with at least $4(s - 1) + 2$ cells is primitive.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Titu Andreescu) Prove that for every nonnegative integer $n$, the number $7^{7^n}+1$ is the product of at least $2n+3$ (not necessarily distinct) primes.",
        "Solution_1": "The proof is by induction. The base is provided by the $n = 0$ case, where $7^{7^0} + 1 = 7^1 + 1 = 2^3$. To prove the inductive step, it suffices to show that if $x = 7^{2m - 1}$ for some positive integer $m$ then $(x^7 + 1)/(x + 1)$ is composite. As a consequence, $x^7 + 1$ has at least two more prime factors than does $x + 1$. To confirm that $(x^7 + 1)/(x + 1)$ is composite, observe that\n\nAlso each factor exceeds 1. It suffices to check the smaller one; $\\sqrt{7x}\\leq x$ gives\n\nHence $(x^7 + 1)/(x + 1)$ is composite and the proof is complete.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Kiran Kedlaya, Sungyoon Kim) Let $ABC$ be an acute triangle with $\\omega$, $\\Omega$, and $R$ being its incircle, circumcircle, and circumradius, respectively.  Circle $\\omega_A$ is tangent internally to $\\Omega$ at $A$ and tangent externally to $\\omega$.  Circle $\\Omega_A$ is tangent internally to $\\Omega$ at $A$ and tangent internally to $\\omega$.  Let $P_A$ and $Q_A$ denote the centers of $\\omega_A$ and $\\Omega_A$, respectively.  Define points $P_B$, $Q_B$, $P_C$, $Q_C$ analogously.  Prove that\n\nwith equality if and only if triangle $ABC$ is equilateral.",
        "Solution_1": "Lemma.\nProof. Note $P_{A}$ and $Q_{A}$ lie on $AO$ since for a pair of tangent circles, the point of tangency and the two centers are collinear.\nLet $\\omega$ touch $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Note $AE=AF=s-a$. Consider an inversion, $\\mathcal{I}$, centered at $A$, passing through $E$, $F$. Since $IE\\perp AE$, $\\omega$ is orthogonal to the inversion circle, so $\\mathcal{I}(\\omega)=\\omega$. Consider $\\mathcal{I}(\\omega_{A})=\\omega_{A}'$. Note that $\\omega_{A}$ passes through $A$ and is tangent to $\\omega_{A}$, hence $\\omega_{A}'$ is a line that is tangent to $\\omega$. Furthermore, $\\omega_{A}'\\perp AO$ because $\\omega_{A}$ is symmetric about $OA$, so the inversion preserves that reflective symmetry. Since it is a line that is symmetric about $AO$, it must be perpendicular to $AO$. Likewise, $\\mathcal{I}(\\Omega_{A})=\\Omega_{A}'$ is the other line tangent to $\\omega$ and perpendicular to $AO$.\nLet $\\omega_{A} \\cap AO=X$ and $\\omega_{A}' \\cap AO=X'$ (second intersection).\nLet $\\Omega_{A} \\cap AO=Y$ and $\\Omega_{A}' \\cap AO=Y'$ (second intersection).\nEvidently, $AX=2AP_{A}$ and $AY=2AQ_{A}$. We want:\n\nby inversion. Note that $\\omega_{A}' || \\Omega_{A}'$, and they are tangent to $\\omega$, so the distance between those lines is $2r=AX'-AY'$. Drop a perpendicular from $I$ to $AO$, touching at $H$. Then $AH=AI\\cos\\angle OAI=AI\\cos\\frac{1}{2}|\\angle B-\\angle C|$. Then $AX', AY' = AI\\cos\\frac{1}{2}|\\angle B-\\angle C|\\pm r$. So $AX'\\cdot AY' = AI^{2}\\cos^{2}\\frac{1}{2}|\\angle B-\\angle C|-r^{2}$.\n\nNote that $\\frac{AI}{r} = \\frac{1}{\\sin\\frac{A}{2}}$. Applying the double angle formulas and $1-\\cos\\angle A=\\frac{2(s-b)(s-c)}{bc}$, we get\nEnd Lemma\nThe problem becomes:\n\nwhich is true because $OI^{2}=R(R-2r)$, equality is when the circumcenter and incenter coincide. As before, $\\angle OAI=\\frac{1}{2}|\\angle B-\\angle C|=0$, so, by symmetry, $\\angle A=\\angle B=\\angle C$. Hence the inequality is true iff $\\triangle ABC$ is equilateral.\nComment: It is much easier to determine $AP_{A}$ by considering $\\triangle IAP_{A}$. We have $AI$, $\\angle IAO$, $IP_{A}=r+r(P_{A})$, and $AP_{A}=r(P_{A})$. However, the inversion is always nice to use. This also gives an easy construction for $w_{A}$ because the tangency point is collinear with the intersection of $w_{A}'$ and $w$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Titu Andreescu) Prove that for each positive integer $n$, there are pairwise relatively prime integers $k_0, k_1 \\dotsc, k_n$, all strictly greater than 1, such that $k_0 k_1 \\dotsm k_n -1$ is the product of two consecutive integers.",
        "Solution_1": "We will prove the problem for each nonnegative integer $n$.  We wish to show that\n\nfor some integer $a$.  We induct on $n$.  For our base case, $n=0$, we may let $a$ be positive integer.\nFor the inductive step, suppose that $k_0, \\dotsc, k_{n-1}$ are pairwise relatively prime integers such that\n\nfor some integer $a$.  Let $k_n = a^2 + 3a+3$.  Evidently, $k_n >1$.  Also,\n\nSince $a^2+a+1 = a(a+1)+1$ is odd and relatively prime to $a+1$, it follows that $a^2+a+1$ and $k_n$ are relatively prime, so $k_n$ is relatively prime to each of $k_0, \\dotsc, k_{n-1}$.  Finally,\n\nThis completes the induction.",
        "Solution_2": "Lemma.  If $p$ is prime such that $p\\equiv 1 \\pmod{3}$, there exists a residue $r$ such that $p \\mid r^2+r+1$.\nProof. Let $g$ be a multiplicative generator of the nonzero integers mod p.  Set $r \\equiv g^{(p-1)/3}$.  Then $r-1 \\not\\equiv 0 \\pmod{p}$, but $r^3-1 \\equiv 0 \\pmod{p}$, so $\\frac{r^3-1}{r-1} \\equiv r^2 + r+1 \\equiv 0 \\pmod{p}$.\nEnd Lemma\nBy Dirichlet's Theorem, there are infinitely many primes congruent to 1 (mod 3).  Let $p_0, \\dotsc, p_n$ be $(n+1)$ such primes, and let $r_0, \\dotsc, r_n$ be respective residues as described in the lemma.  By the Chinese Remainder Theorem, there is a positive integer $a$ that satisfies the relation\n\nfor each integer $i \\in [0,n]$.  Then\n\nNow, for $1 \\le i \\le n$, take $k_i$ to be the greatest power of $p_i$ that divides $a^2 + a +1$, and let $k_0 = (a^2+a+1)/(k_1 \\dotsm k_n)$.  Since all the $k_i$ are pairwise relatively prime and are greater than 1, we are done.",
        "Solution_3": "Firstly, we see that there are $n + 1$ numbers $k_i$. Since $n\\geq1$, there are at least 2 values of $k_i$.\nDefine a relatively prime partition to be a set of relatively prime numbers such that their product is equal to some natural number.\nDefine $P$ to be the greatest possible cardinality of a relatively prime partition for that number.\nLemma 1. All cardinalities of the relatively prime partitions of a number up to $P$ can be attained.\nProof. $P = P - 0$ satisfies the properties. For any $P - n$ which satisfies the properties, we can take any of the 2 $P - n$ numbers and multiply them together. Because they are both relatively prime to all the other numbers, their product is relatively prime to all the other numbers as well, and that results in $P - n - 1$ numbers which satisfy the conditions, unless $P - n = 1$, because there is only one number left. Therefore, all numbers of numbers of relatively prime factors from $1$ to $P - 1$ are attainable, if $P$ is attainable as well.\nEnd Lemma\nLemma 2. $a^2 + a + 1$ can have arbitrarily many prime factors for some value of $a$.\nProof. Let $f(x) = x^2 + x + 1$. Let $a^2 + a + 1 = k$. Then $f(a + k) = a^2 + 2ak + k^2 + a + k + 1 = 2ak + k^2 + 2k$.\nAfter factoring, we get $k(k + 2 + 2a) = (a^2 + a + 1)(a^2 + 3a + 3)$.\n$a^2 + 3a + 3 \\geq a^2 + a + 1$, because $a$ is strictly positive.\nLet their GCD=g.\n$g|2a + 2$\n$g|2a^2 + 2a$\n$g|a^2 - 3a - 3 + a^2 + 3a + 3, g|2a^2$\n$g|2a$, and so $g|2$, but $g|a^2 + a + 3$ which is strictly odd, so g=1.\n$(a^2 + a + 1)(a^2 + 3a + 3)$ therefore must contain prime factors not in $a^2 + a + 1$. Upon repeating this an arbitrary number of times, we get a number of the form $a^2 + a + 1$ which has arbitrarily many distinct prime factors.\nEnd Lemma\nThen $a(a + 1) + 1 = a^2 + a + 1$ would not have $n + 1 = P$ elements in the relatively prime partition for some value of n, and any value of a.\nIt is possible to choose a value of $a$ such that $a^2 + a + 1$ has arbitrarily many unique prime factors, by our second Lemma, and so it is possible for $P$ to be arbitrarily high. By our first Lemma, all numbers up to P are possible values for the cardinality of some relatively prime partition, and so there always exists some number that has an arbitrary number of elements in a relatively prime partition. Because $a^2 + a + 1 = a(a + 1) + 1$, $a(a + 1)$ is the product of 2 consecutive integers, we see that the given statement is true because if $k_0k_1...k_n = a(a + 1) + 1$, then $k_0k_1...k_n - 1 = a(a + 1)$.",
        "Solution_4": "Write the relation to be proved as\n\nThere are infinitely many primes for which $-3$ is a quadratic residue. Let $2 < p_0 < p_1 < \\cdots < p_n$ be such primes. Using the Chinese Remainder Theorem to specify $a$ modulo $p_n$, we can find an integer $a$ such that $(2a + 1)^2 + 3 = 4p_0p_1\\cdots p_nm$ for some positive integer $m$. Grouping the factors of $m$ appropriately with the $p_i$'s, we obtain $(2a + 1)^2 + 3 = 4k_0k_1\\cdots k_n$ with $k_i$ pairwise relatively prime. We then have $k_0k_1\\cdots k_n - 1 = a(a + 1)$, as desired.",
        "Solution_5": "We are supposed to show that for every positive integer $n$, there is a positive integer $x$ such that $x(x + 1) + 1 = x^2 + x + 1$ has at least $n$ distinct prime divisors. We can actually prove a more general statement.\nClaim. Let $P(x) = a_dx^d + \\cdots + a_1x + 1$ be a polynomial of degree $d\\geq 1$ with integer coefficients. Then for every positive integer $n$, there is a positive integer $x$ such that $P(x)$ has at least $n$ distinct prime divisors.\nThe proof follows from the following two lemmas.\nLemma 1. The set\n\nis infinite.\nProof. The proof is analogous to Euclid's proof that there are infinitely many primes. Namely, if we assume that there are only finitely many primes $p_1, p_2, \\ldots, p_k$ in $Q$, then for each integer $m$, $P(mp_1p_2\\cdots p_k)$ is an integer with no prime factors, which must equal $1$ or $-1$. However, since $P$ has degree $d\\geq 1$, it takes each of the values $1$ and $-1$ at most $d$ times, a contradiction.\nEnd Lemma\nLemma 2. Let $p_1, p_2, \\ldots, p_n$, $n\\geq 1$ be primes in $Q$. Then there is a positive integer $x$ such that $P(x)$ is divisible by $p_1p_2\\cdots p_n$.\nProof. For $i = 1, 2, \\ldots, n$, since $p_i\\in Q$ we can find an integer $c_i$ such that $P(x)$ is divisible by $p_i$ whenever $x\\equiv c_i\\pmod{p_i}$. By the Chinese Remainder Theorem, the system of $n$ congruences $x\\equiv c_i\\pmod{p_i}$, $i = 1, 2, \\ldots, n$ has positive integer solutions. For every positive integer $x$ that solves this system, $P(x)$ is divisible by $p_1p_2\\cdots p_n$.\nEnd Lemma",
        "Solution_6": "Very similar to some above solutions.\nLet $k_{0}k_{1}\\dots k_{n}-1=a(a+1)$. We wish to show this equation has a solution $(a,k_{0},\\dots,k_{n})\\in\\mathbb{Z}^{n+2}$ for any $n\\in\\mathbb{N}$. If $a^{2}+a+1$ has at least $n+1$ distinct prime factors, we can set $a^{2}+a+1=p_{1}^{q_{1}}p_{2}^{q_{2}}\\dots p_{m}^{q_{m}}$ for some $m\\geq n$ where $p_{1},\\dots p_{m}$ are distinct primes. Then, we let $k_{i}=p_{i+1}^{q_{i+1}}$ for all $0\\leq i\\leq n-1$ and set $k_{n}=p_{n+1}^{q_{n+1}}\\dots p_{m}^{q_{m}}$. Thus, it remains to show $a^{2}+a+1$ can have at least $n+1$ distinct prime factors. We will do this by showing it can have arbitrarily many.\nUsing an induction-type argument (it isn't exactly induction though), we first show that $a^{2}+a+1$ can have $1$ prime factor by setting $a=2$. Now, assume $a^{2}+a+1$ has $n$ distinct prime factors. We wish to show there exists some $b$ such that $b^{2}+b+1$ has more than $n$ distinct prime factors (we need not show it has $n+1$, since we only wish to demonstrate that the number of prime factors can be arbitrarily large rather than any integer). Set $b=a^{2}$. Then\n\nSuppose $p$ is a positive integer dividing both $a^{2}+a+1$ and $a^{2}-a+1$. Then $p\\mid(a^{2}+a+1)-(a^{2}-a+1)=2a$, and because $(a^{2}+a+1)$ is always odd, we have $2\\nmid p$ so $p\\mid a$. But this yields $p\\mid\\gcd(a,a^{2}+a+1)=1$, so $p=1$ and thus $a^{2}+a+1$ and $a^{2}-a+1$ are coprime. Therefore, as one can make $a^{2}-a+1>1$, we see that $a^{2}-a+1$ has at least one prime factor which does not divide $a^{2}+a+1$ (which has $n$ distinct prime factors). Therefore, $b^{2}+b+1$ has more than $n$ distinct prime factors.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Zuming Feng) Let $ABC$ be an acute, scalene triangle, and let $M$, $N$, and $P$ be the midpoints of $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively. Let the perpendicular bisectors of $\\overline{AB}$ and $\\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $CE$ intersect in point $F$, inside of triangle $ABC$. Prove that points $A$, $N$, $F$, and $P$ all lie on one circle.",
        "Solution_1_.28synthetic.29": "Without Loss of Generality, assume $AB >AC$. It is sufficient to prove that $\\angle OFA = 90^{\\circ}$, as this would immediately prove that $A,P,O,F,N$ are concyclic.\nBy applying the Menelaus' Theorem in the Triangle $\\triangle BFC$ for the transversal $E,M,D$, we have (in magnitude)\n \nHere, we used that $BM=MC$, as $M$ is the midpoint of $BC$. Now, since $EC =EA$ and $BD=DA$, we have\n\nNow, note that $OE$ bisects the exterior $\\angle FED$ and $OD$ bisects exterior $\\angle FDE$, making $O$ the $F$-excentre of $\\triangle FED$. This implies that $OF$ bisects interior $\\angle EFD$, making $OF \\perp AF$, as was required.",
        "Solution_2_.28complex.29": "Let $A=1,B=b,C=c$ where $b,c$ all lie on the unit circle. Then $O$ is 0. As noted earlier, $(FOBC)$ is cyclic. We will find the ghost point $F',$ the second intersection of $OBC$ and $ANP$.\nWe know that these two circles already intersect at $O$ so we can reflect over the line between their centers. The center of $ANPO$ is the midpoint of $AO$ namely $\\frac12$. With the tangent formula and then taking the midpoint, we find that the center of $OBC$ is $\\frac{bc}{b+c}.$ Then we want to find the reflection of 0 over the line through $\\frac12$ and $\\frac{bc}{b+c}.$ Then we get\n\nNow it remains to show $\\angle F'BA=\\angle ABM;$ the other angle equality would follow by symmetry.\nThen we get:\n\nThus $\\measuredangle F'BA=\\measuredangle BAM,$ so $F'=F$ and we're done.\n~cocohearts",
        "Solution_3_.28synthetic.29": "Without loss of generality $AB < AC$. The intersection of $NE$ and $PD$ is $O$, the circumcenter of $\\triangle ABC$.\nLet $\\angle BAM = y$ and $\\angle CAM = z$. Note $D$ lies on the perpendicular bisector of $AB$, so $AD = BD$. So $\\angle FBC = \\angle B - \\angle ABD = B - y$. Similarly, $\\angle FCB = C - z$, so $\\angle BFC = 180 - (B + C) + (y + z) = 2A$. Notice that $\\angle BOC$ intercepts the minor arc $BC$ in the circumcircle of $\\triangle ABC$, which is double $\\angle A$. Hence $\\angle BFC = \\angle BOC$, so $BFOC$ is cyclic.\nLemma. $\\triangle FEO$ is directly similar to $\\triangle NEM$\nProof.\n\nsince $F$, $E$, $C$ are collinear, $BFOC$ is cyclic, and $OB = OC$. Also\n\nbecause $NE\\perp AC$, and $MNP$ is the medial triangle of $\\triangle ABC$ so $AB \\parallel MN$. Hence $\\angle OFE = \\angle ENM$.\nNotice that $\\angle AEN = 90 - z = \\angle CEN$ since $NE\\perp AC$. $\\angle FED = \\angle MEC = 2z$. Then\n\nHence $\\angle FEO = \\angle NEM$.\nHence $\\triangle FEO$ is similar to $\\triangle NEM$ by AA similarity. It is easy to see that they are oriented such that they are directly similar.\nEnd Lemma\nBy the similarity in the Lemma, $FE: EO = NE: EM\\implies FE: EN = OE: EM$. $\\angle FEN = \\angle OEM$ so $\\triangle FEN\\sim\\triangle OEM$ by SAS similarity. Hence\n\nUsing essentially the same angle chasing, we can show that $\\triangle PDM$ is directly similar to $\\triangle FDO$. It follows that $\\triangle PDF$ is directly similar to $\\triangle MDO$. So\n\nHence $\\angle OPF = \\angle ONF$, so $FONP$ is cyclic. In other words, $F$ lies on the circumcircle of $\\triangle PON$. Note that $\\angle ONA = \\angle OPA = 90$, so $APON$ is cyclic. In other words, $A$ lies on the circumcircle of $\\triangle PON$. $A$, $P$, $N$, $O$, and $F$ all lie on the circumcircle of $\\triangle PON$. Hence $A$, $P$, $F$, and $N$ lie on a circle, as desired.",
        "Solution_4_.28synthetic.29": "This solution utilizes the phantom point method. Clearly, APON are cyclic because $\\angle OPA = \\angle ONA = 90$. Let the circumcircles of triangles $APN$ and $BOC$ intersect at $F'$ and $O$.\nLemma. If $A,B,C$ are points on circle $\\omega$ with center $O$, and the tangents to $\\omega$ at $B,C$ intersect at $Q$, then $AP$ is the symmedian from $A$ to $BC$.\nProof. This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.\nEnd Lemma\nIt is easy to see $Q$ (the intersection of ray $OM$ and the circumcircle of $\\triangle BOC$) is colinear with $A$ and $F'$, and because line $OM$ is the diameter of that circle, $\\angle QBO = \\angle QCO = 90$, so $Q$ is the point $Q$ in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that $F'$ satisfies the original construction for $F$, showing $F=F'$; we are done.\n",
        "Solution_5_.28trigonometric.29": "By the Law of Sines, $\\frac {\\sin\\angle BAM}{\\sin\\angle CAM} = \\frac {\\sin B}{\\sin C} = \\frac bc = \\frac {b/AF}{c/AF} = \\frac {\\sin\\angle AFC\\cdot\\sin\\angle ABF}{\\sin\\angle ACF\\cdot\\sin\\angle AFB}$. Since $\\angle ABF = \\angle ABD = \\angle BAD = \\angle BAM$ and similarly $\\angle ACF = \\angle CAM$, we cancel to get $\\sin\\angle AFC = \\sin\\angle AFB$. Obviously, $\\angle AFB + \\angle AFC > 180^\\circ$ so $\\angle AFC = \\angle AFB$.\nThen $\\angle FAB + \\angle ABF = 180^\\circ - \\angle AFB = 180^\\circ - \\angle AFC = \\angle FAC + \\angle ACF$ and $\\angle ABF + \\angle ACF = \\angle A = \\angle FAB + \\angle FAC$. Subtracting these two equations, $\\angle FAB - \\angle FCA = \\angle FCA - \\angle FAB$ so $\\angle BAF = \\angle ACF$. Therefore, $\\triangle ABF\\sim\\triangle CAF$ (by AA similarity), so a spiral similarity centered at $F$ takes $B$ to $A$ and $A$ to $C$. Therefore, it takes the midpoint of $\\overline{BA}$ to the midpoint of $\\overline{AC}$, or $P$ to $N$. So $\\angle APF = \\angle CNF = 180^\\circ - \\angle ANF$ and $APFN$ is cyclic.",
        "Solution_6_.28isogonal_conjugates.29": "Construct $T$ on $AM$ such that $\\angle BCT = \\angle ACF$. Then $\\angle BCT = \\angle CAM$. Then $\\triangle AMC\\sim\\triangle CMT$, so $\\frac {AM}{CM} = \\frac {CM}{TM}$, or $\\frac {AM}{BM} = \\frac {BM}{TM}$. Then $\\triangle AMB\\sim\\triangle BMT$, so $\\angle CBT = \\angle BAM = \\angle FBA$. Then we have\n$\\angle CBT = \\angle ABF$ and $\\angle BCT = \\angle ACF$. So $T$ and $F$ are isogonally conjugate. Thus $\\angle BAF = \\angle CAM$. Then\n$\\angle AFB = 180 - \\angle ABF - \\angle BAF = 180 - \\angle BAM - \\angle CAM = 180 - \\angle BAC$.\nIf $O$ is the circumcenter of $\\triangle ABC$ then $\\angle BFC = 2\\angle BAC = \\angle BOC$ so $BFOC$ is cyclic. Then $\\angle BFO = 180 - \\angle BOC = 180 - (90 - \\angle BAC) = 90 + \\angle BAC$.\nThen $\\angle AFO = 360 - \\angle AFB - \\angle BFO = 360 - (180 - \\angle BAC) - (90 + \\angle BAC) = 90$. Then $\\triangle AFO$ is a right triangle.\nNow by the homothety centered at $A$ with ratio $\\frac {1}{2}$, $B$ is taken to $P$ and $C$ is taken to $N$. Thus $O$ is taken to the circumcenter of $\\triangle APN$ and is the midpoint of $AO$, which is also the circumcenter of $\\triangle AFO$, so $A,P,N,F,O$ all lie on a circle.",
        "Solution_7_.28symmedians.29": "Median $AM$ of a triangle $ABC$ implies $\\frac {\\sin{BAM}}{\\sin{CAM}} = \\frac {\\sin{B}}{\\sin{C}}$.\nTrig ceva for $F$ shows that $AF$ is a symmedian.\nThen $FP$ is a median, use the lemma again to show that $AFP = C$, and similarly $AFN = B$, so you're done.\n",
        "Solution_8_.28inversion.29_.28Official_Solution_.232.29": "Invert the figure about a circle centered at $A$, and let $X'$ denote the image of the point $X$ under this inversion. Find point $F_1'$ so that $AB'F_1'C'$ is a parallelogram and let $Z'$ denote the center of this parallelogram. Note that $\\triangle BAC\\sim\\triangle C'AB'$ and $\\triangle BAD\\sim\\triangle D'AB'$. Because $M$ is the midpoint of $BC$ and $Z'$ is the midpoint of $B'C'$, we also have $\\triangle BAM\\sim\\triangle C'AZ'$. Thus\n\nHence quadrilateral $AB'D'F_1'$ is cyclic and, by a similar argument, quadrilateral $AC'E'F_1'$ is also cyclic. Because the images under the inversion of lines $BDF$ and $CFE$ are circles that intersect in $A$ and $F'$, it follows that $F_1' = F'$.\nNext note that $B'$, $Z'$, and $C'$ are collinear and are the images of $P'$, $F'$, and $N'$, respectively, under a homothety centered at $A$ and with ratio $1/2$. It follows that $P'$, $F'$, and $N'$ are collinear, and then that the points $A$, $P$, $F$, and $N$ lie on a circle.",
        "Solution_9_.28Official_Solution_.231.29": "Let $O$ be the circumcenter of triangle $ABC$. We prove that\n\nIt will then follow that $A, P, O, F, N$ lie on the circle with diameter $\\overline{AO}$. Indeed, the fact that the first two angles in $(1)$ are right is immediate because $\\overline{OP}$ and $\\overline{ON}$ are the perpendicular bisectors of $\\overline{AB}$ and $\\overline{AC}$, respectively. Thus we need only prove that $\\angle AFO = 90^\\circ$.\nWe may assume, without loss of generality, that $AB > AC$. This leads to configurations similar to the ones shown above. The proof can be adapted to other configurations. Because $\\overline{PO}$ is the perpendicular bisector of $\\overline{AB}$, it follows that triangle $ADB$ is an isosceles triangle with $AD = BD$. Likewise, triangle $AEC$ is isosceles with $AE = CE$. Let $x = \\angle ABD = \\angle BAD$ and $y = \\angle CAE = \\angle ACE$, so $x + y = \\angle BAC$.\nApplying the Law of Sines to triangles $ABM$ and $ACM$ gives\n\nTaking the quotient of the two equations and noting that $\\sin\\angle BMA = \\sin\\angle CMA$, we find\n\nBecause $BM = MC$, we have\n\nApplying the Law of Sines to triangles $ABF$ and $ACF$, we find\n\nTaking the quotient of the two equations yields\n\nso by $(2)$,\n\nBecause $\\angle ADF$ is an exterior angle to triangle $ADB$, we have $\\angle EDF = 2x$. Similarly, $\\angle DEF = 2y$. Hence\n\nThus $\\angle BFC = 2\\angle BAC = \\angle BOC$, so $BOFC$ is cyclic. In addition,\n\nand hence, from $(3)$, $\\angle AFB = \\angle AFC = 180^\\circ - \\angle BAC$. Because $BOFC$ is cyclic and $\\triangle BOC$ is isosceles with vertex angle $\\angle BOC = 2\\angle BAC$, we have $\\angle OFB = \\angle OCB = 90^\\circ - \\angle BAC$. Therefore,\n\nThis completes the proof.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Gabriel Carroll) Let $n$ be a positive integer. Denote by $S_n$ the set of points $(x, y)$ with integer coordinates such that\n\nA path is a sequence of distinct points $(x_1 , y_1 ), (x_2 , y_2 ), \\ldots , (x_\\ell, y_\\ell)$ in $S_n$ such that, for $i = 2, \\ldots , \\ell$, the distance between $(x_i , y_i )$ and $(x_{i - 1} , y_{i - 1} )$ is $1$ (in other words, the points $(x_i , y_i )$ and $(x_{i - 1} , y_{i - 1} )$ are neighbors in the lattice of points with integer coordinates). Prove that the points in $S_n$ cannot be partitioned into fewer than $n$ paths (a partition of $S_n$ into $m$ paths is a set $\\mathcal{P}$ of $m$ nonempty paths such that each point in $S_n$ appears in exactly one of the $m$ paths in $\\mathcal{P}$).",
        "Solution_1": "Color all the points in $S_{n}$ red or black such that each row of points starts and ends with a red point, and alternates between red and black for each point in the row. This creates a checkerboard pattern, except that the middle two rows are identical.\nSuppose there is a partition of $m < n$ paths that works. For each of the $m$ paths, we break a path into two separate paths wherever there are consecutive red points. This only happens only with the $n$ red pairs in the middle two rows. We now have $m + l,\\, l \\le n$ paths whose colors are strictly alternating. In a path with alternating colors, there can be at most one more red dot than the number of black dots (which occurs when both endpoints are red). Thus there can be no more than $m + l$ red dots than black dots in $S_n$.\nThere are $2n$ more red points than black points, so\n\nwhich implies that $n \\le m$, contradiction.",
        "Solution_2": "Suppose you have a partition of less than $n$ paths. Then start from $(0,n - 1)$ and work your way down to $(0, - n)$ along the right edge of $S_n$ performing the following algorithm when going from each point.\nIf a path already has the point you're coming from and the point you're going to as adjacent in the path, do nothing.\nIf not the first case, and the point you're coming from and the point you're going to are both endpoints of paths, join them (one less path).\nIf not the first case, and there is exactly one endpoint among the point you're coming from and the point you're going to, remove a segment from the one in the middle of a path (paths +1) and now you have two endpoints; join them (paths -1). If the point in the middle of the path is the one you're coming from, remove the segment that doesn't come from the point that came before it. The number of paths remains the same.\nIf not the first case, and both points are in the middle of a path, which is impossible. Since the two points are in the middle of the path, they are adjacent to two other points. But they are also adjacent to each other, so both are adjacent to at least three points. Then, no two points adjacent to at least three others are travelled consecutively.\nThus, given any configuration we can create the path shown above without increasing the number of paths. The remaining points form $S_{n - 1}$. Now suppose for some positive integer $n$, we have a partition into less than $n$ paths. Apply the algorithm to make the path shown above. Then the remaining points are partitioned into less than $n - 1$ paths. Apply the algorithm again. Repeat until we show that $S_1$ is partitioned into less than $1$ path, contradiction.",
        "Solution_3": "We induct on $n$ to prove that an $n - 1$-path partition is impossible. This can be restated as saying that we cannot fit $2n^2 - n + 1$ edges into the graph with each vertex having degree at most 2. The base case is trivial (2 edges when there only is 1). Suppose it was possible for $n + 1$, and we'll prove it's possible for $n$.\nCall the 'complete' network (graph) of points $G_{n + 1}$ and the $n$-path partition for $n + 1$ is a subgraph of $G_{n + 1}$, $H_{n + 1}$. There are a total of $2(n + 1)^2 - n = 2n^2 + 3n + 2$ edges in $H_{n + 1}$ out of $4n^2 + 4n + 1$ in $G_{n + 1}$ (not hard to prove), and each vertex has maximum order 2. Now consider the complement of $H_{n + 1}$ in $G_{n + 1}$, that is, take all the edges in $G_{n + 1}$ that are not in $H_{n + 1}$ and of course all the vertices. Call this $F_{n + 1}$ We will trim $F_{n + 1}$ down a bit to get $H_n$.\nFirst, call the points in $S_{n + 1}$ but not $S_n$ exterior points, and the points in $S_n$ interior points. $F_{n + 1}$ has $2n^2 + n - 1$ edges. Since the interior points have degree 4 in $G_{n + 1}$, they have degree at least 2 in $F_{n + 1}$, for a total degree of $4n^2 + m$ where $0\\le m\\le 2n - 2$ since the total overall degree is $2(2n^2 + n - 1) = 4n^2 + 2n - 2$. This makes the total degree of the exterior points $2n - 2 - m$. Since the exterior points are not connected in $G_{n + 1}$, these $2n - 2 - m$ edges must connect on exterior point to an interior point.\nI then take away $2n - 2$ edges and the exterior points, yielding $2n^2 - n + 1$ edges and the interior points, which is $H_n$, as follows: First take away those $2n - 2 - m$ interior-exterior edges and shave off all the exterior points. Then the total degree (now just of the interior points) must be $4n^2 - 2n + 2 + 2m$.\nLet the number of 0-degree vertices be $a$, 1-degree vertices be $b$, 2-degree vertices be $c$, 3-degree vertices be $d$, and 4-degree vertices be $e$. Since all the vertices began as 2-degree or higher, $2a + b\\le2n - 2 - m$. Moreover, the total degree is $b + 2c + 3d + 4e = 2(2n^2) + (d + 2e) - (b + 2a) = 4n^2 - 2n + 2 + 2m$. So $d + 2e = b + 2a - (2n - 2 + 2m)\\le (2n - 2 - m) - (2n - 2 - 2m) = m$ So we can use our last $m$ edges to remove any excess degree on vertices with degree 3 or 4, yielding $H_n$.\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "(Gregory Galparin) Let $\\mathcal{P}$ be a convex polygon with $n$ sides, $n\\ge3$. Any set of $n - 3$ diagonals of $\\mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $\\mathcal{P}$ into $n - 2$ triangles. If $\\mathcal{P}$ is regular and there is a triangulation of $\\mathcal{P}$ consisting of only isosceles triangles, find all the possible values of $n$.",
        "Solution": "We label the vertices of $\\mathcal{P}$ as $P_0, P_1, P_2, \\ldots, P_n$. Consider a diagonal $d = \\overline{P_a\\,P_{a+k}},\\,k \\le n/2$ in the triangulation. We show that $k$ must have the form $2^{m}$ for some nonnegative integer $m$.\nThis diagonal partitions $\\mathcal{P}$ into two regions $\\mathcal{Q},\\, \\mathcal{R}$, and is the side of an isosceles triangle in both regions. Without loss of generality suppose the area of $Q$ is less than the area of $R$ (so the center of $P$ does not lie in the interior of $Q$); it follows that the lengths of the edges and diagonals in $Q$ are all smaller than $d$. Thus $d$ must the be the base of the isosceles triangle in $Q$, from which it follows that the isosceles triangle is $\\triangle P_aP_{a+k/2}\\,P_{a+k}$, and so $2|k$. Repeating this process on the legs of isosceles triangle ($\\overline{P_aP_{a+k/2}},\\,\\overline{P_{a+k}P_{a+k/2}}$), it follows that $k = 2^m$ for some positive integer $m$ (if we allow degeneracy, then we can also let $m=0$).\nNow take the isosceles triangle $P_xP_yP_z,\\,0 \\le x < y < z < n$ in the triangulation that contains the center of $\\mathcal{P}$ in its interior; if a diagonal passes through the center, select either of the isosceles triangles with that diagonal as an edge. Without loss of generality, suppose $P_xP_y = P_yP_z$. From our previous result, it follows that there are $2^a$ edges of $P$ on the minor arcs of $P_xP_y,\\, P_yP_z$ and $2^b$ edges of $P$ on the minor arc of $P_zP_x$, for positive integers $a,\\,b$. Therefore, we can write\n so $n$ must be the sum of two powers of $2$.\nWe now claim that this condition is sufficient. Suppose without loss of generality that $a+1 \\ge b$; then we rewrite this as\nBy induction, it follows that we can cover all the desired $n$.\nFor $n = 3,4$, this is trivial. For $k>1$, we construct the diagonals of equal length $\\overline{P_0P_{2^{k-1}}}$ and $\\overline{P_{2^{k-1}+1}P_0}$. This partitions $\\mathcal{P}$ into $3$ regions: an isosceles $\\triangle P_0P_{2^{k-1}}P_{2^{k-1}+1}$, and two other regions. For these two regions, we can recursively construct the isosceles triangles defined above in the second paragraph. It follows that we have constructed $2(2^{k-1}-1) + (1) = 2^k-1 = n-2$ isosceles triangles with non-intersecting diagonals, as desired.\nWe construct the diagonals $\\overline{P_0P_2},\\ \\overline{P_2P_4},\\ \\ldots \\overline{P_{2n-2}P_0}$. This partitions $\\mathcal{P}$ into $n$ isosceles triangles of the form $\\triangle P_{2k}P_{2k+1}P_{2k+2}$, as well as a central regular polygon with $n$ sides. However, we know that there exists a triangulation for the $n$-sided polygon that yields $n-2$ isosceles triangles. Thus, we have created $(n) + (n-2) = 2n-2$ isosceles triangles with non-intersecting diagonals, as desired.\nIn summary, the answer is all $n$ that can be written in the form $2^{a+1} + 2^{b},\\, a,b \\ge 0$. Alternatively, this condition can be expressed as either $n=2^{k},\\, k \\ge 2$ (this is the case when $a+1 = b$) or $n$ is the sum of two distinct powers of $2$, where $1= 2^0$ is considered a power of $2$."
