[ { "Problem": "Prove that the fraction $\\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$.", "Solution_1": "Denoting the greatest common divisor of $a, b$ as $(a,b)$, we use the Euclidean algorithm:\n\nTheir greatest common divisor is 1, so $\\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.", "Solution_2": "Proof by contradiction:\nAssume that $\\dfrac{14n+3}{21n+4}$ is a reducible fraction where $p$ is the greatest common factor of $14n+3$ and $21n + 4$. Thus,\nSubtracting the second equation from the first equation we get $1\\equiv 0\\pmod{p}$ which is clearly absurd.\nHence $\\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.", "Solution_3": "Proof by contradiction:\nAssume that $\\dfrac{14n+3}{21n+4}$ is a reducible fraction.\nIf a certain fraction $\\dfrac{a}{b}$ is reducible, then the fraction $\\dfrac{2a}{3b}$ is reducible, too.\nIn this case, $\\dfrac{2a}{3b} = \\dfrac{42n+8}{42n+9}$.\nThis fraction consists of two consecutives numbers, which never share any factor. So in this case, $\\dfrac{2a}{3b}$ is irreducible, which is absurd.\nHence $\\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.\n", "Solution_4": "We notice that:\nSo it follows that $7n+1$ and $14n+3$ must be coprime for every natural number $n$ for the fraction to be irreducible. Now the problem simplifies to proving $\\frac{7n+1}{14n+3}$ irreducible. We re-write this fraction as:\nSince the denominator $2(7n+1) + 1$ differs from a multiple of the numerator $7n+1$ by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that $\\frac{21n+4}{14n+3}$ is irreducible.\nQ.E.D\n", "Solution_5": "By Bezout's Lemma, $3 \\cdot (14n+3) - 2 \\cdot (21n + 4) = 1$, so the GCD of the numerator and denominator is $1$ and the fraction is irreducible.", "Solution_6": "To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with $\\frac{21n}{14n}$ + $\\frac{4}{3}$.\nSimplifying the fraction, we end up with. $\\frac{3}{2}$. Now combining these fractions with addition as shown before in the problem, we end up with $\\frac{17}{6}$. It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us\n$\\frac{25}{17}$. Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16).\nTrying 2, we end up with $\\frac{46}{31}$. Again, both results are 1 off from being multiples of 15.\nTrying 3, we end up with $\\frac{67}{45}$. Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:\n1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but\n2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us $\\frac{319}{199}$, we keep ending up with results that at the end will only have a common divisor of 1.\n", "Solution_7": "Let $\\gcd (21n+4,14n+3)=a.$ So for some co-prime positive integers $x,y$ we have Multiplying $(1)$ by $2$ and $(2)$ by $3$ and then subtracting $(1)$ from $(2)$ we get We must have $a=1$, since $a$ is an positive integer. Thus, $\\gcd(21n+4,14n+3)=1$ which means the fraction is irreducible, as needed." }, { "Problem": "Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$.", "Solution_1": "Let $N = 100a + 10b+c$ for some digits $a,b,$ and $c$. Then for some $m$. We also have $m=a^2+b^2+c^2$. Substituting this into the first equation and simplification, we get For an integer divisible by $11$, the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by $11$. Thus we get: $b = a + c$ or $b = a + c - 11$.\nCase $1$: Let $b=a+c$. We get Since the right side is even, the left side must also be even. Let $c=2q$ for some $q = 0,1,2,3,4$. Then \nSubstitute $q=0,1,2,3,4$ into the last equation and then solve for $a$.\nWhen $q=0$, we get $a=5$. Thus $c=0$ and $b=5$. We get that $N=550$ which works.\nWhen $q=1$, we get that $a$ is not an integer. There is no $N$ for this case.\nWhen $q=2$, we get that $a$ is not an integer. There is no $N$ for this case.\nWhen $q=3$, we get that $a$ is not an integer. There is no $N$ for this case.\nWhen $q=4$, we get that $a$ is not an integer. There is no $N$ for this case.\nCase $2$: Let $b = a + c - 11$. We get \nNow we test all $c=0\\rightarrow10$. When $c=0,1,2,4,5,6,7,8,9$, we get no integer solution to $a$. Thus, for these values of $c$, there is no valid $N$. However, when $c=3$, we get We get that $a=8$ is a valid solution. For this case, we get $a=8,b=0,c=3$, so $N=803$, and this is a valid value. Thus, the answers are $\\boxed{N=550,803}$.", "Solution_2": "Define a ten to be all ten positive integers which begin with a fixed tens digit.\nWe can make a systematic approach to this:\nBy inspection, $\\dfrac{N}{11}$ must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well.\nFor a given ten, the sum of the squares of the digits of $N$ increases faster than $\\dfrac{N}{11}$, so we can have at most one number in every ten that works.\nWe check the first ten:\n$11*11=121$\n$1^2+2^2+1^2=4$\n$12*11=132$\n$1^2+3^2+2^2=14$\n11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work.\nWe try the second ten:\n$21*11=231$\n$2^2+3^2+1^2=14$\n$22*11=242$\n$2^2+4^2+2^2=24$\nTherefore, no numbers in the second ten work.\nWe continue, to find out that 50 and 73 are the only ones that works.\n$N=50*11=550$, $N=73*11=803$ so there are two $N$ that works.", "Solution_3": "Note that there are only 900 three-digit numbers. Of these, only 81 are divisible by 11. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer. There is plenty of time on the IMO, and listing is highly unlikely to miss cases. We end up getting $\\boxed{N=550,803}$.", "Solution_4": "https://artofproblemsolving.com/community/c6h54826p22231968\n~franzliszt" }, { "Problem": "(Hungary)\nSolve the system of equations:\nwhere $a$ and $b$ are constants. Give the conditions that $a$ and $b$ must satisfy so that $x, y, z$ (the solutions of the system) are distinct positive numbers.", "Solution": "Note that $x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2$, so the first two equations become\nWe note that $(x+y)^2 - z^2 = \\Big[ (x+y)+z \\Big]\\Big[ (x+y)-z\\Big]$, so if $a$ equals 0, then $b$ must also equal 0. We then have $x+y = -z$; $xy = (x+y)^2$. This gives us $x^2 + xy + y^2 = 0$. Mutiplying both sides by $(x-y)$, we have $x^3 - y^3 = 0$. Since we want $x,y$ to be real, this implies $x = y$. But $x^2 + x^2 + x^2$ can only equal 0 when $x=0$ (which, in this case, implies $y,z = 0$). Hence there are no positive solutions when $a = 0$.\nWhen $a \\neq 0$, we divide $(**)$ by $(*)$ to obtain the system of equations\nwhich clearly has solution $x+y = \\frac{a^2 + b^2}{2a}$, $z = \\frac{a^2 - b^2}{2a}$. In order for these both to be positive, we must have positive $a$ and $a^2 > b^2$. Now, we have $x+y = \\frac{a^2 + b^2}{2a}$; $xy = \\left(\\frac{a^2 - b^2}{2a}\\right)^2$, so $x,y$ are the roots of the quadratic $m^2 - \\frac{a^2 + b^2}{2a}m + \\left(\\frac{a^2 - b^2}{2a}\\right)^2$. The discriminant for this equation is\nIf the expressions $(3a^2 - b^2), (3b^2 - a^2)$ were simultaneously negative, then their sum, $2(a^2 + b^2)$, would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when $3a^2 > b^2$ and $3b^2 > a^2$. But we have already replaced the first inequality with the sharper bound $a^2 > b^2$. It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from $\\left(\\frac{a^2 + b^2}{2a}\\right)^2 > \\left(\\frac{a^2 + b^2}{2a}\\right)^2 - \\left(2\\frac{a^2 -b^2}{2a}\\right)^2$ or from Descartes' Rule of Signs). We have now found the solutions to the system, and determined that it has positive solutions if and only if $a$ is positive and $3b^2 > a^2 > b^2$. Q.E.D.\n" }, { "Problem": "Find the smallest natural number $n$ which has the following properties:\n(a) Its decimal representation has 6 as the last digit.\n(b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number $n$.", "Solution_1": "As the new number starts with a $6$ and the old number is $1/4$ of the new number, the old number must start with a $1$.\nAs the new number now starts with $61$, the old number must start with $\\lfloor 61/4\\rfloor = 15$.\nWe continue in this way until the process terminates with the new number $615\\,384$ and the old number $n=\\boxed{153\\,846}$.", "Solution_2": "Let the original number = $10n + 6$, where $n$ is a 5 digit number.\nThen we have $4(10n + 6) = 600000 + n$.\n=> $40n + 24 = 600000 + n$.\n=> $39n = 599976$.\n=> $n = 15384$.\n=> The original number = $\\boxed{153\\,846}$." }, { "Problem": "Find all real roots of the equation\nwhere $p$ is a real parameter.", "Solution": "Assuming $x \\geq 0$, square the equation, obtaining\n.\nIf we have $p + 4 \\geq 4x^2$, we can square again, obtaining\nWe must have $4 - 2p > 0 \\iff p < 2$, so we have\nHowever, this is only a solution when\nso we have $p\\geq 0$ and $p \\leq \\frac {4}{3}$\nand $x = \\frac {4 - p}{2\\sqrt {4 - 2p}}$" }, { "Problem": "(a) Find all positive integers $n$ for which $2^n-1$ is divisible by $7$.\n(b) Prove that there is no positive integer $n$ for which $2^n+1$ is divisible by $7$.", "Solution_1": "We claim $2^n$ is equivalent to $2, 4,$ and $1$ $\\pmod{7}$ for $n$ congruent to $1$, $2$, and $0$ $\\pmod{3}$, respectively.\n(a) From the statement above, only $n$ divisible by $3$ will work.\n(b) Again from the statement above, $2^n$ can never be congruent to $-1$ $\\pmod{7}$, so there are no solutions for $n$.\n", "Solution_1.1": "This solution is clearer and easier to understand.\n(1) Since we know that $2^n-1$ is congruent to 0 (mod 7), we know that $2^n$ is congruent to 8 mod 7, which means $2^n$ is congruent to 1 mod 7.\nExperimenting with the residue of $2^n$ mod 7:\n$n$=1: 2\n$n$=2: 4\n$n$=3: 1 (this is because when $2^n$ is doubled to $2*2^n$, the residue doubles too, but $4*2=8$ is congruent to 1 (mod 7).\n$n$=4: 2\n$n$=5: 4\n$n$=6: 1\nThrough induction, we easily show that this is true since the residue doubles every time you double $2^n$.\nSo, the residue of $2^n$ mod 7 cycles in 2, 4, 1. Therefore, $n$ must be a multiple of 3. Proved.\n(2) According to part (1), the residue of $2^n$ cycles in 2, 4, 1.\nIf $2^n+1$ is congruent to 0 mod 7, then $2^n$ must be congruent to 6 mod 7, but this is not possible due to how $2^n$ mod 7 cycles. Therefore, there is no solution. Proved.\n~hastapasta" }, { "Problem": "Determine all values $x$ in the interval $0\\leq x\\leq 2\\pi$ which satisfy the inequality", "Solution": "We shall deal with the left side of the inequality first ($2\\cos x \\leq \\left| \\sqrt{1+\\sin 2x} - \\sqrt{1-\\sin 2x } \\right|$) and the right side after that.\nIt is clear that the left inequality is true when $\\cos x$ is non-positive, and that is when $x$ is in the interval $[\\pi/2, 3\\pi/2]$. We shall now consider when $\\cos x$ is positive. We can square the given inequality, and the resulting inequality will be true whenever the original left inequality is true. $4\\cos^2{x}\\leq 1+\\sin 2x+1-\\sin 2x-2\\sqrt{1-\\sin^2 2x}=2-2\\sqrt{\\cos^2{2x}}$. This inequality is equivalent to $2\\cos^2 x\\leq 1-\\left| \\cos 2x\\right|$. I shall now divide this problem into cases.\nCase 1: $\\cos 2x$ is non-negative. This means that $x$ is in one of the intervals $[0,\\pi/4]$ or $[7\\pi/4, 2\\pi]$. We must find all $x$ in these two intervals such that $2\\cos^2 x\\leq 1-\\cos 2x$. This inequality is equivalent to $2\\cos^2 x\\leq 2\\sin^2 x$, which is only true when $x=\\pi/4$ or $7\\pi/4$.\nCase 2: $\\cos 2x$ is negative. This means that $x$ is in one of the interavals $(\\pi/4, \\pi/2)$ or $(3\\pi/2, 7\\pi/4)$. We must find all $x$ in these two intervals such that $2\\cos^2 x\\leq 1+\\cos 2x$, which is equivalent to $2\\cos^2 x\\leq 2\\cos^2 x$, which is true for all $x$ in these intervals.\nTherefore the left inequality is true when $x$ is in the union of the intervals $[\\pi/4, \\pi/2)$, $(3\\pi/2, 7\\pi/4]$, and $[\\pi/2, 3\\pi/2]$, which is the interval $[\\pi/4, 7\\pi/4]$. We shall now deal with the right inequality.\nAs above, we can square it and have it be true whenever the original right inequality is true, so we do that. $2-2\\sqrt{\\cos^2{2x}}\\leq 2$, which is always true. Therefore the original right inequality is always satisfied, and all $x$ such that the original inequality is satisfied are in the interval $[\\pi/4, 7\\pi/4]$.\n" }, { "Problem": "Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another.", "Solution_1": "In triangle $ABC$, let $BC=a$, $AC=b$, $AB=c$, $\\angle ABC=\\alpha$, and $\\angle BAC=2\\alpha$. Using the Law of Sines gives that\n\nTherefore $\\cos{\\alpha}=\\frac{a}{2b}$. Using the Law of Cosines gives that\n\nThis can be simplified to $a^2c=b(a^2+c^2-b^2)$. Since $a$, $b$, and $c$ are positive integers, $b|a^2c$. Note that if $b$ is between $a$ and $c$, then $b$ is relatively prime to $a$ and $c$, and $b$ cannot possibly divide $a^2c$. Therefore $b$ is either the least of the three consecutive integers or the greatest.\nAssume that $b$ is the least of the three consecutive integers. Then either $b|b+2$ or $b|(b+2)^2$, depending on if $a=b+2$ or $c=b+2$. If $b|b+2$, then $b$ is 1 or 2. $b$ couldn't be 1, for if it was then the triangle would be degenerate. If $b$ is 2, then $b(a^2+c^2-b^2)=42=a^2c$, but $a$ and $c$ must be 3 and 4 in some order, which means that this triangle doesn't exist. therefore $b$ cannot divide $b+2$, and so $b$ must divide $(b+2)^2$. If $b|(b+2)^2$ then $b|(b+2)^2-b^2-4b=4$, so $b$ is 1, 2, or 4. Clearly $b$ cannot be 1 or 2, so $b$ must be 4. Therefore $b(a^2+c^2-b^2)=180=a^2c$. This shows that $a=6$ and $c=5$, and the triangle has sides that measure 4, 5, and 6.\nNow assume that $b$ is the greatest of the three consecutive integers. Then either $b|b-2$ or $b|(b-2)^2$, depending on if $a=b-2$ or $c=b-2$. $b|b-2$ is absurd, so $b|(b-2)^2$, and $b|(b-2)^2-b^2+4b=4$. Therefore $b$ is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so $b$ cannot be the greatest of the three consecutive integers.\nThis shows that there is exactly one triangle with this property - and it has side lengths of 4, 5, and 6. $\\blacksquare$\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.", "Solution_2": "[Incomplete, please edit]\nIn a given triangle $ABC$, let $A=2B$, $\\implies C=180-3B$, and $\\sin C=\\sin 3B$.\nThen \nHence, $(*)$ $a^2 = b(b+c)$ (with assumptions. This needs clearing up)\nIf $b$ is the shortest side, $(b+2)^2 = b^2 +b(b+1)$\n$\\implies (b-4)(b+1)=0$,\n$\\implies b=4, c=5, a=6$,\nNo other permutation of $a$, $b$ and $c$ in terms of size gives integral values to $(*)$ [show]. So there is only one such triangle.", "Solution_3": "NO TRIGONOMETRY!!!\nLet $a, b, c$ be the side lengths of a triangle in which $\\angle C = 2\\angle B.$\nExtend $AC$ to $D$ such that $CD = BC = a.$ Then $\\angle CDB = \\frac{\\angle ACB}{2} = \\angle ABC$, so $ABC$ and $ADB$ are similar by AA Similarity. Hence, $c^2 = b(a+b)$. Then proceed as in Solution 2, as only algebraic manipulations are left." }, { "Problem": "Prove that there are infinitely many natural numbers $a$ with the following property: the number $z = n^4 + a$ is not prime for any natural number $n$.", "Solution": "Suppose that $a = 4k^4$ for some $a$. We will prove that $a$ satisfies the property outlined above.\nThe polynomial $n^4 + 4k^4$ can be factored as follows:\n\nBoth factors are positive, because if the left one is negative, then the right one would also negative, which is clearly false.\nIt is also simple to prove that $n^2 + 2k^2 - 2nk > 1$ when $k > 1$. Thus, for all $k > 2$, $4k^4$ is a valid value of $a$, completing the proof. $\\square$\n~mathboy100\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Let $M$ be a point on the side $AB$ of $\\triangle ABC$. Let $r_1, r_2$, and $r$ be the inscribed circles of triangles $AMC, BMC$, and $ABC$. Let $q_1, q_2$, and $q$ be the radii of the exscribed circles of the same triangles that lie in the angle $ACB$. Prove that", "Solution": "We use the conventional triangle notations.\nLet $I$ be the incenter of $ABC$, and let $I_{c}$ be its excenter to side $c$. We observe that\nand likewise,\nSimplifying the quotient of these expressions, we obtain the result\nThus we wish to prove that\nBut this follows from the fact that the angles $AMC$ and $CBM$ are supplementary.", "Solution_2": "By similar triangles and the fact that both centers lie on the angle bisector of $\\angle{C}$, we have $\\frac{r}{q} = \\frac{s-c}{s} = \\frac{a + b - c}{a + b + c}$, where $s$ is the semi-perimeter of $ABC$. Let $ABC$ have sides $a, b, c$, and let $AM = c_1, MB = c_2, MC = d$. After simple computations, we see that the condition, whose equivalent form is\n\nis also equivalent to Stewart's Theorem\n\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 following assertion is true for $n=3$ and $n=5$, and that it is false for every other natural number $n>2:$\nIf $a_1, a_2,\\cdots, a_n$ are arbitrary real numbers, then $(a_1-a_2)(a_1-a_3)\\cdots (a_1-a_n)+(a_2-a_1)(a_2-a_3)\\cdots (a_2-a_n)+\\cdots+(a_n-a_1)(a_n-a_2)\\cdots (a_n-a_{n-1})\\ge 0.$", "Solution": "Take $a_1 < 0$, and the remaining $a_i = 0$. Then $E_n = a_1(n-1) < 0$ for $n$ even,\nso the proposition is false for even $n$.\nSuppose $n \\ge 7$ and odd. Take any $c > a > b$, and let $a_1 = a$, $a_2 = a_3 = a_4= b$,\nand $a_5 = a_6 = ... = a_n = c$.\nThen $E_n = (a - b)^3 (a - c)^{n-4} < 0$.\nSo the proposition is false for odd $n \\ge 7$.\nAssume $a_1 \\ge a_2 \\ge a_3$.\nThen in $E_3$ the sum of the first two terms is non-negative, because $a_1 - a_3 \\ge a_2 - a3$.\nThe last term is also non-negative.\nHence $E_3 \\ge 0$, and the proposition is true for $n = 3$.\nIt remains to prove $S_5$.\nSuppose $a_1 \\ge a_2 \\ge a_3 \\ge a_4 \\ge a_5$.\nThen the sum of the first two terms in $E_5$ is\n$(a_1 - a_2){(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5)} \\ge 0$.\nThe third term is non-negative (the first two factors are non-positive and the last two non-negative).\nThe sum of the last two terms is:\n$(a_4 - a_5){(a_1 - a_5)(a_2 - a_5)(a_3 - a_5) - (a_1 - a_4)(a_2 - a_4)(a_3 - a_4)} \\ge 0$.\nHence $E_5 \\ge 0$.\nThis solution was posted and copyrighted by e.lopes. The original thread can be fond here: [1]" }, { "Problem": "Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum.", "Solution": "There are $2^{10}-2=1022$ distinct subsets of our set of 10 two-digit numbers. The sum of the elements of any subset of our set of 10 two-digit numbers must be between $10$ and $90+91+92+93+94+95+96+97+98+99 < 10 \\cdot 100 = 1000$. (There are fewer attainable sums.) As $1000 < 1022$, the Pigeonhole Principle implies there are two distinct subsets whose members have the same sum. Let these sets be $A$ and $B$. Now, let $A' = A - (A \\cap B)$ and $B' = B - (A \\cap B)$. Notice $A'$ and $B'$ are disjoint. They are also nonempty because if $A = A \\cap B$ or $B = A \\cap B$, then one of $A$ and $B$ is a subset of the other, so they are either not distinct or have different sums. Therefore $A'$ and $B'$ are disjoint subsets our set of 10 distinct two-digit numbers, which proves the claim. $\\square$" }, { "Problem": "Point $O$ lies on line $g;$ $\\overrightarrow{OP_1}, \\overrightarrow{OP_2},\\cdots, \\overrightarrow{OP_n}$ are unit vectors such that points $P_1, P_2, \\cdots, P_n$ all lie in a plane containing $g$ and on one side of $g.$ Prove that if $n$ is odd, Here $\\left|\\overrightarrow{OM}\\right|$ denotes the length of vector $\\overrightarrow{OM}.$", "Solution": "We prove it by induction on the number $2n+1$ of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the $2n-1$ vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm $\\ge 1$ betwen two with norm $1$. The sum of the two vectors of norm $1$ makes an angle of $\\le\\frac\\pi 2$ with the vector of norm $\\ge 1$, so their sum has norm $\\ge 1$, and we're done.\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Three players $A, B$ and $C$ play the following game: On each of three cards an integer is written. These three numbers $p, q, r$ satisfy $0 < p < q < r$. The three cards are shuffled and one is dealt to each player. Each then receives the number of counters indicated by the card he holds. Then the cards are shuffled again; the counters remain with the players.\nThis process (shuffling, dealing, giving out counters) takes place for at least two rounds. After the last round, $A$ has 20 counters in all, $B$ has 10 and $C$ has 9. At the last round $B$ received $r$ counters. Who received $q$ counters on the first round?", "Solution": "Answer: player $C$.\nLet $n$ be the number of rounds played, then obviously $n (p + q + r) = 20 + 10 + 9 = 39$. So $n$ must be a divisor of 39, i. e. $n \\in \\{ 1, 3, 13, 39 \\}$. But $p \\geq 1,$ $q \\geq p + 1 \\geq 2,$ $r \\geq q + 1 \\geq 3,$ so $p + q + r \\geq 1 + 2 + 3 = 6$ and $n = \\frac{39}{p + q + r} \\leq \\frac{39}{6} < 7$. Also, by condition $n \\geq 2,$ so we conclude to $n = 3$ and $p + q + r = \\frac{39}{n} = 13$.\nAs $B$ received 10 counters including $r$ counters at the last round, $10 \\geq p + p + r \\geq 2 + r,$ so $r \\leq 8$. On the other hand, from number of counters received by $A$ we get $r > \\frac{20}{3} > 6$. So $r \\in \\{ 7, 8 \\}$.\nIf $r = 7,$ from number of counters received by $B$ we get $3 = 10 - r \\in \\{ 2p, p + q, 2q \\},$ and since 3 is odd, we get $p + q = 3$. But then $p + q + r = 3 + 7 = 10 \\neq 13$ - a contradiction.\nSo $r = 8$ and $2 = 10 - r \\in \\{ 2p, p + q, 2q \\}$. On the other hand, $2 \\leq 2p < p + q < 2q,$ so $2 = 2p,$ $p = 1$ and $q = 13 - p - r = 4$. It is easy to check that the only way to distribute the counters for players is $20 = r + r + q,$ $10 = p + p + r$ and $9 = q + q + p,$ so player $C$ received $q$ counters on the first round." }, { "Problem": "Let $x_i, y_i$ $(i=1,2,\\cdots,n)$ be real numbers such that Prove that, if $z_1, z_2,\\cdots, z_n$ is any permutation of $y_1, y_2, \\cdots, y_n,$ then", "Solution": "We can expand and simplify the inequality a bit, and using the fact that $z$ is a permutation of $y$, we can cancel some terms.\n\n\nConsider the pairing $x_1 \\rightarrow y_1$, $x_2 \\rightarrow y_2$, ... $x_n \\rightarrow y_n$. By switching around some of the $y$ values, we have obtained the pairing $x_1 \\rightarrow z_1$, $x_2 \\rightarrow z_2$, ... $x_n \\rightarrow z_n$. Suppose that we switch around two $y$-values, $y_m$ and $y_n$, such that $y_m > y_n$. If $x_m > x_n$, call this a type 1 move. Otherwise, call this a type 2 move.\nType 2 moves only increase the sum of the products of the pairs. The sum is increased by $x_n \\cdot y_m + x_m \\cdot y_n - x_m \\cdot y_m - x_n \\cdot y_n$. This is equivalent to $(x_n - x_m)(y_m - y_n)$, which is clearly nonnegative since $y_m > y_n$ and $x_n > x_m$.\nWe will now consider switching from the $x$-$z$ pairing back to the $x$-$y$ pairing. We will prove that from any pairing of $x$ and $z$ values, you can use just type 2 moves to navigate back to the pairing of $x$ and $y$ values, which will complete the proof.\nSuppose that $x_a$ is the biggest $x$-value that is not paired with its $y$-value in the $x$ and $z$ pairing. Then, switch this $y$-value with the $y$-value currently paired with $x_a$. This is obviously a type 2 move. Continue this process until you reach back to the $x$-$y$ pairing. All moves are type 2 moves, so the proof is complete.\n~mathboy100", "Solution_2": "We can rewrite the summation as \n\nSince $\\sum^n_{i=1} y_i^2 = \\sum^n_{i=1} z_i^2$, the above inequality is equivalent to \n\nWe will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of $\\sum^n_{i=1}x_iz_i$. Obviously, if $x_1 = x_2 = \\ldots = x_n$ or $y_1 = y_2 = \\ldots = y_n$, the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of $z_i$s that are not ordered in the form $z_1 \\ge z_2 \\ge \\ldots \\ge z_n$ such that $\\sum^n_{i=1}x_iz_i$ is at a maximum out of all possible permutations and is greater than the sum $\\sum^n_{i=1}x_iy_i$. This necessarily means that in the sum $\\sum^n_{i=1}x_iz_i$ there exists two terms $x_pz_m$ and $x_qz_n$ such that $x_p > x_q$ and $z_m < z_n$. Notice that \n\nwhich means if we make the terms $x_pz_n$ and $x_qz_m$ instead of the original $x_pz_m$ and $x_qz_n$, we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality \n \nis proved, which is equivalent to what we wanted to prove.\n\n~Imajinary" }, { "Problem": "In a convex quadrilateral (in the plane) with the area of $32 \\text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \\text{ cm}$. Determine all the possible values that the other diagonal can have.", "Solution": "Label the vertices $A$, $B$, $C$, and $D$ in such a way that $AB + BD + DC = 16$, and $\\overline{BD}$ is a diagonal.\nThe area of the quadrilateral can be expressed as $BD \\cdot ( d_1 + d_2 ) / 2$, where $d_1$ and $d_2$ are altitudes from points $A$ and $C$ onto $\\overline{BD}$. Clearly, $d_1 \\leq AB$ and $d_2 \\leq DC$. Hence the area is at most $BD \\cdot ( AB + DC ) / 2 = BD(16-BD) / 2$.\nThe quadratic function $f(x)=x(16-x)/2$ has its maximum for $x=8$, and its value is $f(8)=32$.\nThe area of our quadrilateral is $32$. This means that we must have $BD=8$. Also, equality must hold in both $d_1 \\leq AB$ and $d_2 \\leq DC$. Hence both $\\overline{AB}$ and $\\overline{DC}$ must be perpendicular to $\\overline{BD}$. And in any such case it is clear from the Pythagorean theorem that $AC = 8\\sqrt 2$.\nTherefore the other diagonal has only one possible length: $8\\sqrt 2$." }, { "Problem": "In the interior of a square $ABCD$ we construct the equilateral triangles $ABK, BCL, CDM, DAN.$ Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon.", "Solution": "Just use complex numbers, with $a = 1$, $b = i$, $c = - 1$ and $d = - i$.\nWith some calculations, we have $k = \\frac {\\sqrt {3} - 1}{2}( - 1 - i)$, $l = \\frac {\\sqrt {3} - 1}{2}(1 - i)$, $m = \\frac {\\sqrt {3} - 1}{2}(1 + i)$ and $n = \\frac {\\sqrt {3} - 1}{2}( - 1 + i)$.\nNow it's an easy job to calculate the twelve midpoints and to find out they are all of the form $\\frac {\\sqrt {3} - 1}{2}e^{\\frac {k\\pi}{6}i}$, with $k\\in\\mathbb{N}: 0\\le k\\le 11$, and the result follows.\nThe above solution was posted and copyrighted by Joao Pedro Santos. The original thread for this problem can be found here: [1]" }, { "Problem": "Let $m$ and $n$ be positive integers such that $1 \\le m < n$. In their decimal representations, the last three digits of $1978^m$ are equal, respectively, to the last three digits of $1978^n$. Find $m$ and $n$ such that $m + n$ has its least value.", "Solution": "We have $1978^m\\equiv 1978^n\\pmod {1000}$, or $978^m-978^n=1000k$ for some positive integer $k$ (if it is not positive just do $978^n-978^m=-1000k$). Hence $978^n\\mid 1000k$. So dividing through by $978^n$ we get $978^{m-n}-1=\\frac{1000k}{978^n}$. Observe that $2\\nmid LHS$, so $2\\nmid RHS$. So since $2|| 978^n$, clearly the minimum possible value of $n$ is $3$ (and then $489^n\\mid k$). We will show later that if $n$ is minimal then $m$ is minimal. We have $978^{m-3}-1\\equiv 0\\pmod {125}\\Leftrightarrow 103^{m-3}\\equiv 1\\pmod {125}$. Hence, $m-3\\mid \\varphi(125)\\Rightarrow m-3\\mid 100$. Checking by hand we find that only $m-3=100$ works (this also shows that minimality of $m$ depends on $n$, as claimed above). So $m=103$. Consequently, $m+n=106$ with $\\boxed{(m,n)=(103,3)}$.\nThe above solution was posted and copyrighted by cobbler and Andreas. The original thread for this problem can be found here: [1] and [2]" }, { "Problem": "If $p$ and $q$ are natural numbers so thatprove that $p$ is divisible with $1979$.", "Solution": "We first write\nNow, observe that\nand similarly $\\frac{1}{661}+\\frac{1}{1318}=\\frac{1979}{661\\cdot 1318}$ and $\\frac{1}{662}+\\frac{1}{1317}=\\frac{1979}{662\\cdot 1317}$, and so on. We see that the original equation becomes\nwhere $s=660\\cdot 661\\cdots 1319$ and $r=\\frac{s}{660\\cdot 1319}+\\frac{s}{661\\cdot 1318}+\\cdots+\\frac{s}{989\\cdot 990}$ are two integers. Finally consider $p=1979\\cdot\\frac{qr}{s}$, and observe that $s\\nmid 1979$ because $1979$ is a prime, it follows that $\\frac{qr}{s}\\in\\mathbb{Z}$. Hence we deduce that $p$ is divisible with $1979$.\nThe above solution was posted and copyrighted by Solumilkyu. The original thread for this problem can be found here: [1]" }, { "Problem": "$P$ is a point inside a given triangle $ABC$. $D, E, F$ are the feet of the perpendiculars from $P$ to the lines $BC, CA, AB$, respectively. Find all $P$ for which\nis least.", "Solution": "We note that $BC \\cdot PD + CA \\cdot PE + AB \\cdot PF$ is twice the triangle's area, i.e., constant. By the Cauchy-Schwarz Inequality,\nwith equality exactly when $PD = PE = PF$, which occurs when $P$ is the triangle's incenter, 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": "The function $f(n)$ is defined on the positive integers and takes non-negative integer values. $f(2)=0,f(3)>0,f(9999)=3333$ and for all $m,n:$Determine $f(1982)$.", "Solution_1": "Clearly $f(1) \\ge 1 \\Rightarrow f(m+1) \\ge f(m)+f(1) \\ge f(m)+1$ so $f(9999) \\ge 9999$.Contradiction!So $f(1)=0$.This forces $f(3)=1$.Hence $f(3k+3) \\ge f(3k)+f(3)>f(3k)$ so the inequality $f(3) 0$, so $f(3) = 1$. It follows by induction that $f(3n) >= n$. For $f(3n+3) = f(3) + f(3n)$ + $0$ or $1 = f(3n) + 1$ or $2$. Moreover if we ever get $f(3n) > n$, then the same argument shows that $f(3m) > m$ for all $m > n$. But $f(3.3333) = 3333$, so $f(3n) = n$ for all $n <= 3333$. Now $f(3n+1) = f(3n) + f(1) + 0$ or $1$ = $n$ or $n + 1$. But $3n+1 = f(9n+3) >= f(6n+2) + f(3n+1) >= 3f(3n+1)$, so $f(3n+1) < n+1$. Hence $f(3n+1) = n.$ Similarly, $4f(3n+2) = n$. In particular $f(1982) = 660$.\nThis solution was posted and copyrighted by Tega. The original thread for this problem can be found here: [3]", "Solution_4": "Similar to solution 3.\nProof:\nLemma 1: $f(mn)\\geq nf(m)$\nLet, $P(m,m)$ be assertion.\nSimilarly,we can induct to get $f(nm)\\geq nf(m)$.\nLemma proved.\nThen we see that,\nThen,\nThen we can easily get,by assertion $P(1980,2)$\nHence, $\\boxed{f(1980)=660,661}$.And, we are done. $\\blacksquare$\nThis solution was posted and copyrighted by IMO2019. The original thread for this problem can be found here: [4]\n" }, { "Problem": "Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy the conditions: \n(i) $f(xf(y))=yf(x)$ for all $x,y$; \n(ii) $f(x)\\to0$ as $x\\to \\infty$.", "Solution_1": "Let $x=y=1$ and we have $f(f(1))=f(1)$. Now, let $x=1,y=f(1)$ and we have $f(f(f(1)))=f(1)f(1)\\Rightarrow f(1)=[f(1)]^2$ since $f(1)>0$ we have $f(1)=1$.\nPlug in $y=x$ and we have $f(xf(x))=xf(x)$. If $a=1$ is the only solution to $f(a)=a$ then we have $xf(x)=1\\Rightarrow f(x)=\\frac{1}{x}$. We prove that this is the only function by showing that there does not exist any other $a$:\nSuppose there did exist such an $a\\ne1$. Then, letting $y=a$ in the functional equation yields $f(xa)=af(x)$. Then, letting $x=\\frac{1}{a}$ yields $f(\\frac{1}{a})=\\frac{1}{a}$. Notice that since $a\\ne1$, one of $a,\\frac{1}{a}$ is greater than $1$. Let $b$ equal the one that is greater than $1$. Then, we find similarly (since $f(b)=b$) that $f(xb)=bf(x)$. Putting $x=b$ into the equation, yields $f(b^2)=b^2$. Repeating this process we find that $f(b^{2^k})=b^{2^k}$ for all natural $k$. But, since $b>1$, as $k\\to \\infty$, we have that $b^{2^k}\\to\\infty$ which contradicts the fact that $f(x)\\to0$ as $x\\to \\infty$.", "Solution_2": "Let $x=1$ so If $f(a)=f(b),$ then $af(1)=f(f(a))=f(f(b))=bf(1)\\implies a=b$ because $f(1)$ goes to the real positive integers, not $0.$ Hence, $f$ is injective. Let $x=y$ so \n so $xf(x)$ is a fixed point of $f.$ Then, let $y=1$ so $f(xf(1))=f(x)\\implies f(1)=1$ as $x$ can't be $0$ so $1$ is a fixed point of $f.$ We claim $1$ is the only fixed point of $f.$ Suppose for the sake of contradiction that $a,b$ be fixed points of $f$ so $f(a)=a$ and $f(b)=b.$ Then, setting $x=a,y=b$ in (i) gives so $ab$ is also a fixed point of $f.$ Also, let $x=\\frac{1}{a},y=a$ so so $\\frac{1}{a}$ is a fixed point of $f.$ If $f(a)=a$ with $a>1,$ then $f(a^n)$ is a fixed point of $f$, contradicting (ii). If $f(a)=a$ with $00$ and $z-x-\\epsilon=z-\\frac{1}{3}>0$. If we repeat the process, defining after similar reasoning, we see that ." }, { "Problem": "A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.", "Solution_1": "Let $O$ be the center of the circle mentioned in the problem. Let $T$ be the second intersection of the circumcircle of $CDO$ with $AB$. By measures of arcs, $\\angle DTA = \\angle DCO = \\frac{\\angle DCB}{2} = \\frac{\\pi}{2} - \\frac{\\angle DAB}{2}$. It follows that $AT = AD$. Likewise, $TB = BC$, so $AD + BC = AB$, as desired.", "Solution_2": "Let $O$ be the center of the circle mentioned in the problem, and let $T$ be the point on $AB$ such that $AT = AD$. Then $\\angle DTA = \\frac{ \\pi - \\angle DAB}{2} = \\angle DCO$, so $DCOT$ is a cyclic quadrilateral and $T$ is in fact the $T$ of the previous solution. The conclusion follows.", "Solution_3": "Let the circle have center $O$ and radius $r$, and let its points of tangency with $BC, CD, DA$ be $E, F, G$, respectively. Since $OEFC$ is clearly a cyclic quadrilateral, the angle $COE$ is equal to half the angle $GAO$. Then\nLikewise, $DG = OB - EB$. It follows that\nQ.E.D.", "Solution_4": "We use the notation of the previous solution. Let $X$ be the point on the ray $AD$ such that $AX = AO$. We note that $OF = OG = r$; $\\angle OFC = \\angle OGX = \\frac{\\pi}{2}$; and $\\angle FCO = \\angle GXO = \\frac{\\pi - \\angle BAD}{2}$; hence the triangles $OFC, OGX$ are congruent; hence $GX = FC = CE$ and $AO = AG + GX = AG + CE$. Similarly, $OB = EB + GD$. Therefore $AO + OB = AG + GD + CE + EB$, Q.E.D.", "Solution_5": "This solution is incorrect. The fact that $BC$ is tangent to the circle does not necessitate that $B$ is its point of tangency. -Nitinjan06\nFrom the fact that AD and BC are tangents to the circle mentioned in the problem, we have\n$\\angle{CBA}=90\\deg$\nand\n$\\angle{DAB}=90\\deg$.\nNow, from the fact that ABCD is cyclic, we obtain that\n$\\angle{BCD}=90\\deg$\nand\n$\\angle{CDA}=90\\deg$,\nsuch that ABCD is a rectangle.\nNow, let E be the point of tangency between the circle and CD. It follows, if O is the center of the circle, that\n$\\angle{OEC}=\\angle{OED}=90\\deg$\nSince $AO=EO=BO$, we obtain two squares, $AOED$ and $BOEC$.\nFrom the properties of squares we now have\n$AD+BC=AO+BO=AB$\nas desired.", "Solution_6": "Lemma. Let $I$ be the in-center of $ABC$ and points $P$ and $Q$ be on the lines $AB$ and $BC$ respectively. Then $BP + CQ = BC$ if and only if $APIQ$ is a cyclic quadrilateral.\nSolution. Assume that rays $AD$ and $BC$ intersect at point $P$. Let $S$ be the center od circle touching $AD$, $DC$ and $CB$. Obviosuly $S$ is a $P$-ex-center of $PDB$, hence $\\angle DSI=\\angle DSP = \\frac{1}{2} \\angle DCP=\\frac{1}{2} \\angle A=\\angle DAI$ so DASI is concyclic.", "Solution_1_2": "https://www.youtube.com/watch?v=tM0WhXNCWGU\n", "Solution_2_2": "https://youtu.be/Ormv0y4ZM1E\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 any positive integer not equal to $2, 5$ or $13$. Show that one can find distinct $a,b$ in the set $\\{2,5,13,d\\}$ such that $ab-1$ is not a perfect square.", "Solution_1": "We do casework with modular arithmetic.\n$d\\equiv 0,3 \\pmod{4}: 13d-1$ is not a perfect square.\n$d\\equiv 2\\pmod{4}: 2d-1$ is not a perfect square.\nTherefore, $d\\equiv 1, \\pmod{4}.$ Now consider $d\\pmod{16}.$\n$d\\equiv 1,13 \\pmod{16}: 13d-1$ is not a perfect square.\n$d\\equiv 5,9\\pmod{16}: 5d-1$ is not a perfect square.\nAs we have covered all possible cases, we are done.", "Solution_2": "Proof by contradiction:\nSuppose $p^2=2d-1$, $q^2=5d-1$ and $r^2=13d-1$. From the first equation, $p$ is an odd integer. Let $p=2k-1$. We have $d=2k^2-2k+1$, which is an odd integer. Then $q^2$ and $r^2$ must be even integers, denoted by $4n^2$ and $4m^2$ respectively, and thus $r^2-q^2=4m^2-4n^2=8d$, from which\n can be deduced. Since $m^2-n^2$ is even, $m$ and $n$ have the same parity, so $(m+n)(m-n)$ is divisible by $4$. It follows that the odd integer $d$ must be divisible by $2$, leading to a contradiction. 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": "Let $p_n (k)$ be the number of permutations of the set $\\{ 1, \\ldots , n \\} , \\; n \\ge 1$, which have exactly $k$ fixed points. Prove that\n(Remark: A permutation $f$ of a set $S$ is a one-to-one mapping of $S$ onto itself. An element $i$ in $S$ is called a fixed point of the permutation $f$ if $f(i) = i$.)", "Solution": "The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element $i$ of the set, there are $(n-1)!$ permutations which have $i$ as a fixed point. Therefore\nas desired.\nSlightly Clearer Solution\nFor any $k$, if there are $p_n(k)$ permutations that have $k$ fixed points, then we know that each fixed point is counted once in the product $k \\cdot p_n{k}$. Therefore the given sum is simply the number of fixed points among all permutations of $\\{ 1, \\ldots , n \\}$. However, if we take any $x$ such that $1 \\le x \\le n$ and $x$ is a fixed point, there are $(n-1)!$ ways to arrange the other numbers in the set. Therefore our desired sum becomes $n \\cdot (n-1)! = n!$, so we are done.", "Solution_2": "The probability of any number $i$ where $1\\le i\\le n$ being a fixed point is $\\frac{1}{n}$. Thus, the expected value of the number of fixed points is $n\\times \\frac{1}{n}=1$.\nThe expected value is also $\\sum_{k=0}^{n} \\frac{k \\cdot p_n (k)}{n!}$.