    },
    {
        "Problem": "(Kiran Kedlaya) Three nonnegative real numbers $r_1$, $r_2$, $r_3$ are written on a blackboard. These numbers have the property that there exist integers $a_1$, $a_2$, $a_3$, not all zero, satisfying $a_1r_1 + a_2r_2 + a_3r_3 = 0$. We are permitted to perform the following operation: find two numbers $x$, $y$ on the blackboard with $x \\le y$, then erase $y$ and write $y - x$ in its place. Prove that after a finite number of such operations, we can end up with at least one $0$ on the blackboard.",
        "Solution": "Every time we perform an operation on the numbers on the blackboard $R = \\left < r_1, r_2, r_3 \\right >$, we perform the corresponding operation on the integers $A = \\left < a_1, a_2, a_3 \\right >$ so that $R \\cdot A = 0$ continues to hold.  (For example, if we replace $r_1$ with $r_1 - r_2$ then we replace $a_2$ with $a_1 + a_2$.)\nIt's possible to show we can always pick an operation so that $|A|^2$ is strictly decreasing. Without loss of generality, let $r_3 > r_2 > r_1$ and $a_3$ be positive. Then it cannot be true that both $a_1$ and $a_2$ are at least $\\frac { - a_3}{2}$, or else $a_1r_1 + a_2r_2 + a_3r_3 > 0$. Without loss of generality, let $a_1 < \\frac { - a_3}{2}$. Then we can replace $a_1$ with $a_1 + a_3$ and $r_3$ with $r_3 - r_1$ to make $|A|$ smaller. Since it is a strictly decreasing sequence of positive integers, after a finite number of operations we have $a_3 = 0$. We can now  see that this result holds for $(r_1,r_2,0)$ if and only if it holds for $(1,\\frac{r_2}{r_1},0)$. We can see that $\\frac{r_2}{r_1}$ is a rational number given that $a_3$ = 0. It is a well known result of the euclidean algorithm that if we continue to perform these operations, $r_1$ or $r_2$ will eventually be 0.\n\n"
    },
    {
        "Problem": "(Sam Vandervelde) At a certain mathematical conference, every pair of mathematicians are either friends or strangers. At mealtime, every participant eats in one of two large dining rooms. Each mathematician insists upon eating in a room which contains an even number of his or her friends. Prove that the number of ways that the mathematicians may be split between the two rooms is a power of two (i.e., is of the form $2^k$ for some positive integer $k$).",
        "Solution_1_.28linear_algebra.29": "Make the obvious re-interpretation as a graph. Let $f : G \\to \\{0,1\\}$ be an indicator function with $f(v) = 0$ if a vertex is in the first partition and $f(v) = 1$ otherwise (this corresponds, in the actual problem, to putting a mathematician in the first or second room). Then look at $f$ as a function into the field with two elements, $F_2$. Let $V$ be the vector space of all such functions. Define the linear operator $A : V \\to V$ as\n\nwhere $\\sim$ denotes adjacency. (Note that we can also think of $A$ as a matrix, which is essentially the adjacency matrix where the diagonal is changed to be $1$ whenever the degree is odd; in more technical terms, it is the Laplacian of $G$ over $F_2$). Then $f$ is a valid partition iff $(Af)(v) = d(v)$, where $d$ is the degree of $v$, for all $v$ (this is taken over $F_2$). So we want all solution to $Af = d$. Note that if $Af = d, Ag = d$, then $A(f - g) = 0$, so $f - g$ is in the nullspace. Thus in particular the number of solutions, if non-zero, is the size of the nullspace of $A$, which is $2^{dim(Null(A))}$ by considering all linear combinations of any basis of $Null(A)$ over $F_2$. Also let $i$ be such that $i(v) = 1$ for all $v$. Then clearly $Ai = 0$, so $dim(Null(A)) > 0$, establishing that the number of ways to do this is $2^k$, $k > 0$. Thus we need only prove the existence of a solution.\nSince we can add a new vertex connected to exactly one previously existing vertex without changing the problem, without loss of generality all vertices have odd degree. Then we want to show that $i \\in Col(A)$. But it is a well-known fact in linear algebra that $Col(A) = Null(A^T)^{\\perp} = Null(A)^{\\perp}$ since $A$ is symmetric. Thus we need to show that if $f \\in Null(A)$, $f$ is perpendicular to $i$ and we will be done. So let $f \\in Null(A)$. Take the submatrix of $A$ consisting of the rows and columns $v$ such that $f(v) = 1$. Then, since $f \\in Null(A)$, the sum of each row in this submatrix must be $0$ in $F_2$. Thus the total number of $1$s in this submatrix is even. But since it is symmetric, the total number of $1$s off of the diagonal is even, so the total number of $1$s on the diagonal is even. But since every vertex has odd degree, the entire diagonal of $A$ consists of $1$s, so this says that the size of the diagonal of this submatrix is even. But this is also the number of $v$ such that $f(v) = 1$, so $f(v) = 1$ for an even number of $v$, thus is perpendicular to $i$, and we have our result.",
        "Solution_2_.28group_theory.29": "Define an order to be a set of instructions, one instruction given to each mathematician.  Each mathematician is told to either move or to stay (we can think of this as stay is $0$, and move is $1$).  Now take some good configuration.  Consider the set of all orders which, when performed on this configuration, give us another good configuration.  Note the identity order, $(0, 0, \\ldots, 0)$, is in this set. We claim this set is an abelian group under composition.\nProof: Clearly each is its own inverse, there is the identity, and the operation is clearly associative and commutative (because it's equivalent to addition of n-dimensional vectors $\\mod{2}$).  So it suffices to show this set of orders is closed under composition.\nConsider any mathematician.  If, in one of these orders, he is told to stay, then the number of his friends who are told to move must be even.  Similarly, if he is told to move, then the number of his friends who are told to stay must be even.  Now just consider two orders $A$ and $B$ and you can show that in $A \\times B$ the same property will hold using parity.\nNow that we've shown it is a group (which we will call $G$), we'll prove it has order two.  Let $I$ be the identity.\nLet $\\{I, T_1\\} = H_1$, where $T$ is some element of $G$.  Now pick an element $T_2$ of $G$ which is not in $H_1$.  Notice that because the elements of $H_1$ are distinct, the elements of $\\{T_2X|X \\in H_1\\}$ are distinct (if two elements of that set were the same, multiply by each on the right by $T_2^{ - 1}$ and you have a contradiction).  Now notice that for any $A, B \\in H_1$, if we were to have $T_2A = B$, then $T_2 = BA^{ - 1} \\in H_1$.  Therefore, $H_1$ and $\\{T_2X|X \\in H_1\\}$ are disjoint and of the same size.  Moreover, the product of any element in the first group and any element in the second group is a member of the second group.  Therefore, these two groups together form a group of order $2^2$.  Call this $H_2$. You progressively build larger and larger subgroups of $G$ until you get to $G$ itself, whose order must then be a power of two.  Therefore, the number of good configurations of the mathematicians was a power of two.\nThis, of course, was all assuming $I$ existed and was in the group.",
        "Solution_3": "Let $n$ be the number of participants at the conference. We proceed by induction on $n$.\nIf $n = 1$, then we have one participant who can eat in either room; that gives us a total of $2 = 2^1$ options.\nLet $n\\geq 2$. The case in which some participant, $P$, has no friends is trivial. In this case, $P$ can eat in either of the two rooms, so the total number of ways to split $n$ participants is twice as many as the number of ways to split $(n - 1)$ participants besides the participant $P$. By induction, the latter number is a power of two, $2^k$, hence the number of ways to split $n$ participants is $2\\times 2^k = 2^{k+1}$, also a power of two. So we assume from here on that every participant has at least one friend.\nWe consider two different cases separately: the case when some participant has an odd number of friends, and the case when each participant has an even number of friends.\nClaim: Some participant, $Z$, has an odd number of friends.\nRemove $Z$ from consideration and for each pair $(X,Y)$ of $Z$'s friends, reverse the relationship between $X$ and $Y$ (from friends to strangers or vice versa).\nClaim. The number of possible seatings is unchanged after removing $Z$ and reversing the relationship between $X$ and $Y$ in each pair $(X,Y)$ of $Z$'s friends.\nProof of the claim. Suppose we have an arrangement prior to $Z$'s departure. By assumption, $Z$ has an even number of friends in the room with him.\nIf this number is 0, the room composition is clearly still valid after $Z$ leaves the room.\nIf this number is positive, let $X$ be one of $Z$'s friends in the room with him. By assumption, person $X$ also has an even number of friends in the same room. Remove $Z$ from the room; then $X$ will have an odd number of friends left in the room, and there will be an odd number of $Z$'s friends in this room besides $X$. Reversing the relationship between $X$ and each of $Z$'s friends in this room will therefore restore the parity to even.\nThe same reasoning applies to any of $Z$'s friends in the other dining room. Indeed, there will be an odd number of them in that room, hence each of them will reverse relationships with an even number of individuals in that room, preserving the parity of the number of friends present.\nMoreover, a legitimate seating without $Z$ arises from exactly one arrangement including $Z$, because in the case under consideration, only one room contains an even number of $Z$'s friends.\nEnd Claim\nThus, we have to double the number of seatings for $(n - 1)$ participants which is, by the induction hypothesis, a power of 2. Consequently, for $n$ participants we will get again a power of 2 for the number of different arrangements.\nCase 2: Each participant has an even number of friends.\nIn this case, each valid split of participants in two rooms gives us an even number of friends in either room.\nLet $(A,B)$ be any pair of friends. Remove this pair from consideration and for each pair $(C,D)$, where $C$ is a friend of $A$ and $D$ is a friend of $B$, change the relationship between $C$ and $D$ to the opposite; do the same if $C$ is a friend of $B$ and $D$ is a friend of $A$. Note that if $C$ and $D$ are friends of both $A$ and $B$, their relationship will be reversed twice, leaving it unchanged.\nConsider now an arbitrary participant $X$ different from $A$ and $B$ and choose one of the two dining rooms. [Note that in the case under consideration, the total number of participants is at least 3, so such a triplet $(A, B; X)$ can be chosen.] Let $A$ have $m$ friends in this room and let $B$ have $n$ friends in this room; both $m$ and $n$ are even. When the pair $(A, B)$ is removed, $X$'s relationship will be reversed with either $n$, or $m$, or $m + n - 2k$ (for $k$ the number of mutual friends of $A$ and $B$ in the chosen room), or 0 people within the chosen room (depending on whether he/she is a friend of only $A$, only $B$, both, or neither). Since $m$ and $n$ are both even, the parity of the number of $X$'s friends in that room will be therefore unchanged in any case.\nAgain, a legitimate seating without $A$ and $B$ will arise from exactly one arrangement that includes the pair $(A, B)$: just add each of $A$ and $B$ to the room with an odd number of the other's friends, and then reverse all the relationships between a friend of $A$ and a friend of $B$. In this way we create a one-to-one correspondence between all possible seatings before and after the $(A,B)$ removal.\nSince the number of arrangements for $n$ participants is twice as many as that for $(n - 2)$ participants, and that number for $(n - 2)$ participants is, by the induction hypothesis, a power of 2, we get in turn a power of 2 for the number of arrangements for $n$ participants. The problem is completely solved.",
        "Solution_4_.28proving_only_existence_of_solution.29": "By induction on the number of mathematicians, assuming solution exists for $n\\leq k$ mathematicians. For $n=k+1$  mathematicians, if each of them has even degree then we're done. Otherwise, let $v$ be a node with odd degree, and denote its neighbors by $N(v)$. We form a new graph $G'$ by deleting $v$ and flipping all edges between members of $N(v)$. That is, for each pair of $a$ and $b$ in $N(v)$, they're connected in $G'$ if and only if they're not in $G$. By induction we know a solution $S'$ exists for $G'$, and it is easily verifiable that the similar solution $S$ exists in $G$ by adding $v$ to the side in $S'$ with even number of neighbors in $N(v)$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page."
    },
    {
        "Problem": "Given circles $\\omega_1$ and $\\omega_2$ intersecting at points $X$ and $Y$, let $\\ell_1$ be a line through the center of $\\omega_1$ intersecting $\\omega_2$ at points $P$ and $Q$ and let $\\ell_2$ be a line through the center of $\\omega_2$ intersecting $\\omega_1$ at points $R$ and $S$.  Prove that if $P, Q, R$ and $S$ lie on a circle then the center of this circle lies on line $XY$.",
        "Solution_1": "Let $\\omega_3$ be the circumcircle of $PQRS$, $r_i$ to be the radius of $\\omega_i$, and $O_i$ to be the center of the circle $\\omega_i$, where $i \\in \\{1,2,3\\}$. Note that $SR$ and $PQ$ are the radical axises of $O_1$ , $O_3$ and $O_2$ , $O_3$ respectively. Hence, by power of a point(the power of $O_1$ can be expressed using circle $\\omega_2$ and $\\omega_3$ and the power of $O_2$ can be expressed using circle $\\omega_1$ and $\\omega_3$),\n\n\nSubtracting these two equations yields that $O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2$, so $O_3$ must lie on the radical axis of $\\omega_1$ , $\\omega_2$.\n~AopsUser101"
    },
    {
        "Problem": "Let $n$ be a positive integer.  Determine the size of the largest subset of $\\{ - n, - n + 1, \\ldots , n - 1, n\\}$ which does not contain three elements $a, b, c$ (not necessarily distinct) satisfying $a + b + c = 0$.",
        "Solution_1": "Let $S$ be a subset of $\\{-n,-n+1,\\dots,n-1,n\\}$ of largest size satisfying $a+b+c\\neq 0$ for all $a,b,c\\in S$. First, observe that $0\\notin S$. Next note that $|S|\\geq \\lceil n/2\\rceil$, by observing that the set of all the odd numbers in $\\{-n,-n+1,\\dots,n-1,n\\}$ works. To prove that $|S|\\leq \\lceil n/2\\rceil$, it suffices to only consider even $n$, because the statement for $2k$ implies the statement for $2k-1$ as well. So from here forth, assume $n$ is even.\nFor any two sets $A$ and $B$, denote by $A+B$ the set $\\{a+b\\mid a\\in A,b\\in B\\}$, and by $-A$ the set $\\{-a\\mid a\\in A\\}$. Also, let $A_+$ denote $A\\cap\\{1,2,\\dots\\}$ and $A_-$ denote $A\\cap\\{-1,-2,\\dots\\}$. First, we present a lemma:\nLemma 1: Let $A$ and $B$ be two sets of integers. Then $|A+B|\\geq|A|+|B|-1$.\nProof: Write $A=\\{a_1,\\dots,a_n\\}$ and $B=\\{b_1,\\dots,b_m\\}$ where $a_1<\\dots<a_n$ and $b_1<\\dots<b_m$. Then $a_1+b_1,a_1+b_2,\\dots,a_1+b_m,a_2+b_m,\\dots,a_n+b_m$ is a strictly increasing sequence of $n+m-1$ integers in $A+B$.\nNow, we consider two cases:\nCase 1: One of $n,-n$ is not in $S$. Without loss of generality, suppose $-n\\notin S$. Let $U=\\{-n+1,-n+2,\\dots,n-2,n-1,n\\}$ (a set with $2n$ elements), so that $S\\subseteq U$ by our assumption. Now, the condition that $a+b+c\\neq 0$ for all $a,b,c\\in S$ implies that $-S\\cap(S_++S_-)=\\emptyset$. Since any element of $S_++S_-$ has absolute value at most $n-1$, we have $S_++S_-\\subseteq \\{-n+1,-n+2,\\dots,n-2,n-1\\}\\subseteq U$. It follows that $S_++S_-\\subseteq U\\setminus -S$, so $|S_++S_-|\\leq 2n-|S|$. However, by Lemma 1, we also have $|S_++S_-|\\geq |S_+|+|S_-|-1=|S|-1$.  Therefore, we must have $|S|-1\\leq 2n-|S|$, or $2|S|\\leq 2n+1$, or $|S|\\leq n$.\nCase 2: Both $n$ and $-n$ are in $S$. Then $n/2$ and $-n/2$ are not in $S$, and at most one of each of the pairs $\\{1,n-1\\},\\{2,n-2\\},\\ldots,\\{\\frac n2-1,\\frac n2+1\\}$ and their negatives are in $S$. This means $S$ contains at most $2+(n-2)=n$ elements.\nThus we have proved that $|S|\\leq n=\\lceil n/2\\rceil$ for even $n$, and we are done.",
        "Solution_2": "Let $Z_n$ be the set of subsets satisfying the $a+b+c \\neq 0$ condition for $n$, and let $s_n$ be the largest size of a set in $Z_n$. Let $n = 2k$ if $n$ is even, and $n = 2k-1$ if $n$ is odd. We note that $s_n \\ge 2k$ due to the following constuction:  or all of the odd numbers in the set. Then the sum of any three will be odd and thus nonzero.\nLemma 1: $s_{n+1} \\le s_{n} + 2$. If $M_{n+1} \\in Z_{n+1}, |M_{n+1}| = s_{n+1}$, then we note that $M_{n+1} \\setminus \\{-2n+1, 2n-1\\} \\in Z_n$, so $|M_{n+1}| - 2 \\le s_n$. $\\square$\nLemma 2: $s_{2k} = s_{2k-1}$. Suppose, for sake of contradiction, that $M_{2k} \\in Z_{n+1}$ and $M_{2k} > s_{2k-1}$. Remove $2k,-2k$ from $M_{2k}$, and partition the rest of the elements into two sets $P_{2k-1}, Q_{2k-1}$, where $P$ and $Q$ contain all of the positive and negative elements of $M_{2k}$, respectively. (obviously $0 \\not \\in M_{2k}$, because $0+0+0 = 0$). WLOG, suppose $|P_{2k-1}| \\ge |Q_{2k-1}|$. Then $|P_{2k-1}| + |Q_{2k-1}| \\ge s_{2k-1} - 1 \\ge 2k - 1$. We now show the following two sub-results:\n\n\nThis is sufficient, because the only two elements that may be in $M_{2k}$ that are not in $P_{2k-1}, Q_{2k-1}$ are $2k$ and $-2k$; for $M_{2k} > s_{2k-1}$, we must either have $|P_{2k-1}| + |Q_{2k-1}| = s_{2k-1} - 1 \\ge 2k-1$ and both $2k,-2k \\in M_{2k}$ [but by pigeonhole $|P_{2k-1}| \\ge k$, see sub-lemma (A)], or $|P_{2k-1}| + |Q_{2k-1}| = s_{2k-1} \\ge 2k$, and $2k \\in M_{2k}$, in which case by (A) we must have $|Q_{2k-1}| < k$, violating (B).\n(A): Partition $\\{1,2,\\ldots,2k-1\\}$ into the $k$ sets $\\{k\\}, \\{1,2k-1\\}, \\{2, 2k-2\\}, \\cdots, \\{k-1, k+1\\}$. Because $(k+\\delta) + (k-\\delta) - 2k = 0$, then if any of those sets are within $P_{2k-1}$, $-2k \\not \\in M_{2k}$. But by Pigeonhole at most $k-1$ elements may be in $P_{2k-1}$, contradiction.\n(B): We prove this statement with another induction. We see that the statement easily holds true for $k=1$ or $2$, so suppose it is true for $k-1$, but [for sake of contradiction] false for $k$. Let $P_{2k-3} = P_{2k-1} \\setminus \\{2k-2, 2k-1\\}$, and similarly for $Q_{2k-3}$. Again WLOG $|P_{2k-3}| \\ge |Q_{2k-3}|$. Then we have $|P_{2k-1}| + |Q_{2k-1}| \\le |P_{2k-3}| + |Q_{2k-3}| + 4$.\nSo we have $\\begin{cases} s_1 &= 2 \\\\ s_{2k} &= s_{2k-1} \\\\ s_{2k-1} &\\le s_{2k-2} + 2 \\\\  \\end{cases}$ and by induction, that $s_n \\le \\boxed{2\\left\\lceil \\frac n2 \\right \\rceil} = 2k$, which we showed is achievable above."
    },
    {
        "Problem": "We define a chessboard polygon to be a polygon whose sides are situated along lines of the form $x = a$ or $y = b$, where $a$ and $b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping $1 \\times 2$ rectangles. Finally, a tasteful tiling is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a $3 \\times 4$ rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.\na) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.\nb) Prove that such a tasteful tiling is unique.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it."
    },
    {
        "Problem": "For $n \\ge 2$ let $a_1$, $a_2$, ..., $a_n$ be positive real numbers such that\nProve that $\\text{max}(a_1, a_2, ... ,a_n) \\le  4 \\text{min}(a_1, a_2, ... , a_n)$.",
        "Solution": "Assume without loss of generality that $a_1 \\geq a_2 \\geq \\cdots \\geq a_n$. Now we seek to prove that $a_1 \\le 4a_n$.\nBy the Cauchy-Schwarz Inequality, \nSince $a_1 \\ge a_n$, clearly $(a_1 - {a_n \\over 4}) > 0$, dividing yields:\n\nas desired.\nAlternative Solution (by Deng Tianle, username: Leole)\nAssume without loss of generality that $a_1 \\geq a_2 \\geq \\cdots \\geq a_n$.\nUsing the Cauchy–Bunyakovsky–Schwarz inequality and the inequality given, \n(Note that $n-2 \\ge 0$ since $n \\ge 2$ as given!)\nThis implies that $3a_n -\\frac{3a_1}{4} \\ge 0 \\iff 4a_n \\ge a_1$ as desired."
    },
    {
        "Problem": "Trapezoid $ABCD$, with $\\overline{AB}||\\overline{CD}$, is inscribed in circle $\\omega$ and point $G$ lies inside triangle $BCD$.  Rays $AG$ and $BG$ meet $\\omega$ again at points $P$ and $Q$, respectively.  Let the line through $G$ parallel to $\\overline{AB}$ intersect $\\overline{BD}$ and $\\overline{BC}$ at points $R$ and $S$, respectively.  Prove that quadrilateral $PQRS$ is cyclic if and only if $\\overline{BG}$ bisects $\\angle CBD$.",
        "Solution_1": "We will use directed angles in this solution. Extend $QR$ to $T$ as follows:\nIf:\nNote that \nThus, $BTRG$ is cyclic.\nAlso, note that $GSCP$ is cyclic because  depending on the configuration.\nNext, we have $T, G, C$ are collinear since\nTherefore,  so $PQRS$ is cyclic.\nOnly If:\nThese steps can be reversed.",
        "Solution_2_.28Projective.29": "Extend $QR$ to $T$, and let line $l \\parallel AB$ intersect $\\omega$ at $K$ and another point $V$, as shown:\nIf:\nSuppose that $VP \\cap CB = S'$, and $AC \\cap QV = R'$. Pascal's theorem on the tuple $(V, P, A, C, B, Q)$ implies that the points $S'$, $R'$, and $G = PA \\cap BQ$ are collinear. However, $AC$ and $BD$ are symmetrical with respect to the axis of symmetry of trapezoid $ABCD$, and $TQ$ and $VQ$ are also symmetrical with respect to the axis of symmetry of $ABCD$ (as $Q$ is the midpoint of $\\overset{\\frown}{DC}$, and $TV \\parallel DC$). Since $R = BD \\cap TQ$, $R$ and $R'$ are symmetric with respect to the axis of symmetry of trapezoid $ABCD$. This implies that line $R'G$ is equivalent to line $RG$. Thus, $S'$ lies on line $RG$. However, $S = BC \\cap RG$, so this implies that $S' = S$.\nNow note that $TVPQ$ is cyclic. Since $TV \\parallel RS$, $\\measuredangle VTQ = \\measuredangle SRQ$. However, $\\measuredangle VTQ + \\measuredangle VPQ = 180^{\\circ} = \\measuredangle SRQ + \\measuredangle SPQ$. Therefore, $PQRS$ is cyclic.\nOnly If:\nConsider the same setup, except $Q$ is no longer the midpoint of $\\overset{\\frown}{DC}$. Note that $TV$ must be parallel to $RG$ in order for $PQRS$  to be cyclic. We claim that $S' = S$ and hope to reach a contradiction. Pascal's theorem on the tuple $(V, P, A, C, B, Q)$ implies that $S'$, $R'$, and $G = PA \\cap BQ$ are collinear. However, there exists a unique point $Q$ such that $HQ$, $AC$, and $RG$ are concurrent. By If, $Q$ must be the midpoint of $\\overset{\\frown}{DC}$ in order for the concurrency to occur; hence, $R' \\notin RS$. Then $R'G \\cap BC = S' \\neq S$, since $RG \\cap BC = S$. However, this is a contradiction, so therefore $TV$ cannot be parallel to $RG$ and $PQRS$ is not cyclic.\nSolution by TheBoomBox77"
    },
    {
        "Problem": "Let $s_1, s_2, s_3, \\ldots$ be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \\cdots.$  Suppose that $t_1, t_2, t_3, \\ldots$ is also an infinite, nonconstant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$.  Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.",
        "Solution": "Suppose the $s_i$ can be represented as $\\frac{a_i}{b_i}$ for every $i$, and suppose $t_i$ can be represented as $\\frac{c_i}{d_i}$. Let's start with only the first two terms in the two sequences, $s_1$ and $s_2$ for sequence $s$ and $t_1$ and $t_2$ for sequence $t$. Then by the conditions of the problem, we have $(s_2 - s_1)(t_2 - t_1)$ is an integer, or $(\\frac{a_2}{b_2} - \\frac{a_1}{b_1})(\\frac{c_2}{d_2} - \\frac{c_1}{d_1})$ is an integer. Now we can set $r = \\frac{b_1 b_2}{d_1 d_2}$, because the least common denominator of $s_2 - s_1$ is $b_1 b_2$ and of $t_2 - t_1$ is $d_1 d_2$, and multiplying or dividing appropriately by $\\frac{b_1 b_2}{d_1 d_2}$ will always give an integer.\nNow suppose we kept adding $s_i$ and $t_i$ until we get to $s_m = \\frac{a_m}{b_m}$ in sequence $s$ and $t_m = \\frac{c_m}{d_m}$ in sequence $t$ so that $(t_m - t_i)(s_m - s_i)$ is an integer for all $i$ with $1 \\le i < m$, where $m$ is a positive integer. At this point, we will have $r$ = $\\frac{\\prod_{n=1}^{m}b_n}{\\prod_{n=1}^{m}d_n}$, because these are the least common denominators of the two sequences up to $m$. As we keep adding $s_i$ and $t_i$, $r$ will always have value $\\frac{\\prod_{n=1}^{m}b_n}{\\prod_{n=1}^{m}d_n}$, and we are done."
    },
    {
        "Problem": "Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter\n$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto\nlines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle\nformed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where\n$O$ is the midpoint of segment $AB$.",
        "Solution_1": "Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$.\nSince $XY$ is a chord of the circle with diameter $AB$,\n$\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$,\nwe conclude $\\angle YAZ = \\angle YBZ = \\delta$.\nTriangles $BQY$ and $APY$ are both right-triangles, and share the\nangle $\\gamma$, therefore they are similar, and so the ratio $PY : YQ = AY : YB$. Now by Thales' theorem the angles $\\angle AXB = \\angle AYB = \\angle AZB$ are all right-angles. Also, $\\angle PYQ$,\nbeing the fourth angle in a quadrilateral with 3 right-angles is\nagain a right-angle.  Therefore $\\triangle PYQ \\sim \\triangle AYB$ and\n$\\angle YQP = \\angle YBA = \\gamma + \\beta$.\nSimilarly, $RY : YS = AY : YB$, and so $\\angle YRS = \\angle YAB = \\alpha + \\delta$.\nNow $RY$ is perpendicular to $AZ$ so the direction $RY$ is $\\alpha$ counterclockwise from the vertical, and since $\\angle YRS = \\alpha + \\delta$ we see that $SR$ is $\\delta$ clockwise from the vertical. (Draw an actual vertical line segment if necessary.)\nSimilarly, $QY$ is perpendicular to $BX$ so the direction $QY$ is $\\beta$ clockwise from the vertical, and since $\\angle YQP$ is $\\gamma + \\beta$ we see that $QY$ is $\\gamma$ counterclockwise from the vertical.\nTherefore the lines $PQ$ and $RS$ intersect at an angle $\\chi = \\gamma + \\delta$. Now by the central angle theorem $2\\gamma = \\angle XOY$\nand $2\\delta = \\angle YOZ$, and so $2(\\gamma + \\delta) = \\angle XOZ$,\nand we are done.\nNote that $RTQY$ is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?",
        "Solution_2": "\nLet $T$ be the foot of the perpendicular from $Y$ to $\\overline{AB}$, let $O$ be the center of the semi-circle.\nSince we have a semi-circle, if we were to reflect it over $\\overline {AB}$, we would have a full circle, with $\\triangle{AXB}$ and $\\triangle{AZB}$ inscribed in it. Now, notice that $Y$ is a point on that full circle, so we can say that $T$ lies on the Simson Line $\\overline{PQ}$ from $Y$ to $\\triangle AXB$ and that it also lies on the Simson line $\\overline {RS}$ from $Y$ to $\\triangle AZB$. Thus, $T$ lies on two distinct lines in a plane, which means that $T=\\overline{PQ}\\cap\\overline{RS}$. Therefore, it suffices to show that $\\angle PTS=\\tfrac{1}{2}\\angle XOZ$.\nSince $m\\angle YTA + m\\angle YPA = 90^\\circ + 90^\\circ = 180^\\circ$ and $m \\angle YTB + m \\angle YSB = 90^\\circ + 90^\\circ = 180^\\circ$, we know that $TAPY$ and $TBSY$ are cyclic quadrilaterals.\nWe use this fact to get  \\\\\nNow note that $\\angle XAY$ is the inscribed angle of minor arc $\\overset{\\huge\\frown}{PY}$, and $\\angle XOY$ is the central angle of minor arc $\\overset{\\huge\\frown}{AB}$, so $\\angle XAY = \\frac{\\overset{\\huge\\frown}{PY}}{2} = \\frac{\\angle XOY}{2}$. Similarly, $\\angle YBZ = \\frac{\\overset{\\huge\\frown}{YZ}}{2}=\\frac{\\angle YOZ}{2}$. Thus we can say\nCombining statements $(1)$ and $(2)$, we can say that $\\angle PTS = \\frac{\\angle XOZ}{2}$, as desired. $\\square$\n~thinker123"
    },
    {
        "Problem": "There are $n$ students standing in a circle, one behind the\nother. The students have heights $h_1 < h_2 < \\ldots < h_n$. If a\nstudent with height $h_k$ is standing directly behind a student\nwith height $h_{k-2}$ or less, the two students are permitted to\nswitch places. Prove that it is not possible to make more than\n$\\binom{n}{3}$ such switches before reaching a position in which\nno further switches are possible.",
        "Solution_1": "We adopt the usual convention that $\\binom{i}{j} = 0$ unless $0 \\le j \\le i$.\nWith this, the binomial coefficients are defined for all integers via the\nrecursion:\nIt is clear that the circle is oriented and all the students are facing in same direction (clockwise or counterclockwise). We'll call this direction forward.\nIn any switch consider the taller student to have moved forward and the shorter student to have remained stationary. No backward motion is allowed. With this definition of forward motion, the first two students with heights $h_1$ and $h_2$ are always stationary, while other students potentially move past them.\nFor $k > 2$, the student with height $h_k$ can never switch places with the student with height $h_{k-1}$, and the former can make at most $k-2$ more forward moves than the latter (when all the students of heights $h_1, \\ldots h_{k-2}$ are between $h_k$ and $h_{k-1}$ in the forward direction).\nTherefore, if the $(k-1)^{\\mathrm{st}}$ student can make $s_{k-1}$ forward steps, the $k^{\\mathrm{th}}$ student can make at most $s_{k-1} + k - 2$ steps. With $s_1 = s_2 = 0$ and $s_3 = 0 + (3-2) = 1$, and a constant second difference of $1$, we quickly see that $s_k = \\binom{k-1}{2}$.\nWith $n$ students in all, the total number of steps is therefore at most $\\sum_{i=3}^{n}\\binom{i-1}{2} = \\binom{n}{3}$. The sum is a telescoping sum since: $\\binom{n}{3} - \\binom{n-1}{3} = \\binom{n-1}{2}.$",
        "Solution_2": "WLOG, let $h_k=k$.  Now, we find the end arrangement with the most switches possible.  We claim that the arrangement will be $n,n-1,n-2,\\dots ,1$, where the left to right direction is the \"backwards\" direction.  To prove this makes the most switches, we show that there is always at least one more switch that can be done for any other arrangement.  This is elementary to show.  There will always be one height $x$ such that the number to its right is $2$ less than $x$, unless every number has $x-1$ to the right of $x$ (other than $2$ and $1$).  The exception occurs at our claim, so our claim is proven.  Now we want to find the maximum ways we can \"undo\" our arrangement.  But undoing a switch is just doing a switch from our arrangement in the opposite direction.  So, the start arrangement with the most possible switches is the reverse of the end arrangement or $1,2,3,\\dots ,n$.\nWe want to find how many switches must be done to get from the start arrangement to the end arrangement.  We start by switching $n$ around until it cannot be switched anymore.  We find that we can switch $n$ $n-2$ times.  When we are switching (or, in other words, moving to the right) the number $k$, we can switch it $k-2$ times before it is to the left of $k-1$.  But then we can also switch each of $k+1,k+2,\\dots ,n$ (in that order) $k-2$ times before they get stopped again.  Let $x=n-k+2$.  Then, the sum of the switches is $\\sum\\limits_{x=2}^{n-1} (n-x)(x-1)=\\sum\\limits_{x=1}^{n} (n-x)(x-1)$.\nRearranging and summing, we get"
    },
    {
        "Problem": "The $2010$ positive numbers $a_1, a_2, \\ldots , a_{2010}$ satisfy\nthe inequality $a_ia_j \\le i+j$ for all distinct indices $i, j$.\nDetermine, with proof, the largest possible value of the product\n$a_1a_2\\cdots a_{2010}$.",
        "Solution": "The largest possible value is"
    },
    {
        "Problem": "Let $ABC$ be a triangle with $\\angle A = 90^{\\circ}$. Points $D$\nand $E$ lie on sides $AC$ and $AB$, respectively, such that $\\angle ABD = \\angle DBC$ and $\\angle ACE = \\angle ECB$. Segments $BD$ and\n$CE$ meet at $I$.  Determine whether or not it is possible for\nsegments $AB, AC, BI, ID, CI, IE$ to all have integer lengths.",
        "Solution": "We know that angle $BIC = 135^{\\circ}$, as the other two angles in triangle $BIC$ add to $45^{\\circ}$. Assume that only $AB, AC, BI$, and $CI$ are integers. Using the Law of Cosines on triangle BIC,\n$BC^2 = BI^2 + CI^2 - 2BI\\cdot CI \\cdot \\cos 135^{\\circ}$. Observing that $BC^2 = AB^2 + AC^2$ is an integer and that $\\cos 135^{\\circ} = -\\frac{\\sqrt{2}}{2},$ we have\nand therefore,\nThe LHS ($\\sqrt{2}$) is irrational, while the RHS is the quotient of the division of two integers and thus is rational. Clearly, there is a contradiction. Therefore, it is impossible for $AB, AC, BI$, and $CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.",
        "Solution_2": "The answer is no.\nSuppose otherwise. It is easy to see (through simple angle chasing) that $\\angle DIC=45^{\\circ}$. Also, since $I$ is the incenter, we have $\\angle IAC = 45^{\\circ}$. Using the Law of Cosines, we have  so that $CD$ is irrational. But $\\triangle IAC \\sim \\triangle DIC$, thus $IC^2=CD\\cdot AC$, implying that $CD$ is rational, contradiction. $\\blacksquare$",
        "Solution_3": "The result can be also proved without direct appeal to trigonometry,\nvia just the angle bisector theorem and the structure of Pythagorean\ntriples. (This is a lot more work).\nA triangle in which all the required lengths are integers exists if and\nonly if there exists a triangle in which $AB$ and $AC$ are\nrelatively-prime integers and the lengths of the segments $BI, ID, CI, IE$ are all rational\n(we divide all the lengths by the $\\gcd(AB, AC)$ or\nconversely multiply all the lengths by the least common multiple\nof the denominators of the rational lengths).\nSuppose there exists a triangle in which the lengths $AB$ and $AC$ are\nrelatively-prime integers and the lengths $IB, ID, CI, IE$ are all rational.\nSince $CE$ is the bisector of $\\angle ACB$, by the angle bisector\ntheorem, the ratio $IB : ID = CB : CD$, and since $BD$ is the\nbisector of $\\angle ABC$, $CB : CD = (AB + BC) : AC$. Therefore,\n$IB : ID = (AB + BC) : AC$. Now $IB : ID$ is by assumption rational,\nso $(AB + BC) : AC$ is rational, but $AB$ and $AC$ are assumed integers\nso $BC : AC$ must also be rational. Since $BC$ is the hypotenuse of\na right-triangle, its length is the square root of an integer,\nand thus either an integer or irrational, so $BC$ must be an integer.\nWith $AB$ and $AC$ relatively-prime, we conclude that the side\nlengths of $\\triangle ABC$ must be a Pythagorean triple: $(2pq, p^2 - q^2, p^2 + q^2)$, with $p > q$ relatively-prime positive integers\nand $p+q$ odd.\nWithout loss of generality, $AC = 2pq, AB = p^2 - q^2, BC = p^2+q^2$.\nBy the angle bisector theorem,\nSince $\\triangle CAE$ is a right-triangle, we have:\nand so $CE$ is rational if and only if $2p^2 + 2q^2$ is a perfect square.\nAlso by the angle bisector theorem,\nand therefore, since $\\triangle DAB$ is a right-triangle, we have:\nand so $BD$ is rational if and only if $p^2 + q^2$ is a perfect square.\nCombining the conditions on $CE$ and $BD$, we see that\n$2p^2+2q^2$ and $p^2+q^2$ must both be perfect squares. If it were so,\ntheir ratio, which is $2$, would be the square of a rational number,\nbut $\\sqrt{2}$ is irrational, and so the assumed triangle cannot exist.",
        "Solution_4": "We proceed by contradiction.\nFTSOC, let $AB, BI, ID$ have integer lengths. Then $BD = BI + ID \\in \\mathbb{Z}$ as well. By trigonometry, \nRearranging we find $\\cos{(\\frac{\\angle ABC}{2})} = \\frac{AB}{BD} \\in \\mathbb{Q}$. As $0 < \\angle ABC < \\frac{\\pi}{2}$ so $0 < \\frac{\\angle ABC}{2} < \\frac{\\pi}{4}$ but there are no possible angles in the interval that result in a rational cosine by Niven's theorem. So we have contradiction.\n~Aaryabhatta1."
    },
    {
        "Problem": "Let $q = \\dfrac{3p-5}{2}$ where $p$ is an odd prime, and let\nProve that if $\\dfrac{1}{p}-2S_q = \\dfrac{m}{n}$ for integers\n$m$ and $n$, then $m-n$ is divisible by $p$.",
        "Solution": "Since $p$ is an odd prime, $p = 2r + 1$, for a suitable positive integer $r$, and consequently $q = 3r - 1$.\nThe partial-fraction decomposition of the general term of $S_q$ is:\ntherefore\nwith $a$ and $b$ positive relatively-prime integers.\nSince $r < p$ and $p$ is a prime, in the final sum all the denominators are relatively prime to $p$, but all the numerators are divisible by $p$, and therefore the numerator $a$ of the reduced fraction $\\frac{a}{b}$ will be divisible by $p$. Since the sought difference $m - n = (b-a) - b = -a$, we conclude that $p$ divides $m-n$ as required."
    },
    {
        "Problem": "A blackboard contains 68 pairs of nonzero integers.  Suppose that for each positive integer $k$ at most one of the pairs $(k, k)$ and $(-k, -k)$ is written on the blackboard.  A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0.  The student then scores one point for each of the 68 pairs in which at least one integer is erased.  Determine, with proof, the largest number $N$ of points that the student can guarantee to score regardless of which 68 pairs have been written on the board.",
        "Solution": "Answer: 43\n\nAttainability: Consider 8 distinct positive numbers. Let there be 5 pairs for each of the numbers including 2 clones of that number. Let there also be 28 pairs that include the negatives of those numbers such that each negative associates with another negative once and exactly once (in graph theoretic terms, a K8). Let x be the number of positives chosen out of these 8 (assume the other chosen numbers are negatives) for erasing; then, the number of points the student scores is $28 + 5x - {x \\choose 2}$, which is maximized at x=5 and x=6, and the maximum value is $43$. Choosing the first 5 numbers as positive and the other three as negative attains this. Hence, 43 is a possible maximum possible score.\n\nBounding: We use expected values. WLOG all the pairs of numbers with both numbers identical have only positive values. Consider flipping a weighted coin whether to choose the positive number or its negation for each positive number; it chooses positive with probability p. The pairs with both numbers same are chosen with probability p, the pairs (k, -k) are chosen with probability 1, and the pairs (x, y) for distinct x and y that sum to a nonzero number are chosen with probability $1-p^2$. We are trying to minimize the expected value, so we can assume that no pairs (k, -k) exist. Let A be the number of (k, k) pairs, and 68-A be the number of (x, y) pairs. The expected number of points scored is $Ap + (68-A)(1-p^2)$. We want to prove this is larger than 42 at all times for some choice of p. If $A < 36$, $1/2$ works for p to give this bound. If $A > 36$, $5/8$ works for p for p to give the desired bound. If $A = 36$, we can use $3/5$ for p to get the desired bound. Hence, in any case, the expected value for the number of points scored across all of these weighted processes is larger than 42, so there exists some case that gives a score of 43. Hence, bounding is complete. We are done with both parts. Q.E.D.\n-Solution by thanosaops"
    },
    {
        "Problem": "Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 + (a + b + c)^2 \\le 4$.  Prove that",
        "Solution_1": "Since\n\nit is natural to consider a change of variables:\n\nwith the inverse mapping given by:\n\nWith this change of variables, the constraint becomes\n\nwhile the left side of the inequality we need to prove is now\nTherefore it remains to prove that\nWe note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done.",
        "Solution_2": "Rearranging the condition yields that\nNow note that\nSumming this for all pairs of $\\{ a,b,c \\}$ gives that\nBy AM-GM. Dividing by $2$ gives the desired inequality.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding 2m to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.",
        "Solution": "Let $\\mathbb{F}_5$ be the field of positive residues modulo 5. We label the vertices of the pentagon clockwise with the residues 0, 1, 2, 3 and 4.\nFor each $i \\in \\mathbb{F}_5$ let $n_i$ be the integer at vertex $i$ and let $r_i \\in \\mathbb{F}_5$ be defined as:\n\nLet $s = \\sum_{i\\in\\mathbb{F}_5} n_i$. A move in the game consists of\n\nfor some vertex $i \\in \\mathbb{F}_5$ and integer $m$. We immediately see that $s$ is an invariant of the game. After our move the new value of $r_i$ is decreased by $3m + 2m \\equiv 0 \\pmod 5$ as a result of the change in the $3n_{i+2}$ and $2n_{i+3}$ terms. So $r_i$ does not change after a move at vertex $i$.\nFor all $i \\in \\mathbb{F}_5$ we have:\nTherefore, the $r_i$ form an arithmetic progression in $\\mathbb{F}_5$ with a difference of $s$. Since $r_k$ is unchanged by a move at vertex $k$, so are all the remaining $r_i$ as the differences are constant.\nProvided $s \\not\\equiv 0 \\pmod 5$, we see that the mapping $i \\mapsto r_i$ is a bijection $\\mathbb{F}_5 \\to \\mathbb{F}_5$ and exactly one vertex will have $r_i \\equiv 0 \\pmod 5$. As $r_i$ is an invariant, a winning vertex must have $r_i \\equiv 0$, since in the final state each $n_k$ with $k \\ne i$ is zero. So, for $s \\not\\equiv 0 \\pmod 5$, if a winning vertex exists, it is the unique vertex with $r_i \\equiv 0$.\nWithout loss of generality, it remains to show that if $r_0 \\equiv 0 \\pmod 5$, then 0 must be a winning vertex. To prove this, we perform the following moves:\n\nWe designate the new state $(n'_0, 0, n'_2, n'_3, 0)$. Since $r_0 \\equiv 0$ is an invariant, and $n'_1 = n'_4 = 0$, we now have $r_0 \\equiv n'_3 - n'_2 = 5p$, for some integer $p$. Our final set of moves is:\n\nNow our chosen vertex 0 is the only vertex with a non-zero value, and since $s$ is invariant, that value is $s$ as required. Since a vertex $i$ with $r_i \\equiv 0 \\pmod 5$ is winnable, and with $s = 2011 \\not\\equiv 0 \\pmod 5$ we always have a unique such vertex, we are done.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Consider the assertion that for each positive integer $n \\ge 2$, the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of 4. Either prove the assertion or find (with proof) a counter-example.",
        "Solution": "We will show that $n = 25$ is a counter-example.\nSince $\\textstyle 2^n \\equiv 1 \\pmod{2^n - 1}$, we see that for any integer $k$, $\\textstyle 2^{2^n} \\equiv 2^{(2^n - kn)} \\pmod{2^n-1}$. Let $0 \\le m < n$ be the residue of $2^n \\pmod n$. Note that since $\\textstyle m < n$ and $\\textstyle n \\ge 2$, necessarily $\\textstyle 2^m < 2^n -1$, and thus the remainder in question is $\\textstyle 2^m$. We want to show that $\\textstyle 2^m \\pmod {2^n-1}$ is an odd power of 2 for some $\\textstyle n$, and thus not a power of 4.\nLet $\\textstyle n=p^2$ for some odd prime $\\textstyle p$. Then $\\textstyle \\varphi(p^2) = p^2 - p$. Since 2 is co-prime to $\\textstyle p^2$, we have\n\nand thus\nTherefore, for a counter-example, it suffices that $\\textstyle 2^p \\pmod{p^2}$ be odd. Choosing $\\textstyle p=5$, we have $\\textstyle 2^5 = 32 \\equiv 7 \\pmod{25}$. Therefore, $\\textstyle 2^{25} \\equiv 7 \\pmod{25}$ and thus\n\nSince $\\textstyle 2^7$ is not a power of 4, we are done.",
        "Solution_2": "Lemma (useful for all situations): If $x$ and $y$ are positive integers such that $2^x - 1$ divides $2^y - 1$, then $x$ divides $y$.\nProof: $2^y \\equiv 1 \\pmod{2^x - 1}$. Replacing the $1$ with a $2^x$ and dividing out the powers of two should create an easy induction proof which will be left to the reader as an Exercise.\nConsider $n = 25$. We will prove that this case is a counterexample via contradiction.\nBecause $4 = 2^2$, we will assume there exists a positive integer $k$ such that $2^{2^n} - 2^{2k}$ divides $2^n - 1$ and $2^{2k} < 2^n - 1$. Dividing the powers of $2$ from LHS gives $2^{2^n - 2k} - 1$ divides $2^n - 1$. Hence, $2^n - 2k$ divides $n$. Because $n = 25$ is odd, $2^{24} - k$ divides $25$. Euler's theorem gives $2^{24} \\equiv 2^4 \\equiv 16 \\pmod{25}$ and so $k \\ge 16$. However, $2^{2k} \\geq 2^{32} > 2^{25} - 1$, a contradiction. Thus, $n = 25$ is a valid counterexample.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Let $P$ be a given point inside quadrilateral $ABCD$.  Points $Q_1$ and $Q_2$ are located within $ABCD$ such that $\\angle Q_1 BC = \\angle ABP$, $\\angle Q_1 CB = \\angle DCP$, $\\angle Q_2 AD = \\angle BAP$, $\\angle Q_2 DA = \\angle CDP$.  Prove that $\\overline{Q_1 Q_2} \\parallel \\overline{AB}$ if and only if $\\overline{Q_1 Q_2} \\parallel \\overline{CD}$.",
        "Solution_1": "Lemma. If $AB$ and $CD$ are not parallel, then $AB, CD, Q_1 Q_2$ are concurrent.\nProof. Let $AB$ and $CD$ meet at $R$. Notice that with respect to triangle $ADR$, $P$ and $Q_2$ are isogonal conjugates (this can be proven by dropping altitudes from $Q_2$ to $AB$, $CD$, and $AD$ or $BC$ depending on where $R$ is). With respect to triangle $BCR$, $P$ and $Q_1$ are isogonal conjugates. Therefore, $Q_1$ and $Q_2$ lie on the reflection of $RP$ in the angle bisector of $\\angle{DRA}$, so $R, Q_1, Q_2$ are collinear. Hence, $AB, CD, Q_1 Q_2$ are concurrent at $R$.\nNow suppose $Q_1 Q_2 \\parallel AB$ but $Q_1 Q_2$ is not parallel to $CD$. Then $AB$ and $CD$ are not parallel and thus intersect at a point $R$. But then $Q_1 Q_2$ also passes through $R$, contradicting $Q_1 Q_2 \\parallel AB$. A similar contradiction occurs if $Q_1 Q_2 \\parallel CD$ but $Q_1 Q_2$ is not parallel to $AB$, so we can conclude that $Q_1 Q_2 \\parallel AB$ if and only if $Q_1 Q_2 \\parallel CD$.",
        "Solution_2": "First note that $\\overline{Q_1 Q_2} \\parallel \\overline{AB}$ if and only if the altitudes from $Q_1$ and $Q_2$ to $\\overline{AB}$ are the same, or $|Q_1B|\\sin \\angle ABQ_1 =|Q_2A|\\sin \\angle BAQ_2$.  Similarly $\\overline{Q_1 Q_2} \\parallel \\overline{CD}$ iff $|Q_1C|\\sin \\angle DCQ_1 =|Q_2D|\\sin \\angle CDQ_2$.\nIf we define $S =\\frac{|Q_1B|\\sin \\angle ABQ_1}{|Q_2A|\\sin \\angle BAQ_2} \\times \\frac{|Q_2D|\\sin \\angle CDQ_2}{|Q_1C|\\sin \\angle DCQ_1}$, then we are done if we can show that S=1.\nBy the law of sines, $\\frac{|Q_1B|}{|Q_1C|}=\\frac{\\sin\\angle Q_1CB}{\\sin\\angle Q_1BC}$ and $\\frac{|Q_2D|}{|Q_2A|}=\\frac{\\sin\\angle Q_2AD}{\\sin\\angle Q_2DA}$.\nSo, $S=\\frac{\\sin \\angle ABQ_1}{\\sin \\angle BAQ_2}\\cdot\\frac{\\sin \\angle CDQ_2}{\\sin \\angle DCQ_1}\\cdot\\frac{\\sin \\angle BCQ_1}{\\sin \\angle CBQ_1}\\cdot\\frac{\\sin \\angle DAQ_2}{\\sin \\angle ADQ_2}$\nBy the terms of the problem, $S=\\frac{\\sin \\angle PBC}{\\sin \\angle PAD}\\cdot\\frac{\\sin \\angle PDA}{\\sin \\angle PCB}\\cdot\\frac{\\sin \\angle PCD}{\\sin \\angle PBA}\\cdot\\frac{\\sin \\angle PAB}{\\sin \\angle PDC}$.  (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.)\nRearranging yields $S= \\frac{\\sin \\angle PBC}{\\sin \\angle PCB}\\cdot\\frac{\\sin \\angle PDA}{\\sin \\angle PAD}\\cdot\\frac{\\sin \\angle PCD}{\\sin \\angle PDC}\\cdot\\frac{\\sin \\angle PAB}{\\sin \\angle PBA}$.\nApplying the law of sines to the triangles with vertices at P yields $S=\\frac{|PC|}{|PB|}\\frac{|PA|}{|PD|}\\frac{|PD|}{|PC|}\\frac{|PB|}{|PA|}=1$."
    },
    {
        "Problem": "Let $A$ be a set with $|A| = 225$, meaning that $A$ has 225 elements.  Suppose further that there are eleven subsets $A_1$, $\\dots$, $A_{11}$ of $A$ such that $|A_i | = 45$ for $1 \\le i \\le 11$ and $|A_i \\cap A_j| = 9$ for $1 \\le i < j \\le 11$.  Prove that $|A_1 \\cup A_2 \\cup \\dots \\cup A_{11}| \\ge 165$, and give an example for which equality holds.",
        "Solution_2": "We will count the number of ordered triples, $(A_i,A_j,x)$, where $1\\le i,j\\le11$ and $x\\in A_i \\cap A_j$. We know this is equal to $990=11\\cdot10\\cdot9$­. We can also find that this is $\\sum_{k=1}^{225}b_k^2-b_k$, where $b_k$ is the number of the $11$ subsets the $k^{\\text{th}}$ element of $A$ is in. Since $\\sum_{k=1}^{225}b_k=45\\cdot11=495$, we know $\\sum_{k=1}^{225}b_k^2=990+495=1485$. Let $n=|A_1 \\cup A_2 \\cup \\dots \\cup A_{11}|$. $n$ is equal to the number of $b_k>0$, where $1\\le k\\le225$. By the QM-AM inequality, we know $\\frac{\\sum_{k=1}^{225}b_k^2}n\\ge\\Bigg(\\frac{\\sum_{k=1}^{225}b_k}n\\Bigg)^2\\implies\\frac{1485}n\\ge\\Big(\\frac{495}n\\Big)^2\\implies n\\ge165$ and that equality occurs when $b_1=b_2=b_3=\\cdots=b_{165}=3,b_{166}=b_{167}=\\cdots=b_{225}=0$.\n$\\square$\nSolution by randomdude10807\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
        "Solution_3": "Define $x_i$ to be the amount of elements in $A$ such that the element is contained exactly $i$ times among the $11$ subsets $A_j$. We are trying to prove that $\\sum_{i=1}^{11}x_i\\ge 165$. Now, note that $\\sum_{i=1}^{11}ix_i$ is equivalent to the total amount of not necessarily distinct elements among the subsets $A_j$, thus $\\sum_{i=1}^{11}ix_i=11|A_j|=495$. Furthermore, note that $\\sum_{i=1}^{11}\\binom{i}{2}x_i$ is equivalent to the total amount of not necessarily distinct pairs of subsets such that the subsets share an element. In other words, each subset pair should be counted $9$ times in this sum, and there are $\\binom{11}{2}$ total distinct pairs of subsets, so $\\sum_{i=1}^{11}\\binom{i}{2}x_i=9*\\binom{11}{2}=495$. Noting that $2\\binom{i}{2}+i=i^2$ for all nonnegative integers, we note\n\nFinally, CS inequality gives us\nas desired.\nOne equality case occurs when $x_i=0$ if $i$ is not $3$, and $x_3=165$. Choose any $165$ elements, and represent each with a unique combination of $3$ of the $11$ subsets ($\\binom{11}{3}=165$). It is then obvious that each subset has $45$ distinct elements, because after each subset is chosen, there are $\\binom{10}{2}=45$ ways to choose the other two subsets to produce a unique element. Similarly, each pair of subsets shares exactly $9$ elements, because after choosing the two subsets, we can choose one of the remaining $9$ subsets to produce the unique element. Hence, proved. $\\blacksquare{}$\n~SigmaPiE"
    },
    {
        "Problem": "Find all integers $n \\ge 3$ such that among any $n$ positive real numbers $a_1$, $a_2$, $\\dots$, $a_n$ with\n\nthere exist three that are the side lengths of an acute triangle.",
        "Solution": "Without loss of generality, assume that the set $\\{a\\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \\le n \\cdot a_1.$ Now set $b_i \\equiv \\frac{a_i}{a_1},$ and since a triangle with sidelengths from $\\{a\\}$ will be similar to the corresponding triangle from $\\{b\\},$ we simply have to show the existence of acute triangles in $\\{b\\}.$ Note that $b_1 = 1$ and for all $i$, $b_i \\le n.$\nNow three arbitrary sidelengths $x$, $y$, and $z$, with $x \\le y \\le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \\longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$'s for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$).\nWe now make another substitution: $c_i \\equiv b_i ^2.$ So $c_1 = 1$ and for all $i$, $c_i \\le n^2.$ Now we examine the smallest possible sets $\\{c\\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$, then the smallest possible set, call it $\\{s_3\\},$ is trivially $\\{1,1,2\\}$, since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\\{s_n\\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\\{s_n\\} = \\{F_0, F_1, ... F_n\\}$, then $\\{s_{n+1}\\} = \\{F_0, F_1, ... F_n, c_{n+1}\\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\\{s_n\\}$ which are $F_{n-1}$ and $F_n$. But these sum to $F_{n+1}$ so $\\{s_{n+1}\\} = \\{F_0, F_1, ... F_{n+1}\\}$ and our induction is complete.\nNow since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\\{c\\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\\{c\\}$ is bounded between $1$ and $n^2$, then the conditions of the problem are met if and only if $F_{n-1} > n^2$. The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\\boxed{\\{n \\ge 13\\}}$.",
        "Solution_2": "Outline:\n1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \\ge 3$.\n2. If the chosen $n$ is such that $F_n \\le n^2$, then choose the sequence $a_n$ such that $a_k = \\sqrt{F_k}$ for $1 \\le k \\le n$. It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \\sqrt{F_a}, y = \\sqrt{F_b}, z = \\sqrt{F_c}$ with $a<b<c$, $x^2 + y^2 = F_a + F_b \\le F_{b-1} + F_b = F_{b+1} \\le F_c = z^2$, x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \\le n^2$ do not work.\n3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \\ge F_1a_1^2, a_3^2 \\ge a_2^2 + a_1^2 \\ge F_2a^2$, and by induction $a_n^2 > F_na_1^2$, a contradiction to the condition's inequality.)\n4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$. It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$. Thus, $\\boxed{n \\ge 13}$ is the desired solution set."
    },
    {
        "Problem": "A circle is divided into 432 congruent arcs by 432 points.  The points are colored in four colors such that some 108 points are colored Red, some 108 points are colored Green, some 108 points are colored Blue, and the remaining 108 points are colored Yellow.  Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.",
        "Solution": "If you rotate the red points 431 times, they will overlap with blue points $108\\times 108$ times, for an average of $\\frac{108\\times 108}{431}$ per rotation. Note that this average is slightly greater than 27. Therefore at some point 28 red points overlap with blue points. In other words, there exist 28 red and blue points such that the convex 28-gons formed by them are congruent.\nRotate these 28 red points 431 times. They will overlap with green points $108\\times 28$ times, for an average of $\\frac{108\\times 28}{431}$ per rotation. This average is slightly greater than 7, so at some point 8 of those red points overlap with green points. In other words, there exist 8 red points, 8 blue points, and 8 green points such that the convex octagons formed by them are congruent.\nRotate these 8 red points 431 times. They will overlap with yellow points $108\\times 8$ times, for an average of $\\frac{108\\cdot 8}{431}$ per rotation. This average is slightly greater than 2, so at some point 3 of those red points will overlap with yellow points. In other words, there exist 3 red points, 3 blue points, 3 green points, and 3 yellow points such that the triangles formed by them are congruent."
    },
    {
        "Problem": "Determine which integers $n > 1$ have the property that there exists an infinite sequence $a_1$, $a_2$, $a_3$, $\\dots$ of nonzero integers such that the equality\n\nholds for every positive integer $k$.",
        "Solution_that_involves_a_non-elementary_result": "(Since Bertrand's is well known and provable using elementary techniques, I see nothing wrong with this-tigershark22)\nFor $n=2$, $|a_1| = 2 |a_2| = \\cdots = 2^m |a_{2^m}|$ implies that for any positive integer $m$, $|a_1| \\ge 2^m$, which is impossible.\nWe proceed to prove that the infinite sequence exists for all $n\\ge 3$.\nFirst, one notices that if we have $a_{xy} = a_x a_y$ for any integers $x$ and $y$, then it is suffice to define all $a_x$ for $x$ prime, and one only needs to verify the equation (*)\n\nfor the other equations will be automatically true.\nTo proceed with the construction, I need the following fact: for any positive integer $m>2$, there exists a prime $p$ such that $\\frac{m}{2} <p \\le m$.\nTo prove this, I am going to use Bertrand's Postulate ([1]) without proof. The Theorem states that, for any integer $n>1$, there exists a prime $p$ such that $n<p\\le 2n-1$. In other words, for any positive integer $m>2$, if $m=2n$ with $n>1$, then there exists a prime $p$ such that $\\frac{m}{2} < p <  m$, and if $m=2n-1$ with $n>1$, then there exists a prime $p$ such that $\\frac{m+1}{2} <p\\le m$, both of which guarantees that for any integer $m>2$, there exists a prime $p$ such that $\\frac{m}{2} <p \\le m$.\n\nGo back to the problem. Suppose $n\\ge 3$. Let the largest two primes not larger than $n$ are $P$ and $Q$, and that $n\\ge P > Q$. By the fact stated above, one can conclude that $2P > n$, and that $4Q = 2(2Q) \\ge 2P > n$. Let's construct $a_n$:\nLet $a_1=1$. There will be three cases: (i) $Q>\\frac{n}{2}$,  (ii) $\\frac{n}{2} \\ge Q > \\frac{n}{3}$, and (iii) $\\frac{n}{3} \\ge Q > \\frac{n}{4}$.\nCase (i): $2Q>n$. Let $a_x = 1$ for all prime numbers $x<Q$, and $a_{xy}=a_xa_y$, then (*) becomes:\n\nCase (ii): $2Q\\le n$ but $3Q > n$. In this case, let $a_2=-1$, and $a_x = 1$ for all prime numbers $2<x<Q$, and $a_{xy}=a_xa_y$, then (*) becomes:\n\nor\n\nCase (iii): $3Q\\le n$. In this case, let $a_2=3$, $a_3=-2$, and $a_x = 1$ for all prime numbers $3<x<Q$, and $a_{xy}=a_xa_y$, then (*) becomes:\n\nor\nIn each case, by Bezout's Theorem, there exists non zero integers $a_P$ and $a_Q$ which satisfy the equation. For all other primes $p > P$, just let $a_p=1$ (or any other non-zero integer).\nThis construction is correct because, for any $k> 1$,\n\nSince Bertrand's Theorem is not elementary, we still need to wait for a better proof.\n--Lightest 21:24, 2 May 2012 (EDT)"
    },
    {
        "Problem": "Find all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ (where $\\mathbb{Z}^+$ is the set of positive integers) such that $f(n!) = f(n)!$ for all positive integers $n$ and such that $m - n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$.",
        "Solution": "By the first condition we have $f(1)=f(1!)=f(1)!$ and $f(2)=f(2!)=f(2)!$, so $f(1)=1$ or $2$ and similarly for $f(2)$.  By the second condition, we have\n\nfor all positive integers $n$.\nSuppose that for some $n \\geq 2$ we have $f(n) = 1$.  We claim that $f(k)=1$ for all $k\\ge n$. Indeed, from Equation (1) we have $f(n+1)!\\equiv 1 \\mod n\\cdot n!$, and this is only possible if $f(n+1)=1$; the claim follows by induction.\nWe now divide into cases:\nCase 1: $f(1)=f(2)=1$\nThis gives $f(n)=1$ always from the previous claim, which is a solution.\nCase 2: $f(1)=2, f(2)=1$\nThis implies $f(n)=1$ for all $n\\ge 2$, but this does not satisfy the initial conditions. Indeed, we would have\n\nand so $2\\mid -1$, a contradiction.\nCase 3: $f(1)=1$, $f(2)=2$\nWe claim $f(n)=n$ always by induction. The base cases are $n = 1$ and $n = 2$. Fix $k > 1$ and suppose that $f(k)=k$.  By Equation (1) we have that\n\nThis implies $f(k+1)<2k$ (otherwise $f(k+1)!\\equiv 0 \\mod k\\cdot k!$). Also we have\n\nso $f(k+1)\\equiv 1 \\mod k$. This gives the solutions $f(k+1)=1$ and $f(k+1)=k+1$. The first case is obviously impossible, so $f(k + 1) = k + 1$, as desired.  By induction, $f(n) = n$ for all $n$.  This also satisfies the requirements.\nCase 4: $f(1)=f(2)=2$\nWe claim $f(n)=2$ by a similar induction. Again if $f(k)=2$, then by (1) we have\n\nand so $f(k+1)<2k$.  Also note that\n\nand\n\nso $f(k+1)\\equiv 2 \\mod k(k-1)$. Then the only possible solution is $f(k+1)=2$. By induction, $f(n) = 2$ for all $n$, and this satisfies all requirements.\nIn summary, there are three solutions: $\\boxed{f(n)=1, f(n)=2, f(n)=n}$."