\nThus, or" }, { "Problem": "Prove that in the set $\\{1,2, \\ldots, 1989\\}$ can be expressed as the disjoint union of subsets $A_i, \\{i = 1,2, \\ldots, 117\\}$ such that\ni.) each $A_i$ contains 17 elements\nii.) the sum of all the elements in each $A_i$ is the same.\n", "Solution": "Let us start pairing numbers in the following fashion, where each pair sums to $1990$:\n$(1, 1990-1), (2, 1990-2), \\ldots (936, 1054)$\nThere are a total of $117*8$ pairs above. Let us start putting these pairs into each of the $117$ subsets starting with the first pair $(1, 1990-1)$ going into $A_1$, $(2, 1990-2)$ into $A_2$ and so on $\\ldots$ with $(k, 1990-k)$ going into $A_{k(mod 117)}$.\nNow we have $16$ numbers present in each of the $117$ subsets all of which have the same total sum. We need $1$ more number to be filled in each subset from the remaining $117$ numbers $(937, 938, ... 1053)$.\nLet us arrange these $117$ numbers in the following manner:\n$995-58, 995-57, \\ldots, 995-1, 995, 995+1, 995+2, \\ldots, 995+57, 995+58$ $==> (1)$\nNow notice that for each number that's off by $x$ from $995$, there's a counter number off by $x$ from $995$ in the opposite direction.\nSo we need to create similar imbalances in the Subsets $A_1$ to $A_{117}$, so that we could add the above $117$ numbers to those imbalances to keep the total sum unchanged.\nNow start swapping the $1st$ number of each pair that we added to subsets $A_1$ to $A_{117}$ to create the above imbalances:\nSwap $1$ from $A_1$ with $59$ from $A_{59}$ - this creates an imbalance of $+58$ and $-58$\nSwap $2$ from $A_2$ with $58$ from $A_{58}$ - this creates an imbalance of $+56$ and $-56$\n$\\ldots$\nSwap $29$ from $A_{29}$ with $31$ from $A_{31}$ - creates an imbalance of $+2$ and $-2$\nLeave $30$ in $A_{30}$ as it is - $0$ imbalance.\nSimilarly,\nSwap $60$ from $A_{60}$ with $117$ from $A_{117}$ - this creates an imbalance of $+57$ and $-57$\nSwap $61$ from $A_{61}$ with $116$ from $A_{116}$ - this creates an imbalance of $+55$ and $-55$\n$\\ldots$\nSwap $88$ from $A_{88}$ with $89$ from $A_{89}$ - this creates an imbalance of $+1$ and $-1$\nNow start adding the $117$ numbers from $(1)$ to the subsets $A_1$ to $A_{117}$ to counter the imbalances $-58$ to $+58$ so that the sum remains unchanged. Notice that each subset now has $17$ elements with total sum = $8*1990 + 995 = 16915$.\n$- Kris17$\n" }, { "Problem": "Chords $AB$ and $CD$ of a circle intersect at a point $E$ inside the circle. Let $M$ be an interior point of the segment $\\overline{EB}$. The tangent line at $E$ to the circle through $D, E$, and $M$ intersects the lines $\\overline{BC}$ and ${AC}$ at $F$ and $G$, respectively.\nIf $\\frac{AM}{AB} = t$, find $\\frac{EG}{EF}$ in terms of $t$.", "Solution_1": "With simple angle chasing, we find that triangles $CEG$ and $MDB$ are similar.\nso, $GE/EC = MD/MB$. (*)\nagain with simple angle chasing, we find that triangles $CEF$ and $AMD$ are similar.\nso, $EF/DM = CE/AM$. (**)\nso, by (*) and (**), we have $GE/EF = MA/MB = t/(t-1)$.\nThis solution was posted and copyrighted by e.lopes. The original thread for this problem can be found here: [1]", "Solution_2": "This problem can be bashed with PoP and Ratio Lemma. Rewriting the given ratio gets $\\frac{MA}{MB}=\\frac{t}{1-t}$.\nBy Ratio Lemma, $\\frac{FB}{FC}=\\frac{BE}{CE} \\cdot \\frac{\\sin{\\angle{FEB}}}{\\sin{\\angle{FEC}}}=\\frac{DE}{AE} \\cdot \\frac{\\sin{\\angle{EDM}}}{\\sin{\\angle{DME}}}=\\frac{DE}{AE} \\cdot \\frac{EM}{DE}=\\frac{ME}{EA}$. Similarly, $\\frac{GA}{GC}=\\frac{ME}{EB}$. We can rewrite these equalities to get $\\frac{AM}{EM}=\\frac{BC}{FB}$ and $\\frac{BM}{EM}=\\frac{AC}{GC}$.\nUsing Ratio Lemma, $\\frac{GE}{\\sin{\\angle{ACD}}}=\\frac{GC}{\\sin{\\angle{GED}}}$ and $\\frac{EF}{\\sin{\\angle{BCD}}}=\\frac{FC}{\\sin{\\angle{FEC}}}$. Since $\\angle{GED}=\\angle{FEC}$, we have $\\frac{FE}{GE}=\\frac{FC}{GC} \\cdot \\frac{\\sin{\\angle{BCD}}}{\\sin{\\angle{ACD}}}$ (eq 1). Note that by Ratio Lemma, $\\frac{\\sin{\\angle{BCD}}}{\\sin{\\angle{ACD}}}=\\frac{CA}{CB} \\cdot \\frac{EB}{EA}$. Plugging this into (eq 1), we get $\\frac{EF}{GE}=\\frac{FC}{GC} \\cdot \\frac{CA}{CB} \\cdot \\frac{EB}{EA}=\\frac{\\frac{EA}{EM} \\cdot FB}{\\frac{EB}{EM} \\cdot GA} \\cdot \\frac{CA}{CB} \\cdot \\frac{EB}{EA}=\\frac{FB}{GA} \\cdot \\frac{CA}{CB}=\\frac{EM}{AM} \\cdot \\frac{MB}{EM}=\\frac{MB}{MA}=\\frac{1-t}{t}$. So $\\frac{EG}{EF}=\\boxed{\\frac{t}{1-t}}$.\nThis solution was posted and copyrighted by AIME12345. The original thread for this problem can be found here: [2]" }, { "Problem": "Find all integers $a$, $b$, $c$ satisfying $1 < a < b < c$ such that $(a - 1)(b -1)(c - 1)$ is a divisor of $abc - 1$.", "Solution": "$1<\\frac{abc-1}{(a-1)(b-1)(c-1)}<\\frac{abc}{(a-1)(b-1)(c-1)}$\n$1<\\frac{abc-1}{(a-1)(b-1)(c-1)}<\\left(\\frac{a}{a-1}\\right) \\left(\\frac{b}{b-1}\\right) \\left(\\frac{c}{c-1}\\right)$\nWith $11$ is an integer. Prove that $f\\left(x\\right)$ cannot be expressed as the product of two non-constant polynomials with integer coefficients.", "Solution": "For the sake of contradiction, assume that $f\\left(x\\right)=g\\left(x\\right)h\\left(x\\right)$ for polynomials $g\\left(x\\right)$ and $h\\left(x\\right)$ in $\\mathbb{R}$. Furthermore, let $g\\left(x\\right)=b_mx^m+b_{m-1}x^{m-1}+\\ldots+b_1x+b_0$ with $b_i=0$ if $i>m$ and $h\\left(x\\right)=c_{n-m}x^{n-m}+c_{n-m-1}x^{n-m-1}+\\ldots+c_1x+c_0$ with $c_i=0$ if $i>n-m$. This gives that $f\\left(x\\right)=\\sum_{i=0}^{n}\\left(\\sum_{j=0}^{i}b_jc_{i-j}\\right)x^i$.\nWe have that $3=b_0c_0$, or $3|b_0c_0$. WLOG, let $3|b_0$ (and thus $3\\not|c_0$). Since $b_0c_1+b_1c_0=0$ and $3$ divides $b_0$ but not $c_0$, we need that $3|b_1$. We can keep on going up the chain until we get that $3|b_{n-2}$. Then, by equating coefficients once more, we get that $b_0c_{n-1}+b_1c_{n-2}+\\ldots+b_{n-2}c_1+b_{n-1}c_0=5$. Taking the equation $\\pmod3$ gives that $b_{n-1}c_0\\equiv2\\pmod3$. This implies that $b_{n-1}\\neq0$. Thus, the degree of $g\\left(x\\right)$ is at least $n-1$. However, because $h\\left(x\\right)$ is a non-constant factor of $f\\left(x\\right)$, we have that the degree of $g\\left(x\\right)$ is at most $n-1$. Thus, the degree of $g\\left(x\\right)$ is $n-1$, implying that the degree of $h\\left(x\\right)$ is $1$.\nFrom this fact, we have that there must exist a rational root of $f\\left(x\\right)$. The only candidates are $1$, $-1$, $3$, and $-3$, though. $f(x)\\equiv x+5x+3\\equiv1\\pmod3$ when $x$ is an integer, so none of these work. Thus, there are no linear factors of $f\\left(x\\right)$.\nIn other words, $f\\left(x\\right)$ cannot be expressed as $g\\left(x\\right)h\\left(x\\right)$ for polynomials $g\\left(x\\right)$ and $h\\left(x\\right)$ in $\\mathbb{R}$. This means that $f\\left(x\\right)$ cannot be expressed as the product of two non-constant polynomials with integer coefficients.\nQ.E.D." }, { "Problem": "Let $A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. The line $XY$ meets $BC$ at $Z$. Let $P$ be a point on the line $XY$ other than $Z$. The line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$, and the line $BP$ intersects the circle with diameter $BD$ at $B$ and $N$. Prove that the lines $AM,DN,XY$ are concurrent.\n", "Solution": "Since $M$ is on the circle with diameter $AC$, we have $\\angle AMC=90$ and so $\\angle MCA=90-A$. We similarly find that $\\angle BND=90$. Also, notice that the line $XY$ is the radical axis of the two circles with diameters $AC$ and $BD$. Thus, since $P$ is on $XY$, we have $PN\\cdot PB=PM\\cdot PC$ and so by the converse of Power of a Point, the quadrilateral $MNBC$ is cyclic. Thus, $90-A=\\angle MCA=\\angle BNM$. Thus, $\\angle MND=180-A$ and so quadrilateral $AMND$ is cyclic. Let the name of the circle $AMND$ be $O$ . Then, the radical axis of $O$ and the circle with diameter $AC$ is line $AM$. Also, the radical axis of $O$ and the circle with diameter $BD$ is line $DN$. Since the pairwise radical axes of 3 circles are concurrent, we have $AM,DN,XY$ are concurrent as desired.", "Solution_2": "Let $AM$ and $PT$ (a subsegment of $XY$) intersect at $Z$. Now, assume that $Z, N, P$ are not collinear. In that case, let $ZD$ intersect the circle with diameter $BD$ at $N'$ and the circle through $D, P, T$ at $N''$.\nWe know that $\\angle AMC = \\angle BND = \\angle ATP = 90^\\circ$ via standard formulae, so quadrilaterals $AMPT$ and $DNPT$ are cyclic. Thus, $N'$ and $N''$ are distinct, as none of them is $N$. Hence, by Power of a Point, However, because $Z$ lies on radical axis $TP$ of the two circles, we have Hence, $ZN'' = ZN'$, a contradiction since $D$ and $D'$ are distinct. We therefore conclude that $Z, N, D$ are collinear, which gives the concurrency of $AM, XY$, and $DN$. This completes the problem.", "Solution_3": "Let $AM$ and $XY$ intersect at $Z$. Because $\\angle AMC = \\angle BND = \\angle APT = 90^\\circ$, we have quadrilaterals $AMPT$ and $DNPT$ cyclic. Therefore, $Z$ lies on the radical axis of the two circumcircles of these quadrilaterals. But $Z$ also lies on radical axis $XY$ of the original two circles, so the power of $Z$ with respect to each of the four circles is all equal to $ZM * ZA$. Hence, $Z$ lies on the radical axis $DN$ of the two circles passing through $D$ and $N$, as desired.\nwhat is $T$ here?" }, { "Problem": "We are given a positive integer $r$ and a rectangular board $ABCD$ with dimensions $|AB|=20$, $|BC|=12$. The rectangle is divided into a grid of $20 \\times 12$ unit squares. The following moves are permitted on the board: one can move from one square to another only if the distance between the centers of the two squares is $\\sqrt{r}$. The task is to find a sequence of moves leading from the square with $A$ as a vertex to the square with $B$ as a vertex.\n(a) Show that the task cannot be done if $r$ is divisible by $2$ or $3$.\n(b) Prove that the task is possible when $r=73$.\n(c) Can the task be done when $r=97$?", "Solution": "First we define the rectangular board in the cartesian plane with centers of the unit squares as integer coordinates and the following coordinates for the squares at the corners of $A$, $B$, $C$, $D$, as follows: $A=(1,1)$, $B=(20,1)$, $C=(20,12)$, $D=(1,12)$\nLet $(x_i,y_i)$ be the coordinates of the piece after move $i$ with $(x_0,y_0)=A=(1,1)$ the initial position of the piece.\n$1 \\le x_i \\le 20$, and also $1 \\le y_i \\le 12$.\nLet $\\Delta x_i = x_i-x_{i-1}$, $\\Delta y_i = y_i-y_{i-1}$\nThen, for any given $r$, we have $(\\Delta x_i)^2+(\\Delta y_i)^2=\\left( \\sqrt{r} \\right)^2=r$ for all $i$\n\nPart (a):\nIn order to find out the conditions for which $r$ is divisible by 2 we are going to look at the following three cases:\n(1) When both $\\Delta x_i$ and $\\Delta y_i$ are divisible by $2$.\n(2) When both $\\Delta x_i$ and $\\Delta y_i$ are odd.\n(3) When one of $\\Delta x_i$ and $\\Delta y_i$ is even and the other one is odd.\nCase (1): Since $\\Delta x_i \\equiv 0\\;(mod \\; 2)$ and $\\Delta y_i \\equiv 0\\;(mod \\; 2)$,\nthen $(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv (0^2+0^2)\\;(mod \\; 2)\\equiv 0\\;(mod \\; 2)$.\nThus, for $r\\equiv 0\\;(mod \\; 2)$, this case is a valid one.\nCase (2): Since $\\Delta x_i \\equiv 1\\;(mod \\; 2)$ and $\\Delta y_i \\equiv 1\\;(mod \\; 2)$,\nthen $(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv (1^2+1^2)\\;(mod \\; 2)\\equiv 0\\;(mod \\; 2)$.\nThus, for $r\\equiv 0\\;(mod \\; 2)$, this case is a valid one.\nCase (3): Since $\\Delta x_i \\equiv 1\\;(mod \\; 2)$ and $\\Delta y_i \\equiv 0\\;(mod \\; 2)$, or $\\Delta x_i \\equiv 0\\;(mod \\; 2)$ and $\\Delta y_i \\equiv 1\\;(mod \\; 2)$,\nthen $(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv (1^2+0^2)\\;(mod \\; 2)\\equiv 1\\;(mod \\; 2)\\not\\equiv 0\\;(mod \\; 2)$.\nThus, for $r\\equiv 0\\;(mod \\; 2)$, this case is NOT a valid one.\nHaving proved that Case (1) and Case (2) are the only valid cases for $r\\equiv 0\\;(mod \\; 2)$ we are going to see what happens for both cases when we start with a square where both coordinates are odd:\nif $(x_{i-1},y_{i-1}) \\equiv (1,1) \\;(mod \\; 2)$,\nthen for case (1): $(x_{i-1}+\\Delta x_i,y_{i-1}+\\Delta y_i) \\equiv (1+0,1+0) \\;(mod \\; 2) \\equiv (1,1) \\;(mod \\; 2)$\nand for case (2): $(x_{i-1}+\\Delta x_i,y_{i-1}+\\Delta y_i) \\equiv (1+1,1+1) \\;(mod \\; 2) \\equiv (0,0) \\;(mod \\; 2)$\nand $(x_{i-1}+2\\Delta x_i,y_{i-1}+2\\Delta y_i) \\equiv (1+2,1+2) \\;(mod \\; 2) \\equiv (1,1) \\;(mod \\; 2)$\nThis means that when $r$ is divisible by two, when starting at $(1,1)$ no matter how many moves you make all $(x_{i},y_{i})$ will either be $\\equiv (0,0) \\;(mod \\; 2)$ or $\\equiv (1,1) \\;(mod \\; 2)$.\nSince the ending coordinate is $(20,1) \\equiv (0,1) \\;(mod \\; 2) \\not\\equiv (1,1) \\;(mod \\; 2)$ and $\\not\\equiv (0,0) \\;(mod \\; 2)$, then the task cannot be done if $r$ is divisible by $2$.\nNow we look at the conditions for which $r$ is divisible by 3 by looking at the following three cases:\n(4) When both $\\Delta x_i$ and $\\Delta y_i$ are divisible by $3$.\n(5) When one of them is not divisible by $3$ and the other one is.\n(6) When neither $\\Delta x_i$ nor $\\Delta y_i$ is divisible by $3$.\nCase (4): Since $\\Delta x_i \\equiv 0\\;(mod \\; 3)$ and $\\Delta y_i \\equiv 0\\;(mod \\; 3)$,\nthen $(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv (0^2+0^2)\\;(mod \\; 3)\\equiv 0\\;(mod \\; 3)$.\nThus, for $r\\equiv 0\\;(mod \\; 3)$, this case is a valid one.\nCase (5): Since either $\\Delta x_i$ or $\\Delta y_i \\equiv \\pm 1\\;(mod \\; 3)$ and the other $\\equiv 1\\;(mod \\; 3)$,\nthen $(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv ((\\pm 1)^2+0^2)\\;(mod \\; 3)\\equiv 1\\;(mod \\; 3)$.\nThus, for $r\\equiv 0\\;(mod \\; 3)$, this case is NOT a valid one.\nCase (6): Since $\\Delta x_i \\equiv \\pm 1\\;(mod \\; 3)$ and $\\Delta y_i \\equiv \\pm1 \\;(mod \\; 3)$,\nthen $(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv ((\\pm 1)^2+(\\pm 1)^2)\\;(mod \\; 3)\\equiv 2\\;(mod \\; 3)\\not\\equiv 0\\;(mod \\; 3)$.\nThus, for $r\\equiv 0\\;(mod \\; 3)$, this case is NOT a valid one.\nHaving proved that Case (4) is the only valid case for $r\\equiv 0\\;(mod \\; 3)$ we are going to see what happens when we start with a square that is $\\equiv (1,1)\\;(mod \\; 3)$\nif $(x_{i-1},y_{i-1}) \\equiv (1,1) \\;(mod \\; 3)$, then $(x_{i-1}+\\Delta x_i ,y_{i-1}+\\Delta y_i) \\equiv (1+0,1+0) \\;(mod \\; 3) \\equiv (1,1) \\;(mod \\; 3)$.\nTherefore starting at any $(x_{i-1},y_{i-1}) \\equiv (1,1) \\;(mod \\; 3)$ no matter how many moves the piece will always land at another square $\\equiv (1,1) \\;(mod \\; 3)$. Since we start at $(1,1)$, then the piece will always land at squares that are $\\equiv (1,1) \\;(mod \\; 3)$.\nSince final square is $(20,1) \\equiv (2,1) \\;(mod \\; 3) \\not\\equiv (1,1) \\;(mod \\; 3)$, then the task cannot be done if $r$ is divisible by $3$.\nIn summary the task cannot be done if $r$ is divisible by $2$ or $r$ is divisible by $3$.\n\nPart (b):\nWhen $r=73$ we have $8^2+3^2=73$\nTherefore the moves can have the following values:\n\nSince we start at $(1,1)$ and we want to end at $(20,1)$ we can write the following $11$ valid moves:\n\nHaving shown that all moves comply with the possible values for $(\\Delta x_i, \\Delta y_i)$ and that all $(x_i, y_i)$ are inside the rectangular grid, then the task is possible when $r=73$.\n\nPart (c):\nWhen $r=97$ we have $9^2+4^2=97$\nTherefore the moves can have the following values:\n\nLet $S$ be a set of the squares of the rectangle with the condition $x_i \\equiv \\left\\lceil \\frac{y_i}{4} \\right\\rceil \\;(mod \\; 2)$ for $(x_i,y_i)$. That is, the reminder of $x_i$ when divided by 2 is the same as the reminder of $\\left\\lceil \\frac{y_i}{4} \\right\\rceil$ when divided by 2 where $\\left\\lceil k \\right\\rceil$ is the smallest integer that is not less than $k$. If that condition is true, then $(x_i,y_i) \\in S$\nNow let's look at a case (c1) where $(x_i,y_i) \\in S$ and $x_i \\equiv 0 \\;(mod \\; 2)$, thus $\\left\\lceil \\frac{y_i}{4} \\right\\rceil \\equiv 0 \\;(mod \\; 2)$:\nc1 subcase 1 : $(x_{i+1},y_{i+1})=(x_i \\pm 9,y_i \\pm 4)$:\n$(x_i \\pm 9) \\equiv (0 \\pm 9)\\;(mod \\; 2) \\equiv 1 \\;(mod \\; 2)$\nand $\\left\\lceil \\frac{y_i \\pm 4}{4} \\right\\rceil \\equiv \\left\\lceil \\frac{y_i}{4} \\right\\rceil \\pm 1 \\;(mod \\; 2)\\equiv (0 \\pm 1) \\;(mod \\; 2)\\equiv 1 \\;(mod \\; 2)$\nsince $x_{i+1} \\equiv \\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\;(mod \\; 2)$, then $(x_{i+1},y_{i+1}) \\in S$ for this subcase.\nc1 subcase 2 : $(x_{i+1},y_{i+1})=(x_i \\pm 4,y_i \\pm 9)$:\nThis move is not allowed because the only values of $1 \\le y_i \\le 12$ such that $\\left\\lceil \\frac{y_i}{4} \\right\\rceil \\equiv 0 \\;(mod \\; 2)$ are $5,6,7,8$. Adding or subtracting 9 from any of these numbers will either make $y_{i+1}<0$ or $y_{i+1}>12$ and will put the piece outside of the board.\nIn summary, for case (c1), $(x_{i+1},y_{i+1}) \\in S$\nNow let's look at a case (c2) where $(x_i,y_i) \\in S$ and $x_i \\equiv 1 \\;(mod \\; 2)$, thus $\\left\\lceil \\frac{y_i}{4} \\right\\rceil \\equiv 1 \\;(mod \\; 2)$:\nc2 subcase 1 : $(x_{i+1},y_{i+1})=(x_i \\pm 9,y_i \\pm 4)$:\n$(x_i \\pm 9) \\equiv (1 \\pm 9)\\;(mod \\; 2) \\equiv 0 \\;(mod \\; 2)$\nand $\\left\\lceil \\frac{y_i \\pm 4}{4} \\right\\rceil \\equiv \\left\\lceil \\frac{y_i}{4} \\right\\rceil \\pm 1 \\;(mod \\; 2)\\equiv (1 \\pm 1) \\;(mod \\; 2)\\equiv 0 \\;(mod \\; 2)$\nsince $x_{i+1} \\equiv \\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\;(mod \\; 2)$, then $(x_{i+1},y_{i+1}) \\in S$ for this subcase.\nc2 subcase 2 : $(x_{i+1},y_{i+1})=(x_i \\pm 4,y_i \\pm 9)$:\n$(x_i \\pm 4) \\equiv (1 \\pm 4)\\;(mod \\; 2) \\equiv 1 \\;(mod \\; 2)$\nNotice that the only values of $1 \\le y_i \\le 12$ such that $\\left\\lceil \\frac{y_i}{4} \\right\\rceil \\equiv 0 \\;(mod \\; 2)$ are $1,2,3,4,9,10,11,12$. Adding 9 to the values of 1,2, or 3 or subtracting 9 from the values of 10, 11, or 12, will result in $\\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\equiv 1 \\;(mod \\; 2)$. However, subtracting 9 from the values of 1,2, or 3, or adding 9 for the values of 10,11, or 12 will put the piece outside the board and not a valid move. Furthermore, for the values of 4 and 9, adding or subtracting 9 from them will result in the piece being outside of the board and not a valid move. Therefore for all valid moves in this subcase which are 9 added to 1, 2, or 3 or 9 subtracted from 10, 11, or 12, the result will be $\\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\equiv 1 \\;(mod \\; 2)$\nsince $x_{i+1} \\equiv \\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\;(mod \\; 2)$, then $(x_{i+1},y_{i+1}) \\in S$ for this subcase.\nIn summary, for case (c2), $(x_{i+1},y_{i+1}) \\in S$\nTherefore if $(x_{i},y_{i}) \\in S$ then $(x_{i+1},y_{i+1}) \\in S$\nNow we look at the starting point: $(1,1)$:\nSince $1 \\equiv 1\\;(mod \\; 2)$ and $\\left\\lceil \\frac{1}{4} \\right\\rceil \\equiv 1\\;(mod \\; 2)$, then $(1,1) \\in S$ and all subsequent squares after that $\\in S$\nNow let's look at the end point: $(20,1)$:\nSince $20 \\equiv 0\\;(mod \\; 2)$ and $\\left\\lceil \\frac{1}{4} \\right\\rceil \\equiv 1\\;(mod \\; 2)$, then $20 \\not\\equiv \\left\\lceil \\frac{1}{4} \\right\\rceil \\;(mod \\; 2)$ and $(20,1) \\not\\in S$\nSince $(20,1) \\not\\in S$, then there are no valid moves starting from $(1,1)$ that will end in $(20,1)$\nTherefore the task cannot be done when $r=97$.\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": "In the plane the points with integer coordinates are the vertices of unit squares. The squares are colored alternatively black and white (as on a chessboard).\nFor any pair of positive integers $m$ and $n$, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths $m$ and $n$, lie along edges of the squares.\nLet $S_{1}$ be the total area of the black part of the triangle and $S_{2}$ be the total area of the white part.\nLet $f(m,n)=|S_{1}-S_{2}|$\n(a) Calculate $f(m,n)$ for all positive integers $m$ and $n$ which are either both even or both odd.\n(b) Prove that $f(m,n) \\le \\frac{1}{2} max\\left\\{ m,n \\right\\}$ for all $m$ and $n$.\n(c) Show that there is no constant $C$ such that $f(m,n) YP \\ge R/2$. But $PC = YC - YP < R - YP \\le R/2$. So $OP > PC$.\nHence $\\angle COP < \\angle OCP$. Let $CE$ be a diameter of the circle, so that $\\angle OCP = \\angle ECB$. But $\\angle ECB = \\angle EAB$ and $\\angle EAB + \\angle BAC = \\angle EAC = 90^\\circ$, since $EC$ is a diameter. Hence $\\angle COP + \\angle BAC < 90^\\circ$.", "Solution_2": "Notice that because $\\angle{PCO} = 90^\\circ - \\angle{A}$, it suffices to prove that $\\angle{POC} < \\angle{PCO}$, or equivalently $PC < PO.$\nSuppose on the contrary that $PC > PO$. By the triangle inequality, $2 PC = PC + PC > PC + PO > CO = R$, where $R$ is the circumradius of $ABC$. But the Law of Sines and basic trigonometry gives us that $PC = 2R \\sin B \\cos C$, so we have $4 \\sin B \\cos C > 1$. But we also have $4 \\sin B \\cos C \\le 4 \\sin B \\cos (B + 30^\\circ) = 2 (\\sin (2B + 30^\\circ) - \\sin 30^\\circ) \\le 2 (1 - \\frac{1}{2}) = 1$ because $\\angle{C} \\ge \\angle{B} + 30^\\circ$, and so we have a contradiction. Hence $PC < PO$ and so $\\angle{PCO} + \\angle{A} < 90^\\circ$, as desired." }, { "Problem": "$S$ is the set of all $(h,k)$ with $h,k$ non-negative integers such that $h + k < n$. Each element of $S$ is colored red or blue, so that if $(h,k)$ is red and $h' \\le h,k' \\le k$, then $(h',k')$ is also red. A type $1$ subset of $S$ has $n$ blue elements with different first member and a type $2$ subset of $S$ has $n$ blue elements with different second member. Show that there are the same number of type $1$ and type $2$ subsets.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $ABC$ be an acute-angled triangle with $AB\\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\\angle BAC$ and $\\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.", "Solution": "\nLet $\\angle{ACB} = a$, $\\angle{CBA} = b$, and $\\angle{ONR} = k$. Call $\\omega$ the circle with diameter $BC$ and $\\odot{AMN}$ the circumcircle of $\\triangle{AMN}$.\nOur ultimate goal is to show that $\\angle{CNR} + \\angle{RMB} = 180^\\circ$. To show why this solves the problem, assume this statement holds true. Call $K$ the intersection point of the circumcircle of $\\triangle{CNR}$ with side $BC$. Then, $\\angle{RKC} = 180^\\circ - \\angle{CNR}$, and $\\angle{RKB} = \\angle{CNR}$. Since $\\angle{RMB} = 180^\\circ - \\angle{CNR}$, $\\angle{RKB} + \\angle{RMB} = 180^\\circ$, implying $K$ also lies on the circumcircle of $\\triangle{BMR}$, thereby solving the problem.\nWe now prove that $\\angle{CNR} + \\angle{RMB} = 180^\\circ$. Note that $ON$ and $OM$ are radii of $\\omega$, so $\\triangle{MON}$ is isosceles. The bisector of $\\angle{MON}$ is thus the perpendicular bisector of $MN$. Since $R$ lies on the bisector of $\\angle{MON}$, $\\angle{ONR} = \\angle{RMO} = k$. Angle computations yield that from $\\angle{AMN} = 180^\\circ - \\angle{BMN} = 180^\\circ - (180^\\circ - \\angle{ACB}) = \\angle{ACB}$ and from $\\angle{BMO} = \\angle{MBO} =$ $\\angle{ABC}$.\nThe bisector of $\\angle{BAC}$ hits $\\odot{AMN}$ at the midpoint of the arc $MN$ not containing $A$. This point must lie on the perpendicular bisector of segment $MN$, which is the bisector of $\\angle{MON}$. It follows that $R$ is indeed the midpoint of arc $MN$, so $A$, $M$, $R$, $N$, are concyclic. Since $\\angle{RAN}$ and $\\angle{RMN}$ subtend the same arc $NR$, $\\angle{RAN}$ = $\\angle{RMN}$. With $AR$ being the bisector of $\\angle{BAC}$, we have We know that $\\angle{RMN} = 180^\\circ - b - k - a$. so we have $180^\\circ - b - k - a = 90^\\circ - \\tfrac{a}{2} - \\tfrac{b}{2}$. Since $\\angle{CNR} = \\angle{CNO} + \\angle{ONR} = a + k$, and $\\angle{RMB} = \\angle{OMB} + \\angle{RMO} = b + k$, we have\nThe problem is solved.\n$\\textbf{NOTE:}$ We have $\\angle{RKB} + \\angle{KBA} + \\angle{BAK} = 180^\\circ \\implies \\angle{BAK} = 180^\\circ - (180^\\circ - \\angle{RMB}) - \\angle{KBA}$. Noting that $180^\\circ - \\angle{RMB} = 180^\\circ - b - k = 180^\\circ - (90^\\circ - \\tfrac{a}{2} + \\tfrac{b}{2}) = 90^\\circ + \\tfrac{a}{2} - \\tfrac{b}{2}$, we then have which is indeed the measure of $\\angle{BAR}$. This implies that $K$ lies on the bisector of $\\angle{BAC}$, and from this, it is clear that $K$ must lie on the interior of segment $BC$. Not proving that $K$ had to lie in the interior of $BC$ was a reason that a large portion of students who submitted a solution received a 1-point deduction." }, { "Problem": "Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1, A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths. Prove that the lines $A_1B_2, B_1C_2$ and $C_1A_2$ are concurrent.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies $\\angle PBA+\\angle PCA = \\angle PBC+\\angle PCB$. Show that $AP \\geq AI$, and that equality holds if and only if $P=I.$", "Solution": "We have\nand similarly \nSince $\\angle PBA + \\angle PCA = \\angle PBC + \\angle PCB$, we have $\\angle PCB - \\angle PCA = \\angle PBA - \\angle PBC.$\nIt follows that Hence, $B,P,I,$ and $C$ are concyclic.\nLet ray $AI$ meet the circumcircle of $\\triangle ABC\\,$ at point $J$. Then, by the Incenter-Excenter Lemma, $JB=JC=JI=JP$.\nFinally, $AP+JP \\geq AJ = AI+IJ$ (since triangle APJ can be degenerate, which happens only when $P=I$), but $JI=JP$; hence $AP \\geq AI$ and we are done.\nBy Mengsay LOEM , Cambodia IMO Team 2015\nlatexed by tluo5458 :)\nminor edits by lpieleanu\n" }, { "Problem": "Real numbers $a_1, a_2, \\dots , a_n$ are given. For each $i$ ($1\\le i\\le n$) define\n\nand let\n.\n(a) Prove that, for any real numbers $x_1\\le x_2\\le \\cdots\\le x_n$,\n\n(b) Show that there are real numbers $x_1\\le x_2\\le x_n$ such that equality holds in (*)\n", "Solution": "Since $d_i=\\max\\{a_j:1\\le j\\le i\\}-\\min\\{a_j:i\\le j\\le n\\}$, all $d_i$ can be expressed as $a_p-a_q$, where $1\\le p\\le i\\le q \\le n$.\nThus, $d$ can be expressed as $a_p-a_q$ for some $p$ and $q$, $1\\le p\\le q\\le n$\n\nLemma) $d\\ge 0$\nAssume for contradiction that $d<0$, then for all $i$, $a_i \\le \\max\\{a_j:1\\le j\\le i\\}\\le \\min\\{a_j:i\\le j\\le n\\}\\le a_{i+1}$\n$a_i\\le a_{i+1}$\nThen, ${a_i}$ is a non-decreasing function, which means, $\\max\\{a_j:1\\le j\\le i\\}=a_i$, and $\\min\\{a_j:i\\le j\\le n\\}\\le a_{i+1}=a_i$, which means, ${d_i}={0}$.\nThen, $d=0$ and contradiction.\n\na)\nCase 1) $d=0$\nIf $d=0$, $\\max\\{|x_i-a_i|:1\\le i\\le n\\}$ is the maximum of a set of non-negative number, which must be at least $0$.\n\nCase 2) $d>0$ (We can ignore $d<0$ because of lemma)\nUsing the fact that $d$ can be expressed as $a_p-a_q$ for some $p$ and $q$, $1\\le p\\le q\\le n$. $x_p\\le x_q$\nAssume for contradiction that $\\max\\{|x_i-a_i|:1\\le i\\le n\\}<\\dfrac{d}{2}$.\nThen, $\\forall x_i$, $|x_i-a_i|<\\dfrac{d}{2}$.\n$|x_p-a_p|<\\dfrac{d}{2}$, and $|x_q-a_q|<\\dfrac{d}{2}$\nThus, $x_p>a_p-\\dfrac{d}{2}$ and $x_qx_q$ --- contradiction ($p\\le q \\rightarrow x_p\\le x_q$).\nThus, $\\max\\{|x_i-a_i|:1\\le i\\le n\\}\\ge\\dfrac{d}{2}$\n\n(b)\nA set of ${x_i}$ where the equality in (*) holds is:\n\nSince $\\max\\{a_j:1\\le j\\le i\\}$ is a non-decreasing function, $x_i$ is non-decreasing.\n\n$\\forall x_i$ :\nLet $a_m=\\max\\{a_j:1\\le j\\le i\\}$, $a_m-a_i\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 $ABC$ has orthocentre $H$. The circle passing through $H$ with centre the midpoint of $BC$ intersects the line $BC$ at $A_1$ and $A_2$. Similarly, the circle passing through $H$ with centre the midpoint of $CA$ intersects the line $CA$ at $B_1$ and $B_2$, and the circle passing through $H$ with centre the midpoint of $AB$ intersects the line $AB$ at $C_1$ and $C_2$. Show that $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ lie on a circle.", "Solution": "Let $M_A$, $M_B$, and $M_C$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. It's not hard to see that $M_BM_C\\parallel BC$. We also have that $AH\\perp BC$, so $AH \\perp M_BM_C$. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that $H$ is on the radical axis of the circles centered at $M_B$ and $M_C$, so $A$ is too. We then have $AC_1\\cdot AC_2=AB_2\\cdot AB_1\\Rightarrow \\frac{AB_2}{AC_1}=\\frac{AC_2}{AB_1}$. This implies that $\\triangle AB_2C_1\\sim \\triangle AC_2B_1$, so $\\angle AB_2C_1=\\angle AC_2B_1$. Therefore $\\angle C_1B_2B_1=180^{\\circ}-\\angle AB_2C_1=180^{\\circ}-\\angle AC_2B_1$. This shows that quadrilateral $C_1C_2B_1B_2$ is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments $C_1C_2$ and $B_1B_2$. However, these are just the perpendicular bisectors of $AB$ and $CA$, which meet at the circumcenter of $ABC$, so the circumcenter of $C_1C_2B_1B_2$ is the circumcenter of triangle $ABC$. Similarly, the circumcenters of $A_1A_2B_1B_2$ and $C_1C_2A_1A_2$ are coincident with the circumcenter of $ABC$. The desired result follows." }, { "Problem": "Let $n$ be a positive integer and let $a_1,\\ldots,a_k (k\\ge2)$ be distinct integers in the set $\\{1,\\ldots,n\\}$ such that $n$ divides $a_i(a_{i+1}-1)$ for $i=1,\\ldots,k-1$. Prove that $n$ doesn't divide $a_k(a_1-1)$.\nAuthor: Ross Atkins, Australia", "Solution": "Let $n=pq$ such that $p\\mid a_1$ and $q\\mid a_2-1$. Suppose $n$ divides $a_k(a_1-1)$. \nNote $q\\mid a_2-1$ implies $(q,a_2)=1$ and hence $q\\mid a_3-1$. Similarly one has $q\\mid a_i-1$ for all $i$'s, in particular, $p\\mid a_1$ and $q\\mid a_1-1$ force $(p,q)=1$. Now $(p,a_1-1)=1$ gives $p\\mid a_k$, similarly one has $p\\mid a_i$ for all $i$'s, that is $a_i$'s satisfy $p\\mid a_i$ and $q\\mid a_i-1$, but there should be at most one such integer satisfies them within the range of $1,2,\\ldots,n$ for $n=pq$ and $(p,q)=1$. A contradiction!!!", "Solution_2": "Let $n = p_1^{b_1}p_2^{b_2} \\cdots p_s^{b_s}$. Then after toying around with the $p_i^{b_i}$ and what they divide, we have that $p_i^{b_i} \\nmid a_k$, and so in particular, $n \\nmid a_k$.\nAssume by way of contradiction that $n \\mid a_k(a_1 - 1)$. Then $n \\mid a_1 - 1$. Now we shift our view towards the $a_i(a_{i + 1} - 1)$. Here each $p_i^{b_i}$ divides $a_i(a_{i + 1} - 1) \\implies a_ia_{i + 1} \\equiv a_i \\pmod{p_i^{b_i}}$. Hence we have the chain of equivalences $a_1a_2 \\equiv a_1 \\pmod{p_i^{b_i}}, a_2a_3 \\equiv a_2 \\pmod{p_i^{b_i}}, \\dots, a_{k - 1}a_k \\equiv a_{k - 1} \\pmod {p_i^{b_i}}$. Now we also have that $p_i^{b_i} \\mid n \\mid a_1 - 1$. Thus $a_1 \\equiv 1 \\pmod{p_i^{b_i}}$. Now plugging this value of $a_1$ modulo $p_i^{b_i}$, we obtain that $a_1 \\equiv a_2 \\equiv a_3 \\equiv \\cdots a_k \\equiv 1 \\pmod{p_i^{b_i}}$. Hence this chain of congruences is also true for $n$ as $p_i$ was arbitrary. However as all the $a_i \\in \\{1, 2, \\dots, n\\}$ we have that not all the $a_i$ are distinct, and so this is a contradiction." }, { "Problem": "Find all function $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for all $x,y\\in\\mathbb{R}$ the following equality holds\n$f(\\left\\lfloor x\\right\\rfloor y)=f(x)\\left\\lfloor f(y)\\right\\rfloor$\nwhere $\\left\\lfloor a\\right\\rfloor$ is greatest integer not greater than $a.$\nAuthor: Pierre Bornsztein, France", "Solution_1": "Put $x=y=0$. Then $f(0)=0$ or $\\lfloor f(0) \\rfloor=1$.\n$\\bullet$ If $\\lfloor f(0) \\rfloor=1$, putting $y=0$ we get $f(x)=f(0)$, that is f is constant. \nSubstituing in the original equation we find $f(x)=0, \\ \\forall x \\in \\mathbb{R}$ or $f(x)=a, \\ \\forall x \\in \\mathbb{R}$, where $a \\in [1,2)$.\n$\\bullet$ If $f(0)=0$, putting $x=y=1$ we get $f(1)=0$ or $\\lfloor f(1) \\rfloor=1$.\nFor $f(1)=0$, we set $x=1$ to find $f(y)=0 \\ \\forall y$, which is a solution.\nFor $\\lfloor f(1) \\rfloor=1$, setting $y=1$ yields $f(\\lfloor x \\rfloor)=f(x), \\ (*)$.\nPutting $x=2, y=\\frac{1}{2}$ to the original we get $f(1)=f(2)\\lfloor f(\\frac{1}{2}) \\rfloor$.\nHowever, from $(*)$ we have $f(\\frac{1}{2})=f(0)=0$, so $f(1)=0$ which contradicts the fact $\\lfloor f(1) \\rfloor=1$.\nSo, $f(x)=0, \\ \\forall x$ or $f(x)=a, \\ \\forall x, \\ a \\in [1,2)$. ( By socrates[1])", "Solution_2": "Substituting $y=0$ we have $f(0) = f(x) [f(0)]$. \nIf $[f0)] \\ne 0$ then $f(x) = \\frac{f(0)}{[f(0)]}$. Then $f(x)$ is constant. Let $f(x)=c$. Then substituting that in (1) we have $c=c[c] \\Rightarrow c(1-[c])=0 \\Rightarrow c=0$, or $[c]=1$. Therefore $f(x)=c$ where $c=0$ or $c \\in [1,2)$\nIf $[f(0)] = 0$ then $f(0)=0$.\nNow substituting $x=1$ we have $f(y)=f(1)[f(y)]$.\nIf $f(1) \\ne 0$ then $[f(y)] = \\frac{f(y)}{f(1)}$ and substituting this in (1) we have $f([x]y)=\\frac{f(x)f(y)}{f(1)}$.\nThen $f([x]y)=f(x[y])$. \nSubstituting $x=1/2, y=2$ we get $f(0)=f(1)$. Then $f(1)=0$, which is a contradiction\nTherefore $f(1)=0$. and then $f(y)=0$ for all $R$\nThen the only solutions are $f(x)=0$ or $f(x)=c$ where $c \\in [1,2)$.( By m.candales [2])", "Solution_3": "Let $y=0$, then $f(0)=f(x)\\left\\lfloor f(0)\\right\\rfloor$.\nCase 1: $\\left\\lfloor f(0)\\right\\rfloor\\neq 0$\nThen $f(x)=\\frac{f(0)}{\\left\\lfloor f(0)\\right\\rfloor}$ is a constant. Let $f(x)=k$, then $k=k\\left\\lfloor k \\right\\rfloor \\Leftrightarrow k=0 \\vee 1\\leq k<2$. It is easy to check that this are solutions.\nCase 2: $\\left\\lfloor f(0)\\right\\rfloor= 0$\nIn this case we conclude that $\\left\\lfloor f(0)\\right\\rfloor= 0\\Rightarrow f(0)=0$\nLemma:If $y$ is such that $0\\leq f(y)<1$, $f(y)=0$\nProof of the Lemma: If $x=1$ we have that $f(\\left\\lfloor x\\right\\rfloor y)=f(y)=f(x)\\left\\lfloor f(y)\\right\\rfloor =0$, as desired.\nLet $0\\leq x<1$, so that we have:\n$0=f(0)=f(\\left\\lfloor x\\right\\rfloor y)=f(x)\\left\\lfloor f(y)\\right\\rfloor\\Rightarrow$\n$\\Rightarrow f(x)=0 \\vee 0\\leq f(y)<1 \\Rightarrow f(x)=0 \\vee f(y)=0$, using the lemma.\nIf $f$ is not constant and equal to $0$, letting $y$ be such that $f(y)\\neq 0$ implies that $f(x)=0, \\forall 0\\leq x<1$.\nNow it's enough to notice that any real number $x$ is equal to $ky$, where $k\\in \\mathbb{Z}$ and $0\\leq y< 1$, so that $f(x)=f(ky)=f(\\left\\lfloor k\\right\\rfloor y)=f(k)\\left\\lfloor f(y)\\right\\rfloor=0$. Since $x$ was arbitrary, we have that $f$ is constant and equal to $0$.\nWe conclude that the solutions are $f(x)=k$, where $k=0 \\vee 1\\leq k<2$.( By Jorge Miranda [3]\n)", "Solution_4": "Clearly $f(\\left\\lfloor x\\right\\rfloor y) = f(\\left\\lfloor \\lfloor x \\rfloor \\right\\rfloor y) = f(\\lfloor x \\rfloor)\\left\\lfloor f(y)\\right\\rfloor$, so $(f(x) - f(\\lfloor x \\rfloor))\\left\\lfloor f(y)\\right\\rfloor = 0$ for all $x,y\\in\\mathbb{R}$.\nIf $\\left\\lfloor f(y)\\right\\rfloor = 0$ for all $y \\in \\mathbb{R}$, then by taking $x=1$ we get $f(y)=f(1)\\left\\lfloor f(y)\\right\\rfloor = 0$, so $f$ is identically null (which checks).\nIf, contrariwise, $\\left\\lfloor f(y_0)\\right\\rfloor \\neq 0$ for some $y_0 \\in \\mathbb{R}$, it follows $f(x) = f(\\lfloor x \\rfloor)$ for all $x \\in \\mathbb{R}$.\nNow it immediately follows $f(x) = f(\\lfloor x \\rfloor \\cdot 1) = f(x)\\lfloor f(1) \\rfloor$, hence $f(x)(1 - \\lfloor f(1) \\rfloor) = 0$.\nFor $x=y_0$ this implies $\\lfloor f(1) \\rfloor = 1$.\nAssume $\\lfloor f(0) \\rfloor=0$; then $1 \\leq f(1) = f\\left ( 2\\cdot \\dfrac {1} {2} \\right ) = f(2)\\left \\lfloor f \\left ( \\dfrac {1} {2} \\right ) \\right \\rfloor = f(2)\\left \\lfloor f \\left ( \\left \\lfloor \\dfrac {1} {2} \\right \\rfloor \\right ) \\right \\rfloor = f(2)\\left \\lfloor f(0) \\right \\rfloor = 0$, absurd.\nTherefore $\\lfloor f(0) \\rfloor \\neq 0$, and now $y=0$ in the given functional equation yields $f(0) = f(x)\\lfloor f(0) \\rfloor$ for all $x \\in \\mathbb{R}$, therefore $f(x) = c \\neq 0$ constant, with $\\lfloor c \\rfloor = \\lfloor f(1) \\rfloor = 1$, i.e. $c \\in [1,2)$ (which obviously checks).( By mavropnevma [4])" }, { "Problem": "Given any set $A = \\{a_1, a_2, a_3, a_4\\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \\leq i < j \\leq 4$ for which $a_i +a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.\nAuthor: Fernando Campos, Mexico", "Solution_1": "Firstly, if we order $a_1 \\ge a_2 \\ge a_3 \\ge a_4$, we see $2(a_3 + a_4) \\ge (a_1+a_2)+(a_3+a_4) = s_A \\geq 0$, so $(a_3, a_4)$ isn't a couple that satisfies the conditions of the problem. Also, $2(a_4 + a_2) = (a_4 + a_4) + (a_2 + a_2) \\ge (a_4+a_3)+(a_2+a_1) = s_A \\ge 0$, so again $(a_2, a_4)$ isn't a good couple. We have in total 6 couples. So $n_A \\leq 6-2=4$.\nWe now find all sets $A$ with $n_A = 4$. If $(a,b)$ and $(c,d)$ are both good couples, and $A=\\{a, b, c, d\\}$, we have $a+b=c+d=s_A/2$.\nSo WLOG $A=\\{a,b,a+x,b-x\\}$ with $x > 0$ and $a < b, b-x, a+x$. It's easy to see $a=a_1$ and since $(a_2, a_4),(a_3,a_4)$ are bad, all couples containing $a$ must be good. Obviously $(a,b)$ and $(a+x,b-x)$ are good ($s_A=2(a+b)$). So we have $2a+x | 2a+2b$ and $a+b-x|2a+2b \\Rightarrow a+b-x|2x$.\nUsing the second equation, we see that if $y=a+b$, $y-x|2x \\Rightarrow yk_1-xk_1=2x \\Rightarrow yk_1 = x(2+k_1) \\Rightarrow y=x((2+k_1)/k_1)$, for some $k_1$ a positive integer.\nSo now we use the first equation to get $2ak_2 + xk_2 = 2y = 2x(2+k_1)/k_1 \\Rightarrow 2ak_2 = x(\\frac{4+2k_1}{k_1}-k_2) \\Rightarrow 2a=x(\\frac{4+2k_1}{k_1k_2} - 1)$, for a natural $k_2$.