    },
    {
        "Problem": "Let $P$ be a point in the plane of triangle $ABC$, and $\\gamma$ a line passing through $P$.  Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\\gamma$ intersect lines $BC$, $AC$, $AB$, respectively.  Prove that $A'$, $B'$, $C'$ are collinear.",
        "Solution": "By the sine law on triangle $AB'P$,\n\nso\n\nSimilarly,\n\nHence,\nSince angles $\\angle AB'P$ and $\\angle CB'P$ are supplementary or equal, depending on the position of $B'$ on $AC$,\n\nSimilarly,\nBy the reflective property, $\\angle APB'$ and $\\angle BPA'$ are supplementary or equal, so\n\nSimilarly,\n\nTherefore,\n\nso by Menelaus's theorem, $A'$, $B'$, and $C'$ are collinear.",
        "Solution_2.2C_Barycentric_.28Modified_by_Evan_Chen.29": "We will perform barycentric coordinates on the triangle $PCC'$, with $P=(1,0,0)$, $C'=(0,1,0)$, and $C=(0,0,1)$. Set $a = CC'$, $b = CP$, $c = C'P$ as usual. Since $A$, $B$, $C'$ are collinear, we will define $A = (p : k : q)$ and $B = (p : \\ell : q)$.\nClaim: Line $\\gamma$ is the angle bisector of $\\angle APA'$, $\\angle BPB'$, and $\\angle CPC'$. \nThis is proved by observing that since $A'P$ is the reflection of $AP$ across $\\gamma$, etc.\nThus $B'$ is the intersection of the isogonal of $B$ with respect to $\\angle P$\nwith the line $CA$; that is,\n\nAnalogously, $A'$ is the intersection of the isogonal of $A$ with respect to $\\angle P$\nwith the line $CB$; that is,\n\nThe ratio of the first to third coordinate in these two points\nis both $b^2pq : c^2k\\ell$, so it follows $A'$, $B'$, and $C'$ are collinear.\n~peppapig_"
    },
    {
        "Problem": "For integer $n \\ge 2$, let $x_1$, $x_2$, $\\dots$, $x_n$ be real numbers satisfying\n\nFor each subset $A \\subseteq \\{1, 2, \\dots, n\\}$, define\n\n(If $A$ is the empty set, then $S_A = 0$.)\nProve that for any positive number $\\lambda$, the number of sets $A$ satisfying $S_A \\ge \\lambda$ is at most $2^{n - 3}/\\lambda^2$.  For what choices of $x_1$, $x_2$, $\\dots$, $x_n$, $\\lambda$ does equality hold?",
        "Solution_1": "For convenience, let $N=\\{1,2,\\dots,n\\}$.\nNote that $2\\sum_{1\\leq i<j\\leq n} x_ix_j=\\left(\\sum_{i=1}^{n}x_i\\right)^2-\\left(\\sum_{j=1}^{n} x_i^2\\right)=-1$, so the sum of the $x_i$ taken two at a time is $-1/2$. Now consider the following sum:\n\nSince $S_A^2\\geq 0$, it follows that at most $2^{n-2}/\\lambda^2$ sets $A\\subseteq N$ have $|S_A|\\geq \\lambda$.\nNow note that $S_A+S_{N/A}=0$. It follows that at most half of the $S_A$ such that $|S_A|\\geq\\lambda$ are positive. This shows that at most $2^{n-3}/\\lambda^2$ sets $A\\subseteq N$ satisfy $S_A\\geq \\lambda$.\nNote that if equality holds, every subset $A$ of $N$ has $S_A\\in\\{-\\lambda,0,\\lambda\\}$. It immediately follows that $(x_1,x_2,\\ldots , x_n)$ is a permutation of $(\\lambda,-\\lambda,0,0,\\ldots , 0)$. Since we know that $\\sum_{i=1}^{n} x_i^2=1$, we have that $\\lambda=1/\\sqrt{2}$.",
        "Solution_2": "Let $x_i=x_1,x_2,...,x_n$ It is evident that $x_i = \\frac{(-1)^i}{\\sqrt{n}}$ for evens because of the second equation and $x_i=\\frac{(-1)^i}{\\sqrt{n-1}}$ for odds(one term will be 0 to maintain the first condition). \nWe may then try and get an expression for the maximum number of sets that satisfy this which occur when $\\lambda = \\frac{1}{\\sqrt{n}}$:\nsince it will be \n\nfor any choice of A we pick, it will have to be greater than $\\frac{1}{\\sqrt{n}}$ which means we can either pick 0 negative $x_m$ or $j-1$ negatives for j positive terms. Since we also have that there are $\\frac{n}{2}$ positive and negative terms for evens. Which then gives us:\n\nand \n\nFor odd values, let it be the same as the last even valued sequence where n is even(i.e. the same as the sequence before it but with an extra 0 in one of the spots). Then, the following is apparent: \nThus, we may say that this holds to be true for all $n \\ge 2$ since $n2^{n-3}$ grows faster than the sum. Note that equality holds when $S_A \\in \\{\\lambda,0,-\\lambda\\}$ for all i which occurs when $x_i= \\frac{1}{\\sqrt{2}},-\\frac{1}{\\sqrt{2}},0,....$ and $\\lambda = \\frac{1}{\\sqrt{2}}$ since $S_A=\\frac{1}{\\sqrt{2}}$ is the only choice for $S_A \\ge \\lambda$\n$\\blacksquare$"
    },
    {
        "Problem": "In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively.  Let $\\omega_A$, $\\omega_B$, $\\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively.  Given the fact that segment $AP$ intersects $\\omega_A$, $\\omega_B$, $\\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$.",
        "Solution_1": "\nIn this solution, all lengths and angles are directed.\nFirstly, it is easy to see by that $\\omega_A, \\omega_B, \\omega_C$ concur at a point $M$ (the Miquel point). Let $XM$ meet $\\omega_B, \\omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have  Thusly  But we claim that $\\triangle XDP \\sim \\triangle PBM$. Indeed,  and \nTherefore, $\\frac{XD}{XP} = \\frac{PB}{PM}$. Analogously we find that $\\frac{XE}{XP} = \\frac{PC}{PM}$ and we are done.\ncourtesy v_enhance, minor clarification by integralarefun",
        "Solution_2": "Diagram\nRefer to the Diagram link.\nBy Miquel's Theorem, there exists a point at which $\\omega_A, \\omega_B, \\omega_C$ intersect. We denote this point by $M.$ Now, we angle chase:\n\nIn addition, we have\n\nNow, by the Ratio Lemma, we have\n (by the Law of Sines in $\\triangle MZY$) (by the Law of Sines in $\\triangle MBC$) by the Ratio Lemma.\nThe proof is complete.",
        "Solution_3": "Use directed angles modulo $\\pi$.\nLemma. $\\angle{XRY} \\equiv \\angle{XQZ}.$\nProof.\nNow, it follows that (now not using directed angles)\n\nusing the facts that $ARY$ and $APB$, $AQZ$ and $APC$ are similar triangles, and that $\\frac{RA}{\\sin \\angle{RXA}} = \\frac{QA}{\\sin \\angle{QXA}}$ equals twice the circumradius of the circumcircle of $AQR$.",
        "Solution_4": "We will use some construction arguments to solve the problem. Let $\\angle BAC=\\alpha,$ $\\angle ABC=\\beta,$ $\\angle ACB=\\gamma,$ and let $\\angle APB=\\theta.$ We construct lines through the points $Q,$ and $R$ that intersect with $\\triangle ABC$ at the points $Q$ and $R,$ respectively, and that intersect each other at $T.$ We will construct these lines such that $\\angle CQV=\\angle ARV=\\theta.$\nNow we let the intersections of $AP$ with $RV$ and $QU$ be $Y'$ and $Z',$ respectively. This construction is as follows.\nWe know that $\\angle BRY'=180^\\circ-\\angle ARY'=180^\\circ-\\theta.$ Hence, we have,\nSince the opposite angles of quadrilateral $RY'PB$ add up to $180^\\circ,$ it must be cyclic. Similarly, we can also show that quadrilaterals $CQZ'P,$ and $AQTR$ are also cyclic.\nSince points $Y'$ and $Z'$ lie on $AP,$ we know that,\n\nand that\nHence, the points $Y'$ and $Z'$ coincide with the given points $Y$ and $Z,$ respectively.\nSince quadrilateral $AQTR$ is also cyclic, we have,\nSimilarly, since quadrilaterals $CQZ'P,$ and $AQTR$ are also cyclic, we have,\n\nand,\nSince these three angles are of $\\triangle TY'Z',$ and they are equal to corresponding angles of $\\triangle ABC,$ by AA similarity, we know that $\\triangle TY'Z'\\sim \\triangle ABC.$\nWe now consider the point $X=\\omega_c\\cap AC.$ We know that the points $A,$ $Q,$ $T,$ and $R$ are concyclic. Hence, the points $A,$ $T,$ $X,$ and $R$ must also be concyclic.\nHence, quadrilateral $AQTX$ is cyclic.\n\nSince the angles $\\angle ART$ and $\\angle AXT$ are inscribed in the same arc $\\overarc{AT},$ we have,\nConsider by this result, we can deduce that the homothety that maps $ABC$ to $TY'Z'$ will map $P$ to $X.$ Hence, we have that,\nSince $Y'=Y$ and $Z'=Z$ hence,\nas required.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "For a positive integer $n\\geq 3$ plot $n$ equally spaced points around a circle.  Label one of them $A$, and place a marker at $A$.  One may move the marker forward in a clockwise direction to either the next point or the point after that.  Hence there are a total of $2n$ distinct moves available; two from each point.  Let $a_n$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$, without repeating a move.  Prove that $a_{n-1}+a_n=2^n$ for all $n\\geq 4$.",
        "Solution_1": "We label the points in clockwise order as $1,2,3,\\ldots,n$, where point $A$ is the same as point $1$.  We start and end at point $1$, and we must cross over it, either by visiting it again, or else by making the move from point $n$ to point $2$. We interpret each of these cases in terms of tiling. In each move, we either move one or two points clockwise, so we can think of each move as a $1\\times 1$ or $1\\times 2$ tile. If the point $1$ is visited in the middle, then the first cycle around the circle can be thought of as a tiling of a $1\\times n$ board, and the second cycle around the circle can also be thought of as a $1\\times n$ board. We place this second board directly below the first board. Therefore, in this first case, we wish to find the number of tilings of two $1\\times n$ boards, and to guarantee that no move is repeated, we cannot have two tiles of the same type lying directly atop each other in the $2\\times n$ board. Suppose there are $u_n$ such tilings. It can easily be computed that $u_3=2$ and $u_4=6$ as shown below.\n\nIn the second case, where we pass by point $1$ by moving from point $n$ to point $2$, we can similarly think about it in terms of tiling two rows of $1\\times n$ boards, but we remove the last square in the first row and the first square in the second row to make sure that we jump from point $n$ to point $2$. Suppose that we can tile such boards in $s_n$ ways. It can be easily computed that $s_3=3$ and $s_4=5$ as shown below.\n\nSince these are the only two possible cases, we see that\nFor the sake of convenience in determining recurrence relations, we define another type of board with two $1\\times n$ boards where a specific corner is removed (without loss of generality, we place this in the lower left hand corner). Let $t_n$ be the number of ways to tile such a board without placeing two of the same type of tile atop each other. Once again, we compute that $t_3=3$ and $t_4=5$ as shown below.\n\nWe can determine reccurence relations for $s_n$, $t_n$, and $u_n$ in terms of each other. For $s_n$, note that a tiling can end in one of the three following ways such that the rest of the board can be tiled without restriction (the placed tiles are shaded).\n\nIn the first, second, and third cases, we see that we can tile the rest of the board in $t_{n-1}$, $t_{n-2}$, and $s_{n-2}$ ways, respectively. Hence for $n\\ge 5$, we see that\n\nFor $t_n$, note that a tiling can end in one of the three following ways such that the rest of the board can be tiled without restriction.\n\nIn the first, second, and third cases, we see that we can tile the rest of the board in $s_{n-1}$, $s_{n-2}$, and $t_{n-2}$ ways, respectively. Hence for $n\\ge 5$, we see that\n\nFor $u_n$, note that a tiling can end in one of the two following ways such that the rest of the board can be tiled without restriction.\nIn either case, we are left with a board with a corner removed, hence we can tile the rest of the board in $t_{n-1}$  ways in each case. Hence for $n\\ge 4$, we see that\n\nSubtracting (3) from (2), we find that\n\nTherefore, if $s_{n-1}=t_{n-1}$, then $s_n=t_n$. Since $s_3=t_3$ and $s_4=t_4$, we see that $s_n=t_n$ for all $n\\ge 3$. Therefore, (2) can be rewritten as\n\nand (4) can be rewritten as\n\nNow by (1), we know that\n\nIn particular, $a_4+a_3=6+5+2+3=2^4$, so the statement is true for $n=4$. Then by (7), and then substituting (5) and (6) (where these are valid for $n\\ge 5$), we find\n\nBut then by (5), and then substituting $4s_{n-3}=2u_{n-2}$ and $2s_{n-2}=u_{n-1}$ (by (6), and these are valid for $n\\ge 6$), we find\n\nWe substitute this into (8), finding that for $n\\ge 6$,\n\nTherefore, if $a_{n-1}+a_{n-2}=2^{n-1}$, then $a_{n}+a_{n-1}=2^n$. We already know that $a_4+a_3=2^4$. Also, we can compute that \n\nHence $a_5=s_5+u_5=21$. So as $a_4=s_4+u_4=11$, we find that $a_5+a_4=2^5$. Then we can use (9) for $n\\ge 6$ to find by induction that $a_{n}+a_{n-1}=2^n$ for all $n\\ge 4$.",
        "Solution_2": "We choose the tilings for the first and second pass around the circle simultaneously. As we tile both passes, our state is defined by whether or not we are in a continuation of a length 2 block on the first and second pass. We can't be in a continuation of a length 2 block on both passes, since that would mean we used the same move on the same block on both passes. Thus we have three states: (non-continuation, non-continuation), (non-continuation, continuation), and (continuation, non-continuation), where the elements of each tuple refer to the first and second pass respectively. Listing out the state transitions (while following the rule that we can't use the same move on both passes, and that a continuation must become a non-continuation), we get the transition matrix $T = \\begin{bmatrix} 0 & 1 & 1 \\\\ 1 & 0 & 1 \\\\ 1 & 1 & 0 \\end{bmatrix} = J-I$, where $J$ is the ones matrix, and $I$ is the identity. We have that $a_n = \\left(T^n\\right)_{11} + \\left(T^n\\right)_{23}$. The first term represents the case where we end the first pass at point $A$, and the second term represents the case where we jump over point $A$ when finishing the first pass. Thus it simply remains to compute $T^n$. We have\n\\begin{align*}\n&\\mathrel{\\phantom{=}} T^n \\\\\n&= (J-I)^n \\\\\n&= \\sum_{k=0}^n J^k (-1)^{n-k} {n \\choose k} \\\\\n&= J \\sum_{k=1}^n 3^{k-1} (-1)^{n-k} {n \\choose k} + (-1)^n I \\\\\n&= \\frac{1}{3}J\\left(2^n-(-1)^n\\right) + (-1)^n I\n\\end{align*}\nThus $a_n = \\left(T^n\\right)_{11} + \\left(T^n\\right)_{23} = \\frac{2}{3}\\left(2^n-(-1)^n\\right) + (-1)^n$. Thus, $a_{n-1}+a_n = 2^n$, as desired.",
        "Solution_3": "We split this into two cases. \nCase 1: The marker lands on $A$ thrice. \nThis implies that $A$ is touched after one full cycle. Place any nodes not touched in the first cycle in the set $K$. \nLemma 1: Any unique configuration of $K$ such that [Work In Progress]\n~SigmaPiE\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Let $n$ be a positive integer.  There are $\\tfrac{n(n+1)}{2}$ marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing $n$ marks.  Initially, each mark has the black side up.  An operation is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line.  A configuration is called admissible if it can be obtained from the initial configuration by performing a finite number of operations.  For each admissible configuration $C$, let $f(C)$ denote the smallest number of operations required to obtain $C$ from the initial configuration.  Find the maximum value of $f(C)$, where $C$ varies over all admissible configurations.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Given positive integers $m$ and $n$, prove that there is a positive integer $c$ such that the numbers $cm$ and $cn$ have the same number of occurrences of each non-zero digit when written in base ten.",
        "Solution_1": "This solution is adopted from the official solution. Both the problem and the solution were suggested by Richard Stong.\nWithout Loss of Generality, suppose $m \\geq n \\geq 1$. By prime factorization of $n$, we can find a positive integer $c_1$ such that $c_1n=10^s n_1$ where $n_1$ is relatively prime to $10$. If a positive $k$ is larger than $s$, then $(10^k c_1 m - c_1 n)= 10^s t$, where $t=10^{k-s} c_1m-n_1>0$ is always relatively prime to $10$.\nChoose a $k$ large enough so that $t$ is larger than $c_1m$. We can find an integer $b\\geq 1$ such that $10^b-1$ is divisible by $t$, and also larger than $10c_1m$. For example, let $b=\\varphi(t)$ and use Euler's theorem. Now, let $c_2=(10^b-1)/t$, and $c=c_1c_2$. We claim that $c$ is the desired number.\nIndeed, since both $c_1m$ and $n_1$ are less than $t$, we see that the decimal expansion of both the fraction $(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)$ and $n_1/t=(c_2n_1)/(10^b-1)$ are repeated in $b$-digit. And we also see that $10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)$, therefore the two repeated $b$-digit expansions are cyclic shift of one another.\nThis proves that $cm$ and $c_2n_1$ have the same number of occurrences of non-zero digits. Furthermore, $cn = c_2c_1n=10^s c_2n_1$ also have the same number of occurrences of non-zero digits with $c_2n_1$.",
        "Solution_2": "This is a rephrasing of the above solution.\nIt is enough to solve the problem when $m,n$ are replaced by $km,kn$ for any positive integer $k$. In particular, by taking $k=2^a5^b$ for appropriate values of $a,b$, we may assume $n=10^sn_1$ where $n_1$ is relatively prime to 10.\nFurthermore, adding or removing trailing zeros from $m$ and $n$ doesn't affect the claim, so we may further assume $\\gcd(n,10)=1$ and that $m$ has a xillion trailing zeros (enough to make $m$ way bigger than $n$, and also so that $m$ has at least one trailing zero).\nFor clarity of exposition, we will also multiply $m,n$ by a small number to make the units digit of $n$ be $1$ (though this is not necessary for the solution to work).\nThe point is that, for any positive integer $X$, most nonzero digits appear the same number of times in $X$ and $X+999\\dots999$ if there are enough $9$s; In particular, if the units digits of $X$ is $1$, then all nonzero digits appear the same number of times as long as there are at least as many $9$s as digits in $X$.\nSo we will pick $c$ to satisfy:\nBecause we made the units of $n$ to be $1$, the second condition is equivalent to making the units digit of $c$ to be $1$.\nThe first condition is equivalent to $(m-n)c=999\\dots999$. Because $m$ has at least one trailing 0, the units digit of $m-n$ is 9, so $\\gcd(m-n,10)=1$ and there is some $c$ so that $(m-n)c=999\\dots999$, and the units digit of $c$ must be $1$ which agrees with the other condition.\nFinally, as $m\\gg n, (m-n)c\\approx mc\\gg nc$ so it is possible to make the number of $9$s more than the number of digits in $nc$.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Let $ABC$ be a triangle.  Find all points $P$ on segment $BC$ satisfying the following property:  If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then",
        "Solution": "Let circle $PAB$ (i.e. the circumcircle of $PAB$), $PAC$ be $\\omega_1, \\omega_2$ with radii $r_1$, $r_2$ and centers $O_1, O_2$, respectively, and $d$ be the distance between their centers.\nLemma. $XY = \\frac{r_1 + r_2}{d} \\sqrt{d^2 - (r_1 - r_2)^2}.$\nProof. Let the external tangent containing $X$ meet $\\omega_1$ at $X_1$ and $\\omega_2$ at $X_2$, and let the external tangent containing $Y$ meet $\\omega_1$ at $Y_1$ and $\\omega_2$ at $Y_2$. Then clearly $X_1 Y_1$ and $X_2 Y_2$ are parallel (for they are both perpendicular $O_1 O_2$), and so $X_1 Y_1 Y_2 X_2$ is a trapezoid.\nNow, $X_1 X^2 = XA \\cdot XP = X_2 X^2$ by Power of a Point, and so $X$ is the midpoint of $X_1 X_2$. Similarly, $Y$ is the midpoint of $Y_1 Y_2$. Hence, $XY = \\frac{1}{2} (X_1 Y_1 + X_2 Y_2).$ Let $X_1 Y_1$, $X_2 Y_2$ meet $O_1 O_2$ s at $Z_1, Z_2$, respectively. Then by similar triangles and the Pythagorean Theorem we deduce that $X_1 Z_1 = \\frac{r_1 \\sqrt{d^2 - (r_1 - r_2)^2}}{d}$ and $\\frac{r_2 \\sqrt{d^2 - (r_1 - r_2)^2}}{d}$. But it is clear that $Z_1$, $Z_2$ is the midpoint of $X_1 Y_1$, $X_2 Y_2$, respectively, so $XY = \\frac{(r_1 + r_2)}{d} \\sqrt{d^2 - (r_1 - r_2)^2},$ as desired.\nLemma 2. Triangles $O_1 A O_2$ and $BAC$ are similar.\nProof. $\\angle{AO_1 O_2} = \\frac{\\angle{PO_1 A}}{2} = \\angle{ABC}$ and similarly $\\angle{AO_2 O_1} = \\angle{ACB}$, so the triangles are similar by AA Similarity.\nAlso, let $O_1 O_2$ intersect $AP$ at $Z$. Then obviously $Z$ is the midpoint of $AP$ and $AZ$ is an altitude of triangle $A O_1 O_2$.Thus, we can simplify our expression of $XY$:\n\nwhere $h_a$ is the length of the altitude from $A$ in triangle $ABC$. Hence, substituting into our condition and using $AB = c, BC = a, CA = b$ gives\n\nUsing $2 a h_a = 4[ABC] = \\sqrt{(a + b + c)(a + b - c)(a - b + c)(-a + b + c)}$ by Heron's Formula (where $[ABC]$ is the area of triangle $ABC$, our condition becomes\n\nwhich by $(a + b + c)(-a + b + c) = (b + c)^2 - a^2$ becomes\n\nLet $PB = x$; then $PC = a - x$. The quadratic in $x$ is\n\nwhich factors as\n\nHence, $PB = \\frac{ab}{b+c}$ or $\\frac{ac}{b+c}$, and so the $P$ corresponding to these lengths are our answer.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Let $a,b,c,d$ be real numbers such that $b-d \\ge 5$ and all zeros $x_1, x_2, x_3,$ and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take.",
        "Solution": "Using the hint we turn the equation into $\\prod_{k=1} ^4 (x_k-i)(x_k+i) \\implies P(i)P(-i) \\implies (b-d-1)^2 + (a-c)^2 \\implies \\boxed{16}$. This minimum is achieved when all the $x_i$ are equal to $1$."
    },
    {
        "Problem": "Let $\\mathbb{Z}$ be the set of integers. Find all functions $f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that  for all $x, y \\in \\mathbb{Z}$ with $x \\neq 0$.",
        "Solution": "Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.\nLemma 1: $f(0) = 0$.\nProof: Assume the opposite for a contradiction. Plug in $x = 2f(0)$ (because we assumed that $f(0) \\neq 0$), $y = 0$. What you get eventually reduces to:\n\nwhich is a contradiction since the LHS is divisible by 2 but not 4.\nThen plug in $y = 0$ into the original equation and simplify by Lemma 1. We get:\n\nThen:\n\nTherefore, $f(x)$ must be 0 or $x^2$.\nNow either $f(x)$ is $x^2$ for all $x$ or there exists $a \\neq 0$ such that $f(a)=0$. The first case gives a valid solution. In the second case, we let $y = a$ in the original equation and simplify to get:\n\nBut we know that $xf(-x) = \\frac{f(x)^2}{x}$, so:\n\nSince $a$ is not 0, $f(2x)$ is 0 for all $x$ (including 0). Now either $f(x)$ is 0 for all $x$, or there exists some $m \\neq 0$ such that $f(m) = m^2$. Then $m$ must be odd. We can let $x = 2k$ in the original equation, and since $f(2x)$ is 0 for all $x$, stuff cancels and we get:\n\nfor $\\mathbf{k \\neq 0}$.\nNow, let $y = m$ and we get:\n\nNow, either both sides are 0 or both are equal to $m^6$. If both are $m^6$ then:\n\nwhich simplifies to:\n\nSince $k \\neq 0$ and $m$ is odd, both cases are impossible, so we must have:\n\nThen we can let $k$ be anything except 0, and get $f(x)$ is 0 for all $x \\equiv 3 \\pmod{4}$ except $-m^2$.  Also since $x^2f(-x) = f(x)^2$, we have $f(x) = 0 \\Rightarrow f(-x) = 0$, so $f(x)$ is 0 for all $x \\equiv 1 \\pmod{4}$ except $m^2$. So $f(x)$ is 0 for all $x$ except $\\pm m^2$. Since $f(m) \\neq 0$, $m = \\pm m^2$. Squaring, $m^2 = m^4$ and dividing by $m$, $m = m^3$. Since $f(m^3) = 0$, $f(m) = 0$, which is a contradiction for $m \\neq 1$. However, if we plug in $x = 1$ with $f(1) = 1$ and $y$ as an arbitrary large number with $f(y) = 0$ into the original equation, we get $0 = 1$ which is a clear contradiction, so our only solutions are $f(x) = 0$ and $f(x) = x^2$.",
        "Solution_2": "Given that the range of f consists entirely of integers, it is clear that the LHS must be an integer and $f(yf(y))$ must also be an integer, therefore $\\frac{f(x)^2}{x}$ is an integer. If $x$ divides $f(x)^2$ for all integers $x \\ne 0$, then $x$ must be a factor of $f(x)$, therefore $f(0)=0$. Now, by setting $y=0$ in the original equation, this simplifies to $xf(-x)=\\frac{f(x)^2}{x}$. Assuming $x \\ne 0$, we have $x^2f(-x)=f(x)^2$. Substituting in $-x$ for $x$ gives us $x^2f(x)=f(-x)^2$. Substituting in $\\frac{f(x)^2}{x^2}$ in for $f(-x)$ in the second equation gives us $x^2f(x)=\\frac{f(x)^4}{x^4}$, so $x^6f(x)=f(x)^4$. In particular, if $f(x) \\ne 0$, then we have $f(x)^3=x^6$, therefore $f(x)$ is equivalent to $0$ or $x^2$ for every integer $x$. Now, we shall prove that if for some integer $t \\ne 0$, if $f(t)=0$, then $f(x)=0$ for all integers $x$. If we assume $f(y)=0$ and $y \\ne 0$ in the original equation, this simplifies to $xf(-x)+y^2f(2x)=\\frac{f(x)^2}{x}$. However, since $x^2f(-x)=f(x)^2$, we can rewrite this equation as $\\frac{f(x)^2}{x}+y^2f(2x)=\\frac{f(x)^2}{x}$, $y^2f(2x)$ must therefore be equivalent to $0$. Since, by our initial assumption, $y \\ne 0$, this means that $f(2x)=0$, so, if for some integer $y \\ne 0$, $f(y)=0$, then $f(x)=0$ for all integers $x$. The contrapositive must also be true, i.e. If $f(x) \\ne 0$ for all integers $x$, then there is no integral value of $y \\ne 0$ such that $f(y)=0$, therefore $f(x)$ must be equivalent for $x^2$ for every integer $x$, including $0$, since $f(0)=0$. Thus, $f(x)=0, x^2$ are the only possible solutions.",
        "Solution_3": "Let's assume $f(0)\\neq 0.$ Substitute $(x,y)=(2f(0),0)$ to get\nThis means that $2(2f(0)-1)$ is a perfect square. However, this is impossible, as it is equivalent to $2\\pmod{4}.$ Therefore, $f(0)=0.$ Now substitute $x\\neq 0, y=0$ to get \nSimilarly, \nFrom these two equations, we can find either $f(x)=f(-x)=0,$ or $f(x)=f(-x)=x^2.$ Both of these are valid solutions on their own, so let's see if there are any solutions combining the two.\nLet's say we can find $f(x)=x^2, f(y)=0,$ and $x,y\\neq 0.$ Then \n (NEEDS FIXING: $f(x)^2/x= x^4/x = x^3$, so the RHS is $0$ instead of $x-x^3$.)\nIf $f(2x)=4x^2,$ then $y^2=\\frac{x-x^3}{4x^2}=\\frac{1-x^2}{4x},$ which is only possible when $y=0.$ This contradicts our assumption. Therefore, $f(2x)=0.$ This forces $x=\\pm 1$ due to the right side of the equation. Let's consider the possibility $f(2)=0, f(1)=1.$ Substituting $(x,y)=(2,1)$ into the original equation yields   which is impossible. So $f(2)=f(-2)=4$ and there are no solutions \"combining\" $f(x)=x^2$ and $f(x)=0.$\nTherefore our only solutions are $\\boxed{f(x)=0}$ and $\\boxed{f(x)=x^2.}$",
        "Solution_4": "Let the given assertion be $P(x, y)$. We try $P(x, 0)$ and get $xf(2c-x)=f(x)^2/x+c$, where $f(0)=c$. We plug in $x=c$ and get $cf(c)=f(c)^2/c+c$. Rearranging and solving for $c^2$ gives us $c^2=\\frac{f(c)^2}{f(c)-1}$. Obviously, the only $c$ that works such that the RHS is an integer is $c=0$, and thus $f(0)=0$.\nWe use this information on assertion $P(x,0)$ and obtain $xf(-x)=f(x^2)/x$, or $f(-x)=\\frac{f(x)^2}{x^2}$. Thus, $f(x)$ is an even function. It follows that $f(x)=0, x^2$ for each $x$. We now prove that $f(x)=x^2$, f(x)=0$ are the only solutions. [in progress] ~SigmaPiE"
    },
    {
        "Problem": "Prove that there exists an infinite set of points  in the plane with the following property: For any three distinct integers $a,b,$ and $c$, points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a+b+c=2014$.",
        "Solution_.28Group_Theory.29": "Consider an elliptic curve with a generator $g$, such that $g$ is not a root of $0$. By repeatedly adding $g$ to itself under the standard group operation, with can build $g, 2g, 3g, \\ldots$ as well as $-g, -2g, -3g, \\ldots$. If we let  then we can observe that collinearity between $P_a$, $P_b$, and $P_c$ occurs only if $P_a + P_b + P_c = 0$ (by definition of the group operation), which is equivalent to $(3a-2014)g + (3b-2014)g + (3c-2014)g = (3a+3b+3c-3*2014)g = 0$, or $3a + 3b + 3c = 3*2014$, or $a + b + c = 2014$. We know that all these points $P_k$ exist because $3k-2014$ is never 0 for integer $k$, so that none of these points need to be point at infinity (the identity element of the group).",
        "Solution_2_.28Function_Theory.29": "Consider letting $P_x$ be the point $(x, f(x))$, where $f(x) = x^3 - 2014x^2$. Then if three points $P_a, P_b, P_c$ are on the same line $y = mx + p$, they must be the solutions to the equation $x^3 - 2014x^2 = mx + p$ (i.e. the intersection of $f$ and the line). By Vieta's Formulas, the sum of the roots to this equation is 2014. Because each value of x corresponds to a different one of a, b, and c by definition of $P_x$, $a + b + c = 2014$. Conversely, if $a + b + c = 2014$, they must be the solutions to $(x-a)(x-b)(x-c) = x^3 - 2014x^2 - mx - p = 0$ for some real $m$ and $p$. Clearly, then, $P_a, P_b, P_c$ must all lie on the line $y = mx + p$. Hence, our setting $P_x = f(x)$ produces a valid infinite set of points.\nNote: We could have let $f(x) = ax^3 - 2014ax^2 + bx + c$, where a, b, and c are arbitrary constants. (a is nonzero.)"
    },
    {
        "Problem": "Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.",
        "Solution": "We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\\boxed{6}$.\nIt is easy to verify that player A can create a 5 in a row.\nLet A place two counters anywhere, and B take off one of them. Then, A should create a \"triangle of hexagons\" by placing two adjacent counters also next to the unremoved one, so that no matter what B does there will be a line of two counters on adjacent hexagons.\nNote: A diagram and more conciseness is required in the following solution. Better maneuvers are always appreciated.\nThen, it is advisable for A to make a 4-in-a-row by placing 2 counters at the end of the line. B to prevent a 5-in-a-row must counter by removing a middle counter (a counter not on the edge of the 4-in-a-row). Player A then could reinstate the threat by placing a counter there and one adjacent to that hexagon (but not adjacent to the other middle counter). It is not in B's best interest to then remove the other middle counter, for then A can \"add\" a counter to both hexagons (adjacent to the removed counter) on the same side of the originally placed hexagon while preserving the threat. Now, we have two different threats B cannot both counter. Thus, assume B does not remove the other middle counter. Eventually, by \"adding\" counters and preserving the threat player A can create a web of hexagons surrounding the chosen middle counter. As we have proven, it will be disastrous for B to then remove the other middle counter; thus, B has to remove the chosen middle counter.\nNow, player A can administer the decisive winning maneuver as follows. Let X be two adjacent counters of the web, one of which is adjacent to the initial edge counter. Let Y be the other two counters that satisfy the condition. The line X should not have more than two counters on it. Player A should place his two counters on the first two hexagons not adjacent to any counter in X (or the web), so that B is forced to remove the outermost counter A played. Then, A should place two counters adjacent to his remaining counter in a direction parallel to line Y. B has to expunge the outermost one; otherwise A can place two counters to complete a 5-in-a-row. Let A place two more counters in the same direction; then, B has to remove the middle counter played. Now, turn in a direction parallel to line X; let A place a counter on the first two spaces adjacent to counters already placed. B will remove the second counter played. A's knockout blow occurs when he reinstates the lost counter and a counter adjacent to that counter and the remaining one he played last turn, and it is easy to verify he is guaranteed a 5-in-a-row. Therefore, A can (after an elongated siege) obtain a 5-in-a-row.\nTo prove that 6-in-a-row is impossible, tile the hexagon board in an A-B-C-A-B-C-A-B-C format. Define an \"A tile\" to be a tile engraved with the letter A. To prevent player A from obtaining a 6 in a row, all player B has to do is to remove counters placed on A tiles by player A. Because no two A tiles are adjacent, player A can play at most one counter on an A tile at a time; thus, there can at most be one counter on A tiles any time of the game. Clearly, then, a 6 in a row, which requires A-B-C-A-B-C (or two counters on A tiles), is impossible, completing the proof."
    },
    {
        "Problem": "Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\\angle BAC$.  Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$.  Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.",
        "Solution_1": "Let $O_1$ be the center of $(AHPC)$, $O$ be the center of $(ABC)$. Note that $(O_1)$ is the reflection of $(O)$ across $AC$, so $AO=AO_1$. Additionally\n\nso $Y$ lies on $(O)$. Now since $XO,OO_1,XO_1$ are perpendicular to $AB,AC,$ and their bisector, $XOO_1$ is isosceles with $XO=OO_1$, and $\\angle XOO_1=180-\\angle A$. Also\n\nBut $YO=OA$ as well, and $\\angle YOX=\\angle AOO_1$, so $\\triangle OYX\\cong \\triangle OAO_1$. Thus $XY=AO_1=AO$.",
        "Solution_2": "Since $AHPC$ is a cyclic quadrilateral, $\\angle AHC = \\angle APC$. $\\angle AHC = 90^\\circ + \\angle ABC$ and $\\angle APC = 90^\\circ + \\angle AYC$, we find $\\angle ABC = \\angle AYC$. That is, $ABYC$ is a cyclic quadrilateral. Let $D$ be mid-point of $\\overline{AB}$. $O, X, D$ are collinear and $OX \\perp AB$. Let $M$ be second intersection of $AP$ with circumcircle of the triangle $ABC$. Let $YP \\cap AC = E$, $YM \\cap AB = F$. Since $M$ is mid-point of the arc $BC$, $OM\\perp BC$. Since $AYMC$ is a cyclic quadrilateral, $\\angle CYM = \\angle CAM = \\angle BAC /2$. Since $Y$ is the orthocenter of triangle $APC$, $\\angle PYC = \\angle CAP = \\angle BAC /2$. Thus, $\\angle PYM = \\angle BAC$ and $AEYF$ is a cyclic quadrilateral. So, $YF \\perp AB$ and $OX \\parallel MY$. We will prove that $XYMO$ is a parallelogram.\nhttps://wiki-images.artofproblemsolving.com//7/7b/Usamo2014-5.png (figure link)\nWe see that $YPM$ is an isosceles triangle and $YM=YP$. Also $XB=XP$ and $\\angle BXP = 2\\angle BAP = \\angle BAC = \\angle PYM$. Then, $BXP \\sim MYP$. By spiral similarity, $BPM \\sim XPY$ and $\\angle XYP = \\angle BMP = \\angle BCA$. Hence, $\\angle XYP = \\angle BCA$, $XY \\perp BC$. Since $OM \\perp BC$, we get $XYMO$ is a parallelogram. As a result, $OM = XY$.\n(Lokman GÖKÇE)"
    },
    {
        "Problem": "Prove that there is a constant $c>0$ with the following property: If $a, b, n$ are positive integers such that $\\gcd(a+i, b+j)>1$ for all $i, j\\in\\{0, 1, \\ldots n\\}$, then",
        "Solution": "The following solution is due to Gabriel Dospinescu and v_Enhance (also known as Evan Chen).\nLet $N = n+1$ and assume $N$ is (very) large. We construct an $N \\times N$ with cells $(i,j)$ where $0 \\le i, j \\le n$ and in each cell place a prime $p$ dividing $\\gcd (a+i, b+j)$.\nThe central claim is at least $50\\%$ of the primes in this table exceed $0.001n^2$. We count the maximum number of squares they could occupy:\n Here the summation runs over primes $p \\le 0.001n^2$.\nLet $r = \\pi(0.001n^2)$ denote the number of such primes. Now we apply the three bounds:  which follows by adding all the primes directly with some computation,  using the harmonic series bound, and  via Prime Number Theorem.  Hence the sum in question is certainly less than $\\tfrac 12 N^2$ for $N$ large enough, establishing the central claim.\nHence some column $a+i$ has at least one half of its primes greater than $0.001n^2$. Because this is greater than $n$ for large $n$, these primes must all be distinct, so $a+i$ exceeds their product, which is larger than  where $c$ is some constant."
    },
    {
        "Problem": "Solve in integers the equation",
        "Solution": "We first notice that both sides must be integers, so $\\frac{x+y}{3}$ must be an integer.\nWe can therefore perform the substitution $x+y = 3t$ where $t$ is an integer.\nThen:\n$(3t)^2 - xy = (t+1)^3$\n$9t^2 + x (x - 3t) = t^3 + 3t^2 + 3t + 1$\n$4x^2 - 12xt + 9t^2 = 4t^3 - 15t^2 + 12t + 4$\n$(2x - 3t)^2 = (t - 2)^2(4t + 1)$\n$4t+1$ is therefore the square of an odd integer and can be replaced with $(2n+1)^2 = 4n^2 + 4n +1$\nBy substituting using $t = n^2 + n$ we get:\n$(2x - 3n^2 - 3n)^2 = [(n^2 + n - 2)(2n+1)]^2$\n$2x - 3n^2 - 3n = \\pm (2n^3 + 3n^2 -3n -2)$\n$x = n^3 + 3n^2 - 1$ or $x = -n^3 + 3n + 1$\nUsing substitution we get the solutions: $(n^3 + 3n^2 - 1, -n^3 + 3n + 1) \\cup (-n^3 + 3n + 1, n^3 + 3n^2 - 1)$",
        "Solution_2": "Let $n = \\frac{x+y}{3}$.\nThus, $x+y = 3n$.\nWe have\n\nSubstituting $n$ for $\\frac{x+y}{3}$, we have\n\nTreating $x$ as a variable and $n$ as a constant, we have\n\nwhich turns into\n\na quadratic equation.\nBy the quadratic formula, \n\nwhich simplifies to\n\nSince we want $x$ and $y$ to be integers, we need $4(n+1)^3 - 27n^2$ to be a perfect square.\nWe can factor the aforementioned equation to be \nfor an integer $k$. \nSince $(n-2)^2$ is always a perfect square, for $(n-2)^2 (4n+1)$ to be a perfect square, $4n + 1$ has to be a perfect square as well.\nSince $4n + 1$ is odd, the square root of the aforementioned equation must be odd as well.\nThus, we have $4n + 1 = a^2$ for some odd $a$.\nThus,  in which by difference of squares it is easy to see that all the possible values for $n$ are just $n = p(p-1)$, where $p$ is a positive integer. \nThus, \nThus, the general form for  for a positive integer $p$.\n(This is an integer since $4(p(p-1)+1)^3 - 27(p(p-1))^2$ is an even perfect square (since $4(p(p-1)+1)$ is always even, as well as $27(p(p-1))^2$ being always even) as established, and $3p(p-1)$ is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by $2$ always an integer.)\nSince $y = 3n - x$, the general form for $y$ is just\n\n(This is an integer since $4(p(p-1)+1)^3 - 27(p(p-1))^2$ is an even perfect square (since $4(p(p-1)+1)$ is always even, as well as $27(p(p-1))^2$ being always even) as established, and $3p(p-1)$ is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by $2$ always an integer, which thus trivially makes  an integer.)\nfor a positive integer $p$.\nThus, our general in integers $(x, y)$ is \n$\\boxed{}$\n-fidgetboss_4000"
    },
    {
        "Problem": "Quadrilateral $APBQ$ is inscribed in circle $\\omega$ with $\\angle P = \\angle Q = 90^{\\circ}$ and $AP = AQ < BP$. Let $X$ be a variable point on segment $\\overline{PQ}$. Line $AX$ meets $\\omega$ again at $S$ (other than $A$). Point $T$ lies on arc $AQB$ of $\\omega$ such that $\\overline{XT}$ is perpendicular to $\\overline{AX}$. Let $M$ denote the midpoint of chord $\\overline{ST}$. As $X$ varies on segment $\\overline{PQ}$, show that $M$ moves along a circle.",
        "Solution_1": "We will use coordinate geometry.\nWithout loss of generality,\nlet the circle be the unit circle centered at the origin, \n,\nwhere $(1-a)^2+b^2=1$.\nLet angle $\\angle XAB=A$, which is an acute angle, $\\tan{A}=t$, then $X=(1-a,at)$.\nAngle $\\angle BOS=2A$, $S=(-\\cos(2A),\\sin(2A))$.\nLet $M=(u,v)$, then $T=(2u+\\cos(2A), 2v-\\sin(2A))$.\nThe condition $TX \\perp AX$ yields: $(2v-\\sin(2A)-at)/(2u+\\cos(2A)+a-1)=\\cot A.$     (E1)\nUse identities $(\\cos A)^2=1/(1+t^2)$,  $\\cos(2A)=2(\\cos A)^2-1= 2/(1+t^2) -1$, $\\sin(2A)=2\\sin A\\cos A=2t^2/(1+t^2)$, we obtain $2vt-at^2=2u+a$.   (E1')\nThe condition that $T$ is on the circle yields $(2u+\\cos(2A))^2+ (2v-\\sin(2A))^2=1$, namely $v\\sin(2A)-u\\cos(2A)=u^2+v^2$.   (E2)\n$M$ is the mid-point on the hypotenuse of triangle $STX$, hence $MS=MX$, yielding $(u+\\cos(2A))^2+(v-\\sin(2A))^2=(u+a-1)^2+(v-at)^2$.   (E3)\nExpand (E3), using (E2) to replace $2(v\\sin(2A)-u\\cos(2A))$ with $2(u^2+v^2)$, and using (E1') to replace $a(-2vt+at^2)$ with $-a(2u+a)$, and we obtain\n$u^2-u-a+v^2=0$, namely $(u-\\frac{1}{2})^2+v^2=a+\\frac{1}{4}$, which is a circle centered at $(\\frac{1}{2},0)$ with radius $r=\\sqrt{a+\\frac{1}{4}}$.\n",
        "Solution_2": "Let the midpoint of $AO$ be $K$. We claim that $M$ moves along a circle with radius $KP$.\nWe will show that $KM^2 = KP^2$, which implies that $KM = KP$, and as $KP$ is fixed, this implies the claim.\n$KM^2 = \\frac{AM^2+OM^2}{2}-\\frac{AO^2}{4}$ by the median formula on $\\triangle AMO$.\n$KP^2 = \\frac{AP^2+OP^2}{2}-\\frac{AO^2}{4}$ by the median formula on $\\triangle APO$.\n$KM^2-KP^2 = \\frac{1}{2}(AM^2+OM^2-AP^2-OP^2)$.\nAs $OP = OT$, $OP^2-OM^2 = MT^2$ from right triangle $OMT$. $(1)$\nBy $(1)$, $KM^2-KP^2 = \\frac{1}{2}(AM^2-MT^2-AP^2)$.\nSince $M$ is the circumcenter of $\\triangle XTS$, and $MT$ is the circumradius, the expression $AM^2-MT^2$ is the power of point $A$ with respect to $(XTS)$. However, as $AX*AS$ is also the power of point $A$ with respect to $(XTS)$, this implies that $AM^2-MT^2=AX*AS$. $(2)$\nBy $(2)$, $KM^2-KP^2 = \\frac{1}{2}(AX*AS-AP^2)$\nFinally, $\\triangle APX \\sim \\triangle ASP$ by AA similarity ($\\angle XAP = \\angle SAP$ and $\\angle APX = \\angle AQP = \\angle ASP$), so $AX*AS = AP^2$. $(3)$\nBy $(3)$, $KM^2-KP^2=0$, so $KM^2=KP^2$, as desired. $QED$",
        "Solution_3.28synthetic.29": "To begin with, we connect $\\overline{AT}$ and we construct the nine-point circle of $\\triangle AST$ centered at $N_9$.\nLemma $1$: $AX \\cdot AS = AP^2$.\nWe proceed on a directed angle chase. We get $\\measuredangle ASP = \\measuredangle AQP = \\measuredangle QPA$, so $\\triangle PAS \\sim \\triangle XAP$ and the desired result follows by side length ratios.\nLemma $2$: The locus of $N_9$ as $X$ moves along $\\overline{PQ}$ is a circle centered about $A$.\nWe add the midpoint of $\\overline{AS}$, $N$, and let the circumradius of $\\triangle AST$ be $R$. Taking the power of $A$ with respect to $(N_9)$, we get \nHence, $AN_9 = \\sqrt{\\frac{1}{4}R^2 + \\frac{1}{2}AP^2}$, which remains constant as $X$ moves.\nNext, consider the homothety of scale factor $\\frac{2}{3}$ about $O$ mapping $N_9$ to $G$. This means that the locus of $G$ is a circle as well.\nFinally, we take a homothety of scale factor $\\frac{3}{2}$ about $A$ mapping $G$ to $M$. Hence, the locus of $M$  is a circle, as desired. - Spacesam"
    },
    {
        "Problem": "Let $S = \\{1, 2, ..., n\\}$, where $n \\ge 1$. Each of the $2^n$ subsets of $S$ is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set $T \\subseteq S$, we then write $f(T)$ for the number of subsets of T that are blue.\nDetermine the number of colorings that satisfy the following condition: for any subsets $T_1$ and $T_2$ of $S$,",
        "Solution": "Define function: $C(T)=1$ if the set T is colored blue, and $C(T)=0$ if $T$ is colored red.\nDefine the $\\text{Core} =\\text{intersection of all } T \\text{ where } C(T)=1$.\nThe empty set is denoted as $\\varnothing$, $\\cap$ denotes intersection, and $\\cup$ denotes union. Let $S_n=\\{n\\}$ are one-element subsets.\nLet $m_{c_k}=\\dbinom{m}{k} = \\frac{m!}{k!(m-k)!}$ denote m choose k.\n(Case I) $f(\\varnothing)=1$. Then for distinct m and k, $f(S_m \\cup S_k)=f(S_m)f(S_k)$, meaning only if $S_m$ and $S_k$ are both blue is their union blue. Namely $C(S_m \\cup S_k)=C(S_m)C(S_k).$\nSimilarly, for distinct $m,n,k$, $f(S_m \\cup S_k \\cup Sn)=f(S_m \\cup S_k)f(S_n)$, $C(S_m \\cup S_k \\cup S_n)=C(S_m)C(S_k)C(S_n)$. This procedure of determination continues to $S$. Therefore, if $T=\\{a_1,a_2, \\cdots a_k\\}$, then $C(T)=C(S_{a_1})C(S_{a_2}) \\cdots C(S_{a_k})$. All colorings thus determined by the free colors chosen for subsets of one single elements satisfy the condition.  There are $2^n$ colorings in this case.\n(Case II.)  $f(\\varnothing)=0$.\n(Case II.1)  $\\text{Core}=\\varnothing$. Then either (II.1.1) there exist two nonintersecting subsets A and B, $C(A)=C(B)=1$, but f$(A)f(B)=0$, a contradiction, or (II.1.2) all subsets has $C(T)=0$, which is easily confirmed to satisfy the condition $f(T_1)f(T_2)=f(T_1 \\cap T_2)f(T_1 \\cup T_2)$.  There is one coloring in this case.\n(Case II.2) Core = a subset of 1 element. WLOG, $C(S_1)=1$. Then $f(S_1)=1$, and subsets containing element 1 may be colored blue. $f(S_1 \\cup S_m)f(S_1\\cup S_n)=f(S_1 \\cup S_m \\cup S_n)$, namely $C(S_1 \\cup S_m \\cup S_n)=C(S_m \\cup S_1)C(S_n \\cup S_1)$. Now S_1 functions as the $\\varnothing$ in case I, with $n-1$ elements to combine into a base of $n-1$ two-element sets, and all the other subsets are determined. There are $2^{n-1}$ colorings for each choice of core. However, there are nC1 = n such cores. Hence altogether there are $n2^{n-1}$ colorings in this case.\n(Case II.3) Core = a subset of 2 elements. WLOG, let $C(S_1 \\cup S_2)=1$. Only subsets containing the core may be colored blue. With the same reasoning as in the preceding case, there are $(nC2)2^{n-2}$ colorings.\n$\\dots$\n(Case II.n+1) Core = S. Then $C(S)=1$, with all other subsets $C(T)=0$, there is $1=\\dbinom{n}{n}2^0$\nCombining all the cases, we have $1+\\left[2^n+\\dbinom{n}{1}2^{n-1}+\\dbinom{n}{2}2^{n-2}+ \\cdots + \\dbinom{n}{n}2^0\\right]=\\boxed{1+3^n}$ colorings. sponsored by ALLEN"
    },
    {
        "Problem": "Steve is piling $m\\geq 1$ indistinguishable stones on the squares of an $n\\times n$ grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions $(i, k), (i, l), (j, k), (j, l)$ for some $1\\leq i, j, k, l\\leq n$, such that $i<j$ and $k<l$. A stone move consists of either removing one stone from each of $(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively,j or removing one stone from each of $(i, l)$ and $(j, k)$ and moving them to $(i, k)$ and $(j, l)$ respectively.\nTwo ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.\nHow many different non-equivalent ways can Steve pile the stones on the grid?",
        "Solution": "Let the number of stones in row $i$ be $r_i$ and let the number of stones in column $i$ be $c_i$. Since there are $m$ stones, we must have $\\sum_{i=1}^n r_i=\\sum_{i=1}^n c_i=m$\nLemma 1: If any $2$ pilings are equivalent, then $r_i$ and $c_i$ are the same in both pilings $\\forall i$.\nProof: We suppose the contrary. Note that $r_i$ and $c_i$ remain invariant after each move, therefore, if any of the $r_i$ or $c_i$ are different, they will remain different.\nLemma 2: Any $2$ pilings with the same $r_i$ and $c_i$ $\\forall i$ are equivalent.\nProof: Suppose piling 1 and piling 2 not the same piling. Call a stone in piling 1 wrong if the stone occupies a position such that there are more stones in that position in piling 1 than piling 2. Similarly define a wrong stone in piling 2. Let a wrong stone be at $(a, b)$ in piling 1. Since $c_b$ is the same for both pilings, we must have a wrong stone in piling 2 at column b, say at $(c, b)$, such that $c\\not = a$. Similarly, we must have a wrong stone in piling 1 at row c, say at $(c, d)$ where $d \\not = b$. Clearly, making the move $(a,b);(c,d) \\implies (c,b);(a,d)$ in piling 1 decreases the number of wrong stones in piling 1. Therefore, the number of wrong stones in piling 1 must eventually be $0$ after a sequence of moves, so piling 1 and piling 2 are equivalent.\nLemma 3: Given the sequences $g_i$ and $h_i$ such that $\\sum_{i=1}^n g_i=\\sum_{i=1}^n h_i=m$ and $g_i, h_i\\geq 0 \\forall i$, there is always a piling that satisfies $r_i=g_i$ and $c_i=h_i$ $\\forall i$.\nProof: We take the lowest $i$, $j$, such that $g_i, h_j >0$ and place a stone at $(i, j)$, then we subtract $g_i$ and $h_j$ by $1$ each, until $g_i$ and $h_i$ become $0$ $\\forall i$, which will happen when $m$ stones are placed, because $\\sum_{i=1}^n g_i$ and $\\sum_{i=1}^n h_i$ are both initially $m$ and decrease by $1$ after each stone is placed. Note that in this process $r_i+g_i$ and $c_i+h_i$ remains invariant, thus, the final piling satisfies the conditions above.\nBy the above lemmas, the number of ways to pile is simply the number of ways to choose the sequences $r_i$ and $c_i$ such that $\\sum_{i=1}^n r_i=\\sum_{i=1}^n c_i=m$ and $r_i, c_i \\geq 0 \\forall i$. By stars and bars, the number of ways is $\\binom{n+m-1}{m}^{2}$.\nSolution by Shaddoll"
    },
    {
        "Problem": "Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.",
        "Solution_1": "Note:  This solution is definitely not what the folks at MAA intended, but it works!\nLook at the statement $a^4+b^4=e^5$.  This can be viewed as a special case of Beal's Conjecture (http://en.wikipedia.org/wiki/Beal%27s_conjecture), stating that the equation $A^x+B^y=C^z$ has no solutions over positive integers for $gcd(a, b, c) = 1$ and $x, y, z > 2$. This special case was proven in 2009 by Michael Bennet, Jordan Ellenberg, and Nathan Ng, as $(x, y, z) = (2, 4, n)$.  This case $a^4+b^4=e^5$ is obviously contained under that special case, so $a$ and $b$ must have a common factor greater than $1$.\nCall the greatest common factor of $a$ and $b$ $f$.  Then $a = f \\cdot a_1$ for some $a_1$ and likewise $b = f \\cdot b_1$ for some $b_1$.  Then consider the quantity $ac+bd$.\n$ac+bd = f \\cdot a_1 \\cdot c + f \\cdot b_1 \\cdot d = f\\cdot(a_1 \\cdot c + b_1 \\cdot d)$.\nBecause $c$ and $d$ are both positive, $(a_1 \\cdot c + b_1 \\cdot d) > 1$, and by definition $f > 1$, so $ac+bd$ is composite.\n~BealsConjecture",
        "Solution_2": "A more conventional approach, using proof by contradiction:\nWithout loss of generality, assume $a>d$\nSince $a^4 +b^4=c^4+d^4$, It is obvious that $b<c$\nWe construct the equation $(a^4 +b^4)c^2d^2-(c^4+d^4)a^2b^2=(a^2c^2-b^2d^2)(a^2d^2-b^2c^2)$ (the right side is the factorization of the left) Which factors into: $e^5(cd-ab)(cd+ab)=(ac-bd)(ac+bd)(ad-bc)(ad+bc)$ (by using $a^4 +b^4=c^4+d^4=e^5$ and factoring)\nIf $ac-bd$ or $ad-bc$ equal zero, then $a:b$ equals $c:d$ or $d:c$ respectively, which is impossible because $a^4 +b^4=c^4+d^4$ and because $a,b,c,d,e$ are distinct. Therefore the right side of the equation above is non-zero and the left side must be divisible by $ac+bd$\nIf $ac+bd$ were prime,\nWe have $ac+bd-(cd+ab)=(a-d)(c-b)>0$\nWhich means that  $ac+bd>cd+ab>cd-ab$ and neither $cd+ab$ nor $cd-ab$ can be multiples of $ac+bd$ meaning $e$ must be a multiple of $ac+bd$ and $e=k(ac+bd)$ for some integer $k$\nBut clearly this is impossible since $(k(ac+bd))^5>a^4+b^4$.\nTherefore, by contradiction, $ac+bd$ is composite."
    },
    {
        "Problem_6": "Consider $0<\\lambda<1$, and let $A$ be a multiset of positive integers. Let $A_n=\\{a\\in A: a\\leq n\\}$. Assume that for every $n\\in\\mathbb{N}$, the set $A_n$ contains at most $n\\lambda$numbers. Show that there are infinitely many $n\\in\\mathbb{N}$ for which the sum of the elements in $A_n$ is at most $\\frac{n(n+1)}{2}\\lambda$. (A multiset is a set-like collection of elements in which order is ignored, but repetition of elements is allowed and multiplicity of elements is significant. For example, multisets $\\{1, 2, 3\\}$ and $\\{2, 1, 3\\}$ are equivalent, but $\\{1, 1, 2, 3\\}$ and $\\{1, 2, 3\\}$ differ.)",
        "Solution": "Proposed by mengmeng142857.\nWe prove this by contradiction. Suppose that the number of times that an integer $a$ appears in $A$ is $k_a$. Let the sum of the elements in $A_n$ be $S_n=\\sum_{i=1}^{n}i\\cdot k_i$. If there are only finitely many $n\\in\\mathbb{N}$ such that $S_n\\leq\\frac{n(n+1)}{2}\\lambda$, then by the Well Ordering Principle, there must be a largest $m\\in\\mathbb{N}$ such that $S_m\\leq\\frac{m(m+1)}{2}\\lambda$, and for all $n>m$, $S_n>\\frac{n(n+1)}{2}\\lambda$.\nNow, observe that for some $n>m$,\n\nAlso, \n$\\frac{S_{n}}{n}>\\frac{n+1}{2}\\lambda$, \n$\\frac{S_{n-1}}{n(n-1)}>\\frac{1}{2}\\lambda$, \n$\\frac{S_{n-2}}{(n-1)(n-2)}>\\frac{1}{2}\\lambda$, \n$\\hdots$\n$\\frac{S_{m+1}}{(m+2)(m+1)}>\\frac{1}{2}\\lambda$, \n$\\frac{S_{m}}{(m+1)m}\\leq \\frac{1}{2}\\lambda$.\nNow, for any $n\\in\\mathbb{N}$, we let $\\frac{S_n}{(n+1)n}-\\frac{1}{2}\\lambda=d_n$, then\n\nLet $\\sum_{i=1}^{m}d_i=-c$ (where $c$ is positive, otherwise we already have a contradiction), then \n\nDividing both sides by $n$, $\\lambda-c/n<\\frac{\\sum_{i=1}^{n}k_i}{n}\\leq \\lambda$. Taking the limit yields\n$\\lim_{n\\to\\infty}{\\frac{\\sum_{i=1}^{n}k_i}{n}}=\\lambda$.\nObserve that this would imply that\n\nHowever, as $k_n$ is an integer, it is impossible for it to converge to $\\lambda$. This yields the desired contradiction."
    },
    {
        "Problem": "Let $X_1, X_2, \\ldots, X_{100}$ be a sequence of mutually distinct nonempty subsets of a set $S$. Any two sets $X_i$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$, that is, $X_i\\cap X_{i+1}=\\emptyset$ and $X_i\\cup X_{i+1}\\neq S$, for all $i\\in\\{1, \\ldots, 99\\}$. Find the smallest possible number of elements in $S$.",
        "Solution_1": "The answer is that $|S| \\ge 8$.\nFirst, we provide a inductive construction for $S = \\left\\{ 1, \\dots, 8 \\right\\}$. Actually, for $n \\ge 4$ we will provide a construction for $S = \\left\\{ 1, \\dots, n \\right\\}$ which has $2^{n-1} + 1$ elements in a line. (This is sufficient, since we then get $129$ for $n = 8$.) The idea is to start with the following construction for $|S| = 4$: Then inductively, we do the following procedure to move from $n$ to $n+1$: take the chain for $n$ elements, delete an element, and make two copies of the chain (which now has even length). Glue the two copies together, joined by $\\varnothing$ in between. Then place the element $n+1$ in alternating positions starting with the first (in particular, this hits $n+1$). For example, the first iteration of this construction gives: \nNow let's check $|S| \\ge 8$ is sufficient. Consider a chain on a set of size $|S| = 7$. (We need $|S| \\ge 7$ else $2^{|S|} < 100$.) Observe that there are sets of size $\\ge 4$ can only be neighbored by sets of size $\\le 2$, of which there are $\\binom 71 + \\binom 72 = 28$. So there are $\\le 30$ sets of size $\\ge 4$. Also, there are $\\binom 73 = 35$ sets of size $3$. So the total number of sets in a chain can be at most $30 + 28 + 35 = 93 < 100$.",
        "Solution_2": "My proof that $|S|\\ge 8$ is basically the same as the one above. Here is another construction for $|S| = 8$ that I like because it works with remainders and it's pretty intuitive. The basic idea is to assign different subsets to different remainders when divided by particular numbers, and then to use the Chinese Remainder Theorem to show that all of the subsets are distinct. The motivation for this comes from the fact that we want $X_i$ and $X_{i+1}$ to always be disjoint, so remainders are a great way to systematically make that happen, since $i$ and $i+1$ do not have the same remainder modulo any positive integer greater than $1.$ Anyway, here is the construction:\nLet $S = \\left\\{ {1, 2, ..., 8} \\right\\}.$ For $i = 1, 2, ..., 98,$ we will choose which elements of the set $\\left\\{ {1, 2, 3, 4} \\right\\}$ belong to $X_i$ based on the remainder of $i$ modulo $9,$ and we will choose which elements of the set $\\left\\{ {5, 6, 7, 8} \\right\\}$ belong to $X_i$ based on the remainder of $i$ modulo $11.$ We do this as follows:\n\n\nFinally, we specially define $X_{99} = \\left\\{ {1, 2, 3} \\right\\}$ and $X_{100} = \\left\\{ {5, 6, 7} \\right\\}.$\nIt is relatively easy to see that this configuration satisfies all of the desired conditions. We see that $X_{98} = \\left\\{ {4, 7} \\right\\},$ so $X_{98}$ and $X_{99}$ are disjoint, as are $X_{99}$ and $X_{100}.$ The remainder configuration above takes care of the rest, so any two consecutive sets are disjoint. Then, by the Chinese Remainder Theorem, no two integers from $1$ to $98$ have the same combination of residues modulo $9$ and modulo $11,$ so all of the sets $X_i$ are distinct for $i = 1, 2, ..., 98.$ It is also easy to verify that none of these match $X_{99}$ or $X_{100},$ since they all have at most two elements of $\\left\\{ {1, 2, 3, 4} \\right\\}$ and at most two elements of $\\left\\{ {5, 6, 7, 8} \\right\\},$ whereas $X_{99}$ and $X_{100}$ do not satisfy this; hence all of the sets are distinct. Finally, notice that, for any pair of consecutive sets, at least one of them has at most $3$ elements, while the other has at most $4.$ Thus, their union always has at most $7$ elements, so $X_i\\cup X_{i+1}\\neq S$ for all $i = 1, 2, ..., 99.$\nAll of the conditions are satisfied, so this configuration works. We thus conclude that $\\text{min}\\left(\\left|S\\right|\\right) = 8.$\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Prove that for any positive integer $k,$\n\nis an integer.",
        "Solution_1": "Define $v_p(N)$ for all rational numbers $N$ and primes $p$, where if $N=\\frac{x}{y}$, then $v_p(N)=v_p(x)-v_p(y)$, and $v_p(x)$ is the greatest power of $p$ that divides $x$ for integer $x$. Note that the expression(that we're trying to prove is an integer) is clearly rational, call it $N$.\n$v_p(N)=\\sum_{i=1}^\\infty \\left\\lfloor \\frac{k^{2}}{p^{i}} \\right\\rfloor+\\sum_{j=0}^{k-1} \\sum_{i=1}^\\infty \\left\\lfloor \\frac{j}{p^{i}}\\right\\rfloor-\\sum_{j=k}^{2k-1} \\sum_{i=1}^\\infty \\left\\lfloor \\frac{j}{p^{i}} \\right\\rfloor$, by Legendre. Clearly, $\\left\\lfloor{\\frac{x}{p}}\\right\\rfloor={\\frac{x-r(x,p)}{p}}$, and $\\sum_{i=0}^{k-1} r(i,m)\\leq \\sum_{i=k}^{2k-1} r(i,m)$, where $r(i,m)$ is the remainder function(we take out groups of $m$ which are just permutations of numbers $1$ to $m$ until there are less than $m$ left, then we have $m$ distinct values, which the minimum sum is attained at $0$ to $k-1$). Thus, $v_p(N)=\\sum_{m=p^{i}, i\\in \\mathbb{N}_{+}}-\\frac{k^{2}}{m}+\\left\\lfloor{\\frac{k^{2}}{m}}\\right\\rfloor-\\frac{\\sum_{i=0}^{k-1} r(i,m)-\\sum_{i=k}^{2k-1} r(i,m)}{m} \\geq \\sum_{m=p^{i}, i\\in \\mathbb{N}} \\left\\lceil -\\frac{k^{2}}{m}+\\lfloor{\\frac{k^{2}}{m}}\\rfloor\\right\\rceil \\geq 0$, as the term in each summand is a sum of floors also and is clearly an integer.",
        "Solution_2_.28Controversial.29": "Consider an $k\\times k$ grid, which is to be filled with the integers $1$ through $k^2$ such that the numbers in each row are in increasing order from left to right, and such that the numbers in each column are in increasing order from bottom to top. In other words, we are creating an $k\\times k$ standard Young tableaux.\nThe Hook Length Formula is the source of the controversy, as it is very powerful and trivializes this problem. The Hook Length Formula states that the number of ways to create this standard Young tableaux (call this $N$ for convenience) is:\n\nNow, we do some simple rearrangement:\n\n\nThis is exactly the expression given in the problem! Since the expression given in the problem equals the number of distinct $k\\times k$ standard Young tableaux, it must be an integer, so we are done.",
        "Solution_3_.28Induction.29": "Define  Clearly, $A(1) = 1$ and $A(2) = 2.$\nThen  Lots of terms cancel, and we are left with  The numerator has $2k+1$ consecutive positive integers, so one of them must be divisible by $(2k+1).$ Also, there are $2k$ terms left, $k$ of which are even. We can choose one of these to cancel out the $2$ in the denominator. Therefore, the ratio between $A(k+1)$ and $A(k)$ is an integer. By our inductive hypothesis, $A(k)$ is an integer. Therefore, $A(k+1)$ is as well, and we are done."