\nFinally, we obtain $k_1 | 4+2k-1 \\Rightarrow k_1 | 4 \\Rightarrow k_1=$ 1, 2 or 4. We divide in cases:\nCASE I: $k_1=1$. \nSo $y=3x$ and $2a=x((\\frac{6}{k_2}) -1)$. But $a < b-x \\Rightarrow 2a < y-x=2x \\Rightarrow (6/k_2) - 1 < 2 \\Rightarrow k_2 > 2 \\Rightarrow k_2 =$3, 4,5 or 6. $k_2=6$ implies $a=0$, impossible. $a=x$ when $k_2=3$. We easily see $b=3x=3x$ and $A=\\{x, 3x, 3x-x, 2x\\}$, impossible since $3x-x=2x$. When $k_2=4$, $a=x/4=y/12$, and we get $\\{11a, a, 5a, 7a\\}$.Uf $k_2=5$, $a=x/5=y/15$ and we get $\\{a, 14a, 6a, 9a\\}$.\nCASE II and III:$k_1=$2, 4. Left to the reader.\nANSWER: $\\{11a, a, 5a, 7a\\}$,$\\{a, 11a, 19a, 29a\\}$, for any positive integer $a$.\n(Note: The above solution looks generally correct, but the actual answer should be $\\{11a, a, 5a, 7a\\}$,$\\{a, 11a, 19a, 29a\\}$. You can check that $\\{a, 14a, 6a, 9a\\}$ doesn't actually work. -Someone who didn't write up the above solution but solved the problem in a similar way)", "Solution_2_.28Avoids_the_error_of_the_first.29": "Similarly to the first solution, let us order the terms in set $A$ by $0 < a_1 < a_2 < a_3 < a_4$ since it is given that these four integers must be distinct and positive. Obviously, $s_a > 0$ from this. We know that pair $(a_3, a_4)$ cannot work because $2(a_3 + a_4) > a_1 + a_2 + a_3 + a_4 = s_A.$ Additionally, pair $(a_2, a_4)$ cannot work for the same reason: $2(a_2 + a_4) = a_4 + a_4 + a_2 + a_2 > s_A.$\nHence, since there were only ${4 \\choose 2} = 6$ pairs $(i, j)$ in total, and two pairs are guaranteed not to work, we are to find sets $A$ satisfying $n_A = 6 - 2 = 4.$ Due to our making $a_4$ the largest term of set $A$, the only way to have a pair $(a_i, a_4)$ that exists such that $a_i + a_4 | s_A$ is to make $i$ equal to 1 and find an $a_1$ such that $\\frac{s_A}{a_1 + a_4} = 2.$ When this happens, we know that $\\frac{s_A}{a_2 + a_3} = 2$, as well. It clearly follows that we need pairs $(a_1, a_2)$ and $(a_1, a_3)$ that exist such that both $a_1 + a_2$ and $a_1 + a_3$ divide $s_A.$\nTo satisfy conditions $\\frac{s_A}{a_1 + a_4} = 2$ and $\\frac{s_A}{a_2 + a_3} = 2$, choose a set $A$ with four distinct positive integer elements $a - x, a, b, b + x$ with $0 < a - x < a < b < b + x.$ Obviously, $x < a < b$ for our set to consist solely of positive integers. The problem becomes to find $x, a,$ and $b$ such that $2a - x | 2(a + b)$ and $(a + b) - x | 2a + 2b.$\nWe realize that $x$ can be any positive even integer. This allows $2a - x$ and $2(a + b)$ to have a common factor of 2, simplifying our problem a bit. Hence, $a - \\frac{x}{2}$ must divide $(a + b).$ Now, we can look at the other condition, $(a + b) - x | 2a + 2b$, to solve this problem. To find $a, b, x$ satisfying $(a + b) - x | 2a + 2b$, it suffices to define a positive integer $n$ that makes $n(a + b - x) = 2(a + b) \\Rightarrow 0 = (n - 2)a + (n - 2)b - nx \\Rightarrow (n - 2)(a + b) = nx.$ Note that $x$ must be less than $\\frac{a + b}{2}.$ If $x \\ge \\frac{a + b}{2}$, $(n - 2)(a + b) = nx$ would have solutions $x, a,$ and $b$ that will make $a - x < 0$, since $a < \\frac{a + b}{2}$ due to $a < b.$ This is a contradiction with the condition that each element in set $A$ has to be positive. Despite this, We move on to some cases.\n\nThus, the only possible sets $A$ we have for satisfying the maximum $n_A$, which is 4, are $A = \\{p, 5p, 7p, 11p\\}$ and $\\{p, 11p, 19p, 29p\\}$, with $p$ being any positive integer since it is obvious that any $a$ that is a positive multiple of 5 or 11 makes $\\frac{a}{5}$ and $\\frac{a}{11}$ positive integers, respectively." }, { "Problem": "Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST$.", "Solution": "First, $BK = BM$ because $BK$ and $BM$ are both tangents from $B$ to the excircle $J$. Then $BJ \\bot KM$. Call the $X$ the intersection between $BJ$ and $KM$. Similarly, let the intersection between the perpendicular line segments $CJ$ and $LM$ be $Y$. We have $\\angle XBM = \\angle XBK = \\angle FBA$ and $\\angle XMB = \\angle XKB$. We then have, $\\angle XBM + \\angle XBK + \\angle XMB + \\angle XKB = \\angle MBK + \\angle XMB + \\angle XKB$ $= \\angle MBK + \\angle KMB + \\angle MKB = 180^{\\circ}$. So $\\angle XBM = 90^{\\circ} - \\angle XMB$. We also have $180^{\\circ} = \\angle FBA + \\angle ABC + \\angle XBM = 2\\angle XBM + \\angle ABC = 180^{\\circ} - 2\\angle XMB$ $+ \\angle ABC$. Then $\\angle ABC = 2\\angle XMB$. Notice that $\\angle XFM = 90^{\\circ} - \\angle XMB - \\angle BMF = 90^{\\circ} - \\angle XMB - \\angle YMC$. Then, $\\angle ACB = 2\\angle YMC$. $\\angle BAC = 180^{\\circ} - \\angle ABC - \\angle ACB = 180^{\\circ} - 2(\\angle XMB + \\angle YMC)$ $= 2(90^{\\circ} - (\\angle XMB + \\angle YMC) = 2\\angle XFM$. Similarly, $\\angle BAC = 2\\angle YGM$. Draw the line segments $FK$ and $GL$. $\\triangle FXK$ and $\\triangle FXM$ are congruent and $\\triangle GYL$ and $\\triangle GYM$ are congruent. Quadrilateral $AFJL$ is cyclic because $\\angle JAL = \\frac{\\angle BAC}{2} = \\angle XFM = \\angle JFL$. Quadrilateral $AFKJ$ is also cyclic because $\\angle JAK = \\frac{\\angle BAC}{2} = \\angle XFM = \\angle XFK = \\angle JFK$. The circumcircle of $\\triangle AFJ$ also contains the points $K$ and $J$ because there is a circle around the quadrilaterals $AFJL$ and $AFKJ$. Therefore, pentagon $AFKJL$ is also cyclic. Finally, quadrilateral $AGLJ$ is cyclic because $\\angle JAL = \\frac{\\angle BAC}{2} = \\angle YGM = \\angle YGL = \\angle JGL$. Again, $\\triangle AJL$ is common in both the cyclic pentagon $AFKJL$ and cyclic quadrilateral $AGLJ$, so the circumcircle of $\\triangle AJL$ also contains the points $F$, $K$, and $G$. Therefore, hexagon $AFKJLG$ is cyclic. Since $\\angle AKJ$ and $\\angle ALJ$ are both right angles, $AJ$ is the diameter of the circle around cyclic hexagon $AFKJLG$. Therefore, $\\angle AFJ$ and $\\angle AGJ$ are both right angles. $\\triangle BFS$ and $\\triangle BFA$ are congruent by ASA congruency, and so are $\\triangle CGT$ and $\\triangle CGA$. We have $SB = AB$, $TC = AC$, $BM = BK$, and $CM = CL$. Since $AK$ and $AL$ are tangents from $A$ to the circle $J$, $AK = AL$. Then, we have $AK = AL$, which becomes $AB + BK = AC + CL$, which is $SB + BM = TC + CM$, or $SM = TM$. This means that $M$ is the midpoint of $ST$.\nQED\n--Aopsqwerty 21:19, 19 July 2012 (EDT)", "Solution_2": "For simplicity, let $A, B, C$ written alone denote the angles of triangle $ABC$, and $a$, $b$, $c$ denote its sides.\nLet $R$ be the radius of the A-excircle. Because $CM = CL$, we have $CML$ isosceles and so $\\angle{CML} = \\dfrac{\\angle{C}}{2}$ by the Exterior Angle Theorem. Then because $\\angle{FBS} = 90^\\circ - \\dfrac{B}{2}$, we have $\\angle{BFM} = \\dfrac{\\angle{A}}{2}$, again by the Exterior Angle Theorem.\nNotice that $\\angle{BJM} = \\dfrac{\\angle{B}}{2}$ and $\\angle{CJM} = \\dfrac{\\angle{C}}{2}$, and so\n\nafter converting tangents to sine and cosine. Thus,\n\nIt follows that $BM = a \\sin \\dfrac{B}{2} \\cos \\dfrac{C}{2} \\sec \\frac{A}{2}$. By the Law of Sines on triangle $BFM$ and $ABC$ and the double-angle formula for sine, we have\n\nTherefore, triangle $BFA$ is congruent to a right triangle with hypotenuse length $c$ and one angle of measure $90^\\circ - \\dfrac{B}{2}$ by SAS Congruence, and so $\\angle{BFA} = 90^\\circ$. It then follows that triangles $BFS$ and $BFA$ are congruent by $ASA$, and so $AF = FS$. Thus, $J$ lies on the perpendicular bisector of $AS$. Similarly, $J$ lies on the perpendicular bisector of $AT$, and so $J$ is the circumcenter of $ATS$. In particular, $J$ lies on the perpendicular bisector of $ST$, and so, because $JM$ is perpendicular to $ST$, $M$ must be the midpoint of $ST$, as desired.\n--Suli 17:53, 8 February 2015 (EST)", "Solution_3": "Same as Solution 2, except noticing that (letting $s = \\dfrac{a + b + c}{2}$ be the semi-perimeter):\n--Suli 18:21, 8 February 2015 (EST)", "Solution_4": "As before in Solution 2, we find that $\\angle{JFL} = \\dfrac{\\angle{A}}{2}.$ But it is clear that $AJ$ bisects $\\angle{KAL}$, so $\\angle{JAL} = \\dfrac{\\angle{A}}{2} = \\angle{JFL}$ and hence $AFJL$ is cyclic. In particular, $\\angle{AFJ} = \\angle{ALJ} = 90^\\circ$, and continue as in Solution 2." }, { "Problem": "Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1,m_2,...,m_k$ (not necessarily different) such that\n$1+\\frac{2^k-1}{n}=(1+\\frac{1}{m_1})(1+\\frac{1}{m_2})...(1+\\frac{1}{m_k})$.", "Solution": "We prove the claim by induction on $k$.\nBase case: If $k = 1$ then $1 +\\frac{2^1-1}{n} = 1 + \\frac{1}{n}$, so the claim is true for all positive integers $n$.\nInductive hypothesis: Suppose that for some $m \\in \\mathbb{Z}^{+}$ the claim is true for $k = m$, for all $n \\in \\mathbb{Z}^{+}$.\nInductive step: Let $n$ be arbitrary and fixed. Case on the parity of $n$:\n[Case 1: $n$ is even] $1 + \\frac{2^{m+1} - 1}{n} = \\left( 1 + \\frac{2^{m} - 1}{\\frac{n}{2}} \\right) \\cdot \\left( 1 + \\frac{1}{n + 2^{m+1} - 2} \\right)$\n[Case 2: $n$ is odd] $1 + \\frac{2^{m+1} - 1}{n} = \\left( 1 + \\frac{2^{m}-1}{\\frac{n+1}{2}} \\right) \\cdot \\left( 1 + \\frac{1}{n} \\right)$\nIn either case, $1 + \\frac{2^{m+1} - 1}{n} = \\left( 1 + \\frac{2^m - 1}{c} \\right) \\cdot \\left( 1 + \\frac{1}{a_{m+1}} \\right)$ for some $c, a_{m+1} \\in \\mathbb{Z}^+$.\nBy the induction hypothesis we can choose $a_1, ..., a_m$ such that $\\left( 1 + \\frac{2^m - 1}{c} \\right) = \\prod_{i=1}^{m} (1 + \\frac{1}{a_i})$.\nTherefore, since $n$ was arbitrary, the claim is true for $k = m+1$, for all $n$. Our induction is complete and the claim is true for all positive integers $n$, $k$." }, { "Problem": "Let $a_0 f(n+1)$; i.e., $f$ is monotonic decreasing. Therefore, because $f(0) > 0$, there exists a unique $N$ such that $f(N-1) > 0 \\ge f(N)$. In other words,\n\n\nThis rearranges to give\n\nDefine $g(n) = a_0 + a_1 + \\dots + a_n - n a_n$. Then because $a_{n+1} > a_n$, we have\n\nTherefore, $g$ is also monotonic decreasing. Note that $g(N+1) = a_0 + a_1 + \\dots + a_{N+1} - (N+1) a_{N+1} \\le 0$ from our inequality, and so $g(k) \\le 0$ for all $k > N$. Thus, the given inequality, which requires that $g(n) > 0$, cannot be satisfied for $n > N$, and so $N$ is the unique solution to this inequality.\n--Suli 22:38, 7 February 2015 (EST)" }, { "Problem": "We say that a finite set $\\mathcal{S}$ in the plane is balanced \nif, for any two different points $A$, $B$ in $\\mathcal{S}$, there is\na point $C$ in $\\mathcal{S}$ such that $AC=BC$. We say that\n$\\mathcal{S}$ is centre-free if for any three points $A$, $B$, $C$ in\n$\\mathcal{S}$, there is no point $P$ in $\\mathcal{S}$ such that\n$PA=PB=PC$.", "Solution": "Part (a): We explicitly construct the sets $\\mathcal{S}$. For\nodd $n$, $\\mathcal{S}$ can be taken to be the vertices of \nregular polygons $P_n$ with $n$ sides: given any two vertices $A$ and\n$B$, one of the two open half-spaces into\nwhich $AB$ divides $P_n$ contains an odd number of $k$ of vertices of\n$P_n$. The $((k+1)/2)^{th}$ vertex encountered while moving from $A$\nto $B$ along the circumcircle of $P_n$ is therefore equidistant from\n$A$ and $B$.\nIf $n \\geq 4$ is even, choose $m\\geq 0$ to be the largest\ninteger such that Hence\n$x < 4\\pi/3 < 2\\pi$. Consider a circle \n$K$ with centre $O$, and let\n$A_1, \\ldots, A_{n-1}$ be distinct points placed counterclockwise\n(say) on $K$ such that $\\angle A_iOA_{i+1}=\\pi/3/2^m$ (for\n$i=1,\\ldots,n-2$). Hence for any line\n$OA_i$, there is a line $OA_j$ such that $\\angle A_iOA_j=\\pi/3$\n(using the facts that\n$2\\pi > x=\\angle A_1OA_{n-1} \\geq 2\\pi/3$, and $n-1$ odd). Thus\n$O$, $A_i$ and $A_j$ form an equilateral triangle. In other words, for\narbitrary $A_i$, there exists $A_j$ equidistant to\n$O$ and $A_i$. Also given any $i,j$ such that\n$1 \\leq i, j \\leq n-1$, $O$ is equidistant to $A_i$ and $A_j$. Hence\nthe $n$ points $O, A_1, \\ldots, A_{n-1}$ form a balanced set.\nPart (b): Note that if $n$ is odd, the set $\\mathcal{S}$ of\nvertices of a regular polygon $P_n$ of $n$ sides forms a balanced set\n(as above) and a centre-free set (trivially, since the centre of the\ncircumscribing circle of $P_n$ does not belong to $\\mathcal{S}$).\nFor $n$ even, we prove that a balanced, centre free set consisting of $n$\npoints does not exist. Assume that\n$\\mathcal{S}=\\{A_i: 1\\leq i \\leq n\\}$ is centre-free. Pick an\narbitrary $A_i \\in \\mathcal{S}$, and let $n_i$ be the number of\ndistinct non-ordered pairs of points $(A_j,A_k)$ ($j\\neq k$) to\nwhich $A_i$ is equidistant. Any \ntwo such pairs are disjoint (for, if there were two such pairs $(A_r,A_s)$\nand $(A_r, A_t)$ with $r, s, t$ distinct, then $A_i$ would be\nequidistant to $A_r$, $A_s$, and $A_t$, violating the centre-free\nproperty). Hence $n_i \\leq (n-2)/2$ \n(we use the fact that $n$ is even here), which means\n$\\sum_i n_i \\leq n(n-2)/2 = n(n-1)/2 -n/2$. Hence there are at least\n$n/2$ non-ordered pairs $(A_j, A_k)$ such that no point in\n$\\mathcal{S}$ is equidistant to $A_j$ and $A_k$." }, { "Problem": "Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \\ldots$ for $n \\geq 0$ as\nDetermine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.", "Solution": "First we observe the following:\nWhen we start with $a_0=3$, we get $a_1=6$, $a_2=9$, $a_3=3$ and the pattern $3,6,9$ repeats.\nWhen we start with $a_0=6$, we get $a_1=9$, $a_2=3$, $a_3=6$ and the pattern $3,6,9$ repeats.\nWhen we start with $a_0=9$, we get $a_1=3$, $a_2=6$, $a_3=9$ and the pattern $3,6,9$ repeats.\nWhen we start with $a_0=12$, we get $a_1=15$, $a_2=15$,..., $a_8=36$, $a_9=6$, $a_{10}=9$, $a_{11}=3$ and the pattern $3,6,9$ repeats.\nWhen this pattern $3,6,9$ repeats, this means that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$ and that number $A$ is either $3,6,$ or $9$.\nWhen we start with any number $a_0\\not\\equiv 0\\; mod\\; 3$, we don't see a repeating pattern.\nTherefore the claim is that $a_0=3k$ where $k$ is a positive integer and we need to prove this claim.\nWhen we start with $a_0=3k$, the next term if it is not a square is $3k+3$, then $3k+6$ and so on until we get $3k+3p$ where $p$ is an integer and $(k+p)=3q^2$ where $q$ is an integer. Then the next term will be $\\sqrt{9q^2}=3q$ and the pattern repeats again when $q=k$ or when $q=3$ or $6$.\nIn order for these patterns to repeat, any square in the sequence need to be a multiple of 3.\nTo try the other two cases where $a_0\\not\\equiv 0\\; mod\\; 3$, we can try $a_0=3k\\pm 1$ then the next terms will be in the form $3k+3p\\pm 1 = 3(k+p) \\pm 1$.\nWhen $3(k+p) \\pm 1$ is a square, it will not be a multiple of $3$ because $3(k+p) \\pm 1$ is not a multiple of $3$ and $3(k+p) \\pm 1 \\ne 9q^2$ because $3(k+p) \\pm 1 \\equiv \\pm 1\\; mod\\; 3$ and $q^2$ would have to be $\\frac{(k+p)}{3} \\pm \\frac{1}{9}$ which is not an integer even if $k+p$ is a multiple of $3$.\nTherefore the pattern doesn't repeat for any of the other cases where $a_0=3k\\pm 1$ and only repeats when $a_0\\equiv 0\\; mod\\; 3$\nSo, the answer to this problem is $a_0=3k\\;\\forall k \\in \\mathbb{Z}^{+}$ and $A=3,6,$ or $9$.\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": "Let $\\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.", "Solution": "http://wiki-images.artofproblemsolving.com/5/5d/FB_IMG_1531446409131.jpg\nThe diagram is certainly not to scale, but the argument is sound (I believe) and involves re-ordering the construction as specified in the original problem so that an identical state of affairs results, yet in so doing differently it is made clear that the line segments in question are parallel.\nConstruct a right-angled triangle ABC'. Select an arbitrary point H along the segment BC', and from point H select an arbitrary point F such that the segment HF is perpendicular to the segment AB. Mark the distance from the intersection of HF and AB to B at B\" (i.e., HF is a perpendicular bisector). It follows that the triangle B\"FB is isosceles. Construct an isosceles triangle FHG. Mark the distance of AB\" along AC' at C\". (From here, a circle can be constructed according to the sets of points A, B, F, and A, B, G. Points F and G may be repositioned to allow for these circles to coincide; also, point H may be repositioned so that point C falls on the coinciding circle, understood that HG is the other perpendicular bisector.)\nAssign the angle BAC the value α. Hence, the angle FHG has the value 180°-α, and the angle HFG (also, HGF) has the value α/2. Assign the angle BFH the value β. Hence, the angle B'FB\" has the value β-α/2. Consequently, the angle FB'B\" has the value 180°-(β-α/2)-(90°-β) = 90°+α/2, and so too its vertical angle BB'G. As the triangle B\"AC\" is isosceles, and its subtended angle has the value α, the angles BB\"C\" and CC\"B\" both have the value 90°+α/2. It follows therefore that segments B\"C\" and FG are parallel.\n(N.B. Points D and E, as given in the wording of the original problem, have been renamed B\" and C\" here.)", "Solution_2": "The essence of the proof is using a rhombus formed by the perpendicular bisectors of the segments $BD, CE, AB$ and $AC$ and the parallelism of its diagonal and the base of the triangle formed by the perpendiculars from one vertex of the rhombus.\nThe perpendicular bisectors of the segments $BD, CE, AB$ and $AC$ intersect at points $O$ (the circumcenter $ABC), H, H',$ and $Q.$ Let $AD = AE = 2x, AC = 2b.$ The distance from the line $HO$ to point $C$ is $b,$ from $QH'$ to $C$ is $b - x.$ Therefore, the distance between lines $HO$ and $QH'$ is $x.$ Similarly, the distance between the lines $HQ$ and $OH'$ is $x.$\nThe quadrilateral $OHQH'$ is formed by the intersection of two pairs of equidistant lines $\\implies$ $OHQH'$ is a parallelogram with equal heights $\\implies$ $OHQH'$ is a rhombus.\nLet $I$ and $I'$ be the feet of the heights of the rhombus from the vertex $O.$ In isosceles triangles $\\triangle ADE$ and $\\triangle OII'$ the sides are parallel, hence the bases of these triangles are parallel, $DE||II'.$\nThe feet of the heights $I$ and $I'$ divide the sides of the rhombus in the same ratio, which means that the diagonal $HH'$ of the rhombus is parallel to the segments $II' || DE.$\nThe angles between the diagonal of the rhombus and its sides are the same, so $\\angle HQO = \\angle H'QO.$\nThe lines $QF$ and $QG$ are symmetrical with respect to the diameter $OQ,$ so $QF = QG, QF' = QG'.$\nThe homothety of triangles $QHH', QFG,$ and $QF'G'$ centered at $Q$ implies $FG || F'G' || HH'.$\nTherefore, these three lines are parallel to $DE.$\nNote that triangle $ABC$ may be obtuse.\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "Let $\\mathbb{Z}$ be the set of integers. Determine all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that, for all\nintegers $a$ and $b$,", "Solution_1": "Let us substitute $0$ in for $a$ to get\nNow, since the domain and range of $f$ are the same, we can let $x = f(b)$ and $f(0)$ equal some constant $c$ to get\n\nTherefore, we have found that all solutions must be of the form $f(x) = 2x + c.$\nPlugging back into the original equation, we have: $4a + c + 4b + 2c = 4a + 4b + 2c + c$ which is true. Therefore, we know that $f(x) = 2x + c$ satisfies the above for any integral constant c, and that this family of equations is unique.\n(This solution does not work though because we don't know that $f$ is surjective)", "Solution_2": "We plug in $a=-b=x$ and $a=-b=x+k$ to get \n\n\nrespectively.\nSetting them equal to each other, we have the equation and moving \"like terms\" to one side of the equation yields Seeing that this is a difference of outputs of $f,$ we can relate this to slope by dividing by $2k$ on both sides. This gives us which means that $f$ is linear. (Functional equations don't work like that unfortunately)\nLet $f(x)=mx+n.$ Plugging our expression into our original equation yields $2ma+2mb+3n=m^2a+m^2b+mn+n,$ and letting $b$ be constant, this can only be true if $2m=m^2 \\implies m=0,2.$ If $m=0,$ then $n=0,$ which implies $f(x)=0.$ However, the output is then not all integers, so this doesn't work. If $m=2,$ we have $f(x)=2x+n.$ Plugging this in works, so the answer is $f(x)=2x+c$ for some integer $c.$", "Solution_3:_The_only_one_that_actually_works": "The only solutions are $f(x)=0, 2x+c.$ For some integer $c.$\nObviously these work. We prove these are the only linear solutions. Plug $a=0$ and $b=0$ separately to get that $f(2x)=2f(x)-f(0).$ Plug $(0, a+b)$ to see $f(0)+f(a+b)=f(a)+f(b),$ and subtracting $2f(0)$ from both sides shows $f(a)-f(0)$ to be additive thus linear by Cauchy since this is on integers. Thus, $f(a)$ is linear, and so we are done since it can be easily shown that $0$ and $2x+c$ are the only linear solutions by plugging $mx+n$ into the equation.\n~ LLL2019 (first to post correct solution on here)", "Solution_4:_This_works_as_well": "We claim the only solutions are $f\\equiv0$ and $f(x)=2x+c$ for some integer $c$., which obviously work. Plugging in $(0,n)$ and $(1,n-1)$ give $f(0)+2f(n)=f(f(n))=f(2)+2f(n-1)$, so $f(n)-f(n-1)=\\frac{f(2)-f(0)}2$. Since this difference is constant and $f\\colon\\mathbb Z\\rightarrow\\mathbb Z$, we must have $f$ is linear (finite differences or induction). It is easy to see the only linear solutions are those specified above. $\\blacksquare$\n~ Ezra Guerrero" }, { "Problem": "Consider the convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$. The following ratio equalities hold:\n Prove that the following three lines meet in a point: the internal bisectors of angles $\\angle ADP$ and $\\angle PCB$ and the perpendicular bisector of segment $\\overline{AB}$.", "Solution_2_.28Three_perpendicular_bisectors.29": "The essence of the proof is the replacement of the bisectors of angles by the perpendicular bisectors of the sides of the cyclic pentagon.\nLet $O$ be the circumcenter of $\\triangle ABP, \\angle PAD = \\alpha, OE$ is the perpendicular bisector of $AP,$ and point $E$ lies on $AD.$ Then\n$\\hspace{33mm} ABPE$ is cyclic.\n\nthe bisector of the $\\angle ADP$ is the perpendicular bisector of the side $EP$ of the cyclic $ABPE$ that passes through the center $O.$\nA similar reasoning can be done for $OF,$ the perpendicular bisector of $BP.$\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "Let $n \\geq 100$ be an integer. Ivan writes the numbers $n, n+1, \\ldots, 2 n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.", "Solution_1": "If we can guarantee that there exist $3$ cards such that every pair of them sum to a perfect square, then we can guarantee that one of the piles contains $2$ cards that sum to a perfect square. Assume the perfect squares $p^2$, $q^2$, and $r^2$ satisfy the following system of equations: \n\nwhere $a$, $b$, and $c$ are numbers on three of the cards. Solving for $a$, $b$, and $c$ in terms of $p$, $q$, and $r$ tells us that $a = \\frac{p^2 + r^2 - q^2}{2}$, $b=\\frac{p^2 + q^2 - r^2}{2}$, and $c=\\frac{q^2 + r^2 - p^2}{2}$. We can then substitute $p^2 = (2e-1)^2$, $q^2 = (2e)^2$, and $r^2 = (2e+1)^2$ to cancel out the $2$s in the denominatior, and simplifying gives $a = 2e^2 + 1$, $b = 2e(e-2)$, and $c = 2e(e+2)$. Now, we have to prove that there exists three numbers in these forms between $n$ and $2n$ when $n \\ge 100$. Notice that $b$ will always be the least of the three and $c$ will always be the greatest of the three. So it is sufficient to prove that there exists numbers in the form $2e(e-2)$ and $2e(e+2)$ between $n$ and $2n$.\nFor two numbers in the form of $2e(e-2)$ and $2e(e+2)$ to be between $n$ and $2n$, the inequalities\n\nmust be satisfied. We can then expand and simplify to get that\n\nThen, we can complete the square on the left sides of both inequalities and isolate $e$ to get that\n\nNotice that $e$ must be an integer, so there must be an integer between $\\sqrt{1 + n} - 1$ and $\\sqrt{1 + \\frac{n}{2}} + 1$. If $\\sqrt{1 + n} - 1$ and $\\sqrt{1 + \\frac{n}{2}} + 1$ differ by at least $1$, then we can guarantee that there is an integer between them (and those integers are the possible values of $e$). Setting up the inequality $\\sqrt{1 + n} - \\sqrt{1 + \\frac{n}{2}} - 2 \\ge 1$ and solving for $n$ tells us that $n \\in [107, \\infty)$ always works. Testing the remaining $7$ numbers ($100$ to $106$) manually tells us that there is an integer between $\\sqrt{1 + n} - 1$ and $\\sqrt{1 + \\frac{n}{2}} + 1$ when $n \\ge 100$. Therefore, there exists a triplet of integers $(a,b,c)$ with $a, b, c \\in \\{n, n+1, ..., 2n\\}$ when $n \\ge 100$ such that every pair of the numbers sum to a perfect square. By the pigeonhole principle, we know that $2$ of the numbers must be on cards in the same pile, and hence, when $n \\ge 100$, there will always be a pile with $2$ numbers that sum to a perfect square. $\\square$\n~Mathdreams", "Solution_2": "For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that\np+q = x^2 and q+r = y^2, p+r = z^2.\nby equation 1\nif we add (2) and (3) to (1),\n(1) + (2) + (3) =>\nAt this time 100 ≤ n, so let's put n = 100 to this\nwhere\nx = 16, y = 18, z = 20 fits perfectly\ntherefore the minimum of n fits the proposition so the proposition is true\n~Mathhyhyhy" }, { "Problem": "The Bank of Oslo issues two types of coin: aluminium (denoted A) and bronze (denoted B). Marianne has $n$ aluminium coins and $n$ bronze coins, arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k\\le 2n$, Marianne repeatedly performs the following operation: she identifies the longest chain containing the $k^{th}$ coin from the left, and moves all coins in that chain to the left end of the row. For example, if $n = 4$ and $k = 4$, the process starting from the ordering AABBBABA would be\nAABBBABA → BBBAAABA → AAABBBBA → BBBBAAAA → BBBBAAAA → ...\nFind all pairs $(n, k)$ with $1 \\le k \\le 2n$ such that for every initial ordering, at some moment during the process, the leftmost $n$ coins will all be of the same type.", "Solution": "https://www.youtube.com/watch?v=nYD-qIOdi_c [Video contains solutions to all day 1 problems]\nhttps://youtu.be/KHn3xD6wS3A\n[Video contains problem 1 discussion]\nWe call a chain basic when it is the largest possible for the coins it consists of. Let \\(A=[i,j]\\) be the basic chain with the \\(i\\)-th and \\(j\\)-th coins being the first and last, respectively.\nClaim:\nProof:\nFor \\(k < n\\), it is easy to see that the arrangement \\(A\\ldots AB\\ldots BA\\) remains the same.\nFor \\(k > \\lceil \\frac{3n}{2} \\rceil = 2n - \\lfloor \\frac{n}{2} \\rfloor\\), we obtain the arrangement \\(A\\ldots AB\\ldots BA\\ldots AB\\ldots B\\), where each basic chain consists of \\(\\lfloor \\frac{n}{2} \\rfloor, \\lceil \\frac{n}{2} \\rceil, \\lceil \\frac{n}{2} \\rceil, \\lfloor \\frac{n}{2} \\rfloor\\) coins, respectively. Since the number of coins in the last chain is \\(\\geq \\lfloor \\frac{n}{2} \\rfloor\\), it follows that \\(k\\) is greater than the number of the remaining coins, or in other words, it is always contained in the last chain.\nHowever, we have a loop:\nWe will prove that in any other case, the number of basic chains decreases by a constant, which proves the claim.\nFor \\(k \\in B=[l, m]\\), where \\(l > 1\\) and \\(m < 2n\\), the basic chains \\(B_1=[l{'}, l-1]\\) and \\(B_2=[m+1, m']\\) merge into one, and we are done since it is impossible to increase.\nFor \\(k \\in C=[1, l]\\), where \\(n + 1 > l \\geq k \\geq n\\), it holds that \\(l = n\\), which is what we need to prove.\nFor \\(k \\in D=[m, 2n]\\), we will prove that the basic chains are of quantity 2 or 3 (two are obtained with one move):\nIndeed, if there are at least 4 basic chains, from the beginning of the pigeonhole, we have at least one chain with a number of coins < \\(\\lfloor \\frac{n}{2} \\rfloor + 1 \\leq 2n - k + 1\\). Therefore, \\(k\\) does not belong to this chain when it is the last, and then the number of basic chains decreases, which completes the proof. $\\blacksquare$\nIn conclusion, such pairs are \\((n, k)\\), where \\(k \\in \\{n, n+1, \\ldots, \\lceil \\frac{3n}{2} \\rceil\\}\\)." }, { "Problem": "Determine all composite integers $n>1$ that satisfy the following property: if $d_1,d_2,\\dots,d_k$ are all the positive divisors of $n$ with $1=d_1q$, the divisor is larger than $q$. In the second case, all divisors divisible by $q^2$ are also larger than $q$, and otherwise are of the form $p^x \\cdot q^1$ or $p^x$ for some nonnegative integer $x$. If the divisor is of the form $p^x$, then it is a power of $p$. If it is of the form $p^x \\cdot q^1$, the smallest of these factors is $p^0 \\cdot q^1 = q$. Therefore, (in the case where $2$ or more primes divide $n$) the ordered tuple of divisors is of the form $(1,\\, p,\\, p^2 \\dots,\\, p^a,\\, q \\dots,\\, n)$ for some integer $a\\geq 1$, since after each divisor $p^x$, the next smallest divisor is either $p^{x+1}$ or simply $q$.\nIf $a\\geq 2$, the condition fails. This is because $p^{a-1} \\nmid p^a + q$, since $p^a$ is divisible by $p^{a-1}$, but $q$ is not since it is a prime different from $p$. If $a=1$, then $p^{a-1}=p^0=1$, which does divide $q$. Therefore $a$ must equal $1$ for the condition to be satisfied in this case. However, we know that the ordered list of divisors satisfies $d_i \\cdot d_{k+1-i}=n$, meaning since the first $3$ divisors are $(1, p, q)$, then the last $3$ divisors are $(\\frac{n}{q}, \\frac{n}{p}, n)$, so $(\\frac{n}{q})$ must divide $(\\frac{n}{p} + n)$. But $\\frac{n}{q}$ is divisible by $p$, so $\\frac{n}{p} + n$ must also be divisible by $p$, but since $a=1$ $\\frac{n}{p}$ is and $n$ isn't.\nWhen $n=p^x$, it is easy to verify this works for all primes $p$ and all $x\\geq 2$, since $p^y \\vert p^{y+1} + p^{y+2}$, and the divisors are ordered as ${1,\\, p,\\, p^2…\\, p^x}$.\n~SpencerD.", "Solution_2": "Similar argument as above but restated.\nLet $n = p_1^{e_1} p_2^{d_2} \\dots p_k^{e_k}$, with $p_1 < p_2 < \\dots < p_k$ primes.\nIf $k=1$, it's easy to verify that this property holds, since $p^l \\mid p^{l+1} + p^{l+2}$.\nSuppose $k \\geq 2$. All divisors of $n$ which contain a factor of $p_k$ for $k \\geq 2$ is at least as large as $p_2$. So, the only divisors which are possibly less than $p_2$ are of the form $p_1^{j}$. Also, by construction $p_1$ is less than $p_2$. Putting it all together, this says that the smallest factors are $(1, p_1, \\dots, p_1^j, p_2)$ for some $j \\geq 1$.\nIf $j \\geq 2$, then $p_1^{j-1} \\nmid p_1^j + p_2$, since the LHS has a factor of $p_1$ and the RHS does not since.\nIf $j = 1$, then the largest three factors are $(\\frac{n}{p_2}, \\frac{n}{p_1}, n)$. But $\\frac{n}{p_2}$ and $n$ have a factor of $p_1^{e_1}$, while $\\frac{n}{p_1}$ only has a factor of $p_1^{e_1 - 1}$, so $\\frac{n}{p_2} \\nmid \\frac{n}{p_1} + n$.\nHence the prime powers are the only composite numbers which satisfy the property." }, { "Problem": "For what real values of $x$ is\ngiven (a) $A=\\sqrt{2}$, (b) $A=1$, (c) $A=2$, where only non-negative real numbers are admitted for square roots?", "Solution_1": "The square roots imply that $x\\ge \\frac{1}{2}$.\nSquare both sides of the given equation:\nAdd the first and the last terms to get:\nMultiply the middle terms, and use $(a + b)(a - b) = a^2 - b^2$ to get:\nSince the term inside the square root is a perfect square, and by factoring 2 out, we get\n\nUse the property that $\\sqrt{x^2}=|x|$ to get\nCase I: If $x \\le 1$, then $|x-1| = 1 - x$, and the equation reduces to $A^2 = 2$. This is precisely part (a) of the question, for which the valid interval is now $x \\in \\left[ \\frac{1}{2}, 1 \\right]$\nCase II: If $x > 1$, then $|x-1| = x - 1$ and we have\n\nwhich simplifies to\nThis tells there that there is no solution for (b), since we must have $A^2 \\ge 2$\nFor (c), we have $A = 2$, which means that $A^2 = 4$, so the only solution is $x=\\frac{3}{2}$.\n~flamewavelight (Expanded)\n~phoenixfire (edited)\n", "Solution_2": "Note that the equation can be rewritten to \n\ni.e., $\\sqrt{2x-1}+1 + |\\sqrt{2x-1}-1|=A\\sqrt{2}$.\nCase I: when $2x-1\\ge 1$ (i.e., $x\\ge 1$), the equation becomes $2\\sqrt{2x-1}=\\sqrt{2}A$. For (a), we have $x=1$; for (b) we have $x=\\frac{3}{4}$; for (c) we have $x=\\frac{3}{2}$. Since $x\\ge 1$, (b) $x=\\frac{3}{4}$ is not what we want.\nCase II: when $0\\le 2x-1 <1$ (i.e., $1/2\\le x <1$), the equation becomes $2=\\sqrt{2}A$, which only works for (a) $A=\\sqrt{2}$.\nIn summary, any $x \\in \\left[\\frac{1}{2}, 1\\right]$ is a solution for (a); there is no solution for (b); there is one solution for (c), which is $x=\\frac{3}{2}$.\n~zhaoxiayu\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "For what values of the variable $x$ does the following inequality hold:\n", "Solution": "Set $x = -\\frac{1}{2} + \\frac{a^2}{2}$, where $a\\ge0$.\n$\\frac{4\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)^2}{\\left(1-\\sqrt{1+2\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)}\\right)^2}<2\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)+9$\nAfter simplifying, we get\n$(a+1)^2b$. Then for all integers $0 \\le k \\le n$, $x_n x_k a^n b^k \\ge x_n x_k b^n a^k$, with equality only when $k=n$ or $x_k = 0$. (In particular, we have strict inequality for $k=n-1$.) In summation, this becomes\n\nor\n\nwhich is equivalent to\n\nThis implies\n\nOn the other hand, if $a=b$, then evidently $A_{n-1}/A_n = B_{n-1}/B_n$, and if $a < b$, then by what we have just shown, $A_{n-1}/A_n > B_{n-1}/B_n$. Hence $A_{n-1}/A_n < B_{n-1}/B_n$ if and only if $a>b$, 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": "Consider a convex polyhedron $P_1$ with nine vertices $A_1, A_2, \\cdots, A_9;$ let $P_i$ be the polyhedron obtained from $P_1$ by a translation that moves vertex $A_1$ to $A_i(i=2,3,\\cdots, 9).$ Prove that at least two of the polyhedra $P_1, P_2,\\cdots, P_9$ have an interior point in common.", "Solution": "WLOG let $A_1$ be the origin $0$.\nTake any point $A_i$, then $P_i=A_i+P_1$, lies in $2 P_1$, the polyhedron $P_1$ stretched by the factor $2$ on $P_1=0$.\nMore general: take any $p,q$ in any convex shape $S$. Then $p+q \\in 2S$.\nProve: since $S$ is convex, $\\frac{p+q}{2} \\in S$, thus $p+q \\in 2S$.\nNow all these nine polyhedrons lie inside $2 P_1$. Let $V$ be the volume of $P_1$.\nThen some polyhedrons with total sum of volumes $9V$ lie in a shape of volume $8V$, thus they must overlap, meaning that they have an interior point in common.\nThe above solution was posted by ZetaX. The original thread for this problem can be found here: [1]" }, { "Problem": "Prove that if $n \\geq 4$, every quadrilateral that can be inscribed in a circle can be dissected into $n$ quadrilaterals each of which is inscribable in a circle.", "Solution": "Our initial quadrilateral will be $ABCD$.\nFor $n=4$, we do this:\nTake $E\\in AB,F\\in CD,\\ EF\\|AD$ with $E,F$ sufficiently close to $A,D$ respectively. Take $U\\in AD,V\\in EF$ such that $AEVU$ is an isosceles trapezoid, with $V$ close enough to $F$ (or $U$ close enough to $D$) that we can find a circle passing through $U,D$ (or $F,V$) which cuts the segments $UV,DF$ in $X,Y$. Our four cyclic quadrilaterals are $BCFE,\\ AEVU,\\ VFYX,\\ YXUD$.\nFor $n\\ge 5$ we do the exact same thing as above, but now, since we have an isosceles trapezoid, we can add as many trapezoids as we want by dissecting the one trapezoid with lines parallel to its bases.\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]" }, { "Problem": "Determine whether or not there exists a finite set $M$ of points in space not lying in the same plane such that, for any two points $A$ and $B$ of $M$; one can select two other points $C$ and $D$ of $M$ so that lines $AB$ and $CD$ are parallel and not coincident.", "Solution": "In order to solve this problem we can start by finding at least one finite set $M$ that satisfies the condition.\nWe start by defining our first set $M_{8}$ with the vertices of a cube of side $k$ as follows:\n$M_{8} = \\{ (0,0,0), (k,0,0), (k,k,0), (0,k,0), (0,0,k), (k,0,k), (k,k,k), (0,k,k) \\}$\nSince all the faces of this cube have a parallel face, then any two points on one face will have corresponding 2 points on the opposite face that is parallel. However we have four diagonals on this cube that do not have two points that are parallel to any of these diagonals.\nBy doing a reflection of the points on the $z=k$ plane along the $xy$-plane these four diagonals will have their respective parallel diagonals on the $z \\le 0$ space.\nBut now we have four more diagonals on the set of two cubes that do not have a parallel line. That is, diagonal $(0,0,-k) \\rightarrow (k,k,k)$ does not have a parallel line and neither do the other three.\nBy doing a reflection of the points on the $x=k$ plane along the $yz$-plane these new four diagonals will have their respective parallel diagonals on the $x \\le 0$ space.\nBut now we have four more diagonals on the set of 4 cubes that do not have a parallel line. That is, diagonal $(-k,0,-k) \\rightarrow (k,k,k)$ does not have a parallel line and neither do the other three.\nBy doing a reflection of the points on the $y=k$ plane along the $xz$-plane these new four diagonals will have their respective parallel diagonals on the $y \\le 0$ space.\nThe new 4 longer diagonals will cross the diagonal of two of the cubes and will have a parallel line on one of the other cubes.\nSo, we found a set at least one finite set $M$ that we can define as $M=\\{(x,y,z)\\}$ where $x,y,z \\in \\{-k,0,k\\}$ giving a total of 27 points. Therefore such a set exists.