    },
    {
        "Problem": "Let $\\triangle ABC$ be an acute triangle, and let $I_B, I_C,$ and $O$ denote its $B$-excenter, $C$-excenter, and circumcenter, respectively. Points $E$ and $Y$ are selected on $\\overline{AC}$ such that $\\angle ABY = \\angle CBY$ and $\\overline{BE}\\perp\\overline{AC}.$ Similarly, points $F$ and $Z$ are selected on $\\overline{AB}$ such that $\\angle ACZ = \\angle BCZ$ and $\\overline{CF}\\perp\\overline{AB}.$\nLines $I_B F$ and $I_C E$ meet at $P.$ Prove that $\\overline{PO}$ and $\\overline{YZ}$ are perpendicular.",
        "Solution": "This problem can be proved in the following two steps.\n1. Let $I_A$ be the $A$-excenter, then $I_A,O,$ and $P$ are colinear. This can be proved by the Trigonometric Form of Ceva's Theorem for $\\triangle I_AI_BI_C.$\n2. Show that $I_AY^2-I_AZ^2=OY^2-OZ^2,$ which implies $\\overline{OI_A}\\perp\\overline{YZ}.$ This can be proved by multiple applications of the Pythagorean Thm.",
        "Solution_2": "We find point $T$ on line $YZ,$ we prove that $TY \\perp OI_A$ and state that $P$ is the point $X(24)$ from ENCYCLOPEDIA OF TRIANGLE, therefore $P \\in OI_A.$\nLet $\\omega$ be circumcircle of $\\triangle ABC$ centered at $O.$ \nLet $Y_1,$ and $Z_1$ be crosspoints of $\\omega$ and $BY,$ and $CZ,$ respectively.\nLet $T$ be crosspoint of $YZ$ and $Y_1 Z_1.$\nIn accordance the Pascal theorem for pentagon $AZ_1BCY_1,$ $AT$ is tangent to $\\omega$ at $A.$\nLet $I_A, I_B, I_C$ be $A, B,$ and $C$-excenters of $\\triangle ABC.$\nDenote \n\n\n$V$ is the foot ot perpendicular from $O$ to $AI_A.$\n$I$ is ortocenter of $\\triangle I_A I_B I_C$ and incenter of $\\triangle ABC.$\n$\\omega$ is the Nine–point circle of $\\triangle I_A I_B I_C.$\n$Y_1$ is the midpoint of $II_B, Z_1$ is the midpoint of $II_C$ in accordance with property of Nine–point circle $\\implies$\n\n\n\n\n\n\n \nIn $\\triangle ABC$ segment $YZ$ cross segment $AX \\implies \\frac {AU}{UX} = \\frac {m + nk}{k+1},$ where $n = \\frac {a}{b}, m = \\frac{a}{c}, k=\\frac {b}{c},$\n\nLet $\\triangle II_B I_C$ be the base triangle with orthocenter $I_A,$ center of Nine-points circle $O \\implies OI_A$ be the Euler line of $\\triangle II_B I_C.$\n$\\triangle ABC$ is orthic triangle of $\\triangle II_B I_C,$\n$\\triangle DEF$ is orthic-of-orthic triangle.\n$P$ is perspector of base triangle and orthic-of-orthic triangle.\nTherefore $P$ is point $X(24)$ of ENCYCLOPEDIA OF TRIANGLE CENTERS which lies on Euler line of the base triangle.\n[[1]]\nClaim \nProof\nvladimir.shelomovskii@gmail.com, vvsss"
    },
    {
        "Problem": "Find all functions $f:\\mathbb{R}\\rightarrow \\mathbb{R}$ such that for all real numbers $x$ and $y$,",
        "Solution_1": "Step 1: Set $x = y = 0$ to obtain $f(0) = 0.$\nStep 2: Set $x = 0$ to obtain $f(y)f(-y) = f(y)^2.$\n$\\indent$ In particular, if $f(y) \\ne 0$ then $f(y) = f(-y).$\n$\\indent$ In addition, replacing $y \\to -t$, it follows that $f(t) = 0 \\implies f(-t) = 0$ for all $t \\in \\mathbb{R}.$\nStep 3: Set $x = 3y$ to obtain $\\left[f(y) + 3y^2\\right]f(8y) = f(4y)^2.$\n$\\indent$ In particular, replacing $y \\to t/8$, it follows that $f(t) = 0 \\implies f(t/2) = 0$ for all $t \\in \\mathbb{R}.$\nStep 4: Set $y = -x$ to obtain $f(4x)\\left[f(x) + f(-x) - 2x^2\\right] = 0.$\n$\\indent$ In particular, if $f(x) \\ne 0$, then $f(4x) \\ne 0$ by the observation from Step 3, because $f(4x) = 0 \\implies f(2x) = 0 \\implies f(x) = 0.$ Hence, the above equation implies that $2x^2 = f(x) + f(-x) = 2f(x)$, where the last step follows from the first observation from Step 2.\n$\\indent$ Therefore, either $f(x) = 0$ or $f(x) = x^2$ for each $x.$\n$\\indent$ Looking back on the equation from Step 3, it follows that $f(y) + 3y^2 \\ne 0$ for any nonzero $y.$ Therefore, replacing $y \\to t/4$ in this equation, it follows that $f(t) = 0 \\implies f(2t) = 0.$\nStep 5: If $f(a) = f(b) = 0$, then $f(b - a) = 0.$\n$\\indent$ This follows by choosing $x, y$ such that $x - 3y = a$ and $3x - y = b.$ Then $x + y = \\tfrac{b - a}{2}$, so plugging $x, y$ into the given equation, we deduce that $f\\left(\\tfrac{b - a}{2}\\right) = 0.$ Therefore, by the third observation from Step 4, we obtain $f(b - a) = 0$, as desired.\nStep 6: If $f \\not\\equiv 0$, then $f(t) = 0 \\implies t = 0.$\n$\\indent$ Suppose by way of contradiction that there exists an nonzero $t$ with $f(t) = 0.$ Choose $x, y$ such that $f(x) \\ne 0$ and $x + y = t.$ The following three facts are crucial:\n$\\indent$ 1. $f(y) \\ne 0.$ This is because $(x + y) - y = x$, so by Step 5, $f(y) = 0 \\implies f(x) = 0$, impossible.\n$\\indent$ 2. $f(x - 3y) \\ne 0.$ This is because $(x + y) - (x - 3y) = 4y$, so by Step 5 and the observation from Step 3, $f(x - 3y) = 0 \\implies f(4y) = 0 \\implies f(2y) = 0 \\implies f(y) = 0$, impossible.\n$\\indent$ 3. $f(3x - y) \\ne 0.$ This is because by the second observation from Step 2, $f(3x - y) = 0 \\implies f(y - 3x) = 0.$ Then because $(x + y) - (y - 3x) = 4x$, Step 5 together with the observation from Step 3 yield $f(3x - y) = 0 \\implies f(4x) = 0 \\implies f(2x) = 0 \\implies f(x) = 0$, impossible.\n$\\indent$ By the second observation from Step 4, these three facts imply that $f(y) = y^2$ and $f(x - 3y) = \\left(x - 3y\\right)^2$ and $f(3x - y) = \\left(3x - y\\right)^2.$ By plugging into the given equation, it follows that\n But the above expression miraculously factors into $\\left(x + y\\right)^4$! This is clearly a contradiction, since $t = x + y \\ne 0$ by assumption. This completes Step 6.\nStep 7: By Step 6 and the second observation from Step 4, the only possible solutions are $f \\equiv 0$ and $f(x) = x^2$ for all $x \\in \\mathbb{R}.$ It's easy to check that both of these work, so we're done.",
        "Solution_2": "Step 1: $x=y=0 \\implies f(0)=0$\nStep 2: $x=0 \\implies f(y)f(-y)=f(y)^{2}$. Now, assume $y \\not = 0$. Then, if $f(y)=0$, we substitute in $-y$ to get $f(y)f(-y)=f(-y)^{2}$, or $f(y)=f(-y)=0$. Otherwise, we divide both sides by $f(y)$ to get $f(y)=f(-y)$. If $y=0$, we obviously have $f(0)=f(0)$. Thus, the function is even.\n.\nStep 3: $y=-x \\implies 2f(4x)(f(x)-x^{2})=0$. Thus, $\\forall x$, we have $f(4x)=0$ or $f(x)=x^{2}$.\nStep 4: We now assume $f(x) \\not = 0$, $x\\not = 0$. We have $f(\\frac{x}{4})=\\frac{x^{2}}{16}$. Now, setting $x=y=\\frac{x}{4}$, we have $f(\\frac{x}{2})=\\frac{x^{2}}{4}$ or $f(\\frac{x}{2})=0$. The former implies that $f(x)=0$ or $x^{2}$. The latter implies that $f(x)=0$ or $f(x)=\\frac{x^{2}}{2}$. Assume the latter. $y=-2x \\implies -\\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\\frac{x^{2}}{2}$. Clearly, this implies that $f(x)$ is negative for some $m$. Now, we have $f(\\frac{m}{4})=\\frac{m^{2}}{16} \\implies f(\\frac{m}{2})=0,\\frac{m^{2}}{4} \\implies f(m) \\geq 0$, which is a contradiction. Thus, $\\forall x$$f(x)=0$ or $f(x)=x^{2}$.\nStep 5: We now assume $f(x)=0$, $f(y)=y^{2}$ for some $x,y \\not = 0$. Let $m$ be sufficiently large integer, let $z=|4^{m}x|$ and take the absolute value of $y$(since the function is even). Choose $c$ such that $3z-c=y$. Note that we have $\\frac{c}{z}$~$3$ and $\\frac{y}{z}$~$0$. Note that $f(z)=0$. Now, $x=z, y=c \\implies$ LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to $(z+c)^{4}$~$256z^{4}$. Now if $f(z-3c)=0$, the second term of the LHS/RHS clearly ~0 as $m \\to \\infty$. if $f(z-3c)=0$, then we have LHS/RHS ~ $0$, otherwise, we have LHS/RHS~$\\frac{8^{2}\\cdot 3z^{4}}{256z^{4}}$~$\\frac{3}{4}$, a contradiction, as we're clearly not dividing by $0$, and we should have LHS/RHS=1.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$ such that $M\\in\\overline{AB},$ $Q\\in\\overline{AC},$ and $N, P\\in\\overline{BC}.$ Let $S$ be the intersection of lines $MN$ and $PQ.$ Denote by $\\ell$ the angle bisector of $\\angle MSQ.$\nProve that $\\overline{OI}$ is parallel to $\\ell,$ where $O$ is the circumcenter of triangle $ABC,$ and $I$ is the incenter of triangle $ABC.$",
        "Solution_1": "Let  $D$ be the intersection of line $AI$ and the circumcircle of $\\Delta ABC$ (other than $A$), then $OD\\perp BC$. Let $R$ be the point such that $NPQR$ is a rhombus. It follows that $OD\\perp QR$.\nSince $AM=AQ$, $AI\\perp MQ$, or $DI\\perp MQ$. It follows that $\\angle ODI=\\angle RQM$.\nSince $BO=OD$, $MA=AQ$, $\\angle BOD=\\angle MAQ$, it follows that $\\Delta BOD\\sim\\Delta MAQ$, so $AQ/MQ=OD/BD$.\nIt is given that $AQ=NP=RQ$, and by basic properties of the incenter, $ID=BD$. Therefore, $RQ/MQ=OD/ID$, so $\\Delta RQM\\sim\\Delta ODI$.\nSince the rotation between the two triangles in 90 degrees, $OI\\perp MR$. However, $l$ is parallel to the bisector of $MNR$, which is perpendicular to $MR$, so we are done.",
        "Solution_2": "Write $\\angle{JKL} = K$ for all $J,K,L$ chosen as distinct vertices of triangle $ABC$. Define $a, b, c$ as sides opposite to angles $A, B$, and $C$, respectively. Place the triangle in the Euclidean plane with $A$ at the origin and $C$ on the positive x-axis. Assume without loss of generality that C is acute.\nConsider the sides of the pentagon as vectors and note that\nDefine $\\delta$ and $\\gamma$ as the angles made between the positive x-axis and $\\overrightarrow{MN}$ and $\\overrightarrow{QP}$, respectively. Considering the x and y coordinates of the vectors in $(1)$, it follows that\nSuppose $\\sin C - \\sin A = 0$. Then $A = C$, and the triangle is isosceles. In this case, it is clear by symmetry that $\\overline{OI}$ is vertical. Further, since point $S$ exists, $\\delta \\neq \\gamma$, so $\\delta + \\gamma = 180$ and $\\overrightarrow{MN} + \\overrightarrow{QP}$ must be vertical as well.\nFor the remainder of the proof, assume $\\sin C \\neq \\sin A$. Note that  whenever $x, y \\in \\mathbb{R}$ and $\\sin x \\neq \\sin y$. Note further that the slope of the line defined by the vector formed by summing vectors $(\\cos x, \\sin x)$ and $(\\cos y, \\sin y)$ is this expression. Since $\\ell$ is parallel to $\\overrightarrow{MN} + \\overrightarrow{QP}$, the slope of $\\ell$ can be formed by dividing expressions in $(2)$ and $(3)$ and inverting the sign:\nDetermine the coordinates of $I$ by drawing perpendiculars from $I$ to the sides and vertices of the triangle. By exploiting congruence between pairs of right triangles that share a vertex, one can partition $a,b,c$ into $p+q, p+r, q+r$ where $p,q,r$ are bases of these triangles that lie on the sides of triangle $ABC$. From here it is clear that $I = \\left(\\frac{b + c – a}{2} , \\frac{b + c – a}{2}\\tan(A / 2)\\right)$.\nTo find the coordinates of $O$, note that $\\angle{OJK} + \\angle{OJL} = \\angle{J}$ and that $\\angle {OJK} = \\angle{OKJ}$ in any acute triangle $JKL$. It easily follows that $\\angle{OJK} = 90 – L$. Note also that the perpendicular from $O$ to $\\overline{JK}$ bisects $\\overline{JK}$. Hence,  if triangle $ABC$ is acute.\nIf triangle $JKL$ is obtuse at $\\angle J$, then it can be similarly shown that $\\angle{OKL} = \\angle{OLK} = J – 90$ but that the remaining angles of this form are still $90-L$ and $90-K$. It easily follows that $(5)$ holds if $\\angle A$ is obtuse. If $\\angle B$ is obtuse then $\\angle OAC = B – 90$ and the $y$ coordinate of $O$ is $-\\frac{b}{2} \\tan{B-90}$. From this, $(5)$ follows in this case as well.\nWe can conclude the slope of $\\overline {OI}$ is  by the Law of Sines and rearrangement.\nSetting $(6) = (4)$ is equivalent to\nSince $\\tan(A/2) = \\frac{\\sin A}{1 + \\cos A}$, this equation is equivalent to\nThis equation is equivalent to  which is evident.",
        "Solution_3": "Let $A', B',$ and $C'$ be the arc midpoints of $BC, CA, AB,$ respectively. Let $E$ be crosspoint of $AI$ and $B'C'.$\nTherefore $O$ is the circumcenter of triangle $A'B'C'.$\nPoints $A, I, E,$ and $A'$ are collinear.\n$\\angle A'EB' = \\frac {\\overset{\\Large\\frown} {AC'}+\\overset{\\Large\\frown} {B'C} +\\overset{\\Large\\frown} {CA'}}{2} = 90^\\circ \\implies AA' \\perp B'C'$\n$\\implies I$ is orthocenter of $\\triangle A'B'C' \\implies$\n$IO$ is the Euler line of $\\triangle A'B'C'.$\nLet $G$ be the centroid of $\\triangle A'B'C' \\implies G$ lies on line $IO$\n$\\implies \\overline {OG} = \\frac {\\overline{OA'} + \\overline{OB'} + \\overline{OC'}}{3}$ is paraller to $\\overline{OI}.$\n$\\overline{OB'} \\perp \\overline{AC} \\implies \\overline{OB'} \\perp \\overline{QA}.$\nSimilarly $\\overline{OC'} \\perp \\overline{AM}, \\overline{OA'} \\perp \\overline{NP},$ rotation from $\\overline{OA'}$ to $\\overline{NP}$, from $\\overline{OB'}$ to $\\overline{QA},$ and from $\\overline{OC'}$ to $\\overline{AM}$ is in clockwise direction, $|\\overline{QA}|=| \\overline{AM}| =|\\overline{NP}|, |\\overline{OA'}| = |\\overline{OB'}| = |\\overline{OC'}| \\implies$\n$\\overline{QA} + \\overline{AM} + \\overline{NP} = \\overline{QM} + \\overline{NP}$ is perpendicular to $|\\overline{OI}.$\n$|\\overline{MN}| = |\\overline{QP}|$ therefore in accordance with Claim $\\overline{MN} + \\overline{QP}$ is parallel to $\\overline{OI}.$\nThis sum is parallel to $\\ell,$, so we are done.\nClaim\nLet $|\\overline{BA}|= |\\overline{CD}|, \\overline{CB} + |\\overline{BA} + |\\overline{AD} = |\\overline{CD}.$\nThen $(\\overline{BA} + \\overline{CD}) \\perp (\\overline{CB} + \\overline{AD}).$\nProof\n$(\\overline{BA} + \\overline{CD}) \\cdot (\\overline{CB} + \\overline{AD}) = (\\overline{BA} + \\overline{CD}) \\cdot (\\overline{CB} + \\overline{CD} – \\overline{CB} – \\overline{BA}) = |\\overline{CD}|^2 – |\\overline{BA}|^2 = 0.$\nvladimir.shelomovskii@gmail.com, vvsss"
    },
    {
        "Problem": "Integers $n$ and $k$ are given, with $n\\ge k\\ge 2.$ You play the following game against an evil wizard.\nThe wizard has $2n$ cards; for each $i = 1, ..., n,$ there are two cards labeled $i.$ Initially, the wizard places all cards face down in a row, in unknown order.\nYou may repeatedly make moves of the following form: you point to any $k$ of the cards. The wizard then turns those cards face up. If any two of the cards match, the game is over and you win. Otherwise, you must look away, while the wizard arbitrarily permutes the $k$ chosen cards and turns them back face-down. Then, it is your turn again.\nWe say this game is $\\textit{winnable}$ if there exist some positive integer $m$ and some strategy that is guaranteed to win in at most $m$ moves, no matter how the wizard responds.\nFor which values of $n$ and $k$ is the game winnable?",
        "Solution_2": "We claim that the game is winnable if and only if $n>k$. Suppose after the first step, the cards $1$ to $n$ are shuffled around. Notice that we have $n$ cards that we don't know the position of (which are all cards from $1$ to $n$). Now, suppose we pick $p$ known cards. Note that the $p$ cards are all different(since the known cards are the cards from $1$ to $n$), and there is still a possibility that the other cards from the unknown cards complement and cause $1$ to $n$. Therefore, we are in the same state as before, and the game is unwinnable.\nNow, suppose $n>k$. Denote the ith card counting from the left. We pick cards $1$ to $k$, keeping track of the set of values of the cards. Then, we pick cards $2$ to $k+1$, adding the value of the $k+1$th card into the set of value of cards. We keep doing this, until we pick cards $2n-k$ to $2n-1$, at which point we know the exact number on the $2n$th card. Now, we go back to $1$ through $k$, and repeat this process, until we reveal the $2n-1$th card(unless we win during the process). This process terminates only when there are less or equal to $k$ cards that we don't know the exact numbers on or if we somehow win, clearly, as otherwise we're still revealing new information by picking cards from $1$ through $k$. Note that we now know the exact values on $2n-k$ of the cards. By the Pigeonhole Principle, since $k<n$, $2$ of them are the same, and we pick those $2$ cards and $k-2$ other random cards and we win.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Prove that there are infinitely many distinct pairs $(a,b)$ of relatively prime positive integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$.",
        "Solution_1": "Let $n=a+b$. Since $gcd(a,b)=1$, we know $gcd(a,n)=1$. We can rewrite the condition as\nAssume $a$ is odd. Since we need to prove an infinite number of pairs exist, it suffices to show that infinitely many pairs with $a$ odd exist.\nThen we have\nWe know by Euler's theorem that $a^{\\varphi(n)} \\equiv 1 \\mod{n}$, so if $2a-n=\\varphi(n)$ we will have the required condition.\nThis means $a=\\frac{n+\\varphi(n)}{2}$. Let $n=2p$ where $p$ is a prime, $p\\equiv 1\\mod{4}$. Then $\\varphi(n) = 2p*\\left(1-\\frac{1}{2}\\right)\\left(1-\\frac{1}{p}\\right) = p-1$, so \nNote the condition that $p\\equiv 1\\mod{4}$ guarantees that $a$ is odd, since $3p-1 \\equiv 2\\mod{4}$\nThis makes $b = \\frac{p+1}{2}$. Now we need to show that $a$ and $b$ are relatively prime. We see that\n\n\nBy the Euclidean Algorithm.\nTherefore, for all primes $p \\equiv 1\\mod{4}$, the pair $\\left(\\frac{3p-1}{2},\\frac{p+1}{2}\\right)$ satisfies the criteria, so infinitely many such pairs exist.",
        "Solution_2": "Take $a=2n-1, b=2n+1, n\\geq 2$. It is obvious (use the Euclidean Algorithm, if you like), that $\\gcd(a,b)=1$, and that $a,b>1$.\nNote that\n\n\nSo\n\nSince $a+b=4n$, all such pairs work, and we are done.",
        "Solution_3": "Let $x$ be odd where $x>1$. We have $x^2-1=(x-1)(x+1),$ so $x^2-1 \\equiv 0 \\pmod{2x+2}.$ This means that  $x^{x+2}-x^x \\equiv 0 \\pmod{2x+2},$ and since x is odd, $x^{x+2}+(-x)^x \\equiv 0 \\pmod{2x+2},$ or $x^{x+2}+(x+2)^x \\equiv 0 \\pmod{2x+2},$ as desired.",
        "Solution_4": "I claim that the ordered pair $(2^{n} - 1, 2^{n} + 1)$ satisfies the criteria for all $n \\geq 2.$ \nProof:\nIt is easy to see that the order modulo $2^{n+1}$ of $(2^{n} - 1)^k$ is $2,$ since $(2^{n} - 1)^2 = 2^{2n} - 2 \\cdot 2^{n} + 1 \\equiv 1 \\mod 2^{n+1}.$, and we can assert and prove similarly for the order modulo $2^{n+1}$ of $(2^{n} + 1)^k.$ Thus, the remainders modulo $2^{n+1}$ that the sequences of powers of $2^{n} - 1$ and $2^{n} + 1$ generate are $1$ for even powers and $2^{n} - 1$ and $2^{n} + 1$ for odd powers, respectively. Since $2^{n} + 1$ and $2^{n} - 1$ are both odd for $n \\geq 2,$  Since there are infinitely many powers of $2$ and since all ordered pairs $(2^n - 1, 2^n+1)$ contain relatively prime integers, we are done.\n$\\boxed{}$\n-fidgetboss_4000"
    },
    {
        "Problem": "Let $ABC$ be a scalene triangle with circumcircle $\\Omega$ and incenter $I.$ Ray $AI$ meets $BC$ at $D$ and $\\Omega$ again at $M;$ the circle with diameter $DM$ cuts $\\Omega$ again at $K.$ Lines $MK$ and $BC$ meet at $S,$ and $N$ is the midpoint of $IS.$ The circumcircles of $\\triangle KID$ and $\\triangle MAN$ intersect at points $L_1$ and $L.$ Prove that $\\Omega$ passes through the midpoint of either $IL_1$ or $IL.$",
        "Solution": "Let $X$ be the point on circle $\\Omega$ opposite $M$. This means $\\angle MAX = 90^\\circ, BC \\perp XM.$\n$\\angle XKM = \\angle DKM = 90^\\circ \\implies$ the points $X, D,$ and $K$ are collinear.\nLet $D' = BC \\cap XM \\implies DD' \\perp XM \\implies$\n$S$ is the orthocenter of $\\triangle DMX \\implies$ the points $X, A,$ and $S$ are collinear.\nLet $\\omega$ be the circle centered at $S$ with radius $R = \\sqrt {SK \\cdot SM}.$\nWe denote $I_\\omega$ inversion with respect to $\\omega.$\nNote that the circle $\\Omega$ has diameter $MX$ and contain points $A, B, C,$ and $K.$\n$I_\\omega (K) = M \\implies$ circle $\\Omega \\perp \\omega \\implies C = I_\\omega (B), X = I_\\omega (A).$\n$I_\\omega (K) = M \\implies$ circle $KMD \\perp \\omega \\implies D' = I_\\omega (D) \\in KMD \\implies$\n$\\angle DD'M = 90^\\circ \\implies$  the points $X, D',$ and $M$ are collinear.\nLet $F \\in AM, MF = MI.$ It is well known that $MB = MI = MC \\implies$\n$\\Theta = BICF$ is circle  centered at $M.$ $C = I_\\omega (B) \\implies \\Theta \\perp \\omega.$\nLet $I' =  I_\\omega (I ) \\implies I' \\in \\Theta \\implies \\angle II'M =  90^\\circ.$\n$I' =  I_\\omega (I ), X =  I_\\omega (A ) \\implies AII'X$ is cyclic.\n$\\angle XI'I = \\angle XAI =  90^\\circ \\implies$  the points $X, I' ,$ and $F$ are collinear.\n$I'IDD'$ is cyclic $\\implies \\angle I'D'M = \\angle I'D'C + 90^\\circ =  \\angle I'ID + 90^\\circ,$\n$\\angle XFM = \\angle I'FI = 90^\\circ – \\angle I'IF = 90^\\circ – \\angle I'ID  \\implies$\n$\\angle XFM +  \\angle I'D'M = 180^\\circ \\implies I'D'MF$ is cyclic.\nTherefore point $F$ lies on  $I_\\omega (IDK).$\n$FA \\perp SX, SI' \\perp FX \\implies I$ is orthocenter of $\\triangle FSX.$\n$N$ is midpoint $SI, M$ is midpoint $FI, I$ is orthocenter of $\\triangle FSX, A$ is root of height $FA \\implies AMN$ is the nine-point circle of $\\triangle FSX \\implies I' \\in AMN.$\nLet $N' = I_\\omega (N) \\implies R^2 = SN \\cdot SN' = SI \\cdot  SI' \\implies$\n\n$\\angle XN'I' = \\angle XSI' = 90^\\circ – \\angle AXI' = \\angle IFX \\implies N'XIF$ is cyclic.\nTherefore point $F$ lies on  $I_\\omega (AMN) \\implies I_\\omega(F) = L \\implies$\nThe points $F, L,$ and $S$ are collinear, $AXFL$ is cyclic.\nPoint $I$ is orthocenter $\\triangle FSX \\implies XI \\perp SF, \\angle ILS = \\angle SI'F = 90^\\circ$\n$\\implies$ The points $X, I, E,$ and $L$ are collinear.\n$AXFL$ is circle $\\implies AI \\cdot IF = IL \\cdot XI\\implies$\n$AI \\cdot \\frac {IF}{2} = \\frac {IL}{2} \\cdot IX \\implies AI \\cdot IM = EI \\cdot IX \\implies AEMX$ is cyclic.\n\nvladimir.shelomovskii@gmail.com, vvsss"
    },
    {
        "Problem": "Let $\\mathbf{Z}$ denote the set of all integers. Find all real numbers $c > 0$ such that there exists a labeling of the lattice points $( x, y ) \\in \\mathbf{Z}^2$ with positive integers for which: only finitely many distinct labels occur, and for each label $i$, the distance between any two points labeled $i$ is at least $c^i$.\n",
        "Solution_1_.28Evan_Chen.2C_Calvin_Deng.29": "See page 11 of this PDF:\nhttps://web.evanchen.cc/exams/USAMO-2017-notes.pdf\n",
        "Solution_2_.28INCOMPLETE.29": "For $c\\le 1,$ we can label every lattice point $1.$ For $c\\le \\sqrt[4]{2},$ we can make a \"checkerboard\" labeling, i.e. label $(x, y)$ with $1$ if $x+y$ is even and $2$ if $x+y$ is odd. One can easily verify that these labelings satisfy the required conditions. Therefore, a labeling as desired exists for all $0 < c\\le \\sqrt[4]{2}.$\nAn iterated version of the checkerboard labeling can actually work for all values $c < \\sqrt{2}.$ For convenience, define the original lattice grid to be the set of all lattice points in the coordinate plane. Define a modified lattice grid of size $x$ to be a structure similar to the lattice points on the coordinate plane, but with the minimum separation between any two points equaling $x$ (as opposed to $1$).\nOn the first step, assign a label $1$ to half of the points in a checkerboard arrangement. One can see that the points that have not yet been labeled form a modified lattice grid of size $\\sqrt{2}$ (this lattice grid is also rotated by $45^{\\circ}$ from the original lattice grid). At this point, for the second step, assign a label $2$ to half of the points, again in a checkerboard arrangement. At this point, the points that have not yet been labeled form a modified lattice grid of size $2$ (and again, it is rotated $45^{\\circ}$ from the modified lattice grid after the first step). One then continues in this fashion. For the $N^{\\text{th}}$ step, the points we are labeling are separated by at least $\\sqrt{2}\\times\\left(\\sqrt{2}\\right)^{N-1} = \\left(\\sqrt{2}\\right)^N > c^N,$ so we know that our labeling at each step is acceptable.\nAfter the $N^{\\text{th}}$ step (where $N$ is a natural number), the points that have not yet been labeled form a modified lattice grid with size $\\left(\\sqrt{2}\\right)^N.$ Since $c < \\sqrt{2},$ we will eventually have $\\left(\\sqrt{2}\\right)^N > c^{N+1}$ for some sufficiently large $N.$ At this point, we can label all remaining points in the original lattice grid $N+1,$ and this produces a labeling of all of the lattice points in the plane that satisfies all of the conditions. Therefore, a labeling as desired exists for all $c < \\sqrt{2}.$\nWe now prove that no labeling as desired exists for any $c\\ge 2.$ To do this, we will prove that labeling a $2^k$-by-$2^k$ square grid of lattice points requires at least $k+3$ distinct labels for all natural numbers $k$; hence for a sufficiently large section of the lattice plane the number of distinct labels required grows arbitrarily large, so the entire lattice plane cannot be labeled with finitely many distinct labels. We will prove this using induction.\nFor the base case, $k=1,$ we have four points in a square of side length $1.$ The maximum distance between any two of these points is $\\sqrt{2} < c^1$ for all $c\\ge 2,$ so all four points must have different labels. This completes the base case.\nNow, for the inductive step, suppose that labeling a $2^k$-by-$2^k$ square grid of lattice points requires at least $k+3$ distinct labels for some natural number $k.$ We will now prove that labeling a $2^{k+1}$-by-$2^{k+1}$ square grid of lattice points requires at least $k+4$ distinct labels.\nTake a $2^{k+1}$-by-$2^{k+1}$ square grid of lattice points. Divide this grid into four quadrants, $A, B, C,$ and $D.$ By the inductive hypothesis, $A$ requires at least $k+3$ distinct labels. At least one of these labels must be $k+3$ or greater; take one such label and call it $L.$\nThe largest distance between any two points in the entire grid is $\\sqrt{2}\\left(2^{k+1} - 1\\right) < c^{k+3}$ for all $c\\ge 2.$ Therefore, the label $L$ cannot be used anywhere else in the grid. However, $B, C,$ and $D$ each require at least $k+3$ distinct labels as well by the inductive hypothesis. Thus, they must use at least one label that is not used in $A.$ It follows that the entire grid requires at least $k+4$ distinct labels. This completes the inductive step, and thus we conclude that no labeling as desired exists for any $c\\ge 2.$\nI have heard from others that the actual boundary is $\\sqrt{2}.$ This makes intuitive sense, since the iterated checkerboard labeling outlined above just breaks down at this value (you will be able to get closer and closer to labeling all of the lattice points, but you can never get there, since you will never have $\\left(\\sqrt{2}\\right)^N > c^{N+1}$). The inductive argument above seems fairly loose, so I think that it can be sharpened to bring the upper bound down to $\\sqrt{2},$ but I am not sure yet how exactly to do so. I think the way to do it is to somehow force $2$ new labels (instead of just $1$) each time you double the side length of the square grid.",
        "Solution_3": "We claim that for all $c<\\sqrt{2}$, there exists such a configuration.\nLemma 1a: If there exists a configuration for $c=c_1$, then for any $c_2\\le{c_1}$, there also exists a configuration for $c=c_2$.\nNote that any labeling that can be done for $c=c_1$ exists for $c=c_2$. Thus, we may copy the configuration from $c=c_1$ to $c=c_2$.\nLemma 1b: If there does not exist a configuration for $c=c_1$, then for any $c_2\\ge{c_1}$, there does not exist a configuration for $c=c_2$.\nNote that any labeling which may be done for $c=c_2$ should also be available for $c=c_1$. However, we already stated that $c=c_1$ has no configuration, and so $c=c_2$ doesn't either.\nWe now prove that $c=\\sqrt{2}$ does not work. WLOG, label the origin of our lattice plan with label $1$. WLOG, we can cover every point that we can with label $1$ because not labeling any such possible point does not affect the distance between points that aren't allowed to be labelled.\nName the state of the lattice grid before using label $1$ state $S$. After covering every point possible, we are left with only lattice points that have coordinates that sum to an odd integer. Notice, however, that if we discard the covered points and rotate all the new points by $\\frac{\\pi}{4}$ radians counterclockwise, we end up with the same grid as state $S$, only all the distances are multiplied by scalar $\\sqrt{2}$. However, before placing label $2$s, note that we are in the same state $S$ as before, because our minimum distance is also multiplied by scalar $\\sqrt{2}$. Thus, we will always be in an infinite loop, stuck in state $S$, and thus it is impossible to complete the labeling.\nIt suffices to show that any $c=2^{k}$ works given $k<0.5$. Note that the two smallest distances between any two points in the starting grid are $1$ and $\\sqrt{2}$. When we are currently in state $S$ using label $I$, it is obvious that if $2ki-i\\le{1}$, we can break out of state $S$ (Our effective minimum length would be at most 1, and we can cover every remaining point). Because $k<0.5$, there does exist an $I$ where this occurs. Thus, any $c<\\sqrt{2}$ works, hence proved. $\\blacksquare{}$ ~SigmaPiE"
    },
    {
        "Problem": "Find the minimum possible value of\n\ngiven that $a,b,c,d,$ are nonnegative real numbers such that $a+b+c+d=4$.",
        "Solution": "See here: https://artofproblemsolving.com/community/c5t211539f5h1434574_looks_like_mount_inequality_erupted_\nor:\nhttps://www.youtube.com/watch?v=LSYP_KMbBNc"
    },
    {
        "Problem_1": "Let $a,b,c$ be positive real numbers such that $a+b+c=4\\sqrt[3]{abc}$. Prove that\n",
        "Solution": "WLOG let $a \\leq b \\leq c$. Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get:\nThe last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.",
        "Solution_2": "https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg\n-srisainandan6",
        "Solution_3": "Similarly to Solution 2, we will prove homogeneity but we will use that to solve the problem differently. Let $f(a,b,c)=a+b+c-4\\sqrt[3]{abc}$. Note that $f(a,b,c)=f(ka,kb,kc)$, thus proving homogeneity.\nWLOG, we can scale down all variables such that the lowest one is $1$. WLOG, let this be $a=1$. \nWe now have $1+b+c=4\\sqrt[3]{bc}$, and we want to prove $2bc+2b+2c+4\\ge 1+b^2+c^2.$ Adding $2bc$ to both sides and subtracting $2b+2c$ gives us $4bc+4\\ge 1+ (b+c)(b+c-2)$, or $4bc+3\\ge (b+c)(b+c-2)$. Let $\\sqrt[3]{bc}=x$. Now, we have     By the trivial inequality, this is always true. Since all these steps are reversible, the proof is complete. \n~SigmaPiE"
    },
    {
        "Problem_2": "Find all functions $f:(0,\\infty) \\to (0,\\infty)$ such that\nfor all $x,y,z >0$ with $xyz =1.$\n",