\nAnother way to define this set of points is let $M$ be:\nLet $V$ be a solid cube or right angled parallelepiped\nLet $M_{v}$ be the set of all 8 vertices of $V$\nLet $M_{me}$ be the set of all 12 midpoints of the edges of $V$\nLet $M_{mf}$ be the set of all 6 midpoints of the faces of $V$\nLet $M_{c}$ be the center of $V$\n$M$=$M_{v} \\cup M_{me} \\cup M_{mf} \\cup M_{c}$\nIt is possible that one can construct many other sets of $M$ using regular tetrahedrons with some reflections and with less points than 27, or by translating or rotating or skewing all the the points simultaneously of the finite set $M$ that we defined here. But that is not necessary for this problem because it asks to prove whether there exist a set with the described conditions. By showing that at least one set exists with those conditions, the problem is proved.\n\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": "In the triangle $ABC$, prove that there is a point $D$ on side $AB$ such that $CD$ is the geometric mean of $AD$ and $DB$ if and only if $\\sin{A}\\sin{B} \\leq \\sin^2 (\\frac{C}{2})$.", "Solution": "Let a point $D$ on the side $AB$.\nLet $CF$ the altitude of the triangle $\\triangle ABC$, and $C'$ the symmetric point of $C$ through $F$.\nWe bring a parallel line $L$ from $C'$ to $AB$. This line intersects the ray $CD$ at the point $E$, and we know that $DE=DC$.\nThe distance $d(L,AB)$ between the parallel lines $L$ and $AB$ is $CF$.\nLet $w = (O,R)$ the circumscribed circle of $\\triangle ABC$, and $MM'$ the perpendicular diameter to $AB$, such that $M,C$ are on difererent sides of the line $AB$.\nIn fact, the problem asks when the line $L$ intersects the circumcircle. Indeed:\nSuppose that $DC$ is the geometric mean of $DA,DB$.\n$DA \\cdot DB = DC^{2}\\Rightarrow DA \\cdot DB = DC \\cdot DE$\nThen, from the power of $D$ we can see that $E$ is also a point of the circle $w$.\nOr else, the line $L$ intersects $w \\Leftrightarrow$\n$d(L,AB)\\leq d(M,AB) \\Leftrightarrow$\n$CF \\leq MN,$ where $MN$ is the altitude of the isosceles $\\triangle MAB$.\n$\\Leftrightarrow \\frac{1}{2}CF \\cdot AB \\leq \\frac{1}{2}MN \\cdot AB \\Leftrightarrow$ $(ABC) \\leq (MAB) \\Leftrightarrow$ $\\frac{AB \\cdot BC \\cdot AC}{4R}\\leq \\frac{AB \\cdot MA^{2}}{4R}\\Leftrightarrow$\n$BC \\cdot AC \\leq MA^{2}$\nWe use the formulas:\n$BC = 2R \\cdot \\sin A$\n$AC = 2R \\cdot \\sin B$\nand $\\angle CMA = \\frac{C}{2}\\Rightarrow MA = 2R \\cdot \\sin\\frac{C}{2}$\nSo we have\n$(2R \\cdot \\sin A)(2R \\cdot \\sin B) \\leq (2R \\cdot \\sin\\frac{C}{2})^{2}\\Leftrightarrow$\n$\\sin A \\cdot \\sin B \\leq \\sin^{2}\\frac{C}{2}$\n\n\nFor $(\\Leftarrow)$\nSuppose that $\\sin A \\cdot \\sin B \\leq \\sin^{2}\\frac{C}{2}$\nThen we can go inversely and we find that $d(L,AB)\\leq d(M,AB) \\Leftrightarrow$\nthe line $L$ intersects the circle $w$ (without loss of generality; if $d(L,AB)=d(M,AB)$ then $L$ is tangent to $w$ at $M$)\nSo, if $E \\in L \\cap w$ then for the point $D = CE \\cap AB$ we have $DC=DE$ and $AD \\cdot AB = CD \\cdot DE \\Rightarrow$\n$AD \\cdot AB = CD^{2}$\nThe above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [1]" }, { "Problem": "Let $a_1, a_2, a_3, \\cdots$ be an infinite increasing sequence of positive integers. Prove that for every $p \\geq 1$ there are infinitely many $a_m$ which can be written in the formwith $x, y$ positive integers and $q > p$.", "Solution": "If we can find $p\\ne q$ such that $(a_p,a_q)=1$, we're done: every sufficiently large positive integer $n$ can be written in the form $xa_p+ya_q,\\ x,y\\in\\mathbb N$. We can thus assume there are no two such $p\\ne q$. We now prove the assertion by induction on the first term of the sequence, $a_1$. The base step is basically proven, since if $a_1=1$ we can take $p=1$ and any $q>1$ we want. There must be a prime divisor $u|a_1$ which divides infinitely many terms of the sequence, which form some subsequence $(a_{k_n})_{n\\ge 1},\\ k_1=1$. Now apply the induction hypothesis to the sequence $\\left(\\frac{a_{k_n}}u\\right)_{n\\ge 1}$.\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]" }, { "Problem": "Let $P_{1}(x) = x^{2} - 2$ and $P_{j}(x) = P_{1}(P_{j - 1}(x))$ for $j= 2,\\ldots$ Prove that for any positive integer n the roots of the equation $P_{n}(x) = x$ are all real and distinct.", "Solution": "I shall prove by induction that $P_n(x)$ has $2^n$ distinct real solutions, where $2^{n-1}$ are positive and $2^{n-1}$ are negative. Also, for ever root $r$, $|r|<2$.\nClearly, $P_1(x)$ has 2 real solutions, where 1 is positive and 1 is negative. The absolute values of these two solutions are also both less than 2. This proves the base case.\nNow assume that for some positive integer $k$, $P_k(x)$ has $2^k$ distinct real solutions with absolute values less than 2, where $2^{k-1}$ are positive and $2^{k-1}$ are negative.\nChoose a root $r$ of $P_{k+1}(x)$. Let $P_1(r)=s$, where $s$ is a real root of $P_k(x)$. We have that $-2s_n$ for all $n\\ge 1$.\nWe then have:\n$0 6\\,$ be an integer and $\\,a_{1},a_{2},\\cdots ,a_{k}\\,$ be all the natural numbers less than $n$ and relatively prime to $n$. If\n\nprove that $\\,n\\,$ must be either a prime number or a power of $\\,2$.", "Solution_1": "We use Bertrand's Postulate: for $u\\ge 2$ is a positive integer, there is a prime in the interval $(u,2u)$.\nClearly, $a_2$ must be equal to the smallest prime $p$ which does not divide $n$. If $p=2$, then $n$ is a prime since the common difference $a_{i+1}-a_i$ is equal to $1$, i.e. all positive integers less than $n$ are coprime to $n$. If, on the ther hand, $p=3$, we find $n$ to be a power of $2$: the positive integers less than $n$ and coprime to it are precisely the odd ones. We may thus assume that $p\\ge 5$. Furthermore, since $n>6$, the positive integers less than $n$ and coprime to it cannot be $a_1,a_2$ alone, i.e. $k=\\varphi(n)\\ge 3$.\nBy Bertrand's Postulate, the largest prime less than $p$ is strictly larger than $\\frac{p-1}2$, so it cannot divide $p-1$. We will denote this prime by $q$. We know that $q|n$. It's easy to check that for $p\\le 31$ there is a prime $r$ strictly between $a_2=p,\\ a_3=2p-1$. $rq|n$, so, in particular, $pq32$, the two largest primes $r_11$. Say, $n=2l$. Then, $d$divides $2(l-1)$. If $d$ is odd, $d|l$ and $d|l-1$ forcing $d|1$, contradiction! So, $d=2s$ with $s|l-1$. If $l=2r+1$ for some $r$, then $\\gcd(r,n)=1$. So, again $r=1+d(j-1$ for a $j$. If $s$ is odd, then $s|r$ and thus, $s|1$. Similarly, we find that actually $d|2$ must occur. We finally have, $d=2$. But then all odd numbers less than $n$ are co-prime to $n$. So, $n$ does not have any odd factor i.e. $n=2^k$ for some $k\\in\\mathbb N$.\nThis solution was posted and copyrighted by ngv. The original thread for this problem can be found here: [2]", "Solution_3": "We use Bonse's Inequality: $p_{m+1}^2 < p_1p_2\\cdots p_m$, for all $m \\ge 4$, $p_1 = 2$.\nLet $p$ denote the smallest prime which does not divide $n$.\nCase 1) $p = 2$. This implies $n$ is a prime.\nCase 2) $p = 3$. This implies $n = 2^k$ for some $k \\in \\mathbb{N}$.\nCase 3) $p = 5$. Since $n > 6$, we have that $n > 9$. Therefore $a_3 = 9$, but $\\gcd (n, 9) > 1$.\nCase 4) $p \\ge 7$. Write $n = p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_k^{\\alpha_k}$. We have $(p - 2)^2 < p_1p_2\\cdots p_k \\le n = p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_k^{\\alpha_k}$. Hence $(p-2)^2 = a_i$, a contradiction since $\\gcd(p-2, n) > 1$.\nThis solution was posted and copyrighted by SFScoreLow. The original thread for this problem can be found here: [3]", "Solution_4": "Clearly, $a_1 = 1$. Now let $p$ be the smallest prime number which is not a divisor of $n$. Hence, $a_2 = p$, and the common difference $d = p-1$. Observe that since $a_k = n-1$, we have that, or. This means that all prime factors of $p-1$ are prime factors of $n-2$. However, due to the minimality of $p$, all primes factors of $p-1$ are prime factors of $n$. By the Euclidean algorithm, the only prime factors of $p-1$ are $2$, so $p = 2^{2^k} + 1$ or $p = 2$, by the theory of Fermat primes. We will now do casework on $p$.\nCase 1: $p = 2$\nThis case is relatively easy. $a_2 = 2$, so all numbers less than $n$ must be relatively prime to $n$. However, if $n = pq$, where $p > 1$ and $q > 1$, observe that $p$ is not relatively prime to $n$, so $n$ must be prime for this case to work.\nCase 2: $k = 0$ and $p = 3$.\nThis case is somewhat tricky. Observe that $n$ is even. Since $a_2 = 3$, we have that all numbers less than $n$ which are odd are relatively prime to $n$. However, if $n$ has a odd divisor $p > 1$, clearly $p$ is not relatively prime to $n$, so $n$ must be a power of two.\nCase 3: $k \\ge 1$.\nNote that $3$ must be a divisor of $n$, so $a_k$ cannot be a multiple of $3$ for any $k$. I will contradict this statement. First, I claim that $k = \\phi(n) > 2$ for all integers $n > 6$. $\\phi{n} = 1$ is trivial, so let me show that $\\phi(n) = 2$ has only solutions less than or equal to $6$. Since $\\phi$ is a multiplicative function, observe that $n$ cannot be a multiple of a prime greater than $3$, because for such primes $p$ we have that $\\phi(p^k) = (p-1)p^{k-1} \\ge p-1 > 2$, which is contradiction. Hence, $n$ is just composed of factors of $2$ and $3$. If it divides both, and is of the form $2^m 3^n$ for $m$ and $n$ positive, we get that we need $2^{m} 3^{n-1} = 2$, so $n = 1$ and $m = 1$ in this case. If it is of the form $n = 2^k$, we get $k = 2$. And if it is of the form $3^k$, we get $k = 1$. $3$, $4$, and $6$ are all less than or equal to $6$, so $\\phi(n) > 2$. Now we're almost done. Since $k > 2$, the number $a_3 = 2p-1$ must be in this sequence. However, $a_3 = 2p-1$ is of the form $2^{2^n+1} + 1$, which is a multiple of $3$, and we have contradiction, so we are done.\nThis solution was posted and copyrighted by va2010. The original thread for this problem can be found here: [4]", "Solution_5": "Clearly that $a_1=1$. Let $a_{i+1}-a_i = d \\; (1 \\le i \\le \\varphi(n)-1)$ thenWe can see that if $n$ is a prime then $n$ satisfies the condition. We consider the case when $n$ is a composite number. Let $p$ be the smallest prime divisor of $n$ then $p \\le \\sqrt{n}$. Note that $\\varphi (n) > \\sqrt{n} \\ge p$ for all prime $n>6$ since $p_i^{k_i}-p_i^{k_i-1} \\ge p_i^{k_i/2}$ for all $p \\ge 3$. Therefore, if $p \\nmid d$ then there will exists $1 \\le i \\le \\varphi(n)$ such that $(i-1)d \\equiv -1 \\pmod{p}$ or $p \\mid a_i$, a contradiction since $\\gcd (a_i,n)=1$. Thus, $p \\mid d$. We consider two cases:\nIf $d \\ge 2p$ then $p+1 \\ne a_i \\; (1 \\le i \\le \\varphi(n))$. Hence, $\\gcd (p+1,n)>1$. This means there exists a prime $q>p \\ge 2$ such that $q \\mid p+1$. Since $q>p$ so we get $q=p+1$ or $p=2,q=3$, a contradiction since $n>6$. Thus, $d <2p$ or $d=p$.\nIf $p=d=2$ then all odd numbers $a_i \\le n$ are relatively prime to $n$. This can only happen when $n=2^x \\; (x \\in \\mathbb{N})$.\nIf $p=d >3$ then $p+3 \\ne a_i \\; \\forall 1 \\le i \\le \\varphi(n)$ since $a_i \\equiv 1 \\pmod{p} \\; \\forall 1 \\le i \\le \\varphi(n)$. Thus, $\\gcd (p+3,n)>1$ or there exist a prime $q>p>3$ such that $q \\mid p+3$. Note that $2 \\mid p+3$ so $q \\le \\tfrac{p+3}{2} v$, then there is some $t$ s.t. $u = t^{2}+v$, and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$. Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$, we must have $f(x) = x$ for all x.[/quote]" }, { "Problem": "Let $D$ be a point inside acute triangle $ABC$ such that $\\angle ADB = \\angle ACB+\\frac{\\pi}{2}$ and $AC\\cdot BD=AD\\cdot BC$.\n(a) Calculate the ratio $\\frac{AC\\cdot CD}{AB\\cdot BD}$.\n(b) Prove that the tangents at $C$ to the circumcircles of $\\Delta ACD$ and $\\Delta BCD$ are perpendicular.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "(Nazar Agakhanov, Russia)\nLet $a, b, c$ be positive real numbers such that $abc = 1$. Prove that", "Solution_1": "We make the substitution $x= 1/a$, $y=1/b$, $z=1/c$. Then\n\nSince $(x^2,y^2,z^2)$ and $\\bigl( 1/(y+z), 1/(z+x), 1/(x+y) \\bigr)$ are similarly sorted sequences, it follows from the Rearrangement Inequality that\n\nBy the Power Mean Inequality,\n\nSymmetric application of this argument yields\n\nFinally, AM-GM gives us\n\nas desired. $\\blacksquare$", "Solution_2": "We make the same substitution as in the first solution. We note that in general,\n\nIt follows that $(x,y,z)$ and $\\bigl(x/(y+z), y/(z+x), z/(x+y)\\bigr)$ are similarly sorted sequences. Then by Chebyshev's Inequality,\n\nBy AM-GM, $\\frac{x+y+z}{3} \\ge \\sqrt[3]{xyz}=1$, and by Nesbitt's Inequality,\n\nThe desired conclusion follows. $\\blacksquare$", "Solution_3": "Without clever substitutions:\nBy Cauchy-Schwarz, Dividing by $2(ab+bc+ac)$ gives by AM-GM.", "Solution_3b": "Without clever notation:\nBy Cauchy-Schwarz,\nDividing by $2(ab + bc + ac)$ and noting that $ab + bc + ac \\ge 3(a^2b^2c^2)^{\\frac{1}{3}} = 3$ by AM-GM gives\n\nas desired.", "Solution_4": "After the setting $a=\\frac{1}{x}, b=\\frac{1}{y}, c=\\frac{1}{z},$ and as $abc=1$ so $\\left(\\frac{1}{a}\\cdot\\frac{1}{b}\\cdot\\frac{1} {c}=1\\right)$ concluding $x y z=1 .$\n$\\textsf{Claim}:$\n\nBy Titu Lemma,\n\n\nNow by AM-GM we know that and $xyz=1$ which concludes to $\\implies (x+y+z)\\geq3\\sqrt[3]{1}$\nTherefore we get\nHence our claim is proved ~~ Aritra12", "Solution_5": "Proceed as in Solution 1, to arrive at the equivalent inequality\n\nBut we know that by AM-GM. Furthermore,\n\nby Cauchy-Schwarz, and so dividing by $2(x + y + z)$ gives\n\nas desired.", "Solution_6": "Without clever substitutions, and only AM-GM!\nNote that $abc = 1 \\implies a = \\frac{1}{bc}$. The cyclic sum becomes $\\sum_{cyc}\\frac{(bc)^3}{b + c}$. Note that by AM-GM, the cyclic sum is greater than or equal to $3\\left(\\frac{1}{(a+b)(b+c)(a+c)}\\right)^{\\frac13}$. We now see that we have the three so we must be on the right path. We now only need to show that $\\frac32 \\geq 3\\left(\\frac{1}{(a+b)(b+c)(a+c)}\\right)^\\frac13$. Notice that by AM-GM, $a + b \\geq 2\\sqrt{ab}$, $b + c \\geq 2\\sqrt{bc}$, and $a + c \\geq 2\\sqrt{ac}$. Thus, we see that $(a+b)(b+c)(a+c) \\geq 8$, concluding that $\\sum_{cyc} \\frac{(bc)^3}{b+c} \\geq \\frac32 \\geq 3\\left(\\frac{1}{(a+b)(b+c)(a+c)}\\right)^{\\frac13}$\n", "Solution_7_from_Brilliant_Wiki_.28Muirheads.29_.3D": "https://brilliant.org/wiki/muirhead-inequality/\n", "Solution_8_.28fast_Titu.27s_Lemma_no_substitutions.29": "Rewrite $\\frac{1}{a^3(b+c)} + \\frac{1}{b^3(a+c)} + \\frac{1}{c^3(a+b)}$ as $\\frac{(1/a)^2}{a(b+c)} + \\frac{(1/b)^2}{b(a+c)} + \\frac{(1/c)^2}{c(a+b)}$.\nNow applying Titu's lemma yields $\\frac{(1/a)^2}{a(b+c)} + \\frac{(1/b)^2}{b(a+c)} + \\frac{(1/c)^2}{c(a+b)} \\geq \\frac{(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c})^2}{a(b+c) + b(a+c) + c(a+b)} = \\frac{(ab + bc + ca)^2}{2(ab + bc + ca)} = \\frac{ab + bc + ca}{2}$.\nNow applying the AM-GM inequality on $ab + bc +ca \\geq 3((abc)^2)^{\\frac{1}{3}} = 3$. The result now follows.\nNote: $ab + bc + ca = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$, because $abc = 1$. (Why? Because $a = \\frac{1}{bc}$, and hence $\\frac{1}{a} = bc$).\n~th1nq3r\nScroll all the way down\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Let $P$ be a point inside triangle $ABC$ such that\n\nLet $D$, $E$m be the incenters of triangles $APB$, $APC$, respectively. Show that $AP$, $BD$, $CE$ meet at a point.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "The angle at $A$ is the smallest angle of triangle $ABD$. The points $B$ and $C$ divide the circumcircle of the triangle into two arcs. Let $U$ be an interior point of the arc between $B$ and $C$ which does not contain $A$. The perpendicular bisectors of $AB$ and $AC$ meet the line $AU$ and $V$ and $W$, respectively. The lines $BV$ and $CW$ meet at $T$. Show that.\n$AU=TB+TC$\n", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "In a competition, there are a contestants and b judges, where b ≥ 3 is an odd\ninteger. Each judge rates each contestant as either “pass” or “fail”. Suppose k\nis a number such that, for any two judges, their ratings coincide for at most k\ncontestants. Prove that k/a ≥ (b − 1)/(2b).", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "(Marcin Kuczma, Poland)\nLet $n \\geq 2$ be a fixed integer.\n", "Solution": "The answer is $C=1/8$, and equality holds exactly when two of the $x_i$ are equal to each other and all the other $x_i$ are zero. We prove this by induction on the number of nonzero $x_i$.\nFirst, suppose that at most two of the $x_i$, say $x_a$ and $x_b$, are nonzero. Then the left-hand side of the desired inequality becomes $x_a x_b (x_a^2 + x_b^2)$ and the right-hand side becomes $(x_a + x_b)^4/8$. By AM-GM,\n\nwith equality exactly when $x_a = x_b$, as desired.\nNow, suppose that our statement holds when at most $k$ of the $x_i$ are equal to zero. Suppose now that $k+1$ of the $x_i$ are equal to zero, for $k\\ge 2$. Without loss of generality, let these be $x_1 \\ge \\dotsb \\ge x_k \\ge x_{k+1}$.\nWe define\n\nand for convenience, we will denote $S= \\sum_{i=1}^{k-1} x_i$. We wish to show that by replacing the $x_i$ with the $y_i$, we increase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.\nWe note that\n\nIf we replace $x$ with $y$, then $S(x_k^3 + x_{k+1}^3) + x_k^3 x_{k+1} + x_k x_{k+1}^3$ becomes\n\nbut none of the other terms change. Since $3S > S \\ge a_1 \\ge a_k, a_{k+1}$, it follows that we have strictly increased the right-hand side of the equation, i.e.,\n\nBy inductive hypothesis,\n\nand by our choice of $y_i$,\n\nHence the problem's inequality holds by induction, and is strict when there are more than two nonzero $x_i$, as desired. $\\blacksquare$", "Solution_2_.28elegant.29": "I claim that $C=\\frac{1}{8}$. Let $P_2=\\sum_{i=1}^nx_i^2$ and $S_2=\\sum_{1\\leq i1$. Let us denote\n\nDenote\n\nSince $k(b^3-1) >0$, and $a$ is a positive integer root of $P$, there must be some other root $a'$ of $P$.\nWithout loss of generality, let $a' \\ge a$. Then $a^2 \\le aa' = k(b^3-1)$, so\n\nor\n\nwhich reduces to\n\nIt follows that\n\nor\n\nSince $a$ and $b$ are integers, this can only happen when $2a-b=0$, so $(a,b)$ can be written as $(n,2n)$, and $k = n^2$. It follows that\n\nSince $a'$ is the other root of $P$, it follows that $(a',b)$ also satisfies the problem's condition. Therefore the solutions are exactly the ones given at the solution's start. $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\nIn this problem we can do it by an alternative method\na^2/2ab^2-b^3+1>=1\na^2>=2ab^2-b^3+1\na^2-2ab+b^2>=1/b\n(a-b)^2>=1/b\nThe solutions are a>=2 and b>=1 are all the solutions" }, { "Problem": "Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab + bc + ca = 0$ we have the following relations\n", "Solution_1": "From $b=c=0$, we have $f(a) + f(-a) = 2f(a) \\Longrightarrow f(a) = f(-a)$, so $f$ is even, and all the degrees all of its terms are even. Let $\\text{deg}\\, f(x) = n$\nLet $(a,b,c) = (6x, 3x, -2x)$*; then we have $f(3x) + f(5x) + f(8x) = 2f(7x)$. Comparing lead coefficients, we have $3^n + 5^n + 8^n = 2 \\cdot 7^n$, which cannot be true for $n \\ge 5$. Hence, we have $f(x) = c_1 x^4 + c_2 x^2$. We can easily verify by expanding that all such polynomials work.", "Solution_2": "Let $a = (1 - \\sqrt {3})x$, $b = x$, and $c = (1 + \\sqrt {3})x$. Then it is easy to check that $ab + bc + ca = 0$, so\n\nfor all $x$. Hence, for the coefficient of $x^n$ to be nonzero, we must have $2( - \\sqrt {3})^n + (2 \\sqrt {3})^n = 2 \\cdot 3^n$.\nThis does not hold for $n = 1$, and if $n$ is odd and $n \\ge 3$, then the LHS is irrational and the RHS is a positive integer, so $n$ must be even.\nLet $n = 2m$. Then $2 \\cdot 3^m + 12^m = 2 \\cdot 3^{2m}$, so $2 + 4^m = 2 \\cdot 3^m$. This holds for $m = 1$ and $m = 2$, and $(4/3)^3 = 64/27 > 2$, so $2 + 4^m > 4^m > 2 \\cdot 3^m$ for $m \\ge 3$. Therefore, $P(x)$ must be of the form $a_2 x^2 + a_4 x^4$." }, { "Problem": "Let $a_1, a_2, \\dots$ be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer $n$ the numbers $a_1, a_2, \\dots, a_n$ leave $n$ different remainders upon division by $n$. Prove that every integer occurs exactly once in the sequence.", "Solution": "${a_n}$ satisfies the conditions if and only if ${a_n-a_1}$ does.\nTherefore we can assume that $a_1=0$\nFirst of all, $|a_n|v$). \nLet the set of points $S = \\{A_1, A_2, ... A_k\\}$ be the out-most points of the configuration, such that you could form a convex k-gon, $A_1 A_2 A_3 ... A_k$, that has all of the other points within it.\nIf the set S has at least one blue point, there can be a line that separates the plane into two regions: one only consisting of only a blue point, and one consisting of the rest. For the rest of the blue points, we can draw parallel lines as mentioned before to split them from the red points.\nWe end up with $v$ lines.\nIf the set $S$ has no blue points, there can be a line that divides the plane into two regions: one consisting of two red points, and one consisting of the rest. For the rest of the red points, we can draw parallel lines as mentioned before to split them from the blue points.\nWe end up with $u-1 = v$ lines.\nNow we will show that there are configurations that can not be partitioned with less than $v$ lines.\nConsider the arrangement of these points on a circle so that between every two blue points there are at least one red point (on the circle).\nThere are no less than $2v$ arcs of this circle, that has one end blue and other red (and no other colored points inside the arc) - one such arc on each side of each blue point.\nFor a line partitioning to be good, each of these arcs have to be crossed by at least one line, but one line can not cross more than $2$ arcs on a circle - therefore, this configuration can not be partitioned with less than $v$ lines!\nOur proof is done, and we have our final answer: $2013$." }, { "Problem": "Let $n\\ge2$ be an integer. Consider an $n\\times n$ chessboard consisting of $n^2$ unit squares. A configuration of $n$ rooks on this board is $\\textit{peaceful}$ if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that, for each peaceful configuration of $n$ rooks, there is a $k\\times k$ square which does not contain a rook on any of its $k^2$ squares.", "Solution": "We claim the answer is $k = \\lceil \\sqrt{n}\\rceil - 1$, where $\\lceil n\\rceil$ is the ceiling function of $n$; i.e., the least integer greater than or equal to $n$. Notice that $\\lceil n\\rceil < n + 1$.\nFirst, we shall show that each $n \\times n$ chessboard with a peaceful configuration of $n$ rooks contains a valid $k \\times k$ square. Consider firstly the rook $R$ on the top row of the board. Because $k < n$, there exists a set $C$ of $k$ consecutive columns, one of which contains rook $R$. Consider the $n - k + 1$ $k \\times k$ squares in $C$. Of them, only one contains the rook $R$ on one of its squares. Furthermore, each of the other $k - 1$ rooks in $C$ can only make up to $k$ of the $k \\times k$ squares have it on one of its squares. Therefore, because\n\nby the Pigeonhole Principle there exists a $k \\times k$ square in $C$ not covered by any rook in $C$. Clearly, that square cannot be covered by any rook outside of $C$, and so it is a valid choice.\nIt remains to show that there exists a chessboard without a $(k+1) \\times (k+1)$ square. Indeed, such a board exists. First, place a rook at the upper-left corner of the board. Next, place a rook 1 space to the right and $(k+1)$ spaces down of the first rook. Then, place a rook 1 space to the right and $(k+1)$ spaces down of the second rook, and so on, until the placement of a new rook will be under the lower boundary of the board. In that case, place a rook in the same unoccupied column and in the first unoccupied row, and continue placement of subsequent rooks. This arrangement of rooks is clearly peaceful, and because $(k+1)^2 > n$ it has the property that any $(k+1) \\times (k+1)$ square in the board will be occupied by at least one rook, completing the proof.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Determine all triples of positive integers $(a,b,c)$ such that each of the numbers \n\nis a power of 2.\n(A power of 2 is an integer of the form $2^n$ where $n$ is a non-negative integer ).", "Solution": "The solutions for $(a,b,c)$ are $(2,2,2)$, $(2,2,3)$, $(2,6,11)$, $(3,5,7)$, and\npermutations of these triples.\nThroughout the proof, we assume $a \\leq b \\leq c$,\nso that $ab-c = 2^m$, $ca-b = 2^n$,\n$bc-a=2^p$, with $m \\leq n \\leq p$. Note that $a>1$ since\notherwise $b-c=2^m$, which is impossible. Hence \n$2^n = ac-b \\geq (a-1)c \\geq 2$, i.e., $n$ and $p$ are positive.\nObserve that if $a=b\\geq 3$, we get $a(c-1)=2^n$, so $a$ and $c-1$ are (even and)\npowers of $2$. Hence $c$ is odd and $a^2-c=2^m=1$. Hence $c+1=a^2$ is also a\npower of $2$, which implies $c=3$. But $a=b=c=3$ is not a\nsolution; hence $a=b\\geq 3$ is infeasible. We consider the remaining cases as follows.\nCase 1: $a=2$. We have\n\nFrom the second equation, $b$ is even. From the third equation, if\n$p=1$, then $b=c=2$; if $p>1$, then $c$ is odd, which implies that\n$m=0$. Hence $3b=2^n+2$ (so $n\\geq 2$), \n$3c=2^{n+1}+1$, and $(2^{n-1}+1)(2^{n+1}+1)=9(2^{p-1}+1)$. Hence\n$1\\equiv 9 \\mod 2^{n-1} \\implies n \\leq 4$. Hence $n$ is 2 or 4, and\n$(b,c)$ equals $(2,3)$ or $(6,11)$. Thus the solutions for $(a,b,c)$\nare $(2,2,2)$, $(2,2,3)$ or $(2,6,11)$.\nCase 2: $3\\leq a2$. Hence there are no\nsolutions with $a=4$.\nWe obtain $(a,b,c)=(3,5,7)$ as the only solution with\n$3 \\leq a < b \\leq c$." }, { "Problem": "Find all integers $n$ for which each cell of $n \\times n$ table can be filled with one of the letters $I,M$ and $O$ in such a way that:\nNote. The rows and columns of an $n \\times n$ table are each labelled $1$ to $n$ in a natural order. Thus each cell corresponds to a pair of positive integer $(i,j)$ with $1 \\le i,j \\le n$. For $n>1$, the table has $4n-2$ diagonals of two types. A diagonal of first type consists all cells $(i,j)$ for which $i+j$ is a constant, and the diagonal of this second type consists all cells $(i,j)$ for which $i-j$ is constant.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $\\mathbb{R}$ be the set of real numbers , determine all functions \n$f:\\mathbb{R}\\rightarrow\\mathbb{R}$ such that for any real numbers $x$ and $y$\n", "Solution": "Looking at the equation one can deduce that the functions that will work will be linear. That is, a polynomial of at most a degree of 1.\nThus, $f$ is in the form $f(x)=mx+b$\nTherefore,\n$f((mx+b)(my+b))+m(x+y)+b=mxy+b$\n$f(m^2xy+mb(x+y)+b^2)+m(x+y)+b=mxy+b$\n$m(m^2xy+mb(x+y)+b^2)+b+m(x+y)+b=mxy+b$\n$m^3xy+m^2b(x+y)+mb^2+m(x+y)+b=mxy$\n$m(m^2-1)xy+m(bm+1)(x+y)+b(bm+1)=0$\nTherefore,\n$m(m^2-1)=0$ [Equation 1]\n$m(bm+1)=0$ [Equation 2]\n$b(bm+1)$ [Equation 3]\nFrom [Equation 1] we have, $m=0,\\pm 1$\nFrom [Equation 2] we have, $m=0, bm=-1\\pm 1$\nFrom [Equation 3] we have, $b=0, bm=-1\\pm 1$\nWhen $m=0$, $b=0$, then $f(x)=0$\nWhen $bm=-1$, $b=\\frac{-1}{m}$, then since $m=\\pm 1$, then $b=\\mp 1$\nWhen $bm=-1$, $b=\\frac{-1}{m}$, then since $m=\\pm 1$, then $b=\\mp 1$ then $f(x)=\\pm x \\mp1$ which gives these two functions:\n$f(x)=x-1$ and $f(x)=1-x$, which with $f(x)=0$ provide all the three functions for this problem.\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": "Find all numbers $n \\ge 3$ for which there exists real numbers $a_1, a_2, ..., a_{n+2}$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and \n\nfor $i = 1, 2, ..., n.$", "Solution": "We find at least one series of real numbers for $n = 3,$ for each $n = 3k$ and we prove that if $n = 3k \\pm 1,$ then the series does not exist.\nCase 1\nLet $n = 3.$ We get system of equations\nWe subtract the first equation from the second and get: \nSo $a_2 = – 1 \\implies a_1 = 2, a_3 = – 1.$\nCase 1a\nLet $n = 3k, k={1,2,...}.$ \nReal numbers $a_1 =a_4 =...=2, a_2 = a_3 = a_5=...=-1$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and $a_{i}a_{i+1} + 1 = a_{i+2}$.\nCase 2\nLet $n = 4.$ We get system of equations \n\nWe multiply each equation by the number on the right-hand side and get:\n\nWe multiply each equation by a number that precedes a pair of product numbers in a given sequence $a_1, a_2, a_3, a_4, a_1, a_2.$ So we multiply the equation with product $a_1 a_2$ by $a_4$, we multiply the equation with product $a_4 a_1$ by $a_3$ etc. We get:\n\nWe add all the equations of the first system, and all the equations of the second system. The sum of the left parts are the same! It includes the sum of all the numbers $a_i$ and the sum of the triples of consecutive numbers $a_i a_{i + 1} a_{i + 2}.$ Hence, the sums of the right parts are equal, that is,\n\nIt is known that this expression is doubled\n\nSubstituting into any of the initial equations, we obtain the equation $a_1^2 + 1 = a_1,$ which does not have real roots. Hence, there are no such real numbers.\nCase 2a\nLet $n = 5.$ We get system of equations \n\nWe repeat all steps of Case 2 and get: there are no such real numbers.\nCase 2b\nLet $n = 3k \\pm 1.$ We repeat all steps of cases $2$ and $2a$ and get: there are no such real numbers.\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\\angle PP_1C=\\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\\angle CQ_1Q=\\angle CBA$.\nProve that points $P,Q,P_1$, and $Q_1$ are concyclic.", "Solution": "The essence of the proof is to build a circle through the points $P, Q,$ and two additional points $A_0$ and $B_0,$ then we prove that the points $P_1$ and $Q_1$ lie on the same circle.\nWe assume that the intersection point of $AP$ and $BQ$ lies on the segment $PA_1.$ If it lies on segment $AP,$ then the proof is the same, but some angles will be replaced with additional ones up to $180^\\circ$.\nLet the circumcircle of $\\triangle ABC$ be $\\Omega$. Let $A_0$ and $B_0$ be the points of intersection of $AP$ and $BQ$ with $\\Omega$. Let $\\angle BAP = \\delta.$\n\n$\\angle BAP = \\angle BB_0A_0 = \\delta$ since they intersept the arc $BA_0$ of the circle $\\Omega$.\n$\\angle QPA_0 = \\angle QB_0A_0 \\implies QPB_0A_0$ is cyclic (in circle $\\omega.$)\nLet $\\angle BAC = \\alpha, \\angle AA_0B_0 = \\varphi.$\n$\\angle PP_1C = \\alpha, \\angle BB_0C = \\alpha$ since they intersept the arc $BC$ of the circle $\\Omega.$\nSo $B_0P_1CB_1$ is cyclic.\n$\\angle ACB_0 = \\angle AA_0B_0 = \\varphi$ (they intersept the arc $A_0B_0$ of the circle $\\Omega).$\n$\\angle B_1CB_0 = \\varphi.$\n$\\angle B_1P_1B_0 = \\angle B_1CB_0 = \\varphi$ (since they intersept the arc $B_1B_0$ of the circle $B_0P_1CB_1).$\nHence $\\angle PA_0B_0 = \\angle PP_1B_0 = \\varphi,$ the point $P_1$ lies on $\\omega.$\nSimilarly, point $Q_1$ lies on $\\omega.$\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "The real numbers $a$, $b$, $c$, $d$ are such that $a \\geq b \\geq c \\geq d > 0$ and $a + b + c + d = 1$. Prove that", "Solution": "Using Weighted AM-GM we get\n\n\nSo,\nNow notice that\nSo, we get\nNow, for equality we must have $a=b=c=d=\\frac{1}{4}$\nIn that case we get\n~ftheftics" }, { "Problem": "Show that the inequality \n\nholds for all real numbers $x_1,x_2,\\dots,x_n$.", "Solution": "then, since \nthen, \n\ntherefore we have to prove that \n for every list [Xi], \nand we can describe this to\n \nwe know that\n\ntherefore, \n\n\n--Mathhyhyhy 13:29, 6 June 2023 (EST)" }, { "Problem": "Let $\\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that for each $x \\in \\mathbb{R}^+$, there is exactly one $y \\in \\mathbb{R}^+$ satisfying\n.", "Solution": "https://www.youtube.com/watch?v=nYD-qIOdi_c [Video contains solutions to all day 1 problems]\nhttps://youtu.be/b5OZ62vkF9Y [Video Solution by little fermat]\n\nAnswer: The unique solution is the function \\( f(x) = \\frac{1}{x} \\) for every \\( x \\in \\mathbb{R}^+ \\). This function clearly satisfies the required property since the expression \\( xf(y) + yf(x) = \\frac{x}{y} + \\frac{y}{x} \\) is greater than 2 for every \\( y \\neq x \\) (directly from AM-GM) and equal to 2 (with equality) for the unique value \\( y = x \\).\nProof: Let's consider a solution based on some ideas we encountered in the preparation classes for the Olympiad, specifically involving auxiliary sets and functions with specific properties.\nThe fact that for every \\( x \\in \\mathbb{R}^+ \\), there exists a unique \\( y \\in \\mathbb{R}^+ \\) that satisfies the equation \\( xf(y) + yf(x) \\leq 2 \\) can be equivalently expressed as follows: there exists a well-defined function \\( g: \\mathbb{R}^+ \\to \\mathbb{R}^+ \\) given by \\( g(x) := y \\), where \\( y \\) is the one mentioned above. The well-definedness of this function is evident due to the existence and uniqueness, and it satisfies the equation \\( P(x): \\quad xf(g(x)) + f(x)g(x) \\leq 2 \\) while applying the same property for \\( x \\mapsto g(x) \\) gives another unique \\( y := g(g(x)) \\) such that \\( g(x)f(y) + yf(g(x)) \\leq 2 \\). Therefore, we have \\( xf(y) + yf(x) > 2 \\) for all \\( y \\neq g(x) \\).\nSince this inequality holds for \\( y = x \\) (from \\( xf(y) + yf(x) > 2 \\)), the uniqueness assumption implies that \\( g(g(x)) = x \\), making \\( g \\) an involution (hence bijective).\nGenerally, working with an involution naturally leads us to consider its fixed points, especially since we aim to show that \\( g(x) = x \\) identically (which holds for the solution \\( f(x) = \\frac{1}{x}\\)). Let's define the set of fixed points of \\( g \\) as \\( \\mathcal{S} := \\{ x \\in \\mathbb{R}^+ \\mid g(x) = x \\} \\) and show that \\( \\mathcal{S} = \\mathbb{R}^+ \\) is the entire domain.\nAssume for a contradiction that some \\( x \\notin \\mathcal{S} \\) is not a fixed point, i.e., \\( x \\neq g(x) \\). Then, the inequality \\( 2xf(x) > 2 \\) (derived from \\( y \\mapsto x \\)) holds, implying \\( f(x) > \\frac{1}{x} \\). Similarly, \\( x \\notin \\mathcal{S} \\) implies \\( g(x) \\notin \\mathcal{S} \\) (otherwise \\( g(x) \\in \\mathcal{S} \\) implies \\( x = g(g(x)) = g(x) \\), a contradiction), leading to \\( f(g(x)) > \\frac{1}{g(x)} \\).\nApplying these inequalities to \\( P(x) \\) gives \\( xf(g(x)) + f(x)g(x) < 2 \\), which is clearly a contradiction as \\( \\frac{x}{g(x)} + \\frac{g(x)}{x} \\geqslant 2 \\), e.g., from the AM-GM inequality. Therefore, we must have \\( x \\in \\mathcal{S} \\) for every \\( x \\in \\mathbb{R}^+ \\), i.e., \\( g(x) = x \\).\nSubstituting this relationship into the original equation, we obtain \\( P(x): \\quad xf(x) + f(x)x \\leq 2 \\implies xf(x) \\leq 1 \\implies f(x) \\leq \\frac{1}{x} \\) for every \\( x \\in \\mathbb{R}^+ \\). Applying \\( yf(y) \\leq 1 \\) to the equation \\( xf(y) + yf(x) > 2 \\) (since \\( g(x) = x \\)) yields \\( f(x) > \\frac{2}{y} - \\frac{x}{y^2} \\), and taking the limit \\( y \\to x \\) from either side results in \\( f(x) \\geq \\frac{1}{x} \\).\nCombining the results, we have \\( f(x) \\leq \\frac{1}{x} \\) and \\( f(x) \\geq \\frac{1}{x} \\), implying \\( f(x) = \\frac{1}{x} \\) as desired. \\(\\blacksquare\\)\nNote: This solution is written more extensively and with more details than necessary for a competition, especially since I include comments at certain points to encourage understanding of the ideas and explain the solution. In practice, this idea would take up only a few lines." }, { "Problem": "Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\\Omega$ again at $E \\neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\\omega$. Let $\\omega$ meet $\\Omega$ again at $P \\neq B$. Prove that the line tangent to $\\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\\angle BAC$.", "Solution": "Denote the point diametrically opposite to a point $S$ through $S' \\implies AS'$ is the internal angle bisector of $\\angle BAC$.\nDenote the crosspoint of $BS$ and $AS'$ through $H, \\angle ABS = \\varphi.$\n\nTo finishing the solution we need only to prove that $PH = AH.$\nDenote $F = SS' \\cap AC \\implies \\angle CBS = \\frac {\\overset{\\Large\\frown} {CS}}{2} = \\frac {\\overset{\\Large\\frown} {BS}}{2} = \\frac {\\overset{\\Large\\frown} {AB}}{2} + \\frac {\\overset{\\Large\\frown} {AS}}{2} =$\n$=\\angle FCB + \\varphi \\implies \\angle FBS = \\angle ABS \\implies H$ is incenter of $\\triangle ABF.$\nDenote $T = S'B \\cap SA \\implies SB \\perp TS', S'A \\perp TS \\implies H$ is the orthocenter of $\\triangle TSS'.$\nDenote $G = PS' \\cap AE \\implies \\angle BPG = \\angle BPS' = \\angle BSS' = \\angle BDG \\implies B, L, P, D,$ and $G$ are concyclic.\n$\\angle EBS' = \\varphi, \\angle LBG = \\angle LDG = 90^\\circ = \\angle DBS' \\implies \\angle DBG = \\varphi = \\angle SBF \\implies$\npoints $B, G,$ and $F$ are collinear $\\implies GF$ is symmetric to $AF$ with respect $TF.$\nWe use the lemma and complete the proof.\nLemma 1\nLet acute triangle $\\triangle ABC, AB > AC$ be given.\nLet $H$ be the orthocenter of $\\triangle ABC, BHD$ be the height.\nLet $\\Omega$ be the circle $BCD. BC$ is the diameter of $\\Omega.$\nThe point $E$ is symmetric to $D$ with respect to $AH.$\nThe line $BE$ meets $\\Omega$ again at $F \\neq B$.\nProve that $HF = HD.$\nProof\nLet $\\omega$ be the circle centered at $H$ with radius $HD.$\nThe $\\omega$ meets $\\Omega$ again at $F' \\neq D, HD = HF'.$\nLet $\\omega$ meets $BF'$ again at $E' \\neq F'$.