        "Solution": "Obviously, the output of $f$ lies in the interval $(0,1)$. Define $g:(0,1)\\to(0,1)$ as $g(x)=f\\left(\\frac1x-1\\right)$. Then for any $a,b,c\\in(0,1)$ such that $a+b+c=1$, we have $g(a)=f\\left(\\frac1a-1\\right)=f\\left(\\frac{1-a}a\\right)=f\\left(\\frac{b+c}a\\right)$. We can transform $g(b)$ and $g(c)$ similarly:\n\nLet $x=\\frac ca$, $y=\\frac ab$, $z=\\frac bc$. We can see that the above expression is equal to $1$. That is, for any $a,b,c\\in(0,1)$ such that $a+b+c=1$, $g(a)+g(b)+g(c)=1$.\n(To motivate this, one can start by writing $x=\\frac ab$, $y=\\frac bc$, $z=\\frac ca$, and normalizing such that $a+b+c=1$.)\nFor convenience, we define $h:\\left(-\\frac13,\\frac23\\right)\\to\\left(-\\frac13,\\frac23\\right)$ as $h(x)=g\\left(x+\\frac13\\right)-\\frac13$, so that for any $a,b,c\\in\\left(-\\frac13,\\frac23\\right)$ such that $a+b+c=0$, we have\n\nObviously, $h(0)=0$. If $|a|<\\frac13$, then $h(a)+h(-a)+h(0)=0$ and thus $h(-a)=-h(a)$. Furthermore, if $a,b$ are in the domain and $|a+b|<\\frac13$, then $h(a)+h(b)+h(-(a+b))=0$ and thus $h(a+b)=h(a)+h(b)$.\nAt this point, we should realize that $h$ should be of the form $h(x)=kx$. We first prove this for some rational numbers. If $n$ is a positive integer and $x$ is a real number such that $|nx|<\\frac13$, then we can repeatedly apply $h(a+b)=h(a)+h(b)$ to obtain $h(nx)=nh(x)$. Let $k=6h\\left(\\frac16\\right)$, then for any rational number $r=\\frac pq\\in\\left(0,\\frac13\\right)$ where $p,q$ are positive integers, we have $h(r)=6p*h\\left(\\frac1{6q}\\right)=\\frac{6p}q*h\\left(\\frac16\\right)=kr$.\nNext, we prove it for all real numbers in the interval $\\left(0,\\frac13\\right)$. For the sake of contradiction, assume that there is some $x\\in\\left(0,\\frac13\\right)$ such that $h(x)\\ne kx$. Let $E=h(x)-kx$, then obviously $0<|E|<1$. The idea is to \"amplify\" this error until it becomes so big as to contradict the bounds on the output of $h$. Let $N=\\left\\lceil\\frac1{|E|}\\right\\rceil$, so that $N\\ge2$ and $|NE|\\ge1$. Pick any rational $r\\in\\left(\\frac{N-1}Nx,x\\right)$, so that \nAll numbers and sums are safely inside the bounds of $\\left(-\\frac13,\\frac13\\right)$. Thus\n\nbut picking any rational number $s\\in\\left(N(x-r),\\frac13\\right)$ gives us $|kN(x-r)|<|ks|$, and since $ks=h(s)\\in\\left(-\\frac13,\\frac23\\right)$, we have $kN(x-r)\\in\\left(-\\frac13,\\frac23\\right)$ as well, but since $NE\\ge1$, this means that $h(N(x-r))=kN(x-r)+NE\\notin\\left(-\\frac13,\\frac23\\right)$, giving us the desired contradiction.\nWe now know that $h(x)=kx$ for all $0<x<\\frac13$. Since $h(-x)=-h(x)$ for $|x|<\\frac13$, we obtain $h(x)=kx$ for all $|x|<\\frac13$. For $x\\in\\left(\\frac13,\\frac23\\right)$, we have $h(x)+h\\left(-\\frac x2\\right)+h\\left(-\\frac x2\\right)=0$, and thus $h(x)=kx$ as well. So $h(x)=kx$ for all $x$ in the domain. Since $h(x)$ is bounded by $-\\frac13$ and $\\frac23$, we have $-\\frac12\\le k\\le1$. It remains to work backwards to find $f(x)$.\n- wzs26843545602"
    },
    {
        "Problem_3": "For a given integer $n\\ge 2,$ let $\\{a_1,a_2,…,a_m\\}$ be the set of positive integers less than $n$ that are relatively prime to $n.$ Prove that if every prime that divides $m$ also divides $n,$ then $a_1^k+a_2^k + \\dots + a_m^k$ is divisible by $m$ for every positive integer $k.$\n",
        "Solution_.28NO_FORMATTING.29": "https://maa.org/sites/default/files/pdf/AMC/usamo/2018/2018USAMO.pdf\nThe integer m in the statement of the problem is ϕ(n), where ϕ is the Euler totient function.\nThroughout our proof we write p\ns\n|| m, if s is the greatest power of p that divides m.\nWe begin with the following lemma:\nLemma 1. If p is a prime and p\ns divides n for some positive integer s, then 1k + 2k + · · · + n\nk\nis\ndivisible by p\ns−1\nfor any integer k ≥ 1.\n2018 USAMO – Solutions 4\nProof. Let {a1, a2, . . . , am} be a complete reduced residue set modulo p\ns and m = p\ns−1\n(p − 1).\nFirst we prove by induction on s that for any positive integer k, a\nk\n1 + a\nk\n2 + · · · + a\nk\nm is divisible by\np\ns−1\n. The base case s = 1 is true. Suppose the statement holds for some value of s. Consider the\nstatement for s + 1. Note that\n{a1, . . . , am, ps + a1, . . . , ps + am, . . . , ps\n(p − 1) + a1, . . . , ps\n(p − 1) + am}\nis a complete reduced residue set modulo p\ns+1. Therefore, the desired sum of k-th powers is equal\nto\na\nk\n1 + · · · + a\nk\nm + · · · + (p\ns\n(p − 1) + a1)\nk + · · · + (p\ns\n(p − 1) + am)\nk ≡ p(a\nk\n1 + · · · + a\nk\nm) ≡ 0 (mod p\ns\n),\nwhere we have used the induction hypothesis for the second congruence. This gives the induction\nstep.\nNow we are ready to prove the lemma. Because numbers from 1 to n can be split into blocks of\nconsecutive numbers of length p\ns\n, it is enough to show that 1k + 2k +· · · + (p\ns\n)\nk\nis divisible by p\ns−1\nfor any positive integer k. We use induction on s. The statement is true for s = 1. Assume the\nstatement is true for s − 1. The sum\n1\nk + 2k + · · · + (p\ns\n)\nk = a\nk\n1 + a\nk\n2 + · · · + a\nk\nm + p\nk\n�\n1\nk + 2k + · · · + (p\ns−1\n)\nk\n�\nis divisible by p\ns−1\n, because p\ns−1\n| a\nk\n1 + · · · + a\nk\nm and by the induction hypothesis p\ns−2\n| 1\nk + · · · +\n(p\ns−1\n)\nk\n.\nNow we proceed to prove a second lemma, from which the statement of the problem will immediately\nfollow:\nLemma 2. Suppose p is a prime dividing n. Let {a1, . . . , am} be a complete reduced residue set\nmod n, and define s by p\ns\n|| m. Then p\ns divides a\nk\n1 + · · · + a\nk\nm for any integer k ≥ 1.\nProof. We fix p, and use induction on the number of prime factors of n (counted by multiplicity)\nthat are different from p. If there are no prime factors other than p, then n = p\ns+1\n, m = p\ns\n(p − 1),\nand we proved in Lemma 1 that a\nk\n1 +· · ·+a\nk\nm is divisible by p\ns\n. Now suppose the statement is true\nfor n. We show that it is true for nq, where q is a prime not equal to p.\nCase 1. q divides n. We have p\ns\n|| ϕ(n) and p\ns\n|| ϕ(nq), because ϕ(nq) = qϕ(n). If {a1, a2, . . . , am}\nis a complete reduced residue set modulo n, then\n{a1, . . . , am, n + a1, . . . , n + am, . . . , n(q − 1) + a1, . . . , n(q − 1) + am}\nis a complete reduced residue set modulo nq. The new sum of k-th powers is equal to\na\nk\n1 + · · · + a\nk\nm + · · · + (n(q − 1) + a1)\nk + · · · + (n(q − 1) + am)\nk = mnk\n�\n1\nk + · · · + (q − 1)k\n�\n+\n2018 USAMO – Solutions 5\n�\nk\n1\n�\nn\nk−1\n�\n1\nk−1 + · · · + (q − 1)k−1\n�\n(a1 + · · · + am) + · · · + q(a\nk\n1 + · · · + a\nk\nm).\nThis sum is divisible by p\ns because p\ns\n|| m and p\ns\n| a\nj\n1 + a\nj\n2 + · · · + a\nj\nm for any positive integer j.\nCase 2. q doesn’t divide n. Suppose p\nb\n|| q − 1, where b ≥ 0. Note that ϕ(nq) = ϕ(n)(q − 1), so\np\ns\n|| ϕ(n) and p\ns+b\n|| ϕ(nq). Let {a1, . . . , am} be a complete reduced residue set modulo n. The\ncomplete reduced residue set modulo nq consists of the mq numbers\n{a1, . . . , am, n + a1, . . . , n + am, . . . , n(q − 1) + a1, . . . , n(q − 1) + am}\nwith the m elements {qa1, qa2, . . . , qam} removed.\nThe new sum of k-th powers is equal to\na\nk\n1 + · · · + a\nk\nm + · · · + (n(q − 1) + a1)\nk + · · · + (n(q − 1) + am)\nk − q\nk\n(a\nk\n1 + · · · + a\nk\nm) =\nmnk\n�\n1\nk + · · · + (q − 1)k\n�\n+\n�\nk\n1\n�\nn\nk−1\n�\n1\nk−1 + · · · + (q − 1)k−1\n�\n(a1 + · · · + am) + · · ·\n· · · +\n�\nk\nk − 1\n�\nn (1 + · · · + (q − 1)) (a\nk−1\n1 + · · · + a\nk−1\nm ) + q(a\nk\n1 + · · · + a\nk\nm) − q\nk\n(a\nk\n1 + · · · + a\nk\nm).\nEach term\n�\nk\nj\n�\nn\nk−j\n�\n1\nk−j + · · · + (q − 1)k−j\n�\n(a\nj\n1 + · · · + a\nj\nm),\nfor 0 ≤ j ≤ k − 1, is divisible by p\ns+b because p | n\nk−j\n, p\ns\n| a\nj\n1 + · · · + a\nj\nm, and p\nb−1\n| 1\nk−j + · · · +\n(q − 1)k−j by Lemma 1.\nAlso (q\nk − q)(a\nk\n1 + · · · + a\nk\nm) is divisible by p\ns+b because p\nb\n| q − 1 | q\nk − q and p\ns\n| a\nk\n1 + · · · + a\nk\nm.\nThus p\ns+b divides our sum and our proof is complete.\nRemark. In fact, one can also show the converse statement: if {a1, a2, . . . , am} is as defined in\nthe problem and a\nk\n1 + a\nk\n2 + · · · + a\nk\nm is divisible by m for every positive integer k, then every prime\nthat divides m also divides n.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem_4": "Let $p$ be a prime, and let $a_1, \\dots, a_p$ be integers. Show that there exists an integer $k$ such that the numbers produce at least $\\tfrac{1}{2} p$ distinct remainders upon division by $p$.\n",
        "Solution": "$\\textbf{Lemma: }$ For fixed $i\\neq j,$ where $i, j\\in\\{1, 2, ..., p\\},$ the statement $a_i + ik\\equiv a_j + jk\\text{ (mod } p\\text{)}$ holds for exactly one $k\\in {1, 2, ..., p}.$\n$\\textbf{Proof: }$ Notice that the left side minus the right side is congruent to $(a_i - a_j) + (i - j)k$ modulo $p.$ For this difference to equal $0,$ there is a unique solution for $k$ modulo $p$ given by $k\\equiv (a_j - a_i)(i - j)^{-1}\\text{ (mod } p\\text{)},$ where we have used the fact that every nonzero residue modulo $p$ has a unique multiplicative inverse. Therefore, there is exactly one $k\\in {1, 2, ..., p}$ that satisfies $a_i + ik\\equiv a_j + jk\\text{ (mod } p\\text{)}$ for any fixed $i\\neq j. \\textbf{ End Lemma}$\nSuppose that you have $p$ graphs $G_1, G_2, ..., G_p,$ and graph $G_k$ consists of the vertices $(i, k)$ for all $1\\le i\\le p.$ Within any graph $G_k,$ vertices $(i_1, k)$ and $(i_2, k)$ are connected by an edge if and only if $a_{i_1} + i_1k\\equiv a_{i_2} + i_2k\\text{ (mod } p\\text{)}.$ Notice that the number of disconnected components of any graph $G_k$ equals the number of distinct remainders when divided by $p$ given by the numbers $a_1 + k, a_2 + 2k, ..., a_p + pk.$\nThese $p$ graphs together have exactly one edge for every unordered pair of elements of $\\{1, 2, ..., p\\},$ so they have a total of exactly $\\frac{p(p-1)}{2}$ edges. Therefore, there exists at least one graph $G_k$ that has strictly fewer than $\\frac{p}{2}$ edges, meaning that it has more than $\\frac{p}{2}$ disconnected components. Therefore, the collection of numbers $\\{a_i + ik: 1\\le i\\le p\\}$ for this particular value of $k$ has at least $\\frac{p}{2}$ distinct remainders modulo $p.$ This completes the proof.\n$\\textbf{Note: This is the same problem as 2018 USAJMO Problem 5.}$",
        "Solution_2": "We consider the Lemma and it's proof as in the above solution.\nFor every $k$, let $x_k$ be the number of distinct residues that $a_i + ik$ takes on. Further for each residue $r_i$, let $n_i$ be the number of times it is achieved by an $a_j + jk$ for $i \\in [1, x_k]$. Note that $\\sum_{i = 1}^{x_k} n_i = p.$\nThe number of pairs $(a_i, a_j)$ s.t $(a_i + ik) - (a_j + jk) \\equiv 0$ for a given $k$ is,\n\nThe first equality is given by a well known theorem, which can be proven by C-S or Jensen's.\nSumming over all $k$, the number of times a pair $(a_i, a_j)$ has $(a_i + ik) - (a_j + jk) \\equiv 0$ is,\n\nAlternatively, every pair $(a_i, a_j)$ has a unique $k$ s.t $(a_i + ik) - (a_j + jk) \\equiv 0$ by the Lemma so we double-count as,\n\nBy AM-HM inequality,\n\nSo it is clear that $\\exists k$ s.t $x_k \\ge \\frac{p}{2}$.\n~Aaryabhatta1"
    },
    {
        "Problem_5": "In convex cyclic quadrilateral $ABCD,$ we know that lines $AC$ and $BD$ intersect at $E,$ lines $AB$ and $CD$ intersect at $F,$ and lines $BC$ and $DA$ intersect at $G.$ Suppose that the circumcircle of $\\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\\angle MAC = 90^{\\circ}.$\n",
        "Solution": "\nso $P,E,Q$ are collinear. Furthermore, note that $DQBP$ is cyclic because:\n\nNotice that since $A$ is the intersection of $(EDQ)$ and $(BPE)$, it is the Miquel point of $DQBP$.\nNow define $X$ as the intersection of $BQ$ and $DP$. From Pappus's theorem on $BFPDGQ$ that $A,M,X$ are collinear. It’s a well known property of Miquel points that $\\angle EAX = 90$, so it follows that $MA \\perp AE$, as desired. $\\blacksquare$\n~AopsUser101"
    },
    {
        "Problem_6": "Let $a_n$ be the number of permutations $(x_1, x_2, \\dots, x_n)$ of the numbers $(1,2,\\dots, n)$ such that the $n$ ratios $\\frac{x_k}{k}$ for $1\\le k\\le n$ are all distinct. Prove that $a_n$ is odd for all $n\\ge 1.$\n",
        "Solution": "Write out the mapping of each $k$ to $x_k$ as follows:\nNow, consider what happens when the two rows are swapped (and the top-bottom pairs are reordered so that the top reads (1,2,3,...,n)).  This will result in a new permutation if and only if $x_k = j$ does not imply $x_j = k$ for all $k,j$ in $(1,2,3,\\dots,n)$.  I will denote this new permutation as $(y_1,y_2,y_3,\\dots,y_n)$.  Notice that $(y_1,y_2,y_3,\\dots,y_n)$ is a valid permutation if and only if $(x_1,x_2,x_3,\\dots,x_n)$ is valid, because each fraction $\\frac{y_k}{k}$ is the reciprocal of $\\frac{x_k}{k}$.  This means that we need only consider the parity of number of cases in which $(y_1,y_2,y_3,\\dots, y_n) = (x_1,x_2,x_3,\\dots, x_n)$, as there will be an even number of cases where $(y_1,y_2,y_3,\\dots, y_n) \\ne (x_1,x_2,x_3,\\dots, x_n)$.  Let the number of valid permutations where $(y_1,y_2,y_3,\\dots, y_n) = (x_1,x_2,x_3,\\dots, x_n)$ be $b_n$.\nNotice that the only permutations that have the property $x_{x_k}=k$ (which is an equivalent statement to the one given above) are those that are formed by taking pairs of elements and swapping their positions having the maximum number of pairs possible and having no two pairs both contain the same element.  Some more necessary conditions will be outlined below after we split $n$ into two cases.\nCase 1:  $n$ is odd.  If $n$ is odd then there must be exactly one $k$ such that $x_k = k$.  This yields $b_n=n\\cdot b_{n-1}$* which has the same parity as $b_{n-1}$, so we need only consider the parity of $b_n$ for odd $n$. (*Note: This is wrong, as $b_n$ is only the number of valid involutions for $1,2,...,n$, not $1,2,...,n+1$ with one number removed.)\nCase 2:  $n$ is even.  If $n$ is even then can be no $k$ so that $x_k=k$, or else there must be at least two distinct numbers $k$ and $j$ so that $x_k=k$ and $x_j=j$ which violates the given conditions.  Denote a pair of numbers that are swapped to arrive at the final permutation as the pair $(a,b)$.  Then if a permutation yields an invalid arrangement there must be two pairs $(a,b)$ and $(c,d)$ such that $\\frac{a^2}{b^2} = \\frac{c^2}{d^2}$.  But notice that the two pairs $(a,c)$ and $(b,d)$ will also result in an invalid arrangement. So, there must be an even number of invalid arrangements (Note: This proof doesn't work, as no specific pairing of invalid arrangements is constructed. Indeed, when three pairs all with the same fraction are present, this doesn't work.), meaning the parity we desire is just the number of ways to separate $n$ objects into $\\frac n2$ pairs!  However, this is quite simply just $(n-1)(n-3)(n-5)\\cdots(3)(1)$, which is clearly the product of odd numbers.  So we conclude that there are an odd number of valid permutations $(x_1,x_2,x_3,\\dots,x_n)$.  $\\blacksquare$"
    },
    {
        "Problem": "Let $\\mathbb{N}$ be the set of positive integers. A function $f:\\mathbb{N}\\to\\mathbb{N}$ satisfies the equation for all positive integers $n$. Given this information, determine all possible values of $f(1000)$.",
        "Solution": "Let $f^r(x)$ denote the result when $f$ is applied to $f^{r-1}(x)$, where $f^1(x)=f(x)$.\n$\\hfill \\break \\hfill \\break$\nIf $f(p)=f(q)$, then $f^2(p)=f^2(q)$ and $f^{f(p)}(p)=f^{f(q)}(q)$\n$\\implies p^2=f^2(p)\\cdot f^{f(p)}(p)=f^2(q)\\cdot f^{f(q)}(q)=q^2$\n$\\implies p=\\pm q$\n$\\implies p=q$ since $p,q>0$.\nTherefore, $f$ is injective. It follows that $f^r$ is also injective.\nLemma 1: If $f^r(b)=a$ and $f(a)=a$, then $b=a$.\nProof:\n$f^r(b)=a=f^r(a)$ which implies $b=a$ by injectivity of $f^r$.\nLemma 2: If $f^2(m)=f^{f(m)}(m)=m$, and $m$ is odd, then $f(m)=m$.\nProof:\nLet $f(m)=k$. Since $f^2(m)=m$, $f(k)=m$. So, $f^2(k)=k$. $\\newline f^2(k)\\cdot f^{f(k)}(k)=k^2$.\nSince $k\\neq0$, $f^{f(k)}(k)=k$\n$\\implies f^m(k)=k$\n$\\implies f^{gcd(m, 2)}(k)=k$\n$\\implies f(k)=k$\nThis proves Lemma 2.\nI claim that $f(m)=m$ for all odd $m$.\nOtherwise, let $m$ be the least counterexample.\nSince $f^2(m)\\cdot f^{f(m)}(m)=m^2$, either\n$(1) f^2(m)=k<m$, contradicted by Lemma 1 since $k$ is odd and $f^2(k)=k$.\n$(2) f^{f(m)}(m)=k<m$, also contradicted by Lemma 1 by similar logic.\n$(3) f^2(m)=m$ and $f^{f(m)}(m)=m$, which implies that $f(m)=m$ by Lemma 2.\nThis proves the claim.\nBy injectivity, $f(1000)$ is not odd.\nI will prove that $f(1000)$ can be any even number, $x$. Let $f(1000)=x, f(x)=1000$, and $f(k)=k$ for all other $k$. If $n$ is equal to neither $1000$ nor $x$, then $f^2(n)\\cdot f^{f(n)}(n)=n\\cdot n=n^2$. This satisfies the given property.\nIf $n$ is equal to $1000$ or $x$, then $f^2(n)\\cdot f^{f(n)}(n)=n\\cdot n=n^2$ since $f(n)$ is even and $f^2(n)=n$. This satisfies the given property.\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\\overline{AB}$ satisfying $\\angle APD = \\angle BPC$. Show that line $PE$ bisects $\\overline{CD}$.",
        "Solution": "Let $PE \\cap DC = M$. Also, let $N$ be the midpoint of $AB$.\nNote that only one point $P$ satisfies the given angle condition. With this in mind, construct $P'$ with the following properties:\n(1) $AP' \\cdot AB = AD^2$\n(2) $BP' \\cdot AB = CD^2$\nClaim: $P = P'$\nProof:\nThe conditions imply the similarities $ADP \\sim ABD$ and $BCP \\sim BAC$ whence $\\measuredangle APD = \\measuredangle BDA = \\measuredangle BCA = \\measuredangle CPB$ as desired. $\\square$\nClaim: $PE$ is a symmedian in $AEB$\nProof:\nWe have \n\nas desired. $\\square$\nSince $P$ is the isogonal conjugate of $N$, $\\measuredangle PEA = \\measuredangle MEC = \\measuredangle BEN$. However $\\measuredangle MEC = \\measuredangle BEN$ implies that $M$ is the midpoint of $CD$ from similar triangles, so we are done. $\\square$",
        "Solution_2": "Let $\\omega$ be the circle centered at $A$ with radius $AD.$\nLet $\\Omega$ be the circle centered at $B$ with radius $BC.$\nWe denote $I_\\omega$ and $I_\\Omega$ inversion with respect to $\\omega$ and $\\Omega,$ respectively.\n\n\n\n\nLet $\\theta$ be the circle $ABCD.$\n$I_\\omega (\\theta) = B'C'D,$ straight line, therefore \n$I_\\Omega (\\theta) = A'D'C,$ straight line, therefore\n\n$ABCD$ is cyclic $\\implies \\angle BA'C = \\angle AB'D.$\n points $A'$ and $B'$ are coincide.\nDenote $A' = B' = Q \\in AB.$\nSuppose, we move point $Q$ from $A$ to $B.$ Then $\\angle AQD$ decreases monotonically, $\\angle BQC$ increases monotonically. So, there is only one point where\n$\\hspace{19mm} I_\\omega (CD'P) = AC'D'B$ is cyclic.\n\n$\\hspace{19mm} C'D'CD$ is  trapezoid.\nIt is known that the intersection of the diagonals, intersection point of the lines containing the lateral sides of the trapezoid and the midpoints of two parallel sides are collinear.\nvladimir.shelomovskii@gmail.com, vvsss"
    },
    {
        "Problem": "Let $K$ be the set of all positive integers that do not contain the digit $7$ in their base-$10$ representation. Find all polynomials $f$ with nonnegative integer coefficients such that $f(n)\\in K$ whenever $n\\in K$.",
        "Solution": "I claim the only such polynomials are of the form $f(n)=k$ where $k\\in K$, or $f(n)=an+b$ where $a$ is a power of 10, $b\\in K$, and $b<a$. Obviously, these polynomials satisfy the conditions. We now prove that no other polynomial works. That is, we prove that for any other polynomial $f$ with nonnegative coefficients, there is some $n\\in K$ such that $f(n)\\notin K$.\nWe first prove the result for monomials, polynomials in which only one coefficient is nonzero. This is obvious for constant polynomials $f(n)=k\\notin K$. The next simplest case is $f(n)=an$ with $a$ not a power of 10, and hence $\\lg a$ is irrational. By Dirichlet's approximation theorem, the set of multiples of $\\lg a$ is dense $\\bmod\\ 1$, and thus contains an element with fractional part in the interval $[\\lg 7,\\lg 8)$. In other words, there is a power of $a$ whose decimal expansion starts with a 7. Let $a^x$ be the smallest power of $a$ that is not in $K$. Obviously, $x>0$, so letting $n=a^{x-1}$ completes the proof of this part.\nWe have now proven that for any $a$ that is not a power of 10, there is some $n\\in K$ such that $an\\notin K$. We proceed to the case where $f(n)=an^x$ for $x>1$. This splits into 2 cases. If $ax$ is not a power of 10, then pick $m\\in K$ such that $axm\\notin K$. For any $y$, we have\n\nIf we choose $y$ to be large enough, then the terms in the expression above will not interfere with each other, and the resulting number will contain a 7 in the decimal expansion, and thus not be in $K$.\nOn the other hand, if $ax$ is a power of 10, then $a$ and $x$ are both powers of 10, and $x\\ge10$. Obviously, $\\frac12ax(x-1)$ is not a power of 10. By the previous step, which establishes the result for $x=2$, we can pick $m\\in K$ such that $\\frac12ax(x-1)m^2\\notin K$. Then, for any $y$,\n\nSimilarly, picking a sufficient large $y$ settles this case.\nNow, we extend our proof to general polynomials. If a polynomial $f(n)=a_0+a_1n+a_2n^2+...+a_xn^x$ satisfies the conditions of the problem, then for any $m,y>0$:\n\nSimilarly, choosing $y$ to be sufficiently large results in the terms not interfering with each other. If $f$ contains any monomials that do not satisfy the conditions of the problem, then picking suitable $m$ and sufficiently large $y$ causes $f(m*10^y)$ to not be in $K$. Thus, $f$ is a linear polynomial of the form $ax+b$ where $a$ is 0 or a power of 10, and $b\\in K$. It suffices to rule out those polynomials where $a>0$ and $b>a$. If this is the case, then since the digit of $b$ corresponding to $a$ is not 7, there must be a single-digit number $n$ such that the digit of $f(n)$ corresponding to $a$ is 7. Therefore, we are done.\n-wzs26843545602\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Let $n$ be a nonnegative integer. Determine the number of ways that one can choose $(n+1)^2$ sets $S_{i,j}\\subseteq\\{1,2,\\ldots,2n\\}$, for integers $i,j$ with $0\\leq i,j\\leq n$, such that:\n$\\bullet$ for all $0\\leq i,j\\leq n$, the set $S_{i,j}$ has $i+j$ elements; and\n$\\bullet$ $S_{i,j}\\subseteq S_{k,l}$ whenever $0\\leq i\\leq k\\leq n$ and $0\\leq j\\leq l\\leq n$.",
        "Solution_1": "Note that there are $(2n)!$ ways to choose $S_{1, 0}, S_{2, 0}... S_{n, 0}, S_{n, 1}, S_{n, 2}... S_{n, n}$, because there are $2n$ ways to choose which number $S_{1, 0}$ is, $2n-1$ ways to choose which number to append to make $S_{2, 0}$, $2n-2$ ways to choose which number to append to make $S_{3, 0}$, etc. After that, note that $S_{n-1, 1}$ contains the $n-1$ in $S_{n-1, 0}$ and 1 other element chosen from the 2 elements in $S_{n, 1}$ not in $S_{n-1, 0}$ so there are 2 ways for $S_{n-1, 1}$. By the same logic there are 2 ways for $S_{n-1, 2}$ as well so $2^n$ total ways for all $S_{n-1, j}$, so doing the same thing $n-1$ more times yields a final answer of $(2n)!\\cdot 2^{\\left(n^2\\right)}$.\n-Stormersyle",
        "Solution_2": "There are $\\frac{(2n)!}{2^n}$ ways to choose $S_{0,0}, S_{1,1} .... S_{n,n}$. Since, there are $\\binom{2n}{2}$ ways to choose $S_{1,1}$, and after that, to generate $S_{2,2}$, you take $S_{1,1}$ and add 2 new elements, getting you $\\binom{2n-2}{2}$ ways to generate $S_{2,2}$. And you can keep going down the line, and you get that there are $\\frac{(2n)!}{2^n}$ ways to pick $S_{0,0}, S_{1,1} ... S_{n,n}$ Then we can fill out the rest of the gird. First, let’s prove a lemma.",
        "Solution_3": "Let $C_{j}$ represent the set of sets of the form $S_{ij}$ for $1 \\le i \\le n$, $a_{ij}$ denote $S_{(i+1)j} \\backslash S_{ij}$, and $b_{ij} = S_{i(j+1)} \\backslash S_{ij}$. Begin by considering $C_0$ and $S_{00} = \\emptyset$. Then  given $S_{i0}$ we can create $S_{(i+1)0}$ by adding one element ($a_{i0}$). Using this, the number of ways to form the sequence of $S_{00}, S_{10}, \\dots, S_{n0}$ are $(2n)(2n-1) \\cdots (n+1)$ where we successively add one of the remaining elements of $[2n]$ to get consecutive terms in the sequence.\nNow consider when we are given $C_{j}$ and we need to find $C_{j+1}$. So far, there have been $n + j$ chosen distinct elements (via $S_{nj}$). After finding $C_{j+1}$ we will have $n + j + 1$ distinct elements and so in this process we only add one unique element to sets among $C_{j+1}$. There are $2n - (n+j) = n-j$ ways to chose such a new element called $x$.\nNow notice that $b_{0j}, a_{0(j+1)}, a_{1(j+1)}, \\dots, a_{(n-1)(j+1)}$ is a permutation of $x,  a_{0j}, \\dots, a_{(n-1)j}$ by noting $b_{nj} = x$ and,\n\nFurthermore,\n\nTherefore $b_{ij}, a_{i(j+1)}$ is a permutation of $a_{ij}, b_{(i+1)j}$ for $i < n$. Now let $k$ be the first $i$ such that $b_{ij} = x$. By definition, $b_{(k+1)j}, \\dots = x$. Then the number of ways to order $x,  a_{0j}, \\dots, a_{(n-1)j}$ is $2^{k-1}$ as there are 2 permutations for each pair before $\\{b_{ij}, a_{i(j+1)}\\} = \\{a_{ij}, b_{(i+1)j}\\}$ and each pair after is determined by $b_{ij}= b_{(i+1)j} = k, a_{i(j+1)} = a_{ij}$. For $k  = 0$, the permutation is completely determined so there is one way.\nOverall the number of ways to add the $j+1$'th row is,\nIn total, there are $(2n)(2n-1)\\cdots (n+1)$ ways to find $C_0$ and for each $C_{j}$ there are $2^n(n-j+1)$ ways for $1 \\le j \\le n$. So the answer is,\n~Aaryabhatta1\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Two rational numbers $\\frac{m}{n}$ and $\\frac{n}{m}$ are written on a blackboard, where $m$ and $n$ are relatively prime positive integers. At any point, Evan may pick two of the numbers $x$ and $y$ written on the board and write either their arithmetic mean $\\frac{x+y}{2}$ or their harmonic mean $\\frac{2xy}{x+y}$ on the board as well. Find all pairs $(m,n)$ such that Evan can write $1$ on the board in finitely many steps.\nProposed by Yannick Yao",
        "Solution": "We claim that all odd $m, n$ work if $m+n$ is a positive power of 2.\nProof:\nWe first prove that $m+n=2^k$ works. By weighted averages we have that $\\frac{n(\\frac{m}{n})+(2^k-n)\\frac{n}{m}}{2^k}=\\frac{m+n}{2^k}=1$ can be written, so the solution set does indeed work. We will now prove these are the only solutions.\nAssume that $m+n\\ne 2^k$, so then $m+n\\equiv 0\\pmod{p}$ for some odd prime $p$. Then $m\\equiv -n\\pmod{p}$, so $\\frac{m}{n}\\equiv \\frac{n}{m}\\equiv -1\\pmod{p}$. We see that the arithmetic mean is $\\frac{-1+(-1)}{2}\\equiv -1\\pmod{p}$ and the harmonic mean is $\\frac{2(-1)(-1)}{-1+(-1)}\\equiv -1\\pmod{p}$, so if 1 can be written then $1\\equiv -1\\pmod{p}$ and $2\\equiv 0\\pmod{p}$ which is obviously impossible, and we are done.\n-Stormersyle\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Find all polynomials $P$ with real coefficients such that holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$.",
        "Solution": "If $P(x)=c$ for a constant $c,$ then $\\dfrac{c(x+y+z)}{xyz}=3c$. We have $2c=3c.$ Therefore $c=0.$\nNow consider the case of non-constant polynomials. \nFirst we have  for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$. Both sides of the equality are polynomials (of $x,y,z$). They have the same values on the 2-dimensional surface $2xyz=x+y+z$, except for some 1-dimensional curves in it. By continuity, the equality holds for all points on the surface, including those with $z=0.$ Let $z=0,$ we have $y=-x$ and $x(P(x)-P(-x))=0.$ Therefore $P$ is an even function.\n(Here is a sketch of an elementary proof. Let $z=\\dfrac{x+y}{2xy-1}.$ We have\n\nThis is an equality of rational expressions. By multiplying $(2xy-1)^N$ on both sides for a sufficiently large $N$, they become polynomials, say $A(x,y)=B(x,y)$ for all real $x, y$ with $x\\ne 0, y\\ne 0, x+y\\ne 0$ and $2xy-1\\ne 0.$ For a fixed $x,$ we have two polynomials (of $y$) having same values for infinitely many $y$. They must be identical. Let $y=0,$ we have $x^{N+1}(P(x)-P(-x))=0.$ )\nNotice that if $P(x)$ is a solution,  then is $cP(x)$ for any constant $c.$ For simplicity, we assume the leading coefficient of $P$ is $1$:  where $n$ is a positive even number.\nLet $y=\\dfrac{1}{x}$, $z=x+\\dfrac{1}{x}.$ we have\nSimplify using $P(x)=P(-x),$\nExpand and combine like terms, both sides are of the form\nThey have the same values for infinitely many $x.$ They must be identical. We just compare their leading terms. On the left hand side it is $2nx^{n-1}$. There are two cases for the right hand sides: If $n>2$, it is $x^{n-1}$; If $n=2$, it is $(1+a_0)x.$ It does not work for $n>2.$ When $n=2,$ we have $4=1+a_0.$ therefore $a_0=3.$\nThe solution: $P(x)=c(x^2+3)$ for any constant $c.$\n-JZ"
    },
    {
        "Problem_1": "Let $ABC$ be a fixed acute triangle inscribed in a circle $\\omega$ with center $O$. A variable point $X$ is chosen on minor arc $AB$ of $\\omega$, and segments $CX$ and $AB$ meet at $D$. Denote by $O_1$ and $O_2$ the circumcenters of triangles $ADX$ and $BDX$, respectively. Determine all points $X$ for which the area of triangle $OO_1O_2$ is minimized.\n",
        "Solution": "Let $E$ be midpoint $AD.$ Let $F$ be midpoint $BD \\implies$\n\n$E$ and $F$ are the bases of perpendiculars dropped from $O_1$ and $O_2,$ respectively.\nTherefore  $O_1O_2 \\ge EF =  \\frac {AB}{2}.$\n$\\angle AXC = \\angle ABC (AXBC$ is cyclic) $\\implies \\angle O O_1O_2 = \\angle ABC.$\nSimilarly $\\angle BAC = \\angle O O_2 O_1 \\implies  \\triangle ABC \\sim \\triangle O_2 O_1O.$\nThe area of $\\triangle OO_1O_2$ is minimized if $CX \\perp AB$ because\n\nvladimir.shelomovskii@gmail.com, vvsss"
    },
    {
        "Problem": "An empty $2020 \\times 2020 \\times 2020$ cube is given, and a $2020 \\times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a $1 \\times 1 \\times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:\nWhat is the smallest positive number of beams that can be placed to satisfy these conditions?",
        "Solution": "Take one vertex of the cube as origin and establish 3D coordinates along the cube's edges.\nDefine a beam as $x-dir$ if its long edge is parallel to x-axis. Similarly for $y-dir$ and $z-dir$.\nDefine a beam's location as (direction, ($1 \\times 1$ face's location in 2D coordinate).\nFor example, (y, 2, 4) indicates the beam with vertex (1, 0, 3) and (2, 2020, 4)\nApparently $x$ beam needs the other $x$ or $y$ beams to touch its $xy$ faces, $x$ or $z$ beams to touch its $xz$ faces. Similarly for $y$ and $z$ beam.\nIf there are only 1-dir or 2-dirs beams, it is easy to approve that $2020 \\times 2020$ is the minimal number beams.\n(for example, if only use $x-dir$ and $y-dir$ beams, all the $x-dir$ beams's xz faces can only be touched by $x-dir$ beams. In the other word, $2020 x-dir$ beams will be stacked to meet xz faces touch requirements in that xz layer)\nConsider cases with all $3-dirs$ beams.\nWLOG there is a $x-dir$ beam and it needs $x-dir$ or $y-dir$ beams to touch its $xy$ faces (unless it touches the cube surface).\nAnd this $y-dir$ beam also needs a $x-dir$ or $y-dir$ to touch it's $xy$ faces. And so on until one which touches cube surface. So from $xy$ face perspective, it needs $2020$ beams.\nSimilarly from $xz$ and $yz$ face perspective, it also needs $2020$ and $2020$ beams.\nConsider one beam has four $1 \\times 2020$ faces and it can be counted twice. So there should be at least $2020 \\times 3 \\div 2=3030$ beams.\nHere is one solution with 3030 beams.\n$(x, 1, 1),\\ (y, 1, 2),\\ (z, 2, 2),$\n$\\cdots ,$\n$(x, (2n+1), (2n+1)),\\ (y, (2n+1), (2n+2)),\\ (z, (2n+2), (2n+2)),$\n$\\cdots ,$\n$(x, (2019, 2019)),\\ (y, 2019, 2020),\\ (z, 2020, 2020)$\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Let $p$ be an odd prime. An integer $x$ is called a quadratic non-residue if $p$ does not divide $x - t^2$ for any integer $t$.\nDenote by $A$ the set of all integers $a$ such that $1 \\le a < p$, and both $a$ and $4 - a$ are quadratic non-residues. Calculate the remainder when the product of the elements of $A$ is divided by $p$.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem_4": "Suppose that $(a_1, b_1), (a_2, b_2), \\ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \\le i < j \\le 100$ and $|a_ib_j - a_j b_i|=1$. Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs.",
        "Solution": "Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$:\nWe know this must be true:\nSo\nWe require the maximum conditions for $(a_3, b_3)$\nThen one case can be:\nWe try to do some stuff such as solving for $a_3$ with manipulations:\nWe showed that 3 pairs are a complete graph; however, 4 pairs are not a complete graph. We will now show that:\nThis is clearly impossible because $1$ is not even and also $|a_2b_1-a_1b_2| = 1$.\nThe answer is as follows:\n\n$a_1$ has $0$ subtractions that follow condition while $a_2$ has $1$ and then the rest has $2$.\nThere are $n$ terms, so our answer be $2n-3$ and in case of $n=100$ that means \n~Lopkiloinm",