\nWe use Reim’s theorem for $\\omega, \\Omega$ and lines $CDD$ and $BE'F'$ and get $E'D || BC$\n(this idea was recommended by Leonid Shatunov).\n$AH \\perp BC \\implies AH \\perp E'D \\implies$\nThe point $E'$ is symmetric to $D$ with respect to $AH \\implies E' = E \\implies F' = F \\implies HF = HD.$\nvladimir.shelomovskii@gmail.com, vvsss", "Solution_3": "Identify $\\Omega$ with the unit circle, and let the internal bisector of $\\angle BAC$ meet $\\overleftrightarrow{BS}$ at $Q$ and $\\omega$ again at $T$. We set up so that\n\\begin{align*}\n|a|=|b|=|s|&=1 \\\\\nc &= \\frac{s^2}b \\\\\nt &= -s \\\\\ne = -\\frac{bc}a &= -\\frac{s^2}a \\\\\nd = \\frac{ae(b+s) - bs(a+e)}{ae-bs} &= \\frac{a^2b+abs+as^2-bs^2}{a(b+s)} \\\\\nq = \\frac{at(b+s) - bs(a+t)}{at-bs} &= \\frac{2ab+as-bs}{a+b}\n\\end{align*}\nNow we find the coordinate of $P$. We have $|p|=1$, and\n\nNow $\\frac{d-\\ell}{b-c}, \\frac{b-\\ell}{b-e} \\in \\mathbb{R}$. Thus we have\n\nand so\n\n\n\n\nIt remains to show that $Q$ lies on the tangent to $\\omega$ at $P$. Now let $O$ be the center of $\\omega$. Define the vectors\n\nand\n\\begin{align*}\nd' = d-p &= \\frac{(2s+b-a)(a^2b+abs+as^2-bs^2) - s(b+s)(2ab+as-bs)}{a(b+s)(2s+b-a)} \\\\\n&= \\frac{a^2bs+as^3-bs^3+a^2b^2-a^3b-a^2s^2+abs^2-ab^2s}{a(b+s)(2s+b-a)} \\\\\n&= \\frac{(a-s)(b-a)(ab+s^2)}{a(b+s)(2s+b-a)}\n\\end{align*}\nWe observe that $\\overline{b'} = -\\frac{b'}{bp}$ and $\\overline{d'} = \\frac{d'}{ap}$. Thus if we define $o' = o-p$, we have\n\nMeanwhile, we compute\n\nSo\n\nwhich is pure imaginary. $\\blacksquare$", "Solution_2": "Let $O$ be the circumcenter of $\\triangle ABC$. We proceed with showing that $PH=AH$. Suppose that $PD$ intersects $SS'$ and $\\Omega$ at $Q$ and $F \\ne A$ respectively. Note that\n\nSince $AE \\perp BC$, we have $AE \\perp EF$ and hence $AF$ is a diameter of $\\Omega$. By similar triangles $OQ=\\frac{1}{2}AD$ and therefore\n\nSince $AF$ is a diameter of $\\Omega$, $FP \\perp AP \\implies OH \\perp AP$ and thus $H$ lies on the perpendicular bisector of $AP$. This proves the claim." }, { "Problem": "Let $a,b,c$ be real numbers. Consider the quadratic equation in $\\cos{x}$ :\nUsing the numbers $a,b,c$, form a quadratic equation in $\\cos{2x}$, whose roots are the same as those of the original equation. Compare the equations in $\\cos{x}$ and $\\cos{2x}$ for $a=4, b=2, c=-1$.", "Solution": "Let the original equation be satisfied only for $\\cos{x}=m, \\cos{x}=n$. Then we wish to construct a quadratic with roots $2m^2 -1, 2n^2 -1$.\nClearly, the sum of the roots of this quadratic must be\nand the product of its roots must be\nThus the following quadratic fulfils the conditions:\nNow, when we let $a=4, b=2, c= -1$, our equations are\nand\nThis simplifies to previous equation. \nThe first root of the first equation $\\cos {x} = \\frac{-1 + \\sqrt{5}}{4}$ corresponds to $\\cos {2x} = \\frac{-1 - \\sqrt{5}}{4}$ and the second root of the first equation $\\cos {x} = \\frac{-1 - \\sqrt{5}}{4}$ corresponds to $\\cos {2x} = \\frac{-1 + \\sqrt{5}}{4}$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ an odd integer). Let $\\alpha$ be the acute angle subtending, from $A$, that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:", "Solution": "Using coordinates, let $A=(0,0)$, $B=(b,0)$, and $C=(0,c)$. Also, let $PQ$ be the segment that contains the midpoint of the hypotenuse with $P$ closer to $B$.\n\nThen, $P = \\frac{n+1}{2}B+\\frac{n-1}{2}C = \\left(\\frac{n+1}{2}b,\\frac{n-1}{2}c\\right)$, and $Q = \\frac{n-1}{2}B+\\frac{n+1}{2}C = \\left(\\frac{n-1}{2}b,\\frac{n+1}{2}c\\right)$.\nSo, $\\text{slope}$$(PA)=\\tan{\\angle PAB}=\\frac{c}{b}\\cdot\\frac{n-1}{n+1}$, and $\\text{slope}$$(QA)=\\tan{\\angle QAB}=\\frac{c}{b}\\cdot\\frac{n+1}{n-1}$.\nThus, $\\tan{\\alpha} = \\tan{(\\angle QAB - \\angle PAB)} = \\frac{(\\frac{c}{b}\\cdot\\frac{n+1}{n-1})-(\\frac{c}{b}\\cdot\\frac{n-1}{n+1})}{1+(\\frac{c}{b}\\cdot\\frac{n+1}{n-1})\\cdot(\\frac{c}{b}\\cdot\\frac{n-1}{n+1})}$\n$= \\frac{\\frac{c}{b}\\cdot\\frac{4n}{n^2-1}}{1+\\frac{c^2}{b^2}} = \\frac{4nbc}{(n^2-1)(b^2+c^2)}=\\frac{4nbc}{(n^2-1)a^2}$.\nSince $[ABC]=\\frac{1}{2}bc=\\frac{1}{2}ah$, $bc=ah$ and $\\tan{\\alpha}=\\frac{4nh}{(n^2-1)a}$ as desired.", "Solution_2": "Let $P, Q, R$ be points on side $BC$ such that segment $PR$ contains midpoint $Q$, with $P$ closer to $C$ and (without loss of generality) $AC \\le AB$. Then if $AD$ is an altitude, then $D$ is between $P$ and $C$. Combined with the obvious fact that $Q$ is the midpoint of $PR$ (for $n$ is odd), we have" }, { "Problem": "Solve the equation\n$\\cos^n{x} - \\sin^n{x} = 1$\nwhere $n$ is a given positive integer.\n", "Solution": "Since $cos^2x + sin^2x = 1$, we cannot have solutions with $n\\ne2$ and $0<|cos(x)|,|sin(x)|<1$. Nor can we have solutions with $n=2$, because the sign is wrong. So the only solutions have $sin (x) = 0$ or $cos (x) = 0$, and these are: $x =$ multiple of $\\pi$, and $n$ even; $x$ even multiple of $\\pi$ and $n$ odd; $x$ = even multiple of $\\pi + \\frac{3\\pi}{2}$ and $n$ odd.\n" }, { "Problem": "Consider the cube $ABCDA'B'C'D'$($ABCD$ and $A'B'C'D'$ are the upper and lower bases, respectively, and edges $AA'$, $BB'$, $CC'$, $DD'$ are parallel). The point $X$ moves at constant speed along the perimeter of the square $ABCD$ in the direction $ABCDA$, and the point $Y$ moves at the same rate along the perimeter of the square $B'C'CB$ in the direction $B'C'CBB'$. Points $X$ and $Y$ begin their motion at the same instant from the starting positions $A$ and $B'$, respectively. Determine and draw the locus of the midpoints of the segments $XY$.", "Solution": "First we prove a small lemma: If the particles $X$ and $Y$ move along straight lines at constant velocities, then the locus of the midpoint of $XY$ is also a line. This is rather trivial, since all lines in 3D space may take the parametric form $(at+n, bt+m, ct+p)$, with $t$ being time, and the average of two such lines must also have a linear parametric form.\nThe locus clearly starts at the midpoint of $AB'$, or the center of face $ABB'A'$. As $X$ moves from $A$ to $B$, and $Y$ moves from $C'$ to $C$, both $X$ and $Y$ move along straight lines, so the midpoint of $XY$ traces out a line segment, starting at the midpoint of $AB'$ and ending at the midpoint of $BC'$, or the center of face $BCC'B'$. This concludes the first phase of motion. A quick check reveals that the locus goes nowhere during the second and fourth phases of motion, and only moves backward on the third. Thus the locus is just the segment connecting the centers of sides $ABB'A'$ and $BCC'B'$.\nsolution is actually wrong, it's not segment it's parallelogram- author mixed the movement of points (even if he clearly understands what he is doing- just a little mistake) Here is correct soluton:\nAnswer: the rhombus CUVW, where U is the center of ABCD, V is the center of ABB'A, and W is the center of BCC'B'.\nTake rectangular coordinates with A as (0, 0, 0) and C' as (1, 1, 1). Let M be the midpoint of XY. Whilst X is on AB and Y on B'C', X is (x, 0, 0) and Y is (1, x, 1), so M is (x/2 + 1/2, x/2, 1/2) = x (1, 1/2, 1/2) + (1-x) (1/2, 0, 1/2) = x W + (1-x) V, so M traces out the line VW.\nWhilst X is on BC and Y is on C'C, X is (1, x, 0) and Y is ( 1, 1, 1-x), so M is (1, x/2+1/2, 1/2 - x/2) = x (1, 1, 0) + (1-x) (1, 1/2, 1/2) = x C + (1-x) W, so M traces out the line WC.\nWhilst X is on CD and Y is on CB, X is (1-x, 1, 0) and Y is (1, 1-x, 0), so M is (1-x/2, 1-x/2, 0) = x (1, 1, 0) + (1-x) (1/2, 1/2, 0) = x C + (1-x) U, so M traces out the line CU.\nWhilst X is on DA and Y is on BB', X is (0, 1-x, 0) and Y is (1, 0, x), so M is (1/2, 1/2 - x/2, x/2) = x (1/2, 0, 1/2) + (1-x) (1/2, 1/2, 0) = x V + (1-x) U, so M traces out the line UV." }, { "Problem": "In an $n$-gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation\nProve that $a_1=a_2=\\cdots = a_n$.", "Solution": "Let $a_1 = p_1p_2$, $a_2 = p_2p_3$, etc.\nPlot the $n$-gon on the cartesian plane such that $p_1p_2$ is on the $x$-axis and the entire shape is above the $x$-axis. There are two cases: the number of sides is even, and the number of sides is odd:\n$\\textbf{Case 1: Even}$\nIn this case, the side with the topmost points will be $p_{\\frac{n}{2}+1}p_{\\frac{n}{2}+2}$. To obtain the $y$-coordinate of this top side, we can multiply the lengths of the sides $a_1$, $a_2$, ... $a_{\\frac{n}{2}}$ by the sine of the angle they make with the $x$-axis:\n\nWe can obtain the $y$-coordinate of the top side in a different way by multiplying the lengths of the sides $a_{\\frac{n}{2}+1}$, $a_{\\frac{n}{2}+2}$, ... $a_n$ by the sine of the angle they make with the $x$-axis to get the $\\emph{negated}$ $y$-coordinate of the top side:\n\nIt must be true that $\\textbf{(1)} = \\textbf{(2)}$. This implies that $a_k = a_{k+\\frac{n}{2}}$ for all $1 \\leq k \\leq \\frac{n}{2}$, and therefore $a_1=a_2=\\cdots = a_n$.\n$\\textbf{Case 2: Odd}$\nThis case is very similar to before. We will compute the $y$-coordinate of the top point $p_{frac{n+3}{2}}$ two ways:\n\nIt must be true that $\\textbf{(3)} = \\textbf{(4)}$. Then, we get $a_k = a_{n-k+2}$ for all $2 \\leq k \\leq \\frac{n+1}{2}$. Therefore, $a_2=a_3=\\cdots = a_n$. It is trivial that $a_1$ is then equal to the other values, so $a_1=a_2=\\cdots = a_n$. This completes the proof. $\\square$\n~mathboy100", "Solution_2": "Define the vector $\\vec{v_i}$ to equal $\\cos{\\left(\\frac{2\\pi}{n}i\\right)}\\vec{i}+\\sin{\\left(\\frac{2\\pi}{n}i\\right)}\\vec{j}$. Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length $a_i$ is parallel to $\\vec{v_i}$. We then have that\n\nBut $a_i\\geq a_{n-i}$ for all $i\\leq \\lfloor \\frac{n}{2}\\rfloor$, so\n\nfor all $i\\leq \\lfloor \\frac{n}{2}\\rfloor$. This shows that $a_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)}+a_{n-i}\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)}\\geq 0$, with equality when $a_i=a_{n-i}$. Therefore\n\nThere is equality only when $a_i=a_{n-i}$ for all $i$. This implies that $a_1=a_{n-1}$ and $a_2=a_n$, so we have that $a_1=a_2=\\cdots =a_n$. $\\blacksquare$" }, { "Problem": "A circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Tangents to the circle parallel to the sides of the triangle are contructed. Each of these tangents cuts off a triangle from $\\triangle ABC$. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a,b,c$).", "Solution": "Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let the points of contact of MH,FG,DE with the in circle be X,Y,Z respectively. Then perimeter of BHM = BH+HX+XM+MB=BH+HP+MR+BM=BP+BQ=2(s-b) and similar results follow!\n$[BHM]/[BCA] = \\text{(perimeter of BHM/perimeter of BCA)}^2 ={(2s-2b)/(a+b+c)}^2 ={(c+a-b)/(c+a+b)}^2$\nDenote $[ABC]$ the area of $\\triangle ABC$ and $(ABC)$ the perimeter of $\\triangle ABC$.\nThen $\\frac{[BHM]}{[ABC]} = \\left(\\frac{(BHM)}{(ABC)}\\right)^{2} =\\left(\\frac{c+a-b}{c+a+b}\\right)^{2}$.\nSo $[BHM]=\\left(\\frac{c+a-b}{c+a+b}\\right)^{2}\\cdot [ABC]$.\nWe know, $r_{1}$ is the radius of the incircle of $\\triangle BHM$: $r_{1}= 2 \\cdot \\frac{[BHM]}{(BHM)}$.\nArea of the incircle of $\\triangle BHM$\n\nArea of the incircle of $\\triangle ABC$:$4\\pi (\\frac{[ABC]}{a+b+c})^{2}$.\nSum of the area of the 4 incircles:" }, { "Problem": "Given the tetrahedron $ABCD$ whose edges $AB$ and $CD$ have lengths $a$ and $b$ respectively. The distance between the skew lines $AB$ and $CD$ is $d$, and the angle between them is $\\omega$. Tetrahedron $ABCD$ is divided into two solids by plane $\\varepsilon$, parallel to lines $AB$ and $CD$. The ratio of the distances of $\\varepsilon$ from $AB$ and $CD$ is equal to $k$. Compute the ratio of the volumes of the two solids obtained.", "Solution": "Let the plane meet AD at X, BD at Y, BC at Z and AC at W. Take plane parallel to BCD through WX and let it meet AB in P.\nSince the distance of AB from WXYZ is k times the distance of CD, we have that AX = k·XD and hence that AX/AD = k/(k+1). Similarly AP/AB = AW/AC = AX/AD. XY is parallel to AB, so also AX/AD = BY/BD = BZ/BC.\nvol ABWXYZ = vol APWX + vol WXPBYZ. APWX is similar to the tetrahedron ABCD. The sides are k/(k+1) times smaller, so vol APWX = k3(k+1)3 vol ABCD. The base of the prism WXPBYZ is BYZ which is similar to BCD with sides k/(k+1) times smaller and hence area k2(k+1)2 times smaller. Its height is 1/(k+1) times the height of A above ABCD, so vol prism = 3 k2(k+1)3 vol ABCD. Thus vol ABWXYZ = (k3 + 3k2)/(k+1)3 vol ABCD. We get the vol of the other piece as vol ABCD - vol ABWXYZ and hence the ratio is (after a little manipulation) k2(k+3)/(3k+1)." }, { "Problem": "Let $k, m, n$ be natural numbers such that $m+k+1$ is a prime greater than $n+1.$ Let $c_s=s(s+1).$ Prove that the product is divisible by the product $c_1c_2\\cdots c_n$.", "Solution": "For any $m, n$, $c_m - c_n = m(m+1) - n(n+1) = m^2 + m - n^2 - n = (m+n)(m-n) + m - n = (m+n+1)(m-n)$.\nWe can therefore write the product in the problem as follows:\n\nBut, the product of any $n$ consecutive integers is divisible by $n!$. We can prove this as follows:\n\nTherefore, $\\prod_{a = 1}^n m+a-k$ is divisible by $n!$, and $(m+k+1)\\left(\\prod_{a = 1}^n m+a+k+1\\right)$ is divisible by $(n+1)!$. However, we are told that $m+k+1$ is prime and therefore it is not divisible by any of the numbers $1$ through $n+1$. Therefore, $\\prod_{a = 1}^n m+a+k+1$ is divisible by $(n+1)!$.\nFinally, it is clear that $(c_{m+1}-c_k)(c_{m+2}-c_k)\\cdots (c_{m+n}-c_k)$ is divisible by $n!(n+1)! = (1 \\cdot 2)(2 \\cdot 3)(3 \\cdot 4) \\cdots (n \\cdot (n+1)) =$$c_1c_2\\cdots c_n$. $\\square$\n~mathboy100", "Solution_2": "We have that $c_1c_2c_3...c_n=n!(n+1)!$\nand we have that $c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)$\nSo we have that $(c_{m+1}-c_k)(c_{m+2}-c_k)\\ldots(c_{m+n}-c_k)=\\frac{(m+n-k)!}{(m-n)!}\\frac{(m+n+k+1)!}{(m+k+1)!}$ We have to show that:\n$\\frac{(c_{m+1}-c_k)(c_{m+2}-c_k)\\ldots(c_{m+n}-c_k)}{n!(n+1)!}=\\frac{(m+n-k)!}{(m-n)!n!}\\frac{(m+n+k+1)!}{(m+k)!(n+1)!} \\frac 1{m+k+1}$ is an integer\nBut $\\frac{(m+n-k)!}{(m-n)!n!}=\\binom {m+n-k}n$ is an integer and ${(m+n+k+1)!}{(m+k)!(n+1)!} \\frac 1{m+k+1}=\\binom {m+n+k+1}{n+1}\\frac 1{m+k+1}$ is an integer because $m+k+1|m+n+k+1!$ but does not divide neither $n+1!$ nor $m+n!$ because $m+k+1$ is prime and it is greater than $n+1$ (given in the hypotesis) and $m+n$.\nThe above solution was posted and copyrighted by Simo_the_Wolf. The original thread can be found here: [1]" }, { "Problem": "Consider the system of equations\n\n\n\n\n\nwith unknowns $x_1, x_2, \\cdots, x_n$ where $a, b, c$ are real and $a \\neq 0$. Let $\\Delta = (b - 1)^2 - 4ac$. Prove that for this system\n(a) if $\\Delta < 0$, there is no solution,\n(b) if $\\Delta = 0$, there is exactly one solution,\n(c) if $\\Delta > 0$, there is more than one solution.", "Solution": "Adding the $n$ equations together yields\n\nLet $s_i=ax_i^2+(b-1)x_i+c$.\n(a) If $\\Delta<0$, then there is no solution to the quadratic equation $ax^2+(b-1)x+c$, as the determinant is negative. This implies that either $s_i>0$ for all $i$, or $s_i<0$ for all $i$. In either case the above summation cannot be 0, which implies that there are no solutions to the given system of equations. $\\blacksquare$\n(b) If $\\Delta=0$, then there is exactly one solution to the quadratic equation $ax^2+(b-1)x+c$ (let it be $r$), and either $s_i\\geq 0$ for all $i$, or $s_i\\leq 0$ for all $i$. The only way that the above summation is 0 is if $s_i=0$ for all $i$. As there is exactly one $x_i$ that makes $s_i=0$ (namely $x_i=r$), then the only possible solution to the system of equations is $(x_1,x_2,\\dots , x_n)=(r, r, \\dots , r)$. It's not hard to show that this works, so when $\\Delta=0$ the system of equations has exactly one solution. $\\blacksquare$\n(c) If $\\Delta>1$, then there are exactly two solutions to the quadratic equation $ax^2+(b-1)x+c$. Let the roots be $r_1$ and $r_2$. If $x_i=r_1$ for all $i$, then $ax_i^2+bx_i+c=ar_1^2+br_1+c=r_1=x_{i+1}$. This shows that $(x_1,x_2,\\dots ,x_n)=(r_1,r_1,\\dots, r_1)$ is a solution. We can show that $(x_1,x_2,\\dots ,x_n)=(r_2,r_2,\\dots, r_2)$ is another solution using the same reasoning, which shows that the equation has more than one solution. $\\blacksquare$" }, { "Problem": "For each of $k = 1$, $2$, $3$, $4$, $5$ find necessary and sufficient conditions on $a > 0$ such that there\nexists a tetrahedron with $k$ edges length $a$ and the remainder length $1$.", "Solution": "A plodding question. Take the tetrahedron to be ABCD.\nTake k = 1 and AB to have length a, the other edges length 1. Then we can hinge triangles ACD and BCD about CD to vary AB. The extreme values evidently occur with A, B, C, D coplanar. The least value, 0, when A coincides with B, and the greatest value √3, when A and B are on opposite sides of CD. We rule out the extreme values on the grounds that the tetrahedron is degenerate, thus obtaining 0 < a < √3.\nFor k = 5, the same argument shows that 0 < 1 < √3 a, and hence a > 1/√3.\nFor k = 2, there are two possible configurations: the sides length a adjacent, or not. Consider first the adjacent case. Take the sides length a to be AC and AD. As before, the two extreme cases gave A, B, C, D coplanar. If A and B are on opposite sides of CD, then a = √(2 - √3). If they are on the same side, then a = √(2 + √3). So this configuration allows any a satisfying √(2 - √3) < a < √(2 + √3).\nThe other configuration has AB = CD = a. One extreme case has a = 0. We can increase a until we reach the other extreme case with ADBC a square side 1, giving a = √2. So this configuration allows any a satisfying 0 < a < √2. Together, the two configurations allow any a satisfying: 0 < a < √(2 + √3).\nThis also solves the case k = 4, and allows any a satisfying: a > 1/√(2 + √3) = √(2 - √3).\nFor k = 3, any value of a > 0 is allowed. For a <= 1, we may take the edges length a to form a triangle. For a ≥ 1 we may take a triangle with unit edges and the edges joining the vertices to the fourth vertex to have length a." }, { "Problem": "The real numbers $a_0, a_1, \\ldots, a_n, \\ldots$ satisfy the condition:\nThe numbers $b_{1}, b_{2}, \\ldots, b_n, \\ldots$ are defined by\n(a) Prove that $0 \\leq b_n < 2$ for all $n$.\n(b) given $c$ with $0 \\leq c < 2$, prove that there exist numbers $a_0, a_1, \\ldots$ with the above properties such that $b_n > c$ for large enough $n$.", "Solution": "$b_n = \\sum_{k=1}^{n} \\frac{1 - \\frac{a_{k-1}}{a_{k}} }{\\sqrt{a_k}} = \\sum_{k=1}^{n} \\frac{a_k - a_{k-1}}{a_k\\sqrt{a_k}} = \\sum_{k=1}^{n} (a_k - a_{k-1})\\left(a_k^{-\\dfrac{3}{2}}\\right)$\nLet $X_k$ be the rectangle with the verticies: $(a_{k-1},0)$; $(a_{k},0)$; $(a_{k},a_k^{-\\dfrac{3}{2}})$; $(a_{k-1},a_k^{-\\dfrac{3}{2}})$.\n\nFor all $k \\in \\mathbb{N}$, the area of $X_k$ is $(a_k - a_{k-1})\\left(a_k^{-\\dfrac{3}{2}}\\right)$. Therefore, $b_n = \\sum_{k=1}^{n} [X_k]$\nFor all sequences $\\{ a_k \\}$ and all $k \\in \\mathbb{N}$, $X_k$ lies above the $x$-axis, below the curve $f(x) = x^{-\\dfrac{3}{2}}$, and in between the lines $x = 1$ and $x = a_n$, Also, all such rectangles are disjoint.\nThus, $b_n = \\sum_{k=1}^{n} [X_k] < \\int_{1}^{a_n} x^{-\\dfrac{3}{2}} \\,dx = \\left[-\\dfrac{2}{\\sqrt{x}}\\right]_{1}^{a_n} = 2 - \\dfrac{2}{\\sqrt{a_n}} < 2$ as desired.\nBy choosing $a_k = 1 + k (\\Delta x)$, where $\\Delta x > 0$, $b_n$ is a Riemann sum for $\\int_{1}^{a_n} x^{-\\dfrac{3}{2}} \\,dx$. Thus, $\\lim_{\\Delta x \\to 0^+} b_n = \\int_{1}^{a_n} x^{-\\dfrac{3}{2}} \\,dx$.\nTherefore, $\\lim_{n \\to \\infty} \\left[ \\lim_{\\Delta x \\to 0^+} b_n \\right] = \\lim_{n \\to \\infty} \\int_{1}^{a_n} x^{-\\dfrac{3}{2}} \\,dx$ $= \\lim_{n \\to \\infty} 2 - \\dfrac{2}{\\sqrt{a_n}} = \\lim_{n \\to \\infty} 2 - \\dfrac{2}{\\sqrt{1+n(\\Delta x)}} = 2$.\nSo for any $c \\in [0,2)$, we can always select a small enough $\\Delta x > 0$ to form a sequence $\\{ a_n \\}$satisfying the above properties such that $b_n > c$ for large enough $n$ as desired." }, { "Problem": "Prove that the set of integers of the form $2^k - 3(k = 2; 3; \\cdots)$ contains an infinite subset in which every two members are relatively prime.", "Solution": "Wlog, assume $a_{n}\\ge 3$. Then say $p_{1}, p_{2}, \\ldots, p_{k}\\in \\mathbb{N}$ are all the (pairwise distinct) primes dividing $2^{a_{n}}-3$ and let $a_{n+1}= a_{n}\\cdot \\prod_{i=1}^{k}(p_{i}-1)$. Obviously $p_{i}$ is odd, for any $i \\in \\overline{1,k}$. So $\\prod_{i=1}^{k}p_{i}$ divides $2^{a_{n+1}}-1$, by Fermat's little theorem, and $\\gcd(2^{a_{n+1}}-3, 2^{a_{n}}-3) = 1$. Now, by induction, it follows $\\gcd(2^{a_{m}}-3, 2^{a_{n}}-3) = 1$, for any distinct $m, n \\in \\mathbb{N}$.\nThe above solution was posted and copyrighted by s.tringali. The original thread for this problem can be found here: [1]" }, { "Problem": "Let $a$ and $b$ be real numbers for which the equation\n$x^4 + ax^3 + bx^2 + ax + 1 = 0$\nhas at least one real solution. For all such pairs $(a, b)$, find the minimum value of $a^2 + b^2$.", "Solution": "Substitute $z=x+1/x$ to change the original equation into $z^2+az+b-2=0$. This equation has solutions $z=\\frac{-a \\pm \\sqrt{a^2+8-4b}}{2}$. We also know that $|z|=|x+1/x| \\geq 2$. So,\n\n\n\nRearranging and squaring both sides,\n\n\nSo, $a^2+b^2 \\geq a^2+(2-2|a|)^2 = 5a^2-8|a|+4 = 5(|a|-\\frac{4}{5})^2+\\frac{4}{5}$.\nTherefore, the smallest possible value of $a^2+b^2$ is $\\frac{4}{5}$, when $a=\\pm \\frac{4}{5}$ and $b=\\frac{-2}{5}$.\nBorrowed from [1]" }, { "Problem": "Prove that the number $\\sum^n_{k=0}\\binom{2n+1}{2k+1}2^{3k}$ is not divisible by $5$ for any integer $n\\ge0.$", "Solution": "Everything that follows takes place in $\\mathbb F_5(\\sqrt 2)$, i.e. the field we get by adjoining a root of $x^2-2=0$ to $\\mathbb F_5$, the field with $5$ elements.\nWe have $\\sum_{k=0}^n\\binom{2n+1}{2k+1}2^{3k}=\\sum_{k=0}^n\\binom{2n+1}{2n-2k}3^k=\\sum_{k=0}^n\\binom{2n+1}{2(n-k)}2^{-k}$. Now, this is zero iff it's zero when we multiply it by $2^n$, so we may as well prove that $\\sum_{k=0}^n\\binom{2n+1}{2(n-k)}\\sqrt 2^{2(n-k)}\\ne 0$. The LHS is $\\alpha$ from $(1+\\sqrt 2)^{2n+1}=\\alpha+\\beta\\sqrt 2,\\ \\alpha,\\beta\\in\\mathbb F_5$. We have $(1-\\sqrt 2)^{2n+1}=\\alpha-\\beta\\sqrt 2$, so by multiplying them we get $-1=\\alpha^2-2\\beta^2$. If we were to have $\\alpha=0$, then we would get $1=2\\beta^2,\\ \\beta\\in\\mathbb F_5$, and this is impossible, since it would make $3=2^{-1}$ a square $\\beta^2$ in $\\mathbb F_5$ (i.e. $3$ would be a quadratic residue modulo $5$, and it's not).\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problems": "On the sides of an arbitrary triangle $ABC$, triangles $ABR, BCP, CAQ$ are constructed externally with $\\angle CBP = \\angle CAQ = 45^\\circ, \\angle BCP = \\angle ACQ = 30^\\circ, \\angle ABR = \\angle BAR = 15^\\circ$. Prove that $\\angle QRP = 90^\\circ$ and $QR = RP$.", "Solution_1": "Consider $X$ and $Y$ so that $\\triangle CQA\\sim \\triangle CPX$ and $\\triangle CPB\\sim \\triangle CQY$. Furthermore, let $AX$ and $BY$ intersect at $F$. Now, this means that $\\angle PBC = 45 = \\angle QAC = \\angle PXC$ and $\\angle PCB = 15 = \\angle QCA = \\angle PCX$, so $\\triangle BPC\\cong \\triangle XPC$. Thus, $XC = BC$ and $\\angle BCX = 30 + 30 = 60$, so $\\triangle BCX$ is equilateral. Similarly is $\\triangle ACY$. Yet, it is well-known that the intersection of $BY$ and $AX$, which is $F$, must be the Fermat Point of $\\triangle ABC$ if $\\triangle AYC$ and $\\triangle BXC$ are equilateral. Now, $\\angle PBX = 60 - 45 = 15$. Similarly, $\\angle PXB = 15$, so $\\triangle BRA\\sim \\triangle BPX$, so a spiral similarity maps $\\triangle BPR\\sim \\triangle BXA$. This implies that $\\angle BRP = \\angle FAB$. Similarly, $\\angle ARQ = \\angle FBA$, so $\\angle BRP + \\angle ARQ = \\angle FBA + \\angle FAB = 180 - \\angle AFB = 60$. Then, $\\angle PRQ = (180 - 15 - 15) - 60 = 90$. We also realize that $\\frac {AX}{RP} = \\frac {AB}{RB} = \\frac {AB}{AR} = \\frac {BY}{RQ}$. Now, $Y$ is a rotation of $A$ $60$ degrees around $C$ and $B$ is the same rotation of $X$ around $C$, so $AX$ maps to $YB$ from this rotation, so $AX = YB$. It follows that $RP = RQ$.\nThe above solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [1]", "Solution_2": "Let $\\triangle ABT$ be the equilateral triangle constructed such that $T$ and $R$ are on the same side. $\\triangle RTB \\sim \\triangle PCB \\sim \\triangle QCA$. We have $\\frac {AT}{AC} = \\frac {AR}{AQ}$ from similarity. Also we have $\\angle TAC = \\angle RAQ$ . So $\\triangle ACT \\sim \\triangle ARQ$. Then $\\angle ATC = \\angle ARQ = m$ and $\\frac {AR}{AT} = \\frac {RQ}{TC}$. Similar calculations for $B$. We will have $\\angle BTC = \\angle BRP = 60-m$ and $\\frac {BR}{AT} = \\frac {RP}{TC}$. Also from the question we have $AR = BR$. So $\\angle PRQ = 180 - (60-x) - (60-x) -m -(60-m) = 2x$ and $PR = RQ$.\nThe above solution was posted and copyrighted by xeroxia. The original thread for this problem can be found here: [2]", "Solution_3": "Define $X$ to be a point such that $\\triangle AQC$ is directly similar to $\\triangle AXB$.\nThen, it is trivial to show that $\\angle AXR=60^{\\circ}$ and that $\\angle RXB=45^{\\circ}$; that is, $AX=AR=XR=RB$ and $\\triangle RBX$ is a right isosceles triangle. If we prove that $\\triangle RPQ$ is similar to $\\triangle RBX$, then we will be done.\nAccording to the property of spiral similarity, it suffices to prove that $\\triangle RBP\\sim\\triangle RXQ.$\nSince $\\triangle AQC\\sim \\triangle AXB$, we have $\\triangle AQX\\sim\\triangle ACB$, and this gives $\\angle RBP=60^{\\circ}+\\angle CBA=60^{\\circ}+\\angle QXA=\\angle RXQ$. It remains to prove that $\\frac{BP}{RB}=\\frac{XQ}{RX}$.\nAs $RB=RX$, we must prove that $BP=XQ$. From Law of Sines on $\\triangle BPC$, we have $BP=\\frac{a}{2\\cos 15^{\\circ}}$. Since $\\frac{XQ}{AX}=\\frac{a}{c}$, it remains to prove that $AX=AR=\\frac{c}{2\\cos 15^{\\circ}}$, which is easily verified. We are done. $\\blacksquare$\nThe above solution was posted and copyrighted by fantasylover. The original thread for this problem can be found here: [3]\n" }, { "Problem": "A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$, with the sides parallel to those of the box, then exactly $40$ percent from the volume of the box is occupied. Determine the possible dimensions of the box.", "Solution": "We name a,b,c the sides of the parallelepiped, which are positive integers. We also put\n\nIt is clear that $xyz$ is the maximal number of cubes with sides of length $\\sqrt[3]{2}$ that \ncan be put into the parallelepiped with sides parallels to the sides of the box.\nHence the corresponding volume is $V_2=2\\cdot xyz$. We need $V_2=0.4\\cdot V_1=0.4\\cdot abc$,\nhence \nWe give the values of $x$ and $a/x$ for $a=1,\\dots ,8$. The same table is valid for $b,y$ and $c,z$.\n\nBy simple inspection we obtain two solutions of $(1)$: $\\{a,b,c\\}=\\{2,5,3\\}$ and $\\{a,b,c\\}=\\{2,5,6\\}$.\nWe now show that they are the only solutions.\nWe can assume $\\frac ax\\ge \\frac by \\ge \\frac cz$. So necessarily $\\frac ax\\ge \\sqrt[3]{5}$. Note that\nthe definition of $x$ implies \nhence \nIf $a\\ge 4$ then $x\\ge 3$ and $\\frac ax<\\sqrt[3]2(1+\\frac 1x)\\le \\sqrt[3]2(\\frac 43)<\\sqrt[3]5$ \nsince $2\\cdot \\frac {4^3}{3^3}<5$. So we have only left the cases $a=2$ and $a=3$. But for $a=3$\nwe have $a/x=3/2<\\sqrt[3]5$ and so necessarily $a=2$ and $a/x=2$. \nIt follows\nNote that the definitions of $y,z$ imply \nMoreover we have from (2) and from $b/y\\ge c/z$ that\nIf $b=2$ then $b/y=2$ and we would have $c/z=5/4<\\sqrt[3]2$, which contradicts $(3)$.\nOn the other hand, if $b>5$ then $y>4$ and $\\frac by<\\sqrt[3]2(1+\\frac 1y)\\le \\sqrt[3]2(\\frac 54)<\\sqrt{5/2}$\nsince $2^2\\cdot \\frac {5^6}{4^6}<\\frac{5^3}{2^3}$ as $5^3<2^7$. So we have only left the \ncases $b=3,4,5$. But for $b=3$ we have $b/y=3/2<\\sqrt{5/2}$ and for $b=4$ we have \n$b/y=4/3<\\sqrt{5/2}$ and so necessarily $b=5$ and $b/y=5/3$ ($>\\sqrt{5/2}$)\nSo we arrive finally at $a=2,b=5$ and $c/z=3/2$. If $c\\ge 8$ then $z\\ge 6$ and $\\frac cz<\\sqrt[3]2(1+\\frac 1z)\\le \\sqrt[3]2(\\frac 76)<\\frac 32$ since $2\\cdot \\frac {7^3}{6^3}<\\frac{3^3}{2^3}$. On the other hand, for $c\\le 7$ there are the only two possible values $c=3$ and $c=6$ which yield the known solutions." }, { "Problem": "Let $n$ be a given number greater than 2. We consider the set $V_n$ of all the integers of the form $1 + kn$ with $k = 1, 2, \\ldots$ A number $m$ from $V_n$ is called indecomposable in $V_n$ if there are not two numbers $p$ and $q$ from $V_n$ so that $m = pq.$ Prove that there exist a number $r \\in V_n$ that can be expressed as the product of elements indecomposable in $V_n$ in more than one way. (Expressions which differ only in order of the elements of $V_n$ will be considered the same.)", "Solution": "Lemma: there are $\\infty$ many prime numbers $p \\not\\equiv 1 \\mod n$.\nProof:\nAssume that there are only finitely many of them and let their product be $k$.\nThen all prime factors of $nk - 1$ must be $\\equiv 1 \\mod n$ since this number is coprime with $k$. But that would mean that $- 1 \\equiv nk - 1 \\equiv 1 \\mod n$ which is impossible for $n > 2$.\nNow we can tackle the problem:\nThere are only finetely many residue classes $\\mod n$, thus we can find two primes $p,q$ with $p \\equiv q \\not\\equiv 1 \\mod n$ and coprime with $n$.\nLet $s$ be the smallest positive integer with $p^s \\equiv 1 \\mod n$, then the same property holds for $q$ too.\nNow we have that $p^s, q^s, pq^{s - 1} , p^{s - 1}q \\equiv 1 \\mod n$ are all indecomposable since all their nontrivial divisors are $\\not\\equiv 1 \\mod n$. But the product $p^sq^s = (pq^{s - 1})(p^{s - 1}q)$ gives a number that is represented in two ways.\nThe above solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]" }, { "Problem": "Let $0DA$. Let $O$ be the midpoint of $CD$, and the antipode of $B$ with respect to the circle centered at $O$ with radius $OA$ be $P$. Let $M$ be where this such circle hits $EF$. We first note thatand alsomeaning that since $AD=DF$, $AO=OM$, and $AC=CE$, then SAS similarity gets us that $\\triangle ADF \\sim \\triangle AOM \\sim \\triangle ACE$. This means we have $\\frac{AO}{AM} = \\frac{AC}{AE}$ so $\\frac{AO}{AC} = \\frac{AM}{AE}$ and because $\\angle MAE = \\angle OAC$, then $\\triangle AOC \\sim \\triangle AME$ and similarly, we also have $\\triangle AOD \\sim \\triangle AMF$. This means thatmeaning that $M$ is the midpoint of $EF$. Now, since $P$ is the antipode of $B$, then $\\angle BMP = 90^{\\circ} = \\angle EMP = \\angle FMP$ so $EP=FP$ for all $E$ and $F$ for some fixed $P$ and we are done.\nThis solution was posted and copyrighted by kevinmathz. The original thread for this problem can be found here: [3]" }, { "Problem": "Determine the maximum value of $m^2 + n^2$, where $m$ and $n$ are integers satisfying $m, n \\in \\{ 1,2, \\ldots , 1981 \\}$ and $( n^2 - mn - m^2 )^2 = 1$.", "Solution": "We first observe that since $\\gcd(m,n)=1$, $m$ and $n$ are relatively prime. In addition, we note that $n \\ge m$, since if we had $n < m$, then $n^2 -nm -m^2 = n(n-m) - m^2$ would be the sum of two negative integers and therefore less than $-1$. We now observe\n,\ni.e., $(m,n) = (a,b)$ is a solution iff. $(b, a+b)$ is also a solution. Therefore, for a solution $(m, n)$, we can perform the Euclidean algorithm to reduce it eventually to a solution $(1,n)$. It is easy to verify that if $n$ is a positive integer, it must be either 2 or 1. Thus by trivial induction, all the positive integer solutions are of the form $(F_{n}, F_{n+1})$, where the $F_i$ are the Fibonacci numbers. Simple calculation reveals $987$ and $1597$ to be the greatest Fibonacci numbers less than $1981$, giving $987^2 + 1597^2=3524578$ as the maximal value.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Consider infinite sequences $\\{x_n\\}$ of positive reals such that $x_0=1$ and $x_0\\ge x_1\\ge x_2\\ge\\ldots$.\na) Prove that for every such sequence there is an $n\\ge1$ such that:\nb) Find such a sequence such that for all $n$:", "Solution_1": "By Cauchy, the LHS is at least:\n$\\frac{(x_{0}+...+x_{n-1})^{2}}{x_{1}+...+x_{n}}.$\nIt is clear that $x_{0}=1$ and $x_{n}\\le \\frac{x_{1}+...+x_{n-1}}{n-1}$\nWe thus have that the LHS is at least:\n$(\\frac{1+2(x_{1}+...+x_{n-1})+(x_{1}+...+x_{n-1})^{2}}{x_{1}+...+x_{n-1}})(\\frac{n-1}{n})$\nApplying 2-variable AM-GM to $1$ and $(x_{1}+...+x_{n-1})^{2}$ the LHS is at least $4(\\frac{n-1}{n})$\nWe then simply choose $n$ such that $4(\\frac{n-1}{n}) \\ge 3.999$.\nPart b) is likely to occur only if both the AM-GM and Cauchy equality conditions are close to met, or if\n$1=x_{1}+x_{2}+...$ and $\\frac{x_{0}}{x_{1}}=\\frac{x_{1}}{x_{2}}=...$, respectively. This points to the geometric series with common ratio $\\frac{1}{2}$ as the equality case.\nThis solution was posted and copyrighted by n^4 4. The original thread for this problem can be found here: [1]", "Solution_2": "b) is immediate by taking $x_i = 2^{-i}$.\nFor a), we prove $LHS \\ge 2^{2^0 + 2^{-1}+...+ 2^{2-n}}$, and taking $n\\to \\infty$ will yield the result. (Here $n=1$ gives the empty sum, so $LHS \\ge 2^0=1$, which is trivial).\nThe proof is not hard. Let $s_i$ be the sum of the first $i$ terms, so $s_n = s_{n-2} + \\dfrac{x_{n-2}^2}{x_{n-1}} + \\dfrac{x_{n-1}^2}{x_n} \\ge s_{n-2} + 2^{2^0} x_{n-2} \\sqrt{\\dfrac{x_{n-1}}{x_n}} \\ge s_{n-2} + 2^{2^0} x_{n-2}$. This in turn is $s_{n-3} + \\dfrac{x_{n-3}^2}{x_{n-2}} + 2^{2^0} x_{n-2}\\ge s_{n-3} + 2^{2^0 + 2^{-1}} x_{n-3}$. By repeatedly applying AM-GM to the last two terms of the sum, collecting powers of two, and cancelling variables, we eventually obtain\n$s_0 + \\dfrac{x_0^2}{x_1} + 2^{2^0 + 2^1 + ... + 2^{3-n}}x_1 \\ge 0 + 2^{2^0 + 2^1 + ... + 2^{2-n}}x_0$, as desired.\nThis solution was posted and copyrighted by tastymath75025. The original thread for this problem can be found here: [2]" }, { "Problem": "Let $a$, $b$ and $c$ be positive integers, no two of which have a common divisor greater than $1$. Show that $2abc - ab - bc- ca$ is the largest integer which cannot be expressed in the form $xbc + yca + zab$, where $x$, $y$ and $z$ are non-negative integers.", "Solution_1": "First off, I prove $2abc-ab-bc-ca$ is un-achievable. Also, assume WLOG $a\\ge b\\ge c$.\nAssume $2abc-bc-ca=xbc+yca+zab$, then take the equation $\\pmod{ab}$ to give us $(x+1)bc+(y+1)ac\\equiv 0\\pmod{ab}$. By CRT and $\\gcd(a,b)=1$, we take this equation mod a and mod b to give us $y+1\\equiv 0\\pmod{b}$, and $x+1\\equiv 0\\pmod{a}$ (and using all numbers are relatively prime pairwise). Substituting this back into the original equation gives us $z\\le -1$, contradiction (this is where the $2abc$ part comes in).\nNow, let $2abc-bc-ca-ab+k=xbc+yca+zab$, and if a solution exists where $bc-1\\ge k\\ge 1$ then we are done because we can add $1$ to $x$ always.\nConsider the set $x=a-m$, where $m\\in \\{1,2,\\cdots, a-1, a\\}$ and $y=b-n$ where $n\\in \\{1,2,\\cdots, b-1, b\\}$.\nTherefore, we get $2abc-bc-ca-ab+k=(a-m)bc+(b-n)ca+zab$. This is the same as $bc(m-1)+ca(n-1)+k=(z+1)ab$ after doing some simplifying. By CRT, this must hold mod a and mod b and because $\\gcd(a,b)=1$. Mod $a$ gives us $bc(m-1)+k\\equiv 0\\pmod{a}$, for which a value of $m-1$ obviously exists mod $a$, which can be chosen from the set of values we have assigned for $m$. Similar method shows a value of $n-1$ exists mod $b$ from the set we have given to $n$.\nNow, we already know that $x,y\\ge 0$. Also, the LHS of the equation $bc(m-1)+ca(n-1)+k$ is at least $1$, therefore $z\\ge 0$ and we are done.\nThis solution was posted and copyrighted by binomial-theorem. The original thread for this problem can be found here: [1]", "Solution_2": "First we will prove $2abc-ab-bc-ca$ is unattainable, as such:\nSuppose $xbc+yca+zab=2abc-ab-bc-ca$. Then, taking this mod $a$, we have that $xbc\\equiv 0\\pmod{a}$, so $x\\equiv 0\\pmod{a}$, and $a|x$. Similarly, $b|y$, and $c|z$, so $x\\ge a$, $y\\ge b$, and $z\\ge c$. Thus, $xbc+yca+zac\\ge 3abc$, so $2abc\\ge 3abc$, which is a contradiction.\nNow we will prove all $n>2abc-ab-bc-ca$ is attainable, as such: consider the integer $r$ such that $r\\equiv \\frac{n}{bc} \\pmod{a}$ and $0\\le r\\le a-1$. Rearranging the equation $xbc+yca+zab=n$ yields $\\frac{n-xbc}{a}=yc+zb$, so set $x=r$. We see that $xbc-n=rbc-n\\equiv 0 \\pmod{a}$, so $\\frac{n-xbc}{a}$ is a positive integer (obviously $n-xbc>0)$. Now, note that since $x\\le a-1$, we have that $2abc-ab-ac-bc=(a-1)bc+abc-ab-ac\\ge xbc+abc-ab-ac$, so $n>xbc+abc-ab-ac$. Thus, $n-xbc>abc-ab-ac$ so $\\frac{n-xbc}{a}>bc-b-c$, so by Chicken Mc-Nugget theorem, there exist $b, c$ that satisfy the equation and are now done. $\\blacksquare$\nThis solution was posted and copyrighted by Stormersyle. The original thread for this problem can be found here: [2]\n\n" }, { "Problem": "Given points $O$ and $A$ in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point $X$ in the plane, the circle $C(X)$ has center $O$ and radius $OX+{\\angle AOX\\over OX}$, where $\\angle AOX$ is measured in radians in the range $[0,2\\pi)$. Prove that we can find a point $X$, not on $OA$, such that its color appears on the circumference of the circle $C(X)$.", "Solution": "Let $a_{n},\\ n\\ge 1$ be a sequence of positive reals such that $\\left(\\sum_{n\\ge 1}a_{n}\\right)^{2}<2\\pi\\ (*)$. For each $n\\ge 1$, let $\\mathcal C_{n}$ be the circle centered at $O$ with radius $\\sum_{i=1}^{n}a_{i}$.\nBecause of $(*)$, we can find points $x_{i,j}\\in\\mathcal C_{i},\\ i,j\\ge 1$ such that for all $i,j\\ge 1$ we have $\\mathcal C(x_{i,j})=\\mathcal C_{i+j}$. We now forget about all the other points, and work only with the matrix $M=(x_{i,j})$.\nSuppose we use $n$ colors. There must be one, $c_{1}$, which appears infinitely many times on the first row of $M$, in, say, points $x_{1,j_{1}},x_{1,j_{2}},\\ldots$. Then $c_{1}$ cannot appear on the lines $j_{k}+1$, $k\\ge 1$. Next, there is a color $c_{2}$ which appears infinitely often among the points $x_{j_{1}+1,j_{k}-j_{1}}$, $k\\ge 2$. But then $c_{2}$ cannot appear on the lines $j_{k}+1$ for such $k$. Repeating this procedure, we reach a stage where we have a row of $M$ (infinitely many actually) on which none of our $n$ colors $c_{1},\\ldots,c_{n}$ can appear. This is a contradiction.\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]" }, { "Problem": "For any polynomial $P(x) = a_0 + a_1 x + \\cdots + a_k x^k$ with integer coefficients, the number of coefficients which are odd is denoted by $w(P)$. For $i = 0, 1, \\ldots$, let $Q_i (x) = (1+x)^i$. Prove that if $i_1, i_2, \\ldots , i_n$ are integers such that $0 \\leq i_1 < i_2 < \\cdots < i_n$, then", "Solution": "We first observe that $(1+x)^{2^m} \\equiv 1 + x^{2^m} \\pmod{2}$, so for any polynomial $P$ of degree less than $2^m$, $w(P\\cdot Q_{2^m}) = 2w (P)$.\nLet $k$ be the largest power of 2 that is less than or equal to $i_n$. We proceed by induction on $k$.\nFor $k = 0$, the problem is trivial.\nNow assume that the problem holds for $k-1$. We now have two cases: $i_1 \\ge k$, and $i_1 < k$.\nIn the first case, we note that $w \\left( \\sum_{j=1}^{n}Q_{i_j} \\right) = w \\left( (1+x)^k \\sum_{j=1}^{n}Q_{i_j - k} \\right) = 2 w \\left( \\sum_{j=1}^{n}Q_{i_j - k} \\right)$, which is greater than or equal to $w( Q_{i_1} )$ by assumption.\nIn the second case, we use the division algorithm to note that\nBut by assumption, $w \\left( \\sum_{j=0}^{k-1}a_j x^j \\right) \\ge w( Q_{i_1})$, and for each odd $a_j$, at least one of $(a_j + b_j )$ and $b_j$ must be odd, 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": "To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y<0$, then the following operation is allowed: $x,y,z$ are replaced by $x+y,-y,z+y$ respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps.", "Solution": "The algorithm always stops. Indeed, consider the five numbers on the vertices of the pentagon as forming a $5$-tuple $\\mathbf{x} = (x_1, x_2, x_3, x_4, x_5) \\in \\mathbb{Z}^5$. The sum of the five entries of this $5$-tuple is positive (by assumption), and stays unchanged through the entire process (since our operation does not change the sum); thus, it is always positive. Now, define the function $f : \\mathbb{Z}^5 \\to \\mathbb{Z}$ by\n\nClearly, $f \\geq 0$ always. Now, let $\\mathbf{x}_{\\text{old}} = (x_1, x_2, x_3, x_4, x_5)$ and $\\mathbf{x}_{\\text{new}}$ be two consecutive $5$-tuples obtained in this process. Thus, $\\mathbf{x}_{\\text{new}}$ is obtained from $\\mathbf{x}_{\\text{old}}$ by applying our operation once. Suppose, WLOG, that $y=x_4<0$, and that the last three entries $x_3, x_4, x_5$ of $\\mathbf{x}_{\\text{old}}$ are replaced by $x_3 + x_4, -x_4, x_5 + x_4$ in $\\mathbf{x}_{\\text{new}}$. Let $S = x_1 + x_2 + x_3 + x_4 + x_5$ be the sum of all entries of $\\mathbf{x}_{\\text{old}}$ (or $\\mathbf{x}_{\\text{new}}$). Then, a straightforward computation shows that $f(\\mathbf{x}_{\\text{new}}) - f(\\mathbf{x}_{\\text{old}}) = 2Sx_4 < 0$ (since $S>0$ and $x_4 < 0$). Thus, if the algorithm does not stop, we can find an infinite decreasing sequence of nonnegative integers $f_0>f_1>f_2>\\cdots$ (by applying $f$ to each of the $5$-tuples). This is impossible, so the algorithm must stop. $\\Box$\nThis solution was posted and copyrighted by tc1729. The original thread for this problem can be found here: [1]\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Let $x_1 , x_2 , \\ldots , x_n$ be real numbers satisfying $x_1^2 + x_2^2 + \\cdots + x_n^2 = 1$. Prove that for every integer $k \\ge 2$ there are integers $a_1, a_2, \\ldots a_n$, not all 0, such that $| a_i | \\le k-1$ for all $i$ and", "Solution_1": "We first note that by the Power Mean Inequality, $\\sum_{i=1}^{n} x_i \\le \\sqrt{n}$. Therefore all sums of the form $\\sum_{i=1}^{n} b_i x_i$, where the $b_i$ is a non-negative integer less than $k$, fall in the interval $[ 0 , (k-1)\\sqrt{n} ]$. We may partition this interval into $k^n - 1$ subintervals of length $\\frac{ (k-1)\\sqrt{n} }{k^n - 1}$. But since there are $k^n$ such sums, by the pigeonhole principle, two must fall into the same subinterval. It is easy to see that their difference will form a sum with the desired properties.", "Solution_2": "This solution is very similar to Solution 1 but uses a slightly different approach for the first part. It suffices to find ${a_i}$ where $a_i$ is positive. Let $f(a_1, a_2, ..., a_n) = a_1| x_1|+a_2 |x_2|+\\cdots+a_n |x_n|\\geq0$.\nBy the Cauchy-Schwarz Inequality,\n\nThis implies that $\\left(a_1 x_1+a_2 x_2+...+a_n x_n\\right)^2\\leq \\sqrt{n} (k-1)$, and hence the codomain of $f(a_1, a_2, ..., a_n)$ is $\\left[0, \\sqrt{n} (k-1)\\right]$. The rest of the proof is similar to Solution 1.\n~Iraevid13\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "A function $f$ defined on the positive integers (and taking positive integers values) is given by:\n$\\begin{matrix} f(1) = 1, f(3) = 3 \\\\ f(2 \\cdot n) = f(n) \\\\ f(4 \\cdot n + 1) = 2 \\cdot f(2 \\cdot n + 1) - f(n) \\\\ f(4 \\cdot n + 3) = 3 \\cdot f(2 \\cdot n + 1) - 2 \\cdot f(n), \\end{matrix}$\nfor all positive integers $n.$ Determine with proof the number of positive integers $\\leq 1988$ for which $f(n) = n.$", "Solution_1": "Considering that $f(n)=f(2n)$, the two last equations give :\n$f(4n + 1)-f(4n) = 2(f(2n + 1) - f(2n))$\n$f(4n + 3)-f(4n+2)= 2(f(2n + 1) - f(2n))$\nAnd so, if $n$ is even and $2^{p+1}>n\\geq 2^p>1$, we have $f(n+1)-f(n)=2^p$\nSo, if we have an even $n=\\sum_{i=1}^{k} 2^{a_i}$, where $\\{a_i\\}$ is a strictly increasing sequence with $a_1>0$ ($n$ even) : $f(n+1)=2^{a_k}+f(n)$\nThen $f(n)=f(\\sum_{i=1}^{k} 2^{a_i})$ $=f(\\sum_{i=1}^{k} 2^{a_i-a_1})$ $=2^{a_k-a_1}+f(\\sum_{i=2}^{k} 2^{a_i-a_1})$\nAnd so $f((\\sum_{i=1}^{k} 2^{a_i})+1)=2^{a_k}+2^{a_k-a_1}+f(\\sum_{i=2}^{k} 2^{a_i-a_1})$\nAnd it is easy to conclude that $f(\\sum_{i=1}^{k} 2^{a_i})=\\sum_{i=1}^{k} 2^{a_k-a_i}$ and that applying $f(n)$ means reversing the order of binary representation of n (and this could be also easily shown with induction).\nSo $f(n)=n$ occurs if and only if the binary representation of n is symetrical.\nIt remains to count these \"symetric\" numbers. We have exactly $2^{\\lceil\\frac{m-1}{2}\\rceil}$ such numbers in $[2^m,2^{m+1})$. So :\nWe have exactly 1 such numbers in $[1,2)$\nWe have exactly 1 such numbers in $[2,4)$\nWe have exactly 2 such numbers in $[4,8)$\nWe have exactly 2 such numbers in $[8,16)$\nWe have exactly 4 such numbers in $[16,32)$\nWe have exactly 4 such numbers in $[32,64)$\nWe have exactly 8 such numbers in $[64,128)$\nWe have exactly 8 such numbers in $[128,256)$\nWe have exactly 16 such numbers in $[256,512)$\nWe have exactly 16 such numbers in $[512,1024)$\nSince $1988=B11111000100$, positions 2 to 6 may be any between $00000$ and $11101$, and so :\nWe have exactly 30 such numbers in $[1024,1988]$\nAnd so the requested number is $1+1+2+2+4+4+8+8+16+16+30=92$\nThis solution was posted and copyrighted by pco. The original thread for this problem can be found here: [1]", "Solution_2": "The main claim is following.\nClaim: $f(n)$ is equal to the result when $n$ is written in binary and its digits are reversed.\nProof. Follows directly by induction. $\\blacksquare$\nSo the question asks for the number of binary palindromes which are at most $1988 = 11111000100_2$.\nFor $k = 1, 2, \\dots, 10$ there are $2^{\\left\\lceil k/2 \\right\\rceil-1}$ binary palindromes with $k$ bits (note the first bit must be $1$). For $k=11$, the number of binary palindromes which are also less than $1988$ is $2^5 - 2$ (only $11111011111$ and $11111111111$ are missing).\nHence the final count is\nThis solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]", "Solution_3": "We claim that $f(x)$ is the reversal of the digits of $x$ in binary. We proceed by induction, because given $f(1)=f(2)=1$ and $f(3)=3$, we can determine all other values, and this property holds for $f(1)$, $f(2)$, and $f(3)$.\nWe prove that all operations conserve this property. For $f(2n)=f(n)$, we are adding a zero at the end, thus obviously preserving this property. For $f(4n+1) = 2f(2n+1)-f(n)$, we see that letting $4n+1=\\overline{a_1a_2\\ldots a_k01}_2$ gets that assuming this holds true for $f(2n+1)$ and $f(n)$, thenwhich keeps the property since $\\overline{10a_ka_{k-1}\\ldots a_2a_1}_2$ is the reversal of $\\overline{a_1a_2\\ldots a_k01}_2$ in binary. Similarly, letting $4n+1=\\overline{a_1a_2\\ldots a_k11}_2$, thenwhich keeps the property since $\\overline{11a_ka_{k-1}\\ldots a_2a_1}_2$ is the reversal of $\\overline{a_1a_2\\ldots a_k11}_2$. Due to this, we have proved that this property holds for all $n$ which means we need to find the number of binary palindromes below $1988$. This number is just going to be, for $n<11$-digit numbers, $2^{\\lfloor \\frac {n-1}2 \\rfloor}$ ways, so summing up for less than $11$ digits gets $62$ ways. For $11$-digit numbers, we have $32$ ways but need to subtract a few in which are greater than $1988$; there are two numbers we don't count, namely $2015$ and $2047$. As a result, our answer is $62+30=\\boxed{92}$.\nThis solution was posted and copyrighted by kevinmathz. The original thread for this problem can be found here: [3]" }, { "Problem": "Let $n$ and $k$ be positive integers and let $S$ be a set of $n$ points in the plane such that\ni.) no three points of $S$ are collinear, and\nii.) for every point $P$ of $S$ there are at least $k$ points of $S$ equidistant from $P.$\nProve that:", "Solution_1": "Let $\\{A_i\\}_{i=1}^n$ be our set of points. We count the pairs $(A_i,\\{A_j,A_\\ell\\}),\\ i\\ne j\\ne \\ell\\ne i$ such that $A_iA_j=A_iA_\\ell$. There are $n$ choices for $A_i$, and for each $A_i$ there are $\\ge\\frac{k(k-1)}2$ choices for $\\{A_j,A_\\ell\\}$, so the number of such pairs is $\\ge\\frac{nk(k-1)}2$. On the other hand, there are $\\frac{n(n-1)}2$ choices for $\\{A_j,A_\\ell\\}$, and for each $\\{A_j,A_\\ell\\}$ there are at most two choices for $A_i$ (because we can't have $\\ge 3$ points on the perpendicular bisector of $A_jA_\\ell$). This means that the number of pairs is $\\le n(n-1)$. We have thus found $n(n-1)\\ge\\frac{nk(k-1)}2\\Rightarrow 2(n-1)\\ge k(k-1)\\ (*)$. If $k\\ge \\sqrt{2n}+\\frac 12$, then $k(k-1)\\ge 2n-\\frac 14>2n-2$, contradicting the last inequality in $(*)$, so we're done.\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]", "Solution_2": "So obvious simplification and the fact that $n$ is an integer gives that what we want is $n\\ge\\dbinom{k}{2}+1$\nNow for each point, draw a circle around it that contains at least $k$ points in $S$.\nFor any point $P$ in $S$, let $d_P$ be the number of circles it's on.\nFor any point $O$ in $S$, let $f_O$ be the number of points on the defined circle with center $O$, so $f_O\\ge k$ for all $O$. Since the function $\\dbinom{x}{2}$ is increasing for $x\\ge 1$ an integer, $\\dbinom{f_O}{2}\\ge \\dbinom{k}{2}$ for all $O$. So $E(\\dbinom{f}{2})\\ge \\dbinom{k}{2}$.\nBut notice that since any pair of two points shares at most 2 circles (otherwise we have 3 circles with centers on their perpendicular bisector, which is not allowed), $\\sum_{O\\in S}\\dbinom{f_O}{2}\\le 2\\dbinom{n}{2}$, since the left side counts the number of point-pairs that share a circle (with possible circle multiplicity) in $P$ while the right side upper-bounds the number of point-pairs that can share a circle (with possible circle multiplicity). So now $\\dbinom{k}{2}\\le E(\\dbinom{f}{2})\\le (2/n)((n-1)(n)/2)=n-1$, which is what we want.\nThis solution was posted and copyrighted by antimonyarsenide. The original thread for this problem can be found here: [2]" }, { "Problem": "Determine all integers $n > 1$ such that $\\frac{2^n+1}{n^2}$ is an integer.", "Solution_1": "Let $N = \\{ n\\in\\mathbb{N} : 2^n\\equiv - 1\\pmod{n^2} \\}$ be the set of all solutions and $P = \\{ p\\text{ is prime} : \\exists n\\in N, p|n \\}$ be the set of all prime factors of the solutions.\nIt is clear that the smallest element of $P$ is 3.\nAssume that $P\\ne\\{3\\}$ and let's try to determine the second smallest element $q = \\min (P\\setminus\\{3\\})$.\nLet $n\\in N$ be a multiple of $q$. It is important to notice that $9\\not|n$ (otherwise it is easy to get that any power of $3$ divides $n$, a non-sense). Therefore, $n = 3^t n'$ where $t = 0$ or $1$ and $n'$ does not have prime divisors smaller than $q$.\nSince $2^{2n}\\equiv 1\\pmod{q}$, the multiplicative order $r = ord_q(2)$ of 2 modulo $q$ divides $2n$. Moreover, $r$ must be even, since otherwise we would have $2^n \\equiv 1\\pmod{q}$, a contradiction to the required $2^n \\equiv - 1\\pmod{q}$.\nSince $r < q$ we must have $r = 2$ or $2\\cdot 3 = 6$. But the numbers $2^2 - 1 = 3$ and $2^6 - 1 = 63$ deliver only one new prime factor $7$, implying that $q = 7$. However, in this case $r = ord_7(2) = 3$, a contradiction. This contradiction proves that $P = \\{3\\}$ and thus $N = \\{1,3\\}$.\nThis solution was posted and copyrighted by maxal/orl. The original thread for this problem can be found here: [1]", "Solution_2": "Let $p$ be the smallest prime divisor of $n$. It is easy to check that, $n=3$ is obviously solution. Let $p^{a} || n$ . $p | 2^{2n}-1$ and $p | 2^{p-1} - 1$ (By the fermat's theorem), we obtain that $p=3$. It is also obvious that n is an odd number.\nLemma: For all $n$ positive integers, $2$ is a primitive root modulo $3^{n}$.\n$3^{2a} | 2^{2n} - 1$. Using the lemma, we get that $\\phi(3^{2a}) | 2n$. Using the power of three, we obtain that $3^{2a-1} | 3^{a}$. This is only possible when $a \\geq 2a-1$. So $a=1$. Now $q$ be the second smallest prime divisor of $n$. $(2n,q-1) = 2 , 6$. If this equals to 2, we get $q=3$ which is a contradiction.If $(2n,q-1) = 6$ then $q|63$. We know that $q$ is different from $3$. Hence $q$ must be $7$. But this is impossible since\n$2^{n} +1\\equiv 2\\mod 7$ when $n$ is divisible by $3$. Hence the answer is $n = 3$.\nThis solution was posted and copyrighted by grumpyorum. The original thread for this problem can be found here: [2]", "Solution_3": "Trivially $n=1$ is a solution. Now assume $n\\neq 1$ and define $\\pi(n)$ to be the smallest prime divisor of $n$. Let $\\pi(n)=p\\neq 2$. Then we have:\nNow if $r\\neq 2\\mid n$ then we can't have $r\\mid p-1$ because then $r\\le p-1$ contradiction. Therefore $r=2$ and since $n$ is odd $\\gcd(2n, p-1)=2$. HenceLet $v_3(n)=k$. By lifting the exponent we must haveLet $n=3n_1$. $n_1=1$ is a solution ($2^3+1=3^2=9$). Assume $n_1\\neq 1$ and let $\\pi(n_1)=q\\neq 3$. By Chinese Remainder Theorem since $q\\neq 3$ we have:\nHowever $2^n+1\\equiv 8^{n_1}+1\\equiv 2\\pmod{7}$ contradiction.\nThe solutions are henceforth $n=1, 3$.\nThis solution was posted and copyrighted by binomial-theorem. The original thread for this problem can be found here: [3]", "Solution_4": "et $n>1$ and $p$ be the smallest prime factor of $n$.Then $p|2^{2n}-1,p|2^{p-1}-1 \\Rightarrow p|2^{\\text{gcd}(2n,p-1)}=2^2-1=3\\Rightarrow \\boxed{p=3}$.\nThus $n=3^x*y$ for some positive $x$ and $y$.\nAlso $n^2|2^n+1 \\Rightarrow 2x=v_3(n^2) \\le v_3(2^n+1)=v_3(3)+v_3(n)=x+1 \\Rightarrow \\boxed{x=1}$.\nThus $n=3y$ for some positive integer $y$.\nNow let $y>1$ and $q$ be the smallest prime divisor of $y$.Then we can deduce that $q|2^{\\text{gcd}(2n,q-1)}-1=2^{2\\text{gcd}(n,q-1)}-1=2^6-1=63=7*3^2 \\Rightarrow \\boxed{q=7}$(Note that $\\text{gcd}(n,q-1)$ is $1$ or $3$).\nBut this gives $7|2^n+1$ which is false as $2^n+1$ leaves remainders $1,2,4$ modulo $7$.\nThus $y=1$ and $\\boxed{n=3}$ if $n>1$.\nThus the solutions are $\\boxed{n=1}$ and $\\boxed{n=3}$\nThis solution was posted and copyrighted by sayantanchakraborty. The original thread for this problem can be found here: [4]\nMany more solutions can be found in the thread in Contest Collections." }, { "Problem": "Let $S = \\{1,2,3,\\cdots ,280\\}$. Find the smallest integer $n$ such that each $n$-element subset of $S$ contains five numbers which are pairwise relatively prime.", "Solution_1": "Let's look at this another way. We aim to find the least number of integers we can remove from S to leave a set which does not contain 5 pairwise coprime integers.\nLet $S_p = \\{p^k | p^k \\leq 280\\}$\nAnd more generally $S_{p_1,p_2,...,p_l} = \\{n = p_1^{q_1}p_2^{q_2}....p_l^{q_l}| n \\leq 280 \\}$\nAnd finally $S_1 = \\{1\\}$\nIt is clear that these sets taken over all collections of primes and 1 forms a partition of S.\nIf there are representatives from five sets $S_p$, there are five mutually coprime integers, so all but four of the sets $S_p$ must be completely removed. It is clear that (with the exception of 1, which can clearly be removed first) if $p>q, |S_p| \\leq |S_q|$ so it is worth removing $S_p$ before removing $S_q$ (if we are striving for a minimum).\nFurthermore, if we remove two sets $S_p, S_q$, we must also (to stop there from being 5 mutually coprime integers) remove $S_{p,q}$ and so on, and these have similar ordering relations.\nSo once we have removed everything we need to keeping sizes of sets removed to a minimum, we are left with $S_2,S_3,S_5,S_7$ and the multi-index sets not coprime to all of these.\nIn other words, we have the set of multiples of 2,3,5 and 7, which can be calculated (by lots of inclusion-exclusion) to have cardinality $216$. Therefore, a subset of $S$ of size $217$ must contain 5 coprime integers.\nThis solution was posted and copyrighted by Ilthigore. The original thread for this problem can be found here: [1]\nFirstly note that $\\lfloor\\sqrt{280}\\rfloor = 16$. Therefore, we only need to check primes $p\\le 16$ to check if a number from $1$ to $280$ is prime.", "Solution_2": "I claim the answer is $\\boxed{217}$. First, let's look at a subset of $S$ that consists only of multiples of $2,3,5,7$. It is easy to calculate by PIE that the total number of elements in this set isNow, note that no matter what $5$ element subset we consider of this subset we just made, consisting of multiples of $2,3,5,7$, we will always have at least $2$ of those elements that exist in the subset that share a factor greater than $1$ by the Pigeonhole Principle since $\\left\\lfloor\\tfrac{5}{4}\\right\\rfloor+1=2$. Thus, we know $n>216$ from this. We also know that there are $212$ composite numbers and $4$ prime numbers (obviously just $2,3,5,7$) in this subset of $S$. Now we do casework on the numbers $11,13$ (we don't have to check higher numbers because remember $p\\le 16$!) to find the other composite numbers. We find that all of numbers, $121,143,169,187,209,221,247,253$ fills in the rest of the composite numbers from $1$ to $280$. This gives a total of $8$ numbers when counted. So we can now count up the total as $220$ composite numbers and the remaining must be prime so $59$ prime numbers ($1$ not included as prime or composite). Now suppose $X$ is a subset of our set $S$ such that $|X|=217$ ($|X|$ represents its cardinality or the number of elements in the subset $X$). Also suppose that there is no $5$ element subset of $X$ such that all $5$ elements are relatively prime to each other. This means we would have to have at most $4$ prime numbers and at least $213$ composite numbers to make this subset $X$. This means the set $S-X$ (everything that is in $S$ but not in $X$) has at most $220-213=7$ composite numbers. Now consider the following $8$ sets and notice that at least one of these must be a subset of $X$!:And obviously each of these sets has $5$ elements that are all relatively prime numbers, as desired.\nThis solution was posted and copyrighted by Wave-Particle. The original thread for this problem can be found here: [2]" }, { "Problem": "Consider nine points in space, no four of which are coplanar. Each pair of points is joined by an edge (that is, a line segment) and each edge is either colored blue or red or left uncolored. Find the smallest value of $n$ such that whenever exactly n edges are colored, the set of colored edges necessarily contains a triangle all of whose edges have the same color.", "Solution": "We show that for $n = 32$ we can find a coloring without a monochrome triangle.\nTake two squares $R_1R_2R_3R_4$ and $B_1B_2B_3B_4$. Leave the diagonals of each square uncolored, color the remaining edges of $R$ red and the remaining edges of $B$\nblue. Color blue all the edges from the ninth point $X$ to the red square, and red\nall the edges from $X$ to the blue square. Color $R_iB_j$ red if $i$ and $j$ have the same parity and blue otherwise.\nClearly $X$ is not the vertex of a monochrome square, because if $XY$ and $XZ$ are\nthe same color then, $YZ$ is either uncolored or the opposite color. There is no triangle within the red square or the blue square, and hence no monochrome triangle. It remains to consider triangles of the form $R_iR_jB_k$ and $B_iB_jR_k.$ But if $i$ and $j$ have the same parity, then $R_iR_j$ is uncolored (and similarly $B_iB_j$), whereas if\nthey have opposite parity, then $R_iB_k$ and $R_jB_k$ have opposite colors (and similarly $B_iR_k$ and $B_jR_k$).\nIt remains to show that for $n = 33$ we can always find a monochrome triangle.\nThere are three uncolored edges. Take a point on each of the uncolored edges.\nThe edges between the remaining $6$ points must all be colored. Take one of\nthese, $X.$ At least $3$ of the $5$ edges to $X$, say $XA$, $XB$, $XC$ must be the same color\n(say red). If $AB$ is also red, then $XAB$ is monochrome. Similarly, for $BC$ and $CA.$\nBut if $AB$, $BC$ and $CA$ are all blue, then $ABC$ is monochrome." }, { "Problem": "On an infinite chessboard, a game is played as follows. At the start, $n^2$ pieces are arranged on the chessboard in an $n$ by $n$ block of adjoining squares, one piece in each square. A move in the game is a jump in a horizontal or vertical direction over an adjacent occupied square to an unoccupied square immediately beyond. The piece which has been jumped over is removed. Find those values of $n$ for which the game can end with only one piece remaining on the board.", "Solution": "The solution as described in the video is that all values of $n$ are valid except when $n$ is divisible by 3. He describes why those values are not valid using modular math and why the other ones are by creating an algorithmic pattern of reducing the matrices by taking out pieces in subarrays of 3 by m.\nIn the video he made a mistake when reducing the 5x5 to a 2x2 as someone pointed out in the comments on the video. I made sure my animated gif for when $n=5$ was correct.\nI was going to write a solution to this but since Mr. Math in the video does a really beautiful solution of it I decided to write a code that will output these animated gifs with the cases for n=1 through 10 showing them whether they're valid or not depending on how many pieces are left. I used the algorithm described by Mr. Math in the video along and some recursive functions and some visualization functions. My code can generate the animation for any value of $n$ as long as there is enough memory for the file size and to handle the recursive function. I had to reduce the animated gif file sizes significantly. The original animated give for when $n=13$ was 30MB. I wrote the code in Matlab.\nI don't know how to upload it to this page. If anyone wants the code, you can send me an email and I can gladly share it. Enjoy the animated gifs.\n\n\n\n\n\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": "For any positive integer $k$, let $f(k)$ be the number of elements in the set\n$\\{k + 1, k + 2,\\dots, 2k\\}$ whose base 2 representation has precisely three $1$s.\n", "Solution_1": "a) Surjectivity of f\nFor space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation.\nIt's easy to see that (one has to place two 1s in the less n significative bit of a (n+1)-bit number)\nNow consider $k=2^n + 2$ with $n \\ge 2$ and the corrisponding set $S(2^n+2)=\\{2^n+3,...,2^{n+1},2^{n+1}+1,2^{n+1}+2,2^{n+1}+3,2^{n+1}+4\\}$\nwhose subset $\\{2^n+3,...,2^{n+1}\\}$ contains $f(2^n)$ T-numbers since $S(2^n)=\\{2^n+1,...,2^{n+1}\\}=\\{2^n+1,2^n+2\\} \\cup \\{2^n+3,...,2^{n+1}\\}$ contains $f(2^n)$ T-numbers by definition but $\\{2^n+1,2^n+2\\}$ has none. So $S(2^n+2)=\\{2^n+3,...,2^{n+1}\\} \\cup \\{2^{n+1}+1,2^{n+1}+2,2^{n+1}+3,2^{n+1}+4\\}$ but the last set has the only T-number $2^{n+1}+3$. We conclude that:\nNow consider $k=2^n +2^r + 2^s$ with $n>r>s\\ge 0$ and $n\\ge 2$. \nWe explicitly calculate f(k) for such numbers.\nSo we have to calculate how many T-numbers are in the set $S(k)=\\{2^n +2^r + 2^s+1;...;2^{n+1} +2^{r+1} + 2^{s+1}\\}$.\nThe T-numbers less than $2^{n+1}$ in $S(k)$ are $T_1 = \\{2^n +2^j + 2^i : (j=r \\wedge r>i>s) \\vee (n>j>r \\wedge j>i>0) \\}$ \nwhence $\\# (T_1)= r-s-1 + \\sum_{h=r+1}^{n-1} h$.\nThe T-numbers greater than $2^{n+1}$ in $S(k)$ are $T_2 = \\{2^{n+1} +2^j + 2^i : (j=r+1 \\wedge s+1 \\ge i\\ge 0) \\vee (r\\ge j \\ge 1 \\wedge j>i \\ge 0) \\}$ \nwhence $\\# (T_2)= s+2 + \\sum_{h=1}^{r} h$.\nTherefore $S(k)$ contains $\\# (T_1) + \\# (T_2)=r-s-1 + s+2 + \\sum_{h=1}^{r}h +\\sum_{h=r+1}^{n-1} h = \\sum_{h=1}^{n-1} h + r + 1$ T-numbers and we have:\nSummarizing\nThat's to say that f takes all the values from $f(2^n)$ to $f(2^{n+1})$ for every $n \\ge 2$ and\nthen takes any positive integer value since $f(4)=1$ and $f(2^n)$ becomes arbitrarily large.\nb) one-from-one values\nBy the fact that $f(2^n +2^r + 2^s)=f(2^n) + r +1$ where $r=1,2,...,n-1$ and $r > s \\ge 0$ \nit follows that each of the values $f(2^n) + r +1$ where $r=2,...,n-1$ come from at least two different k because there are at least two different choices for $s$. These values are $f(2^n)+3,f(2^n)+4,...,f(2^n)+n=f(2^{n+1})$.\nThus the only possible one-from-one values are $f(2^n)+1$ that come from $k=2^n + 2$ and $f(2^n)+2$ that come from $k=2^n + 3$.\nIf $k=2^n + 3$ we have $f(k)=f(k+1)$ because k+1 is not T and $2k+1=2^{n+1} + 7$ and $2k+2=2^{n+1} + 8$ are not T so f maps $2^n + 3$ and $2^n + 4$ to $f(2^n)+2$.\nIf $k=2^n + 2$ then $f(k+1)=f(2^n)+2>f(k)=f(2^n)+1>f(k-1)=f(2^n+1)=f(2^n)$. \nThe function f is non-decreasing.\nIt is sufficient to prove that $f(k+1) \\ge f(k)$ but this follows by the fact that if k+1 is T then 2k+2 is T too. \nBy the monotonicity of f $f( 2^n + 2)=f(2^n)+1=\\tbinom n2 +1= n(n-1)/2 + 1$ with $n \\ge 2$ are the only one-from-one values of f.", "Solution_2": "a) Surjectivity of f\nFor space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation and define the sets $S(k)=\\{k+1;k+2;...;2k\\}$.\nSo $f(k)$ is the number of T-numbers in $S(k)$.\nA positive even integer 2n is T iff n is T.(Fundamental theorem of T numbers)(FTT)\nThe function f is non-decreasing.\nIt is sufficient to prove that $f(k+1) \\ge f(k)$. This follows by (FTT) if k+1 is T then 2k+2 is T too; so passing from $S(k)$ to $S(k+1)$ we can loose a T-number $k+1$ but we regain it with $2k+2$ in $S(k+1)$ so $f(k+1)$ cannot be less than $f(k)$. We also have $f(k+1) \\le f(k) + 1$. This would be false if $k+1$ were not T and both $2k+1$ and $2k+2$ were T. But if $2k+2$ were T then $k+1$ would be T too by the (FTT).\nThen we have $f(k) \\le f(k+1) \\le f(k) + 1$ that is $f(k+1)=f(k)$ or $f(k+1)=f(k)+1$.\nBut $f(4)=1$ and $f(2^n)= \\tbinom n2$ so f takes all of the positive integer values because starting from 1 with step of 1 reaches arbitrarily large integer values.\nb) one-from-one values of f\nThe one-from-one values $f(k)$ are such that $f(k)=f(k-1)+1$ and $f(k+1)=f(k)+1$ since f is monotone non-decreasing and has step 1.\nBy the condition $f(k+1)=f(k)+1$ since $k+1$ is T iff $2k+2$ is T we have that $2k+1$ must be T.\nBy the condition $f(k)=f(k-1)+1$ since $k$ is T iff $2k$ is T we have that $2k-1$ must be T.\nThen $2k-1$ and $2k+1$ must have the form $2^{n+1}+2^{r+1}+1$ with $n>r\\ge 0$ since they are odd and $n \\ge 2$ since there is only 1 T-number less than 8 and they must have the same number of bits since their difference is 2 and both are T (the only two binary numbers which differs by 2 and have different number of bits are $2^n+1$ and $2^n-1$ or $2^n$ and $2^n-2$ which are evidently not T). Let $2k+1=2^{n+1}+2^{j+1}+1$ and $2k-1=2^{n+1}+2^{i+1}+1$ with $j>i \\ge 0$. Then $2k+ 1-(2k-1)=2^{j+1}-2^{i+1}=2$ that is $j=1$ and $i=0$.\nWe conclude that $f(k)$ is one-from-one for $k=2^n+2^j=2^n+2$ with $n \\ge 2$.\nSince $f(2^n+1)=f(2^n)=\\tbinom n2$ we have that $f(2^n+2)=\\tbinom n2 +1= n(n-1)/2 + 1$ with $n \\ge 2$ are the only one-from-one values of f." }, { "Problem": "Let $S$ denote the set of nonnegative integers. Find all functions $f$ from $S$ to itself such that\n$f(m+f(n))=f(f(m))+f(n)$ $\\forall m,n \\in S$", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $x_{1}$, $x_{2}$,...,$x_{n}$ be real numbers satisfying the conditions\n$|x_{1}+x_{2}+...+x_{n}|=1$\nand\n$|x_{i}| \\le \\frac{n+1}{2}$, for $i=1,2,...,n$\nShow that there exists a permutation $y_{1}$, $y_{2}$,...,$y_{n}$ of $x_{1}$, $x_{2}$,...,$x_{n}$ such that\n$|y_{1}+2y_{2}+...+ny_{n}|\\le \\frac{n+1}{2}$\n", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "For any positive integer $n$, let $d(n)$ denote the number of positive divisors\nof $n$ (including 1 and $n$ itself). Determine all positive integers $k$ such that\n$d(n^2)/d(n) = k$ for some $n$.", "Solution": "First we must $d$etermine gener$a$l values for $d(n)$:\nLet $n=p1 ^ a1 * p2 ^ a2 * .. * pc ^ ac$, if $d$ is an ar$b$itr$a$ry divisor of $n$ then $d$ must have the same prime factors of $n$, each with an exponent $b_i$ being: $0\\leq b_i\\leq a_i$. \nHence there are $Ai + 1$ choices for each exponent of Pi in the number d => there are $(a_1 + 1)(a_2 + 1)..(a_c + 1)$ such d\n$\\implies d(n) = (a_1 + 1)(a_2+1)..(a_c+1)$ where $a_i$ are exponents of the prime numbers in the prime factorisation of $n$.\n$\\implies d(n^2)/d(n) = {(2a_1 + 1)(2a_2 + 1)..(2a_c + 1)}/{(a_1+1)..(a_c+1)}$\nSo we want to find all integers $k$ that can $b$e represented by the product of fractions of the form $(2n+1)/(n+1)$\nObviously $k$ is odd as the numerator is always odd.\nIt's possible for 1 (1/1) and 3 $(5/3 * 9/5)$, which suggests that it may be possible for all odd integers, which we can show by induction.\n$P(k)$: It's possible to represent $k$ as the product of fractions $(2n+1)/(n+1)$\nBase case: $k = 1: (2(0) + 1) / (0 + 1)$\nNow assume that for $k\\geq 3$ it's possible for all odds < $k$.\nSince $k$ is odd, $k+1 = 2^zy$ where $y$ is odd and $y$ < $k$\nLet there be a number $x$ s.t $k-y=x\\implies k+1 = 2^z(k-x)\\implies (2^zx+1)/(2^z-1)=k$\nAlso consider $k/y$. ISTS $k/y$ can be represented by a product of fractions of the form $2n+1/n+1$ in order to show $k$ can be represented by product of fractions $2n+1/n+1$, since $y$ can be represented in such a manner too.\n$k/y = k/(k-x) = 1/(1 - x/k)$\nUsing our definition of $k$ in terms of $x$:\n$k/y = 1/({1 - {2^z-x}/{2^zx+1}}) = {2^zx+1}/{x+1}$\nAnd that is simply the product of fractions: ${2x+1}/{x+1} * {4x+1}/{2x+1} * .. * {2^zx+1}/{2^{z-1}x}$.\nWe have shown that $k/y$ can be written s.t it's a product of fractions of the form ${2n+!}/{n+1}\\implies k$ can be written in such a way too.\nHence we have shown that all odds less than $k$ satisfies $P(n)\\implies P(k)$ is true. Since we have shown P(1) is true, it must hence be true for all odd integers.\nTherefore, $d(n^2)/d(n) = k\\iff k$ is odd, for some n. I.E all odd $k$ satisfy the condition posed in the question.∎\n\n-dabab_kebab (wrote this solution)" }, { "Problem": "Consider an $n \\times n$ square board, where $n$ is a fixed even positive integer. The board is divided into $n^{2}$ units squares. We say that two different squares on the board are adjacent if they have a common side.\n$N$ unit squares on the board are marked in such a way that every square (marked or unmarked) on the board is adjacent to at least one marked square.\nDetermine the smallest possible value of $N$.\n", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $n \\ge 2$ be a positive integer and $\\lambda$ a positive real number. Initially there are $n$ fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points $A$ and $B$ to the left of $B$, and letting the flea from $A$ jump over the flea from $B$ to the point $C$ so that $\\frac{BC}{AB}=\\lambda$.\nDetermine all values of $\\lambda$ such that, for any point $M$ on the line and for any initial position of the $n$ fleas, there exists a sequence of moves that will take them all to the position right of $M$.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.", "Solution": "For each girl, we know that there is a boy who solved a problem in common with her. Since a girl solves at most six problems, for a girl and her set of six problems there are at least 11 boys who solved a problem in common with the girl such that at least three boys solved that common problem. (This is a simple application of the pigeon hole principle where the problems solved by the girl are the holes and the boys are the pigeons, and the guy/pigeon enters a hole/problem which he solved in common with the girl).\nThe boys have solved a total of at most $21\\times 6$ problems.We view this as a set with repeated elements i.e. if a problem is solved by more than one boy it appears as many times as the number of boys who have solved it. Let this set of problems be A. Clearly, A has size $21\\times 6$.\nWe mark each problem in A which has been solved by at least three boys and a girl. And we mark it the number of times that is the same as the number of girls that has solved it. Since there are a total of $21\\times 11$ marks (since there are at least 11 marks for each girl, by the discussion above), either a problem is marked at least thrice, in which case we're done since then it has been solved by at least three boys and three girls. Or each problem has been marked at most twice. In this case it is clear that more than $21\\times 5$ problems in A have been marked since $21\\times 5 \\times 2<21\\times 11$ (there are a total of $21\\times 11$ marks. This means that there is a boy such that all six of his problems have been marked. But then by our discussion in the first paragraph we know there must a problem that this boy has solved which has been solved by at least three girls. Therefore it must be true that there is a problem such that it has been solved by at least three people of each gender." }, { "Problem": "Find all pairs of positive integers $m,n \\ge 3$ for which here exist infinitely many positive integers $a$ such that\n\nis itself an integer", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Each pair of opposite sides of convex hexagon has the property that the distance between their midpoints is $\\frac{\\sqrt{3}}{2}$ times the sum of their lengths. Prove that the hexagon is equiangular.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Define a \"hook\" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.\n\nDetermine all $m \\times n$ rectangles that can be covered without gaps and without overlaps with hooks such that;\n(a) the rectangle is covered without gaps and without overlaps,\n(b) no part of a hook covers area outside the rectangle.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $x, y, z > 0$ satisfy $xyz\\ge 1$. Prove that", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Determine the least real number $M$ such that the inequality holds for all real numbers $a,b$ and $c$", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size. Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Prove that there are infinitely many positive integers $n$ such that $n^{2} + 1$ has a prime divisor greater than $2n + \\sqrt {2n}$.", "Solutions": "Solution 1\nThe main idea is to take a gaussian prime $a+bi$ and multiply it by a \"twice as small\" $c+di$ to get $n+i$. The rest is just making up the little details.\nFor each sufficiently large prime $p$ of the form $4k+1$, we shall find a corresponding $n$ such that $p$ divides $n^2+1$ and $p>2n+\\sqrt{2n}$. Since there exist infinitely many such primes and, for each of them, $n \\ge \\sqrt{p-1}$, we will have found infinitely many distinct $n$ satisfying the hypothesis.\nTake a prime $p$ of the form $4k+1$ and consider its \"sum-of-two squares\" representation $p=a^2+b^2$, which we know to exist for all such primes. As $a\\ne b$, assume without loss of generality that $b>a$. If $a=1$, then $n=b$ is what we are looking for, and $p=n^2+1 > 2n+\\sqrt{2n}$ as long as $p$ (and hence $n$) is large enough. Assume from now on that $b>a>1$.\nSince $a$ and $b$ are (obviously) co-prime, there must exist integers $c$ and $d$ such that\n\nIn fact, if $c$ and $d$ are such numbers, then $c\\pm ma$ and $d\\mp mb$ work as well for any integer $m$, so we can assume that $c \\in \\left[-\\frac{a}{2}, \\frac{a}{2}\\right]$.