        "Solution_2": "We claim the answer is $197$.\nStudy the points $(0, 0), (a_i, b_i), (a_j, b_j)$. If we let these be the vertices of a triangle, applying shoelace theorem gives us an area of $\\frac{1}{2}|0\\times{b_i}+{a_i}\\times{b_j}+{b_i}\\times{0}-0\\times{a_i}-{b_i}\\times{a_j}-{b_j}\\times{0} = \\frac{1}{2}|a_ib_j - a_j b_i| = \\frac{1}{2}$. Therefore, the triangle formed by the points $(0, 0), (a_i, b_i), (a_j, b_j)$ must have an area of $\\frac{1}{2}$.\nTwo cases follow.\nCase 1: Both $(a_i, b_i), (a_j, b_j)$ have exactly one coordinate equal to $0$. \nHere, one point must be on the $x$ axis and the other on the $y$ axis in order for the triangle to have a positive area. For the area of the triangle to be $\\frac{1}{2}$, it follows that the points must be $(1, 0), (0, 1)$ in some order.\nCase 2: At least one of $(a_i, b_i), (a_j, b_j)$ does not have exactly one coordinate equal to $0$. Define $S[l]$ to be a list of lines such that each line in the list has some two lattice points that, with $(0, 0)$, form a triangle with area $\\frac{1}{2}$. Note that for any such line that passes through such two lattice points, we may trivially generate infinite lattice points on the line that have nonnegative coordinates.\nNote that lines $y=1$ and $x=1$ are included in $S[l]$, because the points $(1, 1), (2, 1)$ serve as examples for $y=1$ and $(1, 1), (1, 2)$ serve as examples for $x=1$. For the optimal construction, include the points $(1, 0)$ and all the points $(0, 1), (0, 2), (0, 3), ... , (0, 99)$, in that order. In this case, every adjacent pair of points would count ($98$), as well as picking $(0, 1)$ and a nonadjacent point ($99$), so this would be $98+99=197$.\nTo prove that this is the maximum, consider the case where some $n$ number of points were neither on $x=1$ nor on $y=1$. In this case, we would be removing $n$ adjacent pairs and $n$ options to choose from after choosing $(0, 0)$, resulting in a net loss of $2n$. By having $n$ points on some other combination of lines in $S[l]$, we would trivially have a maximum gain of $n-1$ pairs of points on the lines such that there are no lattice points between those pairs. Because these points are not on $x=1$ or $y=1$, the altitude from a given point to the line formed by $(0, 0)$ and $(0, 1)$ and $(1, 0)$ is not $1$, and so the area of the triangle cannot be $\\frac{1}{2}$. Thus, by not having all points on lines $x=1$ and $y=1$, we cannot exceed the maximum of $197$. Thus, $\\boxed{197}$ is our answer.\n~SigmaPiE\n"
    },
    {
        "Problem": "A finite set $S$ of points in the coordinate plane is called overdetermined if $|S| \\ge 2$ and there exists a nonzero polynomial $P(t)$, with real coefficients and of degree at most $|S| - 2$, satisfying $P(x) = y$ for every point $(x, y) \\in S$.\nFor each integer $n \\ge 2$, find the largest integer $k$ (in terms of $n$) such that there exists a set of $n$ distinct points that is not overdetermined, but has $k$ overdetermined subsets.",
        "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem_6": "Let $ABCDEF$ be a convex hexagon satisfying $\\overline{AB} \\parallel \\overline{DE}$, $\\overline{BC} \\parallel \\overline{EF}$, $\\overline{CD} \\parallel \\overline{FA}$, andLet $X$, $Y$, and $Z$ be the midpoints of $\\overline{AD}$, $\\overline{BE}$, and $\\overline{CF}$. Prove that the circumcenter of $\\triangle ACE$, the circumcenter of $\\triangle BDF$, and the orthocenter of $\\triangle XYZ$ are collinear.",
        "Solution_1": "Let $M_1$, $M_2$, and $M_3$ be the midpoints of $CE$, $AE$, $AC$ and $N_1$, $N_2$, and $N_3$ be the midpoints of $DF$, $BF$, and $BD$. Also, let $H$ be the orthocenter of $XYZ$. Note that we can use parallel sides to see that $X$, $Z$, and $M_3$ are collinear. Thus we have  by midlines. Applying this argument cyclically, and noting the condition $AB \\cdot DE = BC \\cdot EF = CD \\cdot FA$, $M_1$, $M_2$, $M_3$, $N_1$, $N_2$, $N_3$ all lie on a circle concentric with $(XYZ)$.\nNext, realize that basic orthocenter properties imply that the circumcenter $O_1$ of $(ACE)$ is the orthocenter of $\\triangle M_1M_2M_3$, and likewise the circumcenter $O_2$ of $(BDF)$ is the orthocenter of $\\triangle N_1N_2N_3$.\nThe rest is just complex numbers; toss on the complex plane so that the circumcenter of $\\triangle XYZ$ is the origin. Then we have   \nNote that from the above we have $h=\\frac{o_1+o_2}{2}$, so $H$ is the midpoint of segment $O_1O_2$. In particular, $H$, $O_1$, and $O_2$ are collinear, as required.\n~ Leo.Euler",
        "Solution_2": "We construct two equal triangles, prove that triangle $XYZ$ is the same as medial triangle of both this triangles. We use property of medial triangle and prove that circumcenters of constructed triangles coincide with given circumcenters.\nDenote $A' =  C + E – D, B' = D + F – E, C' =  A+ E – F,$\n$D' =  F+ B – A, E' =  A + C – B, F' =  B+ D – C.$\nThen $A' – D'  =  C + E – D –  ( F+ B – A) = (A + C + E ) – (B+ D + F).$\nDenote  $D' – A' = 2\\vec V.$\nSimilarly we get $B' – E' = F' – C' =  D' – A'  \\implies$\n$\\triangle A'C'E' = \\triangle D'F'B'.$\nThe translation vector maps $\\triangle A'C'E'$ into $\\triangle D'F'B'$ is $2\\vec {V.}$\n$X = \\frac {A+D}{2} =  \\frac { (A+ E – F) + (D + F – E)}{2} =  \\frac {C' + B'}{2} = \\frac {E' + F'}{2},$\nso $X$ is midpoint of $AD, B'C',$ and $E'F'.$ Symilarly $Y$ is the midpoint of $BE, A'F',$ and $C'D', Z$ is the midpoint of $CF, A'B',$ and $D'E'.$\n$Z + V =  \\frac {A' + B'}{2}+ \\frac {D' – A'}{2} =  \\frac {B' + D'}{2} = Z'$ is the midpoint of $B'D'.$\nSimilarly $X' = X + V$  is the midpoint of $B'F',Y'= Y + V$  is the midpoint of $D'F'.$\nTherefore $\\triangle X'Y'Z'$ is the medial triangle of $\\triangle B'D'F'.$\n$\\triangle XYZ$ is $\\triangle X'Y'Z'$ translated on $– \\vec {V}.$\nIt is known (see diagram) that circumcenter of triangle coincide with orthocenter of the medial triangle. Therefore  orthocenter $H$ of $\\triangle XYZ$ is circumcenter of $\\triangle B'D'F'$ translated on $– \\vec {V}.$\nIt is the midpoint of segment $OO'$ connected circumcenters of $\\triangle B'D'F'$ and $\\triangle A'C'E'.$\nAccording to the definition of points $A', C', E',$ quadrangles $ABCE', CDEA',$ and $AFEC'$ are parallelograms. Hence\n\n Power of points A,C, and E with respect circumcircle $\\triangle A'C'E'$ is equal, hence distances between these points and circumcenter of $\\triangle A'C'E'$ are the same. Therefore circumcenter  $\\triangle ACE$ coincide with circumcenter  $\\triangle A'C'E'.$\nSimilarly circumcenter of $\\triangle BDF$ coincide with circumcenter of $\\triangle B'D'F'.$\nvladimir.shelomovskii@gmail.com, vvsss\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
    },
    {
        "Problem": "Let $a$ and $b$ be positive integers. The cells of an $(a+b+1)\\times (a+b+1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.",
        "Solution": "https://www.youtube.com/watch?v=137-z4WahKQ",
        "Solution_2": "We proceed with induction. \nBase case: $a=b=1$. \nWe fill in a 3x3 grid that has at least one bronze and one amber cell. We choose one bronze and one amber cell. Note that if the given bronze and amber cells are not in the same row or column, we are done. However, if they are, consider a cell that is not in the same row or column as either cell. Regardless of the color of the cell, we can match it with an opposing color cell, hence proved.\nInduction case: Given the assertion for $(a, b)$ is true, then $(a+1, b)$ is true as well.\nLabel each cell in the $(a+b+1)(a+b+1)$ grid as either red, blue, or white. There must be exactly $a^2+ab-b$ red cells, $b^2+ab-a$ blue cells, and $(a+b+1)(a+b+1)-(a^2+ab-b)-(b^2+ab-a) = 3(a+b) + 1$ white cells. We show that we can achieve a combination regardless of the amber/bronze of the white cells.\nCase 1: There are three or more white cells present in either the bottommost row or rightmost column.\nLet's ignore the bottom most row and rightmost column, and focus on the $(a+b+1)(a+b+1)$ grid of squares on the topleft. By the induction assumption, we know that this grid must have $a$ amber and $b$ bronze not in the same row or column pairwise. \n[cont]"
    },
    {
        "Problem": "Let $b\\geq2$ and $w\\geq2$ be fixed integers, and $n=b+w$. Given are $2b$ identical black rods and $2w$ identical white rods, each of side length $1$.\nWe assemble a regular $2n$-gon using these rods so that parallel sides are the same color. Then, a convex $2b$-gon $B$ is formed by translating the black rods, and a convex $2w$-gon $W$ is formed by translating the white rods. An example of one way of doing the assembly when $b=3$ and $w=2$ is shown below, as well as the resulting polygons $B$ and $W$.\n\nProve that the difference of the areas of $B$ and $W$ depends only on the numbers $b$ and $w$, and not on how the $2n$-gon was assembled.\n",
        "Solution_1": "First notice that the black rods and the white rods form polygons iff in the original $2n$-gon, if a side is a color $x$, then the side that is parallel to that side in the original $2n$-gon is also the color $x$.\nWe can prove that the difference in areas is only affected by the values of $b$ and $w$ by showing that for any valid arrangement of $2b$ rods and $2w$ rods, we may switch any two adjacent black and white rods(and their \"parallel pairs\"), and end up with the same area difference.\n\nIn the figure above (click to expand), after the switch, we can see that after removing the mutually congruent parts, we are left with two parallelograms from each color. Let $x$, $\\alpha{}$, $y$, and $\\beta{}$ be defined as shown. Notice that if we angle chase, the sides of the other parallelogram are the same, but if the angles of the original $2n$-gon all have measure $2k$, the angles of the new parallelograms are $\\alpha{}+180-2k$ and $\\beta{}+180-2k$, as shown. We must prove that the differences between the areas are the same.\nUsing area formulas, the change in the difference of areas is $x\\sin{\\alpha{}}-y\\sin{\\beta{}}+y\\sin{(2k-\\beta{})}-x\\sin{(2k-\\alpha{})}$, which is equal to $x(\\sin{k}(2\\cos^2{k})-2\\sin{k}\\cos{k}\\cos{\\alpha{}})-y(\\sin{k}(2\\cos^2{k}))+y(\\sin{k}\\cos{k}\\cos{\\beta{}})$, or $2\\cos{k}(x\\sin{(\\alpha{}-k)}+y\\sin{(k-\\beta{}}))$. Since $\\cos{k}$ is not $0$ because $n\\geq{}3$, we are left with proving that $x\\sin{(\\alpha{}-k)}+y\\sin{(k-\\beta{}})=0$.\nNow we rotate the polygon so that the vertex between the two sides that we switched is at the point $(0,0)$, the angle bisector of that vertex is $y=0$, and the black side is in the positive $y$-direction. Now think of all the sides as vectors, all pointing in the clockwise direction of the $2n$-gon.\nNotice the part labeled $a$ in the black polygons. We have that the vector labeled $x$ is really just the sum of all of the vectors in the part labeled $a$ - or all the vectors in the $2n$-gon that are in the positive $y$-direction excluding the one that was interchanged. Also notice that the angle of this vector $x$ has a signed angle of $\\alpha{}-k$ with $y=0$ and has length $x$ - meaning that the vertical displacement of the vector $x$ from $y=0$ is equal to $x\\sin{(\\alpha{}-k)}$! Similarly, we get that the vertical displacement of the vector $y$ is equivalent to $y\\sin{(k-\\beta{}})$.\nAdding these two together, we get that $x\\sin{(\\alpha{}-k)}+y\\sin{(k-\\beta{}})$ is simply the vertical displacement of the sum of the vectors $x$ and $y$. Since the sum of the vectors $x$ and $y$ is equivalent to the sum of the vectors in the positive half of the polygon minus the sum of the black vector that would be switched with the white vector(the leftmost vector in the positive half of the polygon) and the rightmost vector in the positive half(which is the parallel pair of the white vector that would be interchanged later), and we know that this sum happens to have a vertical displacement of $0$, along with the fact that the positive half of the polygon summed together also has a vertical displacement of $0$, we get that the total vertical displacement is $0$, meaning that $x\\sin{(\\alpha{}-k)}+y\\sin{(k-\\beta{}})=0$, and we are done.\n~by @peppapig_\n",
        "Solution_2": "Pick a pair of parallel sides of the regular $2n$-gon and flip their color. Let those two lines be the horizontal. The area of the polygon formed by rods of the original color decreases by a parallelogram with height equal to the sum of the vertical heights of the rods of that color divided by $2$ and base length $1$. Similarly, the area of the polygon formed by rods of the new color increases by the sum of the vertical heights of the rods of the new color divided by $2$. The sum of the heights of all the rod is equal to twice the height of the $2n$-gon, so the difference between the areas of $B$ and $W$ changes by the height of the $2n$-gon, which is fixed when $n$ is fixed.\nIf we pair two pairs different-colored pairs of parallel sides of the $2n$-gon and flip the colors of both pairs, then $b$ and $w$ and the difference in the areas of $B$ and $W$ will remain unchanged. Thus, the difference in the areas of $B$ and $W$ depends only on $b$ and $w$."
    },
    {
        "Problem": "Let $\\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f:\\mathbb{R}_{>0} \\to \\mathbb{R}_{>0}$ such that for all $x,y\\in \\mathbb{R}_{>0}$ we have",
        "Solution": "[WIP]"
    },
    {
        "Problem": "Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares.",
        "Solution_1": "Since $q(p-1)$ is a perfect square and $q$ is prime, we should have $p - 1 = qb^2$ for some positive integer $b$. Let $a^2 = p - q$. Therefore, $q = p - a^2$, and substituting that into the $p - 1 = qb^2$ and solving for $p$ gives\n\nNotice that we also have\n\nand so $b^2 - 1 | a^2 - 1$. We run through the cases\nand each factor is greater than $1$, contradicting the primality of $p$.\nThus, the only solution is $(p, q) = (3, 2)$.",
        "Solution_2": "Let $p-q = a^2$, $pq - q = b^2$, where $a, b$ are positive integers. $b^2 - a^2 = pq - q - (p-q) = pq -p$. So,\n$\\bullet$ For $q=2$, $p = b^2 - a^2 = (b-a)(b+a)$. Then $b-a=1$ and $b+a=p$. $a=\\dfrac{p-1}{2}$ and $p-q = a^2$. Thus, $p - 2 = \\left( \\dfrac{p-1}{2} \\right)^2 \\implies p^2 - 6p + 9 = 0$ and we find $p=3$. Hence $(p,q) = (3,2)$.\n$\\bullet$ For $q=4k+3$, ($k\\geq 0$ integer), by $(1)$, $p(4k+2) = b^2 - a^2$. Let's examine in $\\mod 4$, $b^2 - a^2 \\equiv 2 \\pmod{4}$. But we know that $b^2 - a^2 \\equiv 0, 1 \\text{ or } 3 \\pmod{4}$. This is a contradiction and no solution for $q = 4k + 3$.\n$\\bullet$ For $q=4k+1$, ($k > 0$ integer), by $(1)$, $p(4k) = b^2 - a^2$. Let $k=m\\cdot n$, where $m\\geq n \\geq 1$ and $m, n$ are integers. Since $p>q$, we see $p>4k$. Thus, by $(1)$, $(b-a)(b+a) = 4p\\cdot m \\cdot n$. $b-a$ and $b+a$ are same parity and  $4p\\cdot m \\cdot n$ is even integer. So, $b-a$ and $b+a$ are both even integers. Therefore,\n$\\left\\{ \\begin{array}{rcr} b+a =  & 2pn \\\\ b-a =  & 2m  \\end{array} \\right.$\nor gay sex\n$\\left\\{ \\begin{array}{rcr} b+a =  & 2pm \\\\ b-a =  & 2n  \\end{array} \\right.$\nTherefore, $a=pn - m$ or $a = pm - n$. For each case, $p-q = p - 4mn - 1 < a$. But $p-q = a^2$, this gives a contradiction. No solution for $q = 4k + 1$.\nWe conclude that the only solution is $(p,q) = (3,2)$.\n(Lokman GÖKÇE)"
    },
    {
        "Problem": "A function $f: \\mathbb{R}\\to \\mathbb{R}$ is $\\textit{essentially increasing}$ if $f(s)\\leq f(t)$ holds whenever $s\\leq t$ are real numbers such that $f(s)\\neq 0$ and $f(t)\\neq 0$.\nFind the smallest integer $k$ such that for any 2022 real numbers $x_1,x_2,\\ldots , x_{2022},$ there exist $k$ essentially increasing functions $f_1,\\ldots, f_k$ such that",
        "Solution": "Coming soon."
    },
    {
        "Problem": "There are $2022$ users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.)\nStarting now, Mathbook will only allow a new friendship to be formed between two users if they have at least two friends in common. What is the minimum number of friendships that must already exist so that every user could eventually become friends with every other user?",
        "Solution_1": "No Solution Here Yet!"
    },
    {
        "Problem_2": "Let $\\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f:\\mathbb{R}^{+}\\rightarrow\\mathbb{R}^{+}$ such that, for all $x, y \\in \\mathbb{R}^{+}$,",
        "Solution_1": "Make the following substitutions to the equation:\n1. $(x, 1) \\rightarrow f(x + f(x)) = xf(1) + 2$\n2. $(1, x + f(x)) \\rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4$\n3. $(x, 1 + \\frac{f(1)}{x}) \\rightarrow f(x + f(x) + f(1)) = xf\\biggl(1 + \\frac{f(1)}{x}\\biggr) + 2$\nIt then follows from (2) and (3) that $f(1 + \\frac{f(1)}{x}) = f(1) + \\frac{2}{x}$, so we know that this function is linear for $x > 1$. Substitute $f(x) = ax+b$ and solve for $a$ and $b$ in the functional equation; we find that $f(x) = x + 1 \\forall x > 1$.\nNow, we can let $x > 1$ and $y \\le 1$. Since $f(x) = x + 1$, $xy + f(x) > x > 1$, so $f(xy + f(x)) = xy + x + 2 = xf(y) + 2$. It becomes clear then that $f(y) = y + 1$ as well, so $f(x) = x + 1$ is the only solution to the functional equation.\n~jkmmm3"
    },
    {
        "Problem": "Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.",
        "Solution": "We claim the answer is $(\\frac{n+1}{2})^2$.\nFirst, consider a checkerboard tiling of the board with 4 colors: R, G, B, Y. Number each column from $1$ to $n$ from left to right and each row from $1$ to $n$ from top to bottom. We color a tile R if its row and column are odd, a tile G is its row is even but its column is odd, a tile B if its row and column is even, and a tile Y if its row is odd but its column is even.\nLemma 1: Throughout our moves, the color of the uncolored tile stays an invariant.\nConsider that a domino can either only change rows or can only change columns. Therefore, sliding a domino into the hole and creating a new one has two possible colors. Of these, note that the new hole will always trivially be two tiles away from the old hole, meaning that the parity of both the row and column number stays the same. Thus, the lemma holds.\nLemma 2: There are more red tiles than any other color. \nBecause each color is uniquely defined by the parity of a pair of column and row number, it satisfies to show that given an odd integer $n$, there are more odd positive integers less than or equal to $n$ than even ones. Obviously, this is true, and so red will have more tiles than any other color.\nLemma 3: For any starting configuration $C$ and any blank tile $B$ such that the blank tile's color matches the blank tile's color of $C$, there is no more than one unique configuration $C'$ that can be produced from $C$ using valid moves.\nWe will use proof by contradiction. Assume there exists two different $C'$. We can get from one of these $C'$ to another using moves. However, we have to finish off with the same hole as before. Before the last move, the hole must be two tiles away from the starting hole. However, because the domino we used to move into the blank tile's spot is in the way, that hole must be congruent to the hole produced after the first move. We can induct this logic, and because there is a limited amount of tiles with the same color, eventually we will run out of tiles to apply this to. Therefore, having two distinct $C'$ with the same starting hole $B$ is impossible with some $C$.\nWe will now prove that $(\\frac{n+1}{2})^2$ is the answer. There are $\\frac{n+1}{2}$ rows and $\\frac{n+1}{2}$ columns that are odd, and thus there are $(\\frac{n+1}{2})^2$ red tiles. Given lemma 3, this is our upper bound for a maximum. To establish that $(\\frac{n+1}{2})^2$ is indeed possible, we construct such a $C$:\nIn the first column, leave the first tile up blank. Then, continuously fill in  vertically oriented dominos in that column until it reaches the bottom.\nIn the next $n-1$ columns, place $\\frac{n-1}{2}$ vertically oriented dominos in a row starting from the top. At the bottom row, starting with the first unfilled tile on the left, place horizontally aligned dominos in a row until you reach the right.\nObviously, the top left tile is red. It suffices to show that any red tile may be uncovered. For the first column, one may slide some dominos on the first column until the desired tile is uncovered. For the bottom row, all the first dominos may be slid up, and then the bottom dominos may be slid to the left until the desired red tile is uncovered. Finally, for the rest of the red tiles, the bottom red tile in the same color may be revealed, and then vertically aligned dominos in the same column may be slid down until the desired tile is revealed. Therefore, this configuration may produce $(\\frac{n+1}{2})^2$ different configurations with moves.\nHence, we have proved that $(\\frac{n+1}{2})^2$ is the maximum, and we are done. $\\blacksquare{}$\n~SigmaPiE\n"
    },
    {
        "Problem": "A positive integer $a$ is selected, and some positive integers are written on a board. Alice and Bob play the following game. On Alice's turn, she must replace some integer $n$ on the board with $n+a$, and on Bob's turn he must replace some even integer $n$ on the board with $n/2$. Alice goes first and they alternate turns. If on his turn Bob has no valid moves, the game ends.\nAfter analyzing the integers on the board, Bob realizes that, regardless of what moves Alice makes, he will be able to force the game to end eventually. Show that, in fact, for this value of $a$ and these integers on the board, the game is guaranteed to end regardless of Alice's or Bob's moves.\n",
        "Solution_1": "The contrapositive of the claim is somewhat easier to conceptualize: If it is not guaranteed that the game will end (i.e. the game could potentially last forever), then Bob is not able to force the game to end (i.e. Alice can force it to last forever). So we want to prove that, if the game can potentially last indefinitely, then Alice can force it to last indefinitely.\nClearly, if there is $1$ number on the board initially, all moves are forced. This means the claim is true in this specific case, because if the game \"potentially\" can last forever, this means it must last forever, since the game can only be played in one way. Ergo Alice can \"force\" this to occur because it is guaranteed to occur. Now we look at all cases where there is more than $1$ number on the board.\nCase 1: $v_2 (a)=0$\nThe game lasts forever here no matter what. This is true because, if the game ends, it means the board was in some position $P$, Alice added $a$ to some number on the board, and all the numbers now on the board are odd. If there are only odd numbers on the board in position $P$, Alice will add $a$ to some number on the board, making it even, meaning this cannot have been the state $P$ of the board. If at least one number n on the board is even, Alice can add $a$ to a number other than $n$, meaning there is still at least 1 even number on the board, meaning this also cannot have been the state $P$ of the board. This covers all possible boards when  $v_2 (a)=0$, so we're done.\nCase 2: $v_2 (a)=1$\nIf there is at least one number n on the board that is even, the game can also last forever. On any move, Alice will add $a$ to this number $n$ if and only if $v_2 (n)=1$. This way, the new number $n'= n+a$ satisfies $v_2 (n') \\geq 2$. If Bob does divides $n'$ until $v_2 (n)=1$, Alice will again add $a$ to $n$ resulting in $v_2 (n') \\geq 2$. This means that Alice can always keep an even number on the board for Bob to divide no matter how Bob plays.\nIf there is no even number on the board, then the game can clearly not last forever. No matter what Alice does, Bob will have no even number to divide after her move.\nGeneral Case: $v_2 (a)=x$\nIn general, it seems to be the case that the game can last indefinitely for some given $a$ if and only if there exists some number $n$ on the board such that $v_2 (n) \\geq v_2 (a)$, so we should aim to prove this.\n1. \"If\"\nAlice can apply a similar strategy to the strategy used in case 2. On any move, Alice will add $a$ to $n$ if and only if $v_2 (n)=v_2 (a)$. If she does this addition, then $v_2 (n') \\geq v_2 (a)+1$, keeping an even number on the board. Even if Bob divides $n'$ until $v_2 (n)=v_2 (a)$, Alice will apply the same strategy and keep $v_2 (n') \\geq v_2 (a)+1$. Alice's use of this strategy ensures that there always exists some number n on the board such that $v_2 (n) \\geq v_2 (a)$, ensuring there always exists an even number n on the board.\n2.\"Only If\"\nIf $v_2 (n) < v_2 (a)$ for all n on the board, this means that Alice can never change the value of $v_2 (n)$ for any $n$ on the board. Only Bob can do this, and Bob will subtract $1$ from each $v_2 (n)$ until they are all equal to $0$ (all odd), ending the game.\nWe've shown that the game can last indefinitely iff there exists some number $n$ on the board such that $v_2 (n) \\geq v_2 (a)$, and have shown that Alice can ensure the game lasts forever in these scenarios using the above strategy. This proves the contrapositive, proving the claim.\n~SpencerD."
    },
    {
        "Problem": "Let $n\\geq3$ be an integer. We say that an arrangement of the numbers $1$, $2$, $\\dots$, $n^2$ in a $n \\times n$ table is row-valid if the numbers in each row can be permuted to form an arithmetic progression, and column-valid if the numbers in each column can be permuted to form an arithmetic progression. For what values of $n$ is it possible to transform any row-valid arrangement into a column-valid arrangement by permuting the numbers in each row?",
        "Solution_1": "The answer is all $\\boxed{\\text{prime } n}$."
    },
    {
        "Problem": "Let ABC be a triangle with incenter $I$ and excenters $I_a$, $I_b$, $I_c$ opposite $A$, $B$, and $C$, respectively. Given an arbitrary point $D$ on the circumcircle of $\\triangle ABC$ that does not lie on any of the lines $II_{a}$, $I_{b}I_{c}$, or $BC$, suppose the circumcircles of $\\triangle DIIa$ and $\\triangle DI_bI_c$ intersect at two distinct points $D$ and $F$. If $E$ is the intersection of lines $DF$ and $BC$, prove that $\\angle BAD = \\angle EAC$.",
        "Solution_1": "\nConsider points $G,H,J,K,P,$ and $Q$ such that the intersections of the circumcircle of $\\triangle{}ABC$ with the circumcircle of $\\triangle{}DII_A$ are $D$ and $G$, the intersections of the circumcircle of $\\triangle{}ABC$ with the circumcircle of $\\triangle{}DI_BI_C$ are $D$ and $H$, the intersections of the circumcircle of $\\triangle{}ABC$ with line $\\overline{II_A}$ are $A$ and $J$, the intersections of the circumcircle of $\\triangle{}ABC$ with line $\\overline{I_BI_C}$ are $A$ and $K$, the intersection of lines $\\overline{II_A}$ and $\\overline{BC}$ is $P$, and the intersection of lines $\\overline{I_BI_C}$ and $\\overline{BC}$ is $Q$.\nSince $IBI_AC$ is cyclic, the pairwise radical axes of the circumcircles of $\\triangle{}DII_A,\\triangle{}ABC,$ and $IBI_AC$ concur. The pairwise radical axes of these circles are $\\overline{GD},\\overline{II_A},$ and $\\overline{BC}$, so $G,P,$ and $D$ are collinear. Similarly, since $BCI_BI_C$ is cyclic, the pairwise radical axes of the cirucmcircles of $\\triangle{}DI_BI_C,\\triangle{}ABC,$ and $BCI_BI_C$ concur. The pairwise radical axes of these circles are $\\overline{HD},\\overline{I_BI_C},$ and $\\overline{BC}$, so $H,Q,$ and $D$ are collinear. This means that $-1=(Q,P;B,C)\\stackrel{D}{=}(H,G;B,C)$, so the tangents to the circumcircle of $\\triangle{}ABC$ at $G$ and $H$ intersect on $\\overline{BC}$. Let this intersection be $X$. Also, let the intersection of the tangents to the circumcircle of $\\triangle{}ABC$ at $K$ and $J$ be a point at infinity on $\\overline{BC}$ called $Y$ and let the intersection of lines $\\overline{KG}$ and $\\overline{}HJ$ be $Z$. Then, let the intersection of lines $\\overline{GJ}$ and $\\overline{HK}$ be $E'$. By Pascal's Theorem on $GGJHHK$ and $GJJHKK$, we get that $X,E',$ and $Z$ are collinear and that $E',Y,$ and $Z$ are collinear, so $E',X,$ and $Y$ are collinear, meaning that $E'$ lies on $\\overline{BC}$ since both $X$ and $Y$ lie on $\\overline{BC}$.\n\nConsider the transformation which is the composition of an inversion centered at $A$ and a reflection over the angle bisector of $\\angle{}CAB$ that sends $B$ to $C$ and $C$ to $B$. We claim that this sends $D$ to $E'$ and $E'$ to $D$. It is sufficient to prove that if the transformation sends $G$ to $G'$, then $AE'JG'$ is cyclic. Notice that $\\triangle{}AGB\\sim\\triangle{}ACG'$ since $\\angle{}GAB=\\angle{}G'AC$ and $\\tfrac{AG'}{AC}=\\tfrac{\\frac{AB\\cdot{}AC}{AG}}{AC}=\\tfrac{AB}{AG}$. Therefore, we get that $\\angle{}AG'E'=\\angle{}ABG=\\angle{}AJE'$, so $AE'JG'$ is cyclic, proving the claim. This means that $\\angle{}BAE'=\\angle{}CAD$.\n\nWe claim that $\\angle{}I_BE'I_C+\\angle{}I_BDI_C=180^\\circ$. Construct $D'$ to be the intersection of line $\\overline{AE'}$ and the circumcircle of $\\triangle{}E'I_BI_C$ and let $B'$ and $C'$ be the intersections of lines $\\overline{AC}$ and $\\overline{AB}$ with the circumcircle of $\\triangle{}BI_BI_C$. Since $B'$ and $C'$ are the reflections of $B$ and $C$ over $\\overline{I_BI_C}$, it is sufficient to prove that $A,B',C',D'$ are concyclic. Since $\\overline{B'C},\\overline{D'E'},$ and $\\overline{I_BI_C}$ concur and $D',E',I_B,I_C$ and $I_B,I_C,B',C$ are concyclic, we have that $B',C,D',E'$ are concyclic, so $\\angle{}B'D'A=\\angle{}ACE'=\\angle{}AC'B'$, so $A,B',C',D'$ are concyclic, proving the claim. We can similarly get that $\\angle{}IE'I_A=\\angle{}IDI_A$.\n\nLet line $\\overline{E'J}$ intersect the circumcircle of $\\triangle{}DII_A$ at $G$ and $J'$. Notice that $J$ is the midpoint of $\\overline{II_A}$ and $\\angle{}IE'I_A=\\angle{}IDI_A=\\angle{}IJ'I_A$, so $IE'I_AJ'$ is a parallelogram with center $J$, so $\\tfrac{EJ}{EJ'}=\\tfrac{1}{2}$. Similarly, we get that if line $\\overline{E'K}$ intersects the circumcircle of $\\triangle{}DI_BI_C$ at $H$ and $K'$, we have that $\\tfrac{EK}{EK'}=\\tfrac{1}{2}$, so $\\overline{KJ}\\parallel\\overline{K'J'}$, so $\\angle{}HGJ'=\\angle{}HGJ=\\angle{}HKJ=\\angle{}HK'J'$, so $G,H,J',K'$ are concyclic. Then, the pairwise radical axes of the circumcircles of $\\triangle{}DII_A,\\triangle{}DI_BI_C,$ and $GHJ'K'$ are $\\overline{DF},\\overline{HK'},$ and $\\overline{GJ'}$, so $\\overline{DF},\\overline{HK'},$ and $\\overline{GJ'}$ concur, so $\\overline{DF},\\overline{HK},$ and $\\overline{GJ}$ concur, so $E=E'$. We are then done since $\\angle{}BAE'=\\angle{}CAD$.\n~Zhaom",
        "Solution_2": "Set $\\triangle ABC$ as the reference triangle, and let $D=(x_0,y_0,z_0)$ with homogenized coordinates. To find the equation of circle $DII_A$, we note that $I=(a:b:c)$ (not homogenized) and $I_A=(-a:b:c)$. Thus, for this circle with equation $a^2yz+b^2xz+c^2xy=(x+y+z)(u_1x+v_1y+w_1z)$, we compute that   \nFor circle $DI_BI_C$ with equation $a^2yz+b^2xz+c^2xy=(x+y+z)(u_2x+v_2y+w_2z)$,, we find that   \nThe equation of the radical axis is $ux+vy+wz=0$ with $u=u_1-u_2$, $v=v_1-v_2$, and $w=w_1-w_2$. We want to consider the intersection of this line with line $\\overline{BC}$, so set $x=0$. The equation reduces to $vy+wz=0$. We see that $v=\\frac{2bc^3x_0y_0}{b^2z_0^2-c^2y_0^2}$ and $w=\\frac{2b^3cx_0z_0}{c^2y_0^2-b^2z_0^2}$, so  which is the required condition for $\\overline{AD}$ and $\\overline{AE}$ to be isogonal.\n~MathIsFun286"
    }
]