\nDefine $n=|ac-bd|$ and let's see why this is a good choice. For starters, notice that $(a^2+b^2)(c^2+d^2)=n^2+1$.\nIf $c=\\pm\\frac{a}{2}$, from (1) we see that $a$ must divide $2$ and hence $a=2$. This implies, $d=-\\frac{b-1}{2}$ and $n=\\frac{b(b-1)}{2}-2$. Therefore, \n$\\left(b-\\frac{1}{2}\\right)^2 = 1/4 + 2(n+2) > 2n$, so $b > \\sqrt{2n}+\\frac{1}{2}$. Finally, $p=b^2+2^2 > 2n+\\sqrt{2n}$ and the case $c=\\pm\\frac{a}{2}$ is cleared.\nWe can safely assume now that\n\nAs $b>a>1$ implies $b>2$, we have\n\nso\nTherefore,\nBefore we proceed, we would like to show first that $a+b-1 > \\sqrt{p}$. Observe that the function $x+\\sqrt{p-x^2}$ over $x\\in(2,\\sqrt{p-4})$ reaches its minima on the ends, so $a+b$ given $a^2+b^2=p$ is minimized for $a = 2$, where it equals $2+\\sqrt{p-2^2}$. So we want to show that \nwhich is not hard to show for large enough $p$.\nNow armed with $a+b-1>\\sqrt{p}$ and (2), we get\n\nwhere $u=\\sqrt{p}.$\nFinally, \n\nThe proof is complete.\nSolution 2\nWe begin with a lemma.\nLemma: For every prime $p \\equiv 1 \\mod{4}$, there exists a positive integer $n \\le \\dfrac{p - \\sqrt{p-4}}{2}$ such that $p\\mid n^2 + 1$.\nProof:\nWe know that there must be a solution $x$ to the equation\n\nwith $1 \\le x \\le p-1$. However, if $x$ is a solution then so is $p-x$, so there must be a solution $x \\le p-1$. Let $n$ denote this solution and suppose that $n = \\frac{p-k}{2}$. Then, we have\n\n\nSince $k$ is a positive integer, it follows that $k^2 \\ge p-4$, so we have\n\nas desired. The lemma is proven.\n\nSuppose for sake of contradiction that there are only finitely many integers $n_1,n_2,\\dots,n_k$ that work. Let $p \\equiv 1 \\mod{4}$ be a prime with $p>20$ and such that $p$ is relatively prime to $n_i^2+1$ for all $i$ (the existence of $p$ is guaranteed by the existence of infinitely many primes of the form $4k+1$). Then, we know that there exists an $N$ such that $N\\le \\dfrac{p - \\sqrt{p-4}}{2}$ and $p | N^2 + 1$ (this last condition shows that $N$ is not among $n_1,n_2,\\dots,n_k$. We want to show that\n\n\nBut, we know that $p-2N \\ge \\sqrt{p-4}$, so it suffices to show that\n\n\nOnce again, it suffices to show that\n\nwhich follows from $p>20$.\nThus, $N$ satisfies the required condition and it follows that there exist infinitely many values of $n$ that satisfy the given condition, as desired." }, { "Problem": "Suppose that $s_1,s_2,s_3,\\ldots$ is a strictly increasing sequence of positive integers such that the subsequences\nare both arithmetic progressions. Prove that the sequence $s_1,s_2,s_3,\\ldots$ is itself an arithmetic progression.\nAuthor: Gabriel Carroll, USA", "Solution": "then\ni.s.w.\nput S(3) = b, S(2) = a, S(1) = k\ntherefore, \nevery S(n) is an arithmetic sequence.\nQ.E.D." }, { "Problem": "Find all functions $g:\\mathbb{N}\\rightarrow\\mathbb{N}$ such that $\\left(g(m)+n\\right)\\left(g(n)+m\\right)$ is a perfect square for all $m,n\\in\\mathbb{N}.$\nAuthor: Gabriel Carroll, USA", "Solution": "Suppose such function $g$ exist then:\nLemma 1) $g(m) \\ne g(m+1)$\nbut $\\left(g(m)+m\\right)^2<\\left(g(m+1)+m\\right)\\left(g(m)+m+1\\right)<\\left(g(m)+m+1\\right)^2$.\nLemma 2) $|g(m)-g(m+1)| = 1$ (we have show that it can't be 0)\nThen there must exist a prime number $p$ such that $g(m)$ and $g(m+1)$ are in the same residue class modulo $p$.\nIf $|g(m)-g(m+1)| = p^aq$ where $q$ is not divisible by $p$.\n\n\n\nHence, that number is not a perfect square.\n\nIf $g(m)-g(m+1) = 1$, then $g(x) = -x + k$, $k\\in\\mathbb{N}$.\nIf $g(m)-g(m+1) = -1$, then $g(x) = x + k$, $k\\in\\mathbb{N}$." }, { "Problem": "Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.\nProposed by Alexander A. Polyansky, Russia", "Solution": "Let the excenters opposite $A,B,C$ be $I_a,I_b,I_c$. Let the midpoint of $\\overline{I_bI_c}$ be $M_a$, which lies on $(ABC)$, the nine-point circle of $\\triangle I_aI_bI_c$; analogously define $M_b,M_c$.\n$M_aB=M_aC$ and $BC_1=s-a=B_1C$, so $\\triangle M_aBC_1\\cong\\triangle M_aCB_1$ (SAS), thus $M_a$ is equidistant from $B_1,C_1$, with analogous results for $M_b,M_c$. It follows that the circumcentre of $\\triangle A_1B_1C_1$ is one of $M_a,M_b,M_c$; WLOG, suppose it is $M_a$.\nBy isogonal conjugacy, $I_aA_1,I_bB_1I_cC_1$ concur at the Bevan point $V$ of $\\triangle ABC$. $M_aM_b$ is the common perpendicular bisector of $\\overline{C_1A_1}$ and $\\overline{I_cC}$, so $C_1A_1\\parallel I_cC$. $(A_1C_1M_b)$ is the circle on diameter $\\overline{VB}$, so by Reim's theorem, $V \\in (I_bI_cBC)$.\nHence $\\angle I_cI_aI_b=\\tfrac{1}{2}\\angle I_cVI_b=45^{\\circ}\\implies\\angle CAB=180^{\\circ}-2\\angle I_cI_aI_b=90^{\\circ}$, as required.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Convex quadrilateral $ABCD$ has $\\angle{ABC}=\\angle{CDA}=90^{\\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside $\\triangle{SCT}$ and\nProve that line $BD$ is tangent to the circumcircle of $\\triangle{TSH}.$", "Solution": "Denote $\\angle{HSB}=v$, $\\angle{HTD}=u$, $\\angle{HSC}=s$, $\\angle{HTC}=t$, $\\angle{HCS}=x$, $\\angle{HCT}=y$, $\\angle{AHS}=z$, $\\angle{AHT}=w$. Since $\\angle{CHS}-\\angle{CSB} =90$ and $\\angle{CHT}-\\angle{CTD}=90$, we have $\\angle{CSA}=x+90$, $\\angle{CTA}=y+90$.\nSince $\\angle{CHS} = \\angle{CSB}+90$, the tangent of the circumcircle of $\\triangle{CSH}$ at point $S$ is perpendicular to $SB$; therefore, the circumcenter of $\\triangle{CSH}$ (point $K$) is on $AB$. Similarly, the circumcenter of $\\triangle{CTH}$ (point $L$) is on $AD$. In addition, $KL$ is the perpendicular bisector of $CH$.\nExtend $CB$ to meet circumcircle of $\\triangle{CSH}$ at $P$, and extend $CD$ to meet circumcircle of $\\triangle{CTH}$ at $Q$. Then, since $\\angle{ADC}=\\angle{ABC}=90$, $AD$ and $AB$ are the perpendicular bisector of $CQ$ and $CP$, respectively; hence $A$ is the circumcenter of $\\triangle{PCQ}$. Since $B$ and $D$ are midpoints on $CP$ and $CQ$, $PQ \\parallel BD$; also, $AH \\perp BD$, so $AH \\perp PQ$. Since $A$ is the circumcenter, $AH$ is also the perpendicular bisector of $PQ$. Hence,\nWe have \n\n\nHence, $u+v = -w-z+\\angle{DHC}+\\angle{BHC} = 180 -(w+z)$, or \nSince quadrilaterals $QTHC$ and $PSHC$ are cyclic, we have $\\angle{THQ}=\\angle{TCQ}=90-u$, $\\angle{SHP}=\\angle{SCP}=90-v$; so, \n\n\nHence,\n\nSimilarly,\nNow we apply law of Sines repeatedly on pairs of triangles. \nFor $\\triangle{QCH}$ and $\\triangle{PCH}$, $\\angle{HQC}=\\angle{HTC}=t$, $\\angle{HCQ}=y+z$, $\\angle{HPC}=s$, $\\angle{HCP}=x+w$; hence,\n\nFor $\\triangle{LHK}$, $\\angle{HLK}=\\frac{1}{2}\\angle{HLC}=t$, $\\angle{HKL}=\\frac{1}{2}\\angle{HKC}=s$; hence,\n\nFor $\\triangle{LAK}$, $\\angle{ALK}=90-\\angle{DML}=90-\\angle{CMK}=\\angle{MCH}=y+z$, and similarly, $\\angle{AKL}=w+x$; hence,\n\nCombining $(1), (2), (3)$, we have\n\nTherefore, $TS \\parallel KL$, and $\\angle{ATS} = \\angle{ALK}=y+z$. \nLet the circumcircle of $\\triangle{THS}$ meets $AH$ at $G$. We have,\n\nAnd,\n\nThis proves $HG$ is the diameter of the circle and the center of the circle is on AH. $\\square$\nSolution by $Mathdummy$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Let $P = A_1A_2 \\cdots A_k$ be a convex polygon in the plane. The vertices $A_1,A_2,\\dots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.", "Solution": "Note that $2S$ is always an integer for any lattice polygon, so it remains to show that it is divisible by $n$. It clearly suffices to prove the problem for when $n=p^m$ is a prime power. We proceed using induction on $k$, with the base case of $k=3$ settled by Heron's formula: If $a,b,c$ are the side lengths of the triangle, then the square of the area is $S^2=\\frac{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}{16}$. As $n\\mid a^2,b^2,c^2$, we have that $n^2\\mid S^2\\implies n\\mid S$, as desired.\nFor the inductive step, we claim that there exists a diagonal whose length squared is also divisible by $n$. Then, we may split $P$ into two polygons with less vertices and areas divisible by $n$ by assumption. Let $3\\le i\\le k-1$ be such that $v_p(A_1A_i^2)=q$ is minimized. By Ptolemy's Theorem on cyclic quadrilateral $A_1A_{i-1}A_iA_{i+1}$, we have that $\\sqrt{A_1A_{i-1}^2\\cdot A_iA_{i+1}^2}+\\sqrt{A_1A_{i+1}^2\\cdot A_iA_{i-1}^2}=\\sqrt{A_1A_i^2\\cdot A_{i-1}A_{i+1}^2}$, or $\\sqrt{\\frac{A_1A_{i-1}^2\\cdot A_iA_{i+1}^2}{np^q}}+\\sqrt{\\frac{A_1A_{i+1}^2\\cdot A_iA_{i-1}^2}{np^q}}=\\sqrt{\\frac{A_1A_i^2\\cdot A_{i-1}A_{i+1}^2}{np^q}}$. As we have $p^q\\mid AA_{i-1}^2,AA_{i+1}^2$ and $n\\mid A_iA_{i+1}^2,A_iA_{i-1}^2$, the terms under the square roots on the LHS are integers, so the LHS is an algebraic integer. This implies that the term under the square root on the RHS is also an integer, so $np^q\\mid A_1A_i^2\\cdot A_{i-1}A_{i+1}^2\\implies n\\mid A_{i-1}A_{i+1}^2$, as desired." }, { "Problem": "A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit's starting point, $A_0$, and the hunter's starting point, $B_0$, are the same. After $n-1$ rounds of the game, the rabbit is at point $A_{n-1}$ and the hunter is at point $B_{n-1}$. In the nth round of the game, three things occur in order.\n(i) The rabbit moves invisibly to a point $A_n$ such that the distance between $A_{n-1}$ and $A_n$ is exactly 1.\n(ii) A tracking device reports a point $P_n$ to the hunter. The only guarantee provided by the tracking device is that the distance between $P_n$ and $A_n$ is at most 1.\n(iii) The hunter moves visibly to a point $B_n$ such that the distance between $B_{n-1}$ and $B_n$ is exactly 1.\nIs it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after $10^9$ rounds she can ensure that the distance between her and the rabbit is at most 100?", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from $1$ to $10$\n\nDoes there exist an anti-Pascal triangle with $2018$ rows which contains every integer from $1$ to $1 + 2 + 3 + \\dots + 2018$?", "Solution": "Trivially it is required that every positive integer from $1$ to $1+2+3+\\cdots+2018$ appears exactly once.\nLet $M_n$ denote the maximum number in the $n$th row and let $m_n$ denote the minimum number in the $n$th row.\nNow assume $n\\leq 2017$ and consider the numbers directly below $M_n$. Call these $a$ and $b$ where w.l.o.g. $a>b$. Then $a-b=M_n$. Since $a\\leq M_{n+1}$ and $b\\geq m_{n+1}$, we obtain that $M_{n+1}\\geq M_n+m_{n+1}$.\nThus, for $1\\leq ib$. We have $b=a-l$ and $a\\leq 1+2+3+\\cdots+2018$ so, because $l$ is large, $b\\leq 2018$ so $b$ is small. Thus $b=m_{n+1}$ so $l$ is directly above $m_{n+1}$. Thus there are at most 2 large numbers in the $n$th row.\nThus there are at most 126 large numbers outside the bottom row. Since there are 2019 large numbers, there are at least 1893 large numbers in the bottom row so at most 125 non-large numbers in the bottom row. Now there are 2017 pairs of adjacent large numbers in the bottom row. We remove the pair directly beneath $m_{2017}$ and at most 250 other pairs containing a non-large number. Thus we can find a pair of adjacent large numbers in the bottom row, not directly beneath $m_{2017}$. However, their difference is small and in the 2017th row but not $m_{2017}$, which is a contradiction. Thus there is no such anti-Pascal triangle." }, { "Problem": "A social network has $2019$ users, some pairs of whom are friends. Whenever user $A$ is friends with user $B$, user $B$ is also friends with user $A$. Events of the following kind may happen repeatedly, one at a time:\nThree users $A$, $B$, and $C$ such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends, change their friendship statuses such that $B$ and $C$ are now friends, but $A$ is no longer friends with $B$, and no longer friends with $C$. All other friendship statuses are unchanged.\nInitially, $1010$ users have $1009$ friends each, and $1009$ users have $1010$ friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\\angle DAB= \\angle CAD$. The point $E$ on the segment $AC$ satisfies $\\angle ADE= \\angle BCD$, the point $F$ on the segment $AB$ satisfies $\\angle FDA= \\angle DBC$, and the point $X$ on the line $AC$ satisfies $CX=BX$. Let $O_1$ and $O_2$ be the circumcentres of the triangles $ADC$ and $EXD$ respectively. Prove that the lines $BC$, $EF$, and $O_1 O_2$ are concurrent.", "Solution": "We prove that circles $ACD, EXD$ and $\\Omega_0$ centered at $P$ (the intersection point $BC$ and $EF)$ have a common chord.\nLet $P$ be the intersection point of the tangent to the circle $\\omega_2 = BDC$ at the point $D$ and the line $BC, A'$ is inverse to $A$ with respect to the circle $\\Omega_0$ centered at $P$ with radius $PD.$\nThen the pairs of points $F$ and $E, B$ and $C$ are inverse with respect to $\\Omega_0$, so the points $F, E,$ and $P$ are collinear. Quadrilaterals containing the pairs of inverse points $B$ and $C, E$ and $F, A$ and $A'$ are inscribed, $FE$ is antiparallel to $BC$ with respect to angle $A$ (see $\\boldsymbol{Claim}$).\nConsider the circles $\\omega = ACD$ centered at $O_1, \\omega' = A'BD,$\n$\\omega_1 = ABC, \\Omega = EXD$ centered at $O_2 , \\Omega_1 = A'BX,$ and $\\Omega_0.$\nDenote $\\angle ACB = \\gamma$. Then $\\angle BXC = \\angle BXE = \\pi – 2\\gamma,$\n$\\angle AA'B = \\gamma (AA'CB$ is cyclic), \n$\\angle AA'E = \\pi – \\angle AFE = \\pi – \\gamma (AA'EF$ is cyclic, $FE$ is antiparallel),\n$\\angle BA'E = \\angle AA'E – \\angle AA'B = \\pi – 2\\gamma = \\angle BXE \\implies$\n$\\hspace{13mm}E$ is the point of the circle $\\Omega_1.$\nLet the point $Y$ be the radical center of the circles $\\omega, \\omega', \\omega_1.$ It has the same power $\\nu$ with respect to these circles. The common chords of the pairs of circles $A'B, AC, DT,$ where $T = \\omega \\cap \\omega',$ intersect at this point.\n$Y$ has power $\\nu$ with respect to $\\Omega_1$ since $A'B$ is the radical axis of $\\omega', \\omega_1, \\Omega_1.$\n$Y$ has power $\\nu$ with respect to $\\Omega$ since $XE$ containing $Y$ is the radical axis of $\\Omega$ and $\\Omega_1.$\nHence $Y$ has power $\\nu$ with respect to $\\omega, \\omega', \\Omega.$\nLet $T'$ be the point of intersection $\\omega \\cap \\Omega.$ Since the circles $\\omega$ and $\\omega'$ are inverse with respect to $\\Omega_0,$ then $T$ lies on $\\Omega_0,$ and $P$ lies on the perpendicular bisector of $DT.$ The power of a point $Y$ with respect to the circles $\\omega, \\omega',$ and $\\Omega$ are the same, $DY \\cdot YT = DY \\cdot YT' \\implies$ the points $T$ and $T'$ coincide.\nThe centers of the circles $\\omega$ and $\\Omega$ ($O_1$ and $O_2$) are located on the perpendicular bisector $DT'$, the point $P$ is located on the perpendicular bisector $DT$ and, therefore, the points $P, O_1,$ and $O_2$ lie on a line, that is, the lines $BC, EF,$ and $O_1 O_2$ are concurrent.\n$\\boldsymbol{Claim}$\nLet $AK$ be bisector of the triangle $ABC$, point $D$ lies on $AK.$ The point $E$ on the segment $AC$ satisfies $\\angle ADE= \\angle BCD$. The point $F$ on the segment $AB$ satisfies $\\angle ADF= \\angle CBD.$ Let $P$ be the intersection point of the tangent to the circle $BDC$ at the point $D$ and the line $BC.$ Let the circle $\\Omega_0$ be centered at $P$ and has the radius $PD.$\nThen the pairs of points $F$ and $E, B$ and $C$ are inverse with respect to $\\Omega_0$ and $EF$ and $BC$ are antiparallel with respect to the sides of an angle $A.$\n$\\boldsymbol{Proof}$\nLet the point $E'$ is symmetric to $E$ with respect to bisector $AK, E'L || BC.$ \nSymmetry of points $E$ and $E'$ implies $\\angle AEL = \\angle AE'L.$\n\n\n\n\nSimilarly, we prove that $FL$ and $BC$ are antiparallel with respect to angle $A,$ and the points $L$ in triangles $\\triangle EDL$ and $\\triangle FDL$ coincide. Hence, $FE$ and $BC$ are antiparallel and $BCEF$ is cyclic.\nNote that $\\angle DFE = \\angle DLE – \\angle FDL = \\angle AKC – \\angle CBD$ and \n$\\angle PDE = 180^o – \\angle CDK – \\angle CDP – \\angle LDE = 180^o – (180^o – \\angle AKC – \\angle BCD) – \\angle CBD – \\angle BCD$\n$\\angle PDE = \\angle AKC – \\angle CBD = \\angle DFE,$ so $PD$ is tangent to the circle $DEF.$\n$PD^2 = PC \\cdot PB = PE \\cdot PF,$ that is, the points $B$ and $C, E$ and $F$ are inverse with respect to the circle $\\Omega_0.$\n\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around a circle such that the product of any two neighbours is of the form $x^2 + x + k$ for some positive integer $x$.", "Solution": "https://www.youtube.com/watch?v=nYD-qIOdi_c [Video contains solutions to all day 1 problems]\nhttps://youtu.be/_kF9uXCZ6l4 [Video Solution by little fermat]" }, { "Problem": "For each integer $k \\geqslant 2$, determine all infinite sequences of positive integers $a_1, a_2, \\ldots$ for which there exists a polynomial $P$ of the form $P(x)=x^k+c_{k-1} x^{k-1}+\\cdots+c_1 x+c_0$, where $c_0, c_1, \\ldots, c_{k-1}$ are non-negative integers, such that\n\nfor every integer $n \\geqslant 1$.", "Solution": "https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]\nhttps://www.youtube.com/watch?v=CmJn5FKxpPY [Video contains another solution to problem 3]\nLet $f(n)$ and $g(j)$ be functions of positive integers $n$ and $j$ respectively.\nLet $a_{n}=a_{1}+f(n)$, then $a_{n+1}=a_{1}+f(n+1)$, and $a_{n+k}=a_{1}+f(n+k)$\nLet $P=\\prod_{j=1}^{k}\\left ( a_{n+j} \\right ) = \\prod_{j=1}^{k}\\left ( a_{n}+g(j)) \\right )$\nIf we want the coefficients of $P(a_{n})$ to be positive, then $g(j)\\geq 0$ for all $j$ which will give the following value for $P$:\n$P=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+\\prod_{j=1}^{k} g(j) = P(a_{n})$\nThus for every $j$ and $n$ we need the following:\n$a_{n}+g(j)=a_{n+j}=a_{1}+f(n+j)$\nSolving for $g(j)$ we get:\n$g(j)=a_{1}+f(n+j)-a_{n}=a_{1}+f(n+j)-a_{1}-f(n)$\n$g(j)=f(n+j)-f(n)\\geq 0$ for all $n$ and $j$ because $g(j)$ needs to be greater than or equal to zero for all coefficients to be non-negative.\nThis means that $f(n)$ needs to be increasing with $n$, or staying constant, and also with $f(1)=0$ because $a_{1}=a_{1}+f(1)$.\nIn addition, since we need all coefficients to be integer, then all $f(n)$ and $g(j)$ must also be integers. We also need $g(j)$ to not be dependent of $n$, so in the expression $f(n+j)-f(n)$, the $n$ needs to cancel. This mean that the rate of change for $f(n)$ with respect to $n$ needs to be constant. This can only be achieved with $f(n)$ be the equation of a line with slope being either zero or positive integer.\nSo, we set $f(n)$ to be the equation of a line as $f(n)=mn+b$ with $m$ being the slope with a non-negative value and with $b$ the intercept at $n=0$. We know that $f(1)=0$ so $f(1)=m+b=0$ which means that $b=-m$ and our function becomes $f(n)=mn-m=(n-1)m$. Since $f(n)$ needs to be non-negative integer then $m\\geq 0 \\mid m \\in \\mathbb{Z}$ then $f(n)$ is increasing or constant, with $f(1)=0$\nThen, $g(j)=f(n+j)-f(n)=(n+j-1)m-(n-1)m=jm$\nThis gives:\n$\\prod_{j=1}^{k}\\left ( a_{n}+jm \\right )=P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!m^{k}$\nwith $C_{0}=k!m^{k}$ and coefficients of polynomial $\\geq 0$\nThen, $a_{n}=a_{1}+f(n)$\nWhich provides the solution of all infinite sequences of positive integers as:\n$a_{n}=a_{1}+(n-1)m$, $\\forall m\\geq 0 \\mid m \\in \\mathbb{Z}$ and $a_{1} \\geq 1 \\mid a_{1} \\in \\mathbb{Z}$\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": "Construct a right triangle with a given hypotenuse $c$ such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.", "Solutions": "We denote the catheti of the triangle as $a$ and $b$. We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse. (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)", "Solution_1": "The conditions of the problem require that\nHowever, we notice that twice the area of the triangle $abc$ is $ab$, since $a$ and $b$ form a right angle. However, twice the area of the triangle is also the product of $c$ and the altitude to $c$. Hence the altitude to $c$ must have length $\\frac{c}{4}$. Therefore if we construct a circle with diameter $c$ and a line parallel to $c$ and of distance $\\frac{c}{4}$ from $c$, either point of intersection between the line and the circle will provide a suitable third vertex for the triangle. Q.E.D.", "Solution_2": "We denote the angle between $b$ and $c$ as $\\alpha$. The problem requires that\nor, equivalently, that\nHowever, since $\\frac{a}{c} = \\sin{\\alpha};\\; \\frac{b}{c} = \\cos{\\alpha}$, we can rewrite the condition as\nor, equivalently, as\nFrom this it becomes apparent that $2\\alpha = \\frac{\\pi}{6}$ or $\\frac{5\\pi}{6}$; hence the other two angles in the triangle must be $\\frac{ \\pi }{12}$ and $\\frac{ 5 \\pi }{12}$, which are not difficult to construct. Q.E.D.\nNote. It is not difficult to reconcile these two constructions. Indeed, we notice that the altitude of the triangle is of length $c \\sin{\\alpha}\\cos{\\alpha}$, which both of the solutions set equal to $\\frac{c}{4}$ .", "Solution_3": "If we let the legs be $a$ and $b$ with $a < b$, then $c^2 = a^2 + b^2$. Because $c^2 = 4 a b$ as well, we immediately deduce via some short computations through quadratic formula that $a = (2 - \\sqrt{3})b$ and $a = (2 + \\sqrt{3})b$. Thus, $\\frac{a}{b} = \\tan 15^\\circ$ and $\\frac{a}{b} = \\tan 75^\\circ$, and so one of the angles of the triangle must be $15^\\circ$ and the other must be $75^\\circ$. A $15^\\circ$ angle is easily constructed by bisecting a $30^\\circ$ angle (which is formed by constructing the altitude of an equilateral triangle), and $75^\\circ$ angle is constructed by constructing a 15 degree angle on top of the 60 degree point.", "Solution_4": "We draw perpendicular $h$ on $c$ and denote the angle between the median and $c$ as $\\alpha$. The area of the triangle is $\\frac{ab}{2}$ or $\\frac{ch}{2}$\nWe can write $c$ as $2\\sqrt{ab}$ So we have,\n\nWhich simplifies to be\nOr\n\nSo, we draw $c$ and from it's midpoint, we draw a segment $\\frac{c}{2}$ forming an angle $\\frac{\\pi}{6}$\nJoin the vertices to get the required triangle.", "Solution_5": "This is a geometric solution. We start with some observations in the first paragraph and describe the construction in the next paragraph. Consider any right triangle with hypotenuse c. Construct square with side length c on the opposite side of c from the triangle. Construct 3 more triangles congruent to the original triangle on the outside of each of the three other sides of the square in such a way that a larger square with side length a+b is formed. The area of this larger square is c^2+2ab = 1.5c^2 so a+b = (3/2)^(1/2) *c\nNow go back to the given c. Construct a square with side length c first. Let two adjacent corners of it be M and N. Construct the square's diagonals and let center of the square be O. Construct a circle around O with radius (3/2)^(1/2) * OM. To do so, take OM, first draw a 2:1 right triangle with OM being the short side to obtain a hypothenuse with length 3^(1/2) OM, then draw a 1:1 right triangle from that side to obtain a hypothenuse with length 6^(1/2)*OM. then bisect it for desired length. Next draw the circular locus of points X such that MXO is 45 degrees. To accomplish this, simply find the point on the perpendicular bisector of OM that is 1/(2OM) from OM. There are two such points, pick the one closer to N. Draw circle around this new point going through O and M. The intersection of the two circles is the desired third vertex of the triangle with the given hypotenuse c." }, { "Problem": "Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.", "Solution": "Let $M_a$, $M_b$, and $M_c$ be the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively. Let $H_a$, $H_b$, and $H_c$ be the feet of the altitudes from $A$, $B$, and $C$ to their opposite sides, respectively. Since $\\triangle ABC\\sim\\triangle M_bM_aC$, with $M_bM_a=\\frac12 AB$, the distance from $M_a$ to side $\\overline{AC}$ is $\\frac{h_b}{2}$.\nConstruct $AM_a$ with length $m_a$. Draw a circle centered at $A$ with radius $h_a$. Construct the tangent $l_1$ to this circle through $M_a$. $\\overline{BC}$ lies on $l_1$.\nDraw a circle centered at $M_a$ with radius $\\frac{h_b}{2}$. Construct the tangent $l_2$ to this circle through $A$. $\\overline{AC}$ lies on $l_2$. Then $C=l_1\\cap l_2$.\nConstruct the line $l_3$ parallel to $l_2$ so that the distance between $l_2$ and $l_3$ is $h_b$ and $M_a$ lies between these lines. $B$ lies on $l_3$. Then $B=l_1\\cap l_3$." }, { "Problem": "In the interior of triangle $P_1P_2P_3$ a point $P$ is given. Let $Q_1,Q_2,Q_3$ be the intersections of $PP_1, PP_2,PP_3$ with the opposing edges of triangle $P_1P_2P_3$. Prove that among the ratios $\\frac{PP_1}{PQ_1},\\frac{PP_2}{PQ_2},\\frac{PP_3}{PQ_3}$ there exists one not larger than $2$ and one not smaller than $2$.\n", "Solution_1": "Let $[ABC]$ denote the area of triangle $ABC$.\nSince triangles $P_1P_2P_3$ and $PP_2P_3$ share the base $P_2P_3$, we have $\\frac{[PP_2P_3]}{[P_1P_2P_3]}=\\frac{PQ_1}{P_1Q_1}$.\nSimilarly, $\\frac{[PP_1P_3]}{[P_1P_2P_3]}=\\frac{PQ_2}{P_2Q_2}, \\frac{[PP_1P_2]}{[P_1P_2P_3]}=\\frac{PQ_3}{P_3Q_3}$.\nAdding all of these gives $\\frac{[PP_1P_3]}{[P_1P_2P_3]}+\\frac{[PP_1P_2]}{[P_1P_2P_3]}+\\frac{[PP_2P_3]}{[P_1P_2P_3]}=\\frac{PQ_2}{P_2Q_2}+\\frac{PQ_3}{P_3Q_3}+\\frac{PQ_1}{P_1Q_1}=1$.\nWe see that we must have at least one of the three fractions not greater than $\\frac{1}{3}$, and at least one not less than $\\frac{1}{3}$. These correspond to ratios $\\frac{PP_i}{PQ_i}$ being less than or equal to $2$, and greater than or equal to $2$, respectively, so we are done.", "Solution_2": "Let $K_1=[P_2PP_3], K_2=[P_3PP_1],$ and $K_3=[P_1PP_2].$ Note that by same base in triangles $P_2PP_3$ and $P_2P_1P_3,$ Thus, \nWithout loss of generality, assume $K_1\\leq K_2\\leq K_3.$ Hence, and as desired. $\\blacksquare$" }, { "Problem": "Solve the equation $\\cos^2{x}+\\cos^2{2x}+\\cos^2{3x}=1$.", "Solution": "First, note that we can write the left hand side as a cubic function of $\\cos^2 x$. So there are at most $3$ distinct values of $\\cos^2 x$ that satisfy this equation. Therefore, if we find three values of $x$ that satisfy the equation and produce three different $\\cos^2 x$, then we found all solutions to this cubic equation (without expanding it, which is another viable option). Indeed, we find that $\\frac{\\pi}2$, $\\frac{\\pi}4$, and $\\frac{\\pi}6$ all satisfy the equation, and produce three different values of $\\cos^2 x$, namely $0$, $\\frac12$, and $\\frac34$. So we solve $\\cos^2 x = \\text{each of these}$. Therefore, our solutions are:\n$x = \\frac{(2k+1)\\pi}2,\\, \\frac{(2k+1)\\pi}4,\\, \\frac{(6k+1)\\pi}6,\\, \\frac{(6k+5)\\pi}6 \\quad \\forall k\\in Z$" }, { "Problem": "Find all solutions $x_1,x_2,x_3,x_4,x_5$ of the system\n\nwhere $y$ is a parameter.", "Solution": "Notice: The following words are Chinese.\n首先,我们可以将以上5个方程相加,得到:\n$2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$\n当$x_1+x_2+x_3+x_4+x_5=0$时,因为$x_1,x_2,x_3,x_4,x_5$关于原方程组轮换对称,所以\n$x_1=x_2=x_3=x_4=x_5=0$\n若反之,则方程两边同除以$(x_1+x_2+x_3+x_4+x_5)$,得到$y=2$,显然解为\n$x_1=x_2=x_3=x_4=x_5$\n综上所述,若$y=2$,最终答案为$x_1=x_2=x_3=x_4=x_5$,否则答案为$x_1=x_2=x_3=x_4=x_5=0$\nThe solution in English (translated by Google Translate):\nFirst of all, we can add the five equations to get:\n$2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$\nWhen $x_1+x_2+x_3+x_4+x_5=0$, Because $x_1,x_2,x_3,x_4,x_5$ is symmetric in the original equations,\n$x_1=x_2=x_3=x_4=x_5=0$\nOtherwise, dividing both sides by $(x_1+x_2+x_3+x_4+x_5$, we get $y=2$, and clearly\n$x_1=x_2=x_3=x_4=x_5$\nSummarizing, if $y=2$, then the answer is of the form $x_1=x_2=x_3=x_4=x_5$. Otherwise, $x_1=x_2=x_3=x_4=x_5=0$." }, { "Problem": "Seventeen people correspond by mail with one another - each one with all\nthe rest. In their letters only three different topics are discussed. Each pair\nof correspondents deals with only one of these topics. Prove that there are\nat least three people who write to each other about the same topic.", "Solution": "Lemma: Consider a complete graph with 6 vertices colored with 2 colors. There exists a monochromatic triangle.\nProof: Consider one vertex and all connections leading out from it. Call it $V_1$. It has 5 edges coming out from it. By the Pigeonhole Principle, there are at least 3 of the same color. Call this color red. Call those vertices $V_2$, $V_3$ and $V_4$. If any of the segments $V_2V_3$, $V_2V_4$, or $V_3V_4$ are red, then we have a monochromatic triangle with vertices $V_1$ and the other two that are also red. If they are all the other color, then we have a monochromatic triangle with vertices $V_2$,$V_3$, and $V_4$. $\\blacksquare$\n\nMain Problem: Represent these people as vertices on a connected graph with 17 vertices and colored with 3 colors, one corresponding with each topic. So this problem is reduced to showing that on a connected graph with 17 vertices and colored with three colors, there exists some monochromatic triangle. Look at an arbitrary vertex. Call it $V_1$. Look at the 16 other vertices that it is connected to. By the Pigeonhole Principle, there are at least 6 vertices connected to $V_1$ that are all one color. Call this color 1. If any of the connections inbetween these six vertices are in color 1, then we are done. If none of them are color 1, we know that that there are only 2 colors in those 6 vertices. By Lemma 1, we know that there is a monochromatic triangle in those 6 vertices. 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": "Find all sets of four real numbers $x_1$, $x_2$, $x_3$, $x_4$ such that the sum of any one and the product of the other three is equal to $2$.", "Solution": "Let $P = x_1x_2x_3x_4$ be the product of the four real numbers.\nThen, for $i = 1,2,3,4$ we have: $x_i + \\prod_{j \\neq i}x_j = 2$.\nMultiplying by $x_i$ yields:\n$x^2_i + P = 2x_i \\Longleftrightarrow x^2_i-2x_i+1 = (x_i-1)^2 = 1-P \\Longleftrightarrow x_i = 1 \\pm t$ where $t = \\pm \\sqrt{1-P} \\in \\mathbb{R}$.\nIf $t=0$, then we have $(x_1,x_2,x_3,x_4)=(1,1,1,1)$ which is a solution.\nSo assume that $t \\neq 0$. WLOG, let at least two of $x_i$ equal $1+t$, and $x_1 \\ge x_2 \\ge x_3 \\ge x_4$ OR $x_1 \\le x_2 \\le x_3 \\le x_4$.\nCase I: $x_1 = x_2 = x_3 = x_4 = 1+t$\nThen we have:\n$(1+t)+(1+t)^3 = 2 \\Longleftrightarrow t^3+3t^2+4t = 0 \\Longleftrightarrow t(t^2+3t+4) = 0$\nWhich has no non-zero solutions for $t$.\nCase II: $x_1 = x_2 = x_3 = 1+t$ AND $x_4 = 1-t$\nThen we have:\n$(1-t)+(1+t)^3 = 2 \\Longleftrightarrow t^3+3t^2+2t = 0$ $\\Longleftrightarrow t(t+1)(t+2) = 0 \\Longleftrightarrow t \\in \\{0,-1,-2\\}$\nAND\n$(1+t)+(1-t)(1+t)^2 = 2 (1+t)+(1-t)(1+t)^2 = 2 -t^3-t^2+2t = 0$ $\\Longleftrightarrow -t(t-1)(t+2) = 0 \\Longleftrightarrow t \\in \\{0,1,-2\\}$\nSo, we have $t = -2$ as the only non-zero solution, and thus, $(x_1,x_2,x_3,x_4) = (-1,-1,-1,3)$ and all permutations are solutions.\nCase III: $x_1 = x_2 = 1+t$ AND $x_3 = x_4 = 1-t$\nThen we have:\n$(1-t)+(1-t)(1+t)^2 = 2 \\Longleftrightarrow -t^3-t^2 = 0$ $\\Longleftrightarrow -t^2(t+1) = 0 \\Longleftrightarrow t \\in \\{0,-1\\}$\nAND\n$(1+t)+(1+t)(1-t)^2 = 2 \\Longleftrightarrow t^3-t^2 = 0$ $\\Longleftrightarrow t^2(t-1) = 0 \\Longleftrightarrow t \\in \\{0,1\\}$\nThus, there are no non-zero solutions for $t$ in this case.\nTherefore, the solutions are: $(1,1,1,1)$; $(3,-1,-1,-1)$; $(-1,3,-1,-1)$; $(-1,-1,3,-1)$; $(-1,-1,-1,3)$." }, { "Problem": "Prove that for every natural number $n$, and for every real number $x \\neq \\frac{k\\pi}{2^t}$ ($t=0,1, \\dots, n$; $k$ any integer)", "Solution": "Assume that $\\frac{1}{\\sin{2x}}+\\frac{1}{\\sin{4x}}+\\dots+\\frac{1}{\\sin{2^{n}x}}=\\cot{x}-\\cot{2^{n}x}$ is true, then we use $n=1$ and get $\\cot x - \\cot 2x = \\frac {1}{\\sin 2x}$.\nFirst, we prove $\\cot x - \\cot 2x = \\frac {1}{\\sin 2x}$\nLHS=$\\frac{\\cos x}{\\sin x}-\\frac{\\cos 2x}{\\sin 2x}$\n$= \\frac{2\\cos^2 x}{2\\cos x \\sin x}-\\frac{2\\cos^2 x -1}{\\sin 2x}$\n$=\\frac{2\\cos^2 x}{\\sin 2x}-\\frac{2\\cos^2 x -1}{\\sin 2x}$\n$=\\frac {1}{\\sin 2x}$\nUsing the above formula, we can rewrite the original series as\n$\\cot x - \\cot 2x + \\cot 2x - \\cot 4x + \\cot 4x \\cdot \\cdot \\cdot + \\cot 2^{n-1} x - \\cot 2^n x$\nWhich gives us the desired answer of $\\cot x - \\cot 2^n x$" }, { "Problem": "Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle.\n", "Solution": "Let the edges of one of the faces of the tetrahedron have lengths $a$, $b$, and $c$. Let $d$, $e$, and $f$ be the lengths of the sides that are not adjacent to the sides with lengths $a$, $b$, and $c$, respectively.\nWithout loss of generality, assume that $\\max(a,b,c,d,e,f)=a$. I shall now prove that either $b+f>a$ or $c+e>a$, by proving that if $b+f\\leq a$, then $c+e>a$.\nAssume that $b+f\\leq a$. The triangle inequality gives us that $e+f>a$, so $e$ must be greater than $b$. We also have from the triangle inequality that $b+c>a$. Therefore $e+c>b+c>a$. Therefore either $b+f>a$ or $c+e>a$.\nIf $b+f>a$, then the vertex where the sides of length $a$, $b$, and $f$ meet satisfies the given condition. If $c+e>a$, then the vertex where the sides of length $a$, $c$, and $e$ meet satisfies the given condition. This proves the statement. $\\blacksquare$" }, { "Problem": "A semicircular arc $\\gamma$ is drawn with $AB$ as diameter. $C$ is a point on $\\gamma$ other than $A$ and $B$, and $D$ is the foot of the perpendicular from $C$ to $AB$. We consider three circles, $\\gamma_1, \\gamma_2, \\gamma_3$, all tangent to the line $AB$. Of these, $\\gamma_1$ is inscribed in $\\triangle ABC$, while $\\gamma_2$ and $\\gamma_3$ are both tangent to $CD$ and $\\gamma$, one on each side of $CD$. Prove that $\\gamma_1, \\gamma_2$, and $\\gamma_3$ have a second tangent in common.", "Solution": "Denote the triangle sides $a = BC, b = CA, c = AB$. Let $\\omega$ be the circumcircle of the right angle triangle $\\triangle ABC$ centered at the midpoint $O$ of its hypotenuse $c = AB$. Let $R, S, T$ be the tangency points of the circles $K_1, K_2, K_3$ with the line AB. In an inversion with the center $A$ and positive power $r_A^2 = AC^2 = b^2$ ($r_A$ being the inversion circle radius), the line AB is carried into itself, the circle $\\omega$ is carried into the altitude line $CD$ and the altitude line $CD$ into the circle $\\omega$. This implies that the circle $K_3$ intersecting the inversion circle $A$ is carried into itself, but this is possible only if the circle $K_3$ is perpendicular to the inversion circle $A$. It follows that the tangency point $T$ of the circle $K_3$ is the intersection of the inversion circle $(A, r_A = b)$ with the line $AB$. Similarly, in an inversion with the center B and positive power $r_B^2 = BC^2 = a^2$ ($r_B$ being the inversion circle radius), the line AB is carried into itself, the circle $\\omega$ is carried into the altitude line $CD$ and the altitude line $CD$ into the circle $\\omega$. This implies that the circle $K_2$ intersecting the inversion circle $B$ is carried into itself, but this is possible only if the circle $K_2$ is perpendicular to the inversion circle $B$. It follows that the tangency point S of the circle $K_2$ is the intersection of the inversion circle $(B, r_B = a)$ with the line $AB$.\nThe distance between the tangency points S, T is the equal to ST = AT - AS = AT - (AB - BS) = AC - (AB - BC) = a + b - c. The radius r of the incircle $K_1$ of the right angle triangle $\\triangle ABC$ is equal to\n$r = \\frac{|\\triangle ABC|}{s} = \\frac{ab}{a + b + c} = \\frac{a + b - c}{2} = s - c$\nwhere $|\\triangle ABC|$ and s are the area and semiperimeter of the triangle $\\triangle ABC$, for example, because of an obvious identity\n$(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab$\nor just because the angle $\\angle C = 90^\\circ$ is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then\n$AR' = AT - \\frac{ST}{2} = b - r = b - \\frac{a + b - c}{2} = \\frac{c + b - a}{2} = s - a = AR$\nTherefore, the points $R' \\equiv R$ are identical and the midpoint of the segment ST is the tangency point R of the incircle $K_1$ with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles $K_2, K_3$ are tangent to the incircle $K_1$. Radii $r_2, r_3$ of the circles $K_2, K_3$ are now easily calculated:\n$r_2 = SD = BS - BD = a - \\frac{a^2}{c}$\n$r_3 = TD = AT - AD = b - \\frac{b^2}{c}$\nDenote $I, I_2, I_3$ the centers of the circles $K_1, K_2, K_3$. The line $I_2I_3$ cuts the midline RI of the trapezoid $STI_3I_2$ at the distance from the point R equal to\n$\\frac{SI_2 + TI_3}{2} = \\frac{r_2 + r_3}{2} = \\frac{a + b}{2} - \\frac{a^2 + b^2}{2c} = \\frac{a + b - c}{2} = r = RI$\nAs a result, the centers $I_2, I, I_3$ are collinear (in fact, I is the midpoint of the segment $I_2I_3$). The common center line $I_2I_3$ and the common external tangent AB of the circles $K_1, K_2, K_3$ meet at their common external homothety center $H \\equiv I_2I_3 \\cap AB$ and the other common external tangent of the circles $K_2, K_3$ from the common homothety center H is a tangent to the circle $K_1$ as well.\nThe above solution was posted and copyrighted by yetti. The original thread can be found here: [1]" }, { "Problem": "Find the set of all positive integers $n$ with the property that the set $\\{ n, n+1, n+2, n+3, n+4, n+5 \\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.", "Solution": "The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be $\\{ 1, 2, 3, 4, 5, 6 \\}$, but that does not work because only one of the numbers is a multiple of 5. So there are no such sets.", "Solution_2": "As in the previous solution, none of the six consecutive numbers can be multiples of $7$. This means that together, they take on the values $\\{ 1, 2, 3, 4, 5, 6, \\} \\mod 7$. The product of all the numbers in this set, then, is $-1 \\mod 7$, by Wilson's Theorem. However, $-1$ is not a quadratic residue $\\mod 7$, which means that we cannot partition the original set into two sets of equal product. Thus, no such $n$ exist." }, { "Problem": "All the faces of tetrahedron $ABCD$ are acute-angled triangles. We consider all closed polygonal paths of the form $XYZTX$ defined as follows: $X$ is a point on edge $AB$ distinct from $A$ and $B$; similarly, $Y, Z, T$ are interior points of edges $BC, CD, DA$, respectively. Prove:\n(a) If $\\angle DAB + \\angle BCD \\neq \\angle CDA + \\angle ABC$, then among the polygonal paths, there is none of minimal length.\n(b) If $\\angle DAB + \\angle BCD = \\angle CDA + \\angle ABC$, then there are infinitely many shortest polygonal paths, their common length being $2AC \\sin(\\alpha / 2)$, where $\\alpha = \\angle BAC + \\angle CAD + \\angle DAB$.", "Solution": "Rotate the triangle $BCD$ around the edge $BC$ until $ABCD$ are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting $X$ and $Z$. Therefore, $XYB=ZYC$.\nSumming the four equations like this, we get exactly $\\angle ABC+\\angle ADC=\\angle BCD+\\angle BAD$.\nNow, draw all four faces in the plane, so that $BCD$ is constructed on the exterior of the edge $BC$ of $ABC$ and so on with edges $CD$ and $AD$.\nThe final new edge $AB$ (or rather $A'B'$) is parallel to the original one (because of the angle equation). Call the direction on $AB$ towards $B$ \"right\" and towards $A$ \"left\". If we choose a vertex $X$ on $AB$ and connect it to the corresponding vertex $X'$ on A'B'. This works for a whole interval of vertices $X$ if $C$ lies to the left of $B$ and $D$ and $D$ lies to the right of $A$. It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.\nFinally, regard the sine in half the isosceles triangle $ACA'$ which gives the result with the angles around $C$ instead of $A$, but the role of the vertices is symmetric." }, { "Problem": "A soldier needs to check on the presence of mines in a region having the\nshape of an equilateral triangle. The radius of action of his detector is equal\nto half the altitude of the triangle. The soldier leaves from one vertex of the\ntriangle. What path shouid he follow in order to travel the least possible\ndistance and still accomplish his mission?", "Solution": "Let our triangle be $\\triangle ABC$, let the midpoint of $AB$ be $D$, and let the midpoint of $CD$ be $E$. Let the height of the triangle be $h$. Draw circles around points $B$ and $C$ with radius $\\frac{h}{2}$, and label them $c_1$ and $c_2$. Let the intersection of $BE$ and $c_1$ be $F$.\nThe path that is the solution to this problem must go from $A$ to a point on $c_2$ to a point on $c_1$. Let us first find the shortest possible path. We will then prove that this path fits the requirements.\nThe shortest path is in fact the path from $A$ to $E$ to $F$. We will prove this as follows:\nSuppose that a different point on $c_2$ is the optimal point to go to. Let this point be $P$. Then, the optimal point on $c_1$ would be the intersection of $BP$ and $c_1$ (let this point be $R$). Let the line through $E$ parallel to $AB$ be $l$, and the intersection of $AP$ and $l$ be $Q$. Then, we have\n\nThe second inequality is true due to the triangle inequality, and we can prove the first to be true as follows:\nSuppose we have a point on $l$, $X$. Reflect $B$ across $l$ to get $B'$. Then, $AX + BX = AX + B'X$, which is obviously minimized at the intersection of the two lines, $E$.\nBy $\\textbf{(1)}$, we have\n\nand we have proved the claim.\nThis path easily covers the whole triangle. This is because if you draw a line perpendicular to $AB$ from any point in the triangle, this line will hit the path in a distance less than or equal to $\\frac{h}{2}$. We can compute the length of the path to be\n\n~mathboy100" }, { "Problem": "Consider decompositions of an $8\\times8$ chessboard into $p$ non-overlapping rectangles subject to the following conditions:\n(i) Each rectangle has as many white squares as black squares.\n(ii) If $a_i$ is the number of white squares in the $i$-th rectangle, then $a_1 x$ for $x \\geq 5$, we notice that $S$ must only contain the integers $1,2,3$ and $4$. We can replace any occurrences of $4$ in $S$ by replacing it with a couple of $2$'s, without changing $f(S)$ or $g(S)$, so we may assume that $S$ only contains the integers $1,2$ and $3$. We may further assume that $S$ contains at most one $1$, since any two $1$'s can be replaced by a $2$ without changing $g(S)$, but with an increase in $f(S)$. If $S$ contains exactly one $1$, then it must also contain at least one $2$ (since $1976 \\equiv 2\\;(\\mathrm{mod }\\;3)$). We can then replace this pair of a $1$ and a $2$ with a $3$, thus keeping $g(S)$ constant, and increasing $f(S)$. Now we may assume that $S$ contains only $2$'s and $3$'s.\nNow, as observed in the last solution, any triplet of $2$'s can be replaced by a couple of $3$'s, with $g(S)$ constant, and an increase in $f(S)$. Thus, after repeating this operation, we will be left with at most two $2$'s. Since $g(S) = 1976$, and $1976 \\equiv 2\\;(\\mathrm{mod }\\;3)$, we therefore get that $S$ must have exactly one $2$ (since we already showed it consists only of $2$'s and $3$'s). Thus, we get $\\boxed{f(S) = 2\\cdot 3^{658}}$." }, { "Problem": "Let $a,b,A,B$ be given reals. We consider the function defined byProve that if for any real number $x$ we have $f(x) \\geq 0$ then $a^2 + b^2 \\leq 2$ and $A^2 + B^2 \\leq 1.$", "Solution": "$f(x) = 1-\\sqrt{a^2+b^2}\\sin (x+\\arctan\\frac{a}{b}) - \\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\geq 0$.\n$f(x+\\pi) = 1+\\sqrt{a^2+b^2}\\sin (x+\\arctan\\frac{a}{b}) - \\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\geq 0$\nTherefore, $\\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\leq 1$. Since this identity is true for any real $x$, let the sine term be one, $\\longrightarrow A^2+B^2 \\leq 1$.\nTo get cancellation on the rightmost terms, note $\\sin (x+\\pi/2) = \\cos x, \\sin (x-\\pi/2) = -\\cos x$.\n$f(x+\\pi/4) = 1-\\sqrt{a^2+b^2}\\sin (x+\\pi/4+\\arctan\\frac{a}{b}) - \\sqrt{A^2+B^2}\\cos2x+\\arctan\\frac{A}{B}) \\geq 0$.\n$f(x-\\pi/4) = 1-\\sqrt{a^2+b^2}\\sin (x-\\pi/4+\\arctan\\frac{a}{b}) + \\sqrt{A^2+B^2}\\cos2x+\\arctan\\frac{A}{B}) \\geq 0$.\nLet $x+\\arctan\\frac{a}{b} = y$.\nThen $\\sqrt{a^2+b^2}(\\sin (y+\\pi/4) + \\sin (y-\\pi/4)) \\leq 2$\n$\\sqrt{a^2+b^2} \\leq \\dfrac{2}{\\sqrt{2}(\\sin y)}$. Since it's valid for all real $x$ let $\\sin y = 1$, and we are done.\nThe above solution was posted and copyrighted by aznlord1337. The original thread for this problem can be found here: [1]" }, { "Problem": "In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$", "Solution": "Denote $a = BC, b = AB = AC$ the triangle sides, $r, R$ the triangle inradius and circumradius and $s$, $\\triangle$ the triangle semiperimeter and area. Let $S, T$ be the tangency points of the incircle (I) with the sides $AB, AC$. Let $K, L$ be the midpoints of of $BC, PQ$ and $M$ the midpoint of the arc $BC$ of the circumcircle (O) opposite to the vertex $A$. The isosceles triangles $\\triangle AST \\sim \\triangle APQ \\sim \\triangle ABC$ are all centrally similar with the homothety center $a$. The homothety coefficient for the triangles $\\triangle AST \\sim \\triangle ABC$ is $h_{13} = \\frac{AS}{AB} = \\frac{s - a}{b}$. The homothety coefficient for the triangles $\\triangle AST \\sim \\triangle APQ$ is $h_{12} = \\frac{AS}{AP}$. The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles $\\triangle ASK \\sim \\triangle APM$ are also centrally similar with the homothety center A and the same homothety coefficient as the triangles $\\triangle AST \\sim \\triangle APQ$: $h_{12} = \\frac{AS}{AP} = \\frac{AK}{AM} = \\frac{h}{2R}$, where h = AK is the A-altitude of the triangle $\\triangle ABC$. The homothety coefficient of the triangles $\\triangle APQ \\sim \\triangle ABC$ is then $h_{23} = \\frac{h_{12}}{h_{13}} = \\frac{s - a}{b} \\cdot \\frac{2R}{h}$ Denote $h' = AL$ the $A$-altitude of the triangle $\\triangle APQ$. Then\n$\\frac{h'}{h} = h_{23} = \\frac{s - a}{b} \\cdot \\frac{2R}{h}$\n$KL = h - h' = h - 2R\\ \\frac{s - a}{b}$\nSubstituting $h = \\frac{2 \\triangle}{a}$, $2R = \\frac{ab^2}{2 \\triangle}$ and $s - a = \\frac{\\triangle^2}{s(s - b)^2}$, we get\n$KL = \\frac{2 \\triangle}{a} - \\frac{ab^2}{2 \\triangle} \\cdot \\frac{\\triangle^2}{bs(s - b)^2} =$\n(substituting 2s = 2b + a and $s - b = \\frac a 2$ for an isosceles triangle)\n$= \\frac{\\triangle}{s}\\left(\\frac{2s}{a} - \\frac{ab}{2(s - b)^2}\\right) = \\frac{\\triangle}{s} \\left(\\frac{2b}{a} + 1 - \\frac{2b}{a}\\right) = \\frac{\\triangle}{s} = r$\nwhich means that the point $L \\equiv I$ is identical with the incenter of the triangle $\\triangle ABC$.\nThe above solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]" }, { "Problem": "We consider a point $P$ in a plane $p$ and a point $Q \\not\\in p$. Determine all the points $R$ from $p$ for whichis maximum.", "Solution": "Let $T$ be the orthogonal projection of the point $Q$ on the plane $p$. Then, the line $PT$ is the orthogonal projection of the line $PQ$ on the plane $p$, and thus forms the least angle with the line $PQ$ among all lines through the point $P$ which lie in the plane $p$; hence, $\\measuredangle RPQ\\geq\\measuredangle TPQ$, and equality holds if and only if the point $R$ lies on the ray $PT$ (the only exception is when $PQ\\perp p$; in this case, $P = T$, so the ray $PT$ is undefined, and equality holds for all points $R$ in the plane $p$, since we always have $\\angle RPQ = \\angle TPQ = 90^{\\circ}$).\nFrom $\\measuredangle RPQ\\geq\\measuredangle TPQ$, it follows that $\\frac{\\measuredangle RPQ}{2}\\geq\\frac{\\measuredangle TPQ}{2}$; also, since the angles $\\angle RPQ$ and $\\angle TPQ$ are $180^{\\circ}$, their half-angles $\\frac{\\measuredangle RPQ}{2}$ and $\\frac{\\measuredangle TPQ}{2}$ are < 90°, so that from $\\frac{\\measuredangle RPQ}{2}\\geq\\frac{\\measuredangle TPQ}{2}$ we can conclude that $\\sin\\frac{\\measuredangle RPQ}{2}\\geq\\sin\\frac{\\measuredangle TPQ}{2}$. Equality holds, as in the above, if and only if the point $R$ lies on the ray $PT$ (and, respectively, for all points $R$ in the plane $p$ if $PQ\\perp p$).\nOn the other hand, we obviously have $\\cos\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}\\leq 1$ with equality if and only if $\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}=0^{\\circ}$, i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base $QR$, i. e. if and only if $PR = PQ$, i. e. if and only if the point $R$ lies on the sphere with center $P$ and radius $PQ$.\nNow, applying the Mollweide theorem in triangle $QPR$, we get\n$\\frac{QP+PR}{QR}=\\frac{\\cos\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}}{\\sin\\frac{\\measuredangle RPQ}{2}}$\n$\\leq\\frac{1}{\\sin\\frac{\\measuredangle RPQ}{2}}$ (since $\\cos\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}\\leq 1$)\n$\\leq\\frac{1}{\\sin\\frac{\\measuredangle TPQ}{2}}$ (since $\\sin\\frac{\\measuredangle RPQ}{2}\\geq\\sin\\frac{\\measuredangle TPQ}{2}$),\nand equality holds here if and only if equality holds in both of the inequalities $\\sin\\frac{\\measuredangle RPQ}{2}\\geq\\sin\\frac{\\measuredangle TPQ}{2}$ and $\\cos\\frac{\\measuredangle QRP-\\measuredangle PQR}{2}\\leq 1$ that we have used, i. e. if and only if the point $R$ lies both on the ray $PT$ (this condition should be ignored if $PQ\\perp p$) and on the sphere with center $P$ and radius $PQ$.\nHence, the point $R$ for which the ratio $\\frac{QP+PR}{QR}$ is maximum is the point of intersection of the ray $PT$ with the sphere with center $P$ and radius $PQ$ (or, respectively, it can be any arbitrary point on the intersection of the plane $p$ with the sphere with center $P$ and radius $PQ$ if $PQ\\perp p$).\nThis solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [1]\n" }, { "Problem": "(a) For which values of $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers?\n(b) For which values of $n>2$ is there exactly one set having the stated property?", "Solution": "Let $k = \\prod p_i^{e_i}$ be the greatest element of the set, written in its prime factorization. Then $k$ divides the least common multiple of the other elements of the set if and only if the set has cardinality at least $\\max \\{ p_i^{e_i} \\}$, since for any of the $p_i^{e_i}$, we must go down at least to $k-p_i^{e_i}$ to obtain another multiple of $p_i^{e_i}$. In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.\nFor $n > 3$, we may let $k = \\mbox{lcm} (n-1, n-2) = (n-1)(n-2)$, since all the $p_i^{e_i}$ must clearly be less than $n$ and this product must also be greater than $n$ if $n$ is at least 4. For $n > 4$, we may also let $k = \\mbox{lcm} (n-2, n-3) = (n-2)(n-3)$, for the same reasons. However, for $n = 4$, this does not work, and indeed no set works other than $\\{ 3,4,5,6 \\}$. To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.\nQ.E.D." }, { "Problem": "Prove that if $n$ is a positive integer such that the equation $x^3-3xy^2+y^3=n$ has a solution in integers $x,y$, then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$.", "Solution": "Suppose the equation $x^3-3xy^2+y^3=n$ has solution in integers $(x,y)$ with $y=x+k$. Then, completing the cube yields $(y-x)^3-3x^2y+2x^3$. Using the substitution $y=x+k$ yields $k^3-3kx^2-x^3=n$. Notice that equality directly implies that $(k,-x)$ is also a solution to the original equation. Applying the transformation again yields that $(-x-k,-k)$ is also a solution. We show that these three solutions are indeed distinct: If $(x,y)=(k,-x)$ then $x=k,x+k=-x$ which only has solution $x=k=0$ which implies that $n$ is not a positive integer, a contradiction. Similarly, since the transformation from $(k,-x)$ to $(-x-k,-k)$ and $(-x-k,-k)$ to $(x,y)$ is the same as the transformation from $(x,y)$ to $(k,-x)$, we have that the three solutions are pairwise distinct.\nFor the case $n=2891$, notice that $7|n$. Considering all solutions modulo $7$ of the equation $x^3-3xy^2+y^3\\equiv0\\pmod{7}$ yields only $x\\equiv y\\equiv 0\\pmod{7}$. But, this implies that $7^3$ divides $2891$ which is clearly not true." }, { "Problem": "Let $ABC$ be an equilateral triangle and $\\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$ and $CA$ (including $A$, $B$ and $C$). Determine whether, for every partition of $\\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle. Justify your answer.", "Solution": "The answer is positive; there always a class of the partition will contain the three vertices of a right-angled triangle.\nFirst notice that if there are at least two points of the same color on a side, then all the points in $\\mathcal{E}$ which project orthogonally onto points of that color on the respective side must have the opposite color if we are to have no monochromatic right-angled triangles.\nNow assume we can find a side on which there is at most one point bearing one of the colors (blue, say; we take the two colors to be red and blue). Then, by the observation above, it's obvious that the other two sides contain three blue vertices of a right-angled triangle, and we're done (all the points of those two sides are blue, except maybe for the endpoints which they have in common with the other side and the point which projects orthogonally onto the unique blue point on the third side). We can now assume all three sides contain at least two points of each color. Take $U\\in BC$ s.t. $BC=3BU$. $U$ projects orthogonally onto $V\\in CA$, and $V$ projects orthogonally onto $T\\in AB$. We can easily see that $T$ projects orthogonally onto $U$. Again, by the observation in the previous paragraph, the points $U,V,T$ must have different colors if we are to have no monochromatic right-angled triangle, and this is impossible (we only have two colors).\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]\n" }, { "Problem": "Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.", "Solution": "First, we prove that if $BC$ and $AD$ are parallel then the claim is true: Let $AB$ and $CD$ intersect at $E$ (assume $E$ is closer to $AD$, the other case being analogous). Let $M,N$ be the midpoints of $AB,CD$ respectively. Let the length of the perpendicular from $N$ to $AB$ be $r$. It is known that the length of the perpendicular from $M$ to $CD$ is $\\frac{1}{2}AB$. Let the foot of the perpendicular from $C$ to $AB$ be $H$, and similarly define $G$ for side $CD$. Then, since triangles $MNE$ and $BCE$ are similar, we have $\\frac{CH}{r}=\\frac{BG}{\\frac{1}{2}AB}$. This gives an expression for $r$:\n$r=\\frac{1}{2}AB\\cdot\\frac{CH}{BG}$\nNoticing that $CH=BC\\sin EBC,BG=BC\\sin ECB$ simplifies the expression to\n$r=\\frac{1}{2}AB\\cdot\\frac{\\sin EBC}{\\sin ECB}$\nBy the Law of Sines, $\\frac{\\sin EBC}{\\sin ECB}=\\frac{EC}{EB}$. Since triangles $EDA,ECB$ are similar, we have $\\frac{EC}{EB}=\\frac{CD}{AB}$ and thus we have\n$r=\\frac{1}{2}AB\\cdot\\frac{CD}{AB}=\\frac{1}{2}CD$\nand we are done.\nNow to prove the converse. Suppose we have the quadrilateral with $BC$ parallel to $AD$, and with all conditions satisfied. We shall prove that there exists no point $T$ on $CD$ such that $T$ is a midpoint of a side $CD^\\prime$ of a quadrilateral $ABCD^\\prime$ which also satisfies the condition. Suppose there was such a $T$. Like before, define the points $E,M,N$ for quadrilateral $ABCD$. Let $t$ be the length of the perpendicular from $T$ to $AB$. Then, using similar triangles, $\\frac{ET}{t}=\\frac{EN}{\\frac{1}{2}CD}$. This gives\n$t=\\frac{\\frac{1}{2}CD\\cdot ET}{EN}$\nBut, we must have $t=DT$. Thus, we have\n$DT=\\frac{\\frac{1}{2}CD\\cdot ET}{EN}$\n$\\Rightarrow \\frac{1}{2}CD\\cdot EN+NT\\cdot EN=\\frac{1}{2}CD\\cdot (EN+NT)$\n$\\Rightarrow NT\\cdot EN=\\frac{1}{2}CD\\cdot NT$\nSince $\\frac{1}{2}CD\\ne EN$, we have $NT=0$ as desired.", "Solution_2": "Let $M$ and $N$ be midpoints of $AB$ and $CD$, respectively. Let $[X]$ denote the area of figure $X$.\nSuppose that $AD // BC$ and the circle centered at $M$ with diameter $AB$ touches $CD$ at $T$. We know that midsegment $MN // AD$. It follows that $[AMN] = [DMN]$, so $\\frac{1}{2} AM \\cdot d = \\frac{1}{2} MT \\cdot DN$, where $d$ is the distance from $N$ to $AB$. But we have $MT = MA$, so $DN = d$. It follows that the circle centered at $N$ with diameter $CD$ touches $AB$.\nIf on the other hand we have the circle with diameter $CD$ touching $AB$ at $U$(as well as the circle with diameter $AB$ touching $CD$ at $T$), we must have $\\frac{1}{2} DN \\cdot MT = \\frac{1}{2} AM \\cdot UN$ because of the equivalence of radii in circles. Hence $[ANM] = [DMN]$, so $A$ and $D$ are equidistant from $MN$ (as $MN$ is the common base). Hence $AB // MN$. Similarly $CD // MN$, and so $AB // CD$, as desired." }, { "Problem": "Given a set $M$ of $1985$ distinct positive integers, none of which has a prime divisor greater than $23$, prove that $M$ contains a subset of $4$ elements whose product is the $4$th power of an integer.", "Solution": "We have that $x\\in M\\Rightarrow x=2^{e_1}3^{e_2}\\cdots 19^{e_8}23^{e_9}$. We need only consider the exponents. First, we consider the number of subsets of two elements, such that their product is a perfect square. There are $2^9=512$ different parity cases for the exponents $e_1,e_2,...,e_9$. Thus, we have at least one pair of elements out of $1985$ elements. Removing these two elements yields $1983$ elements. By applying the Pigeon Hole Principle again, we find that there exists another such subset. Continuing on like this yields at least $734$ pairs of elements of $M$ whose product is a perfect square. Let $S$ be the set of the square roots of the products of each pair. Then, by the Pigeon Hole Principle again, there exist at least two elements whose product is a perfect square. Let the elements be $x,y$ and let $x=\\sqrt{ab},y=\\sqrt{cd}$ where $a,b,c,d\\in M$. Then, we have $xy=z^2$ for some $z$ which implies $abcd=z^4$ and the claim is proved." }, { "Problem": "Let $A,B$ be adjacent vertices of a regular $n$-gon ($n\\ge5$) with center $O$. A triangle $XYZ$, which is congruent to and initially coincides with $OAB$, moves in the plane in such a way that $Y$ and $Z$ each trace out the whole boundary of the polygon, with $X$ remaining inside the polygon. Find the locus of $X$.", "Solution": "Let $C \\not = A$ the vertex which is adjacent to $B$. While $XYZ$ moves from $OAB$ to $OBC$, it is easy to see $XYBZ$ is cyclic. Thus $X$ lies on the bisector of $\\angle YBZ = \\angle ABC$. Moreover, $X$ is the intersection of a circle passing through $B$ (the circumcircle of $XYBZ$) and with a fixed radius (the radius is a function of $\\triangle XYZ$). Therefore $X$ varies in a line segment ended in $O$. When $Y$ and $Z$ pass through the other sides, we get as locus $n$ distinct line segments, each passing throught $O$ and contained in $OV$ (but not in $\\vec{OV}$) for some vertex $V$ of the polygon. Each two of these lines are obtained one from another by a rotation with center $O$.\nThis solution was posted and copyrighted by feliz. The original thread for this problem can be found here: [1]" }, { "Problem": "Prove that there is no function $f$ from the set of non-negative integers into itself such that $f(f(n)) = n + 1987$ for every $n$.", "Solution_1": "We prove that if $f(f(n)) = n + k$ for all $n$, where $k$ is a fixed positive integer, then $k$ must be even. If $k = 2h$, then we may take $f(n) = n + h$.\nSuppose $f(m) = n$ with $m \\equiv n \\mod k$. Then by an easy induction on $r$ we find $f(m + kr) = n + kr$, $f(n + kr) = m + k(r+1)$. We show this leads to a contradiction. Suppose $m < n$, so $n = m + ks$ for some $s > 0$. Then $f(n) = f(m + ks) = n + ks$. But $f(n) = m + k$, so $m = n + k(s - 1) \\ge n$. Contradiction. So we must have $m \\ge n$, so $m = n + ks$ for some $s \\ge 0$. But now $f(m + k) = f(n + k(s+1)) = m + k(s + 2)$. But $f(m + k) = n + k$, so $n = m + k(s + 1) > n$. Contradiction.\nSo if $f(m) = n$, then $m$ and $n$ have different residues $\\pmod k$. Suppose they have $r_1$ and $r_2$ respectively. Then the same induction shows that all sufficiently large $s \\equiv r_1 \\pmod k$ have $f(s) \\equiv r_2 \\pmod k$, and that all sufficiently large $s \\equiv r_2 \\pmod k$ have $f(s) \\equiv r_1 \\pmod k$. Hence if $m$ has a different residue $r \\mod k$, then $f(m)$ cannot have residue $r_1$ or $r_2$. For if $f(m)$ had residue $r_1$, then the same argument would show that all sufficiently large numbers with residue $r_1$ had $f(m) \\equiv r \\pmod k$. Thus the residues form pairs, so that if a number is congruent to a particular residue, then $f$ of the number is congruent to the pair of the residue. But this is impossible for $k$ odd.", "Solution_2": "Solution by Sawa Pavlov:\nLet $\\mathbb{N}$ be the set of non-negative integers. Put $A = \\mathbb{N} - f(\\mathbb{N})$ (the set of all $n$ such that we cannot find $m$ with $f(m) = n$). Put $B = f(A)$.\nNote that $f$ is injective because if $f(n) = f(m)$, then $f(f(n)) = f(f(m))$ so $m = n$. We claim that $B = f(\\mathbb{N}) - f(f(\\mathbb{N}))$. Obviously $B$ is a subset of $f(\\mathbb{N})$ and if $k$ belongs to $B$, then it does not belong to $f(f(\\mathbb{N}))$ since $f$ is injective. Similarly, a member of $f(f(\\mathbb{N}))$ cannot belong to $B$.\nClearly $A$ and $B$ are disjoint. They have union $\\mathbb{N} - f(f(\\mathbb{N}))$ which is $\\{0, 1, 2, \\ldots , 1986\\}$. But since $f$ is injective they have the same number of elements, which is impossible since $\\{0, 1, \\ldots , 1986\\}$ has an odd number of elements.", "Solution_3": "Consider the function $g: \\mathbb{Z}_{1987} \\rightarrow \\mathbb{Z}_{1987}$ defined by $g(x) = f(x\\; {\\rm mod }\\; 1987) \\;{\\rm mod }\\; 1987$. Notice that we have $f(k) + 1987 = f(f(f(k))) = f(k + 1987)$, so that $f(x) = f(y) \\;{\\rm mod }\\; 1987$ whenever $x = y \\;{\\rm mod }\\; 1987$, and hence $g$ is well defined.\nNow, we observe that $g$ satisfies the identity $g(g(x)) = x$, for $x \\in Z_{1987}$. Thus, $g$ is an invertible function on a finite set of odd size, and hence must have a fixed point, say $a \\in \\mathbb{Z}_{1987}$. Identifying $a$ with its canonical representative in $\\mathbb{Z}$, we therefore get $f(a) = a + 1987k$ for some non-negative integer $k$.\nHowever, we then have $f(f(a)) = a + 1987$, while $f(f(a)) = f(a + 1987k) = 1987k + f(a) = 2\\cdot1987k + a$ (where we use the identity $f(x + 1987) = f(x) + 1987$ derived above, along with $f(a) = f(a) + 1987$. However, these two equations imply that $k = \\dfrac{1}{2}$, which is a contradiction since $k$ is an integer. Thus, such an $f$ cannot exist.\nNote: The main step in the proof above is that the function $g$ can be shown to have a fixed point. This step works even if 1987 is replaced with any other odd number larger than 1. However, for any even number $c$, $f(x) = x + \\dfrac{c}{2}$ satisfies the condition $f(f(x)) = x + c$.\n--Mahamaya 21:15, 21 May 2012 (EDT)" }, { "Problem": "Show that the solution set of the inequality\n$\\sum_{k=1}^{70}\\frac{k}{x-k}\\ge\\frac{5}{4}$\nis a union of disjoint intervals, the sum of whose length is $1988$.", "Solution": "Consider the graph of $f(x)=\\sum_{k=1}^{70}\\frac{k}{x-k}\\ge\\frac{5}{4}$. On the values of $x$ between $n$ and $n+1$ for $n\\in\\mathbb{N}$ $1\\le n\\le 69$, the terms of the form $\\frac{k}{x-k}$ for $k\\ne n,n+1$ have a finite range. In contrast, the term $\\frac{n}{x-n}$ has an infinite range, from $+\\infty$ to $n$. Similarly, the term $\\frac{n+1}{x-n-1}$ has infinite range from $-n-1$ to $-\\infty$. Thus, since the two undefined values occur at the distinct endpoints, we can deduce that $f(x)$ takes on all values between $+\\infty$ and $-\\infty$ for $x\\in(n,n+1)$. Thus, by the Intermediate Value Theorem, we are garunteed a $n70$, the value of $f(x)$ goes from $+\\infty$ to $0$, since as $x$ increases, all the terms go to $0$. Thus, there exists some $r_{70}>70$ such that $f(r_{70})=\\frac{5}{4}$ and so $f(x)\\ge\\frac{5}{4}$ for $x\\in(70,r_{70})$.\nSo, we have $70$ $r_i$ such that $f(r_i)=\\frac{5}{4}$. There are obviously no other such $r_i$ since $f(x)=\\frac{5}{4}$ yields a polynomial of degree $70$ when combining fractions. Thus, we have that the solution set to the inequality $f(x)\\ge\\frac{5}{4}$ is the union of the intervals $(n,r_n]$ (since if $f(x)<\\frac{5}{4}$ for $x\\in(n,r_n)$ then there would exist another solution to the equation $f(x)=\\frac{5}{4}$.\nThus we have proven that the solution set is the union of disjoint intervals. Now we are to prove that the sum of their lengths is $1988$.\nThe sum of their lengths is $r_1+r_2+\\cdots+r_{70}-(1+2+\\cdots+70)=r_1+r_2+\\cdots+r_{70}-35\\cdot71$. We have that the equation $f(x)=\\frac{5}{4}$ yields a polynomial with roots $r_i$. Thus, opposite of the coeficient of $x^{69}$ divided by the leading coefficient is the sum of the $r_i$. It is easy to see that the coefficient of $x^{69}$ is $-5(1+2+\\cdots+70)-4(1+2+\\cdots+70)=-9\\cdot35\\cdot 71$. Thus, since the leading coefficient is $5$ we have $r_1+r_2+\\cdots+r_{70}=9\\cdot7\\cdot71$. Thus, the sum of the lengths of the intervals is $63\\cdot71-35\\cdot71=28\\cdot71=1988$ as desired." }, { "Problem": "Let $\\mathbb{Q^+}$ be the set of positive rational numbers. Construct a function $f :\\mathbb{Q^+}\\rightarrow\\mathbb{Q^+}$ such that $f(xf(y)) = \\frac{f(x)}{y}$ for all $x, y\\in{Q^+}$.", "Solution": "If we let $y = x = 1$ this implies $f(f(1)) = f(1)$. Plugging $f(1) = y$ in with this new fact. We get $f(f(f(1))) = \\frac{f(1)}{f(1)} = 1$. Using $f(f(1)) = f(1)$ again, we see that $f(f(f(1))) = f(f(1)) = f(1) = 1$. Now plugging in $x = 1$ we get $f(f(y)) = \\frac{1}{y}$. Such a function can not be continuous as if $f(x)$ is increasing or decreasing on some observable interval, $f(f(x))$ will be increasing on that interval but $\\frac{1}{x}$ is decreasing on all positive intervals. This indicates the function is discrete which means we can assign a \"4-chain\" of $f(a) = b$, $f(b) = \\frac{1}{a}$, $f(\\frac{1}{a}) = \\frac{1}{b}$, and $f(\\frac{1}{b}) = a$. It is obvious to see that any function where $f(f(x)) = \\frac{1}{x}$. Must follow such a structure. The problem occurs that we do not know whether our current value of $x$ is a $b$ or an $a$. To make non-trivial progress on this we must split all rational numbers into two sets, $S_1$ and $S_2$, such that if $E \\in S_1$ then $\\frac{1}{E} \\in S_1$. As long as $S_1$ and $S_2$ have the same size, so the bijection $f(\\textrm{an element from }S_1)$ = $(\\textrm{an element from } S_2)$ can hold, there will exist a piece-wise function (possibly with a greater than countable infinity number of pieces) that satisfies f(x).\nTo find one of these functions we can limit our searching by proving that $f(ab) = f(a)f(b)$. To do this simply set $x = \\frac{1}{f(y)}$ yielding:\n\\newline\n$f(1) = f(f(y) \\times \\frac{1}{f(y)}) = \\frac{f(\\frac{1}{f(y)})}{y} = 1$. This means that $f(x) = y$. Since $X,Y \\in Q+$ there will always be an $x$ that suffices this and a corresponding $y$. Thus $f(ab)$ with $b = f(c)$ $\\leftrightarrow$ $c = \\frac{1}{f(b)}$ expands into $f(af(c)) = \\frac{f(a)}{c} = f(a)f(b)$.\nUtilizing $f(ab) = f(a)f(b)$ we can see that if we break-down every ratio into it's prime factorization of the numerator and denominator. Our \"4-chain\" is simply $f(p) = q$, $f(q) = \\frac{1}{p}$, $f(\\frac{1}{p}) = \\frac{1}{q}$, $f(\\frac{1}{q}) = p$ where $p$ and $q$ are unique primes. Thus our 2 sets $S_1$ and $S_2$ is simply 2 sets of equal size containing only primes and their inverse. A good way to do this assuming we aren't given the sequence of all infinite primes (as this would require an explicit formula that does not yet exist) is to split the primes into 2 and 1 mod 6 being in one set with 3 and 5 mod 6 in the other. These sets have equal sizes from a strong form of Dirichlet's theorem on arithmetic progressions.\nThus we have created our two sets and to show that this actually works we will let $x = P_1^{E_{1x}} \\times P_2^{E_{2x}} \\times P_3^{E_{3x}} \\times \\dots$ where $P_i$ is a prime and $E_{ix}$ are integers. Define $y$ similarly. For the sake of simplicity if $i$ is odd then $P_i \\in S_1$ and if $i$ is even $P_i \\in S_2$. Where our \"4-chain\" is $f(p_n) = p_{n+1}$, $f(p_{n+1}) = \\frac{1}{p_n}$, $f(\\frac{1}{p_n}) = \\frac{1}{p_{n+1}}$, $f(\\frac{1}{p_{n+1}}) = p_n$ with $n$ being odd. Then $f(y) = P_2^{E_{1y}} \\times P_1^{-E_{2y}} \\times P_4^{E_{3y}} \\times P_3^{-E_{4y}}\\dots$ (remember $f(ab) = f(a)f(b)$).\nThen $xf(y) = P_1^{E_{1x} - E_{2y}} \\times P_2^{E_{2x} + E_{1y}} \\times P_3^{E_{3x} - E_{4y}} \\times \\dots$\n$f(xf(y)) = P_2^{E_{1x} - E_{2y}} \\times P_1^{-E_{2x} - E_{1y}} \\times P_4^{E_{3x} - E_{4y}} \\times P_3^{-E_{4x} - E_{3y}} \\dots$\n$f(x) = P_2^{E_{1x}} \\times P_1^{-E_{2x}} \\times P_4^{E_{3x}} \\times P_3^{-E_{4x}}\\dots$\n$\\frac{f(x)}{y} = P_2^{E_{1x}- E_{2y}} \\times P_1^{-E_{2x} - E_{1y}} \\times P_4^{E_{3x} - E_{4y}} \\times P_3^{-E_{4x} - E_{3y}}\\dots$\nThus $f(xf(y)) = \\frac{f(x)}{y}$ QED" }, { "Problem": "In the plane let $C$ be a circle, $l$ a line tangent to the circle $C$, and $M$ a point on $l$. Find the locus of all points $P$ with the following property: there exists two points $Q$, $R$ on $l$ such that $M$ is the midpoint of $QR$ and $C$ is the inscribed circle of triangle $PQR$.", "Solution": "Note: This is an alternate method to what it is shown on the video. This alternate method is too long and too intensive in solving algebraic equations. A lot of steps have been shortened in this solution. The solution in the video provides a much faster solution,\nLet $r$ be the radius of the circle $C$.\nWe define a cartesian coordinate system in two dimensions with the circle center at $(0,0)$ and circle equation to be $x^{2}+y{2}=r^{2}$\nWe define the line $l$ by the equation $y=-r$, with point $M$ at a distance $m$ from the tangent and cartesian coordinates $(m,-r)$\nLet $d$ be the distance from point $M$ to point $R$ such that the coordinates for $R$ are $(m+d,-r)$ and thus the coordinates for $Q$ are $(m-d,-r)$\nLet points $S$, $T$, and $U$ be the points where lines $PQ$, $PR$, and $l$ are tangent to circle $C$ respectively.\nFirst we get the coordinates for points $S$ and $T$.\nSince the circle is the incenter we know the following properties:\n$\\left| RU \\right| = \\left| RT \\right|=(m+d)$\nand\n$\\left| QU \\right| = \\left| QS \\right|=(m-d)$\nTherefore, to get the coordinates of point $T=(T_{x},T_{y})$, we solve the following equations:\n$T_{x}^{2}+T_{y}^2=r^{2}$\n$\\left| RT \\right|^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2$\n$(m+d)^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2$\nAfter a lot of algebra, this solves to:\n$T_{x}=\\frac{2r^{2}(m+d)}{(m+d)^{2}+r^{2}}$\n$T_{y}=\\frac{r\\left[ (m+d)^{2}-r^{2} \\right]}{(m+d)^{2}+r^{2} }$\nNow we calculate the slope of the line that passes through $PR$ which is perpendicular to the line that passes from the center of the circle to point $T$ as follows:\n$Slope_{PR}=\\frac{-T_{x}}{T_{y}}=\\frac{-2r(m+d)}{(m+d)^{2}-r^2)}$\nThen, the equation of the line that passes through $PR$ is as follows:\n$Line_{PR}\\colon \\; y+r=\\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\\left( x-(m+d) \\right)$\nNow we get the coordinates of point $S=(S_{x},S_{y})$, we solve the following equations:\n$S_{x}^{2}+S_{y}^2=r^{2}$\n$\\left| QT \\right|^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2$\n$(m-d)^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2$\nAfter a lot of algebra, this solves to:\n$S_{x}=\\frac{2r^{2}(m-d)}{(m-d)^{2}+r^{2}}$\n$S_{y}=\\frac{r\\left[ (m-d)^{2}-r^{2} \\right]}{(m-d)^{2}+r^{2} }$\nNow we calculate the slope of the line that passes through $PQ$ which is perpendicular to the line that passes from the center of the circle to point $S$ as follows:\n$Slope_{PQ}=\\frac{-S_{x}}{S_{y}}=\\frac{-2r(m-d)}{(m-d)^{2}-r^2)}$\nThen, the equation of the line that passes through $PQ$ is as follows:\n$Line_{PQ}\\colon \\; y+r=\\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\\left( x-(m-d) \\right)$\nNow we solve for the coordinates for point $P=(P_{x},P_{y})$ by calculating the intersection of $Line_{PR}$ and $Line_{PQ}$ as follows:\n$\\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\\left( P_{x}-(m+d) \\right)=\\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\\left( P_{x}-(m-d) \\right)$\nSolving for $P_{x}$ we get:\n$P_{x}=\\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}$\nSolving for $P_{y}$ we get:\n$P_{y}=\\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\\left( \\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}-(m-d) \\right)-r$\n$P_{y}=\\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}$\nNow we need to find the limit of $P_{x}$ and $P_{y}$ as $d$ approaches infinity:\n$P_{x_{d \\to \\infty}}=\\lim_{d \\to \\infty} \\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}=0$\n$P_{y_{d \\to \\infty}}=\\lim_{d \\to \\infty} \\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}=r$\nThis means that the locus of $P$ starts at point $(0,r)$ on the circle $C$ but that point is not included in the locus as that is the limit.\nIf we assume that the locus is a ray that starts at $(0,r)$ let's calculate the slope of such ray:\n$Slope_{locus}=\\frac{P_{y}-r}{P_{x}}$\n$Slope_{locus}=\\frac{\\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}-r}{\\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}}$\n$Slope_{locus}=\\frac{-r^{2}}{2m}$\nSince the calculated slope of such locus at any point $P$ is not dependent on $d$ and solely dependent on fixed $r$ and $m$, then this proves the slope is fixed and thus the locus is a ray that starts at $(0,r)$ excluding that point and with a slope of $\\frac{-r^{2}}{2m}$ in the cartesian coordinate system moving upwards to infinity.\nWe can also write the equation of the locus as: $y=\\frac{-r^{2}}{2m}x+r,\\;\\;\\forall y>r\\;$and $x<0$\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": "For three points $A,B,C$ in the plane, we define $m(ABC)$ to be the smallest length of the three heights of the triangle $ABC$, where in the case $A$, $B$, $C$ are collinear, we set $m(ABC) = 0$. Let $A$, $B$, $C$ be given points in the plane. Prove that for any point $X$ in the plane,\n$m(ABC) \\leq m(ABX) + m(AXC) + m(XBC)$.\n", "Solution": "First we prove the claim for all points $X$ within or on the triangle $ABC$: In this specific case, suppose without loss of generality that $a\\ge b\\ge c$. Then the length of the smallest height of $ABC$ is that of the height from vertex $A$. This has a value of $\\frac{2[ABC]}{a}$. Similarly, for triangles $ABX,BCX,CAX$ we have that the length of the smallest height is twice the area of the respective triangle divided by the longest side of that triangle. It is clear that since $a$ is the longest side of $ABC$, no two points within $ABC$ have distance exceeding $a$. Thus, since $X$ is within $ABC$, the longest side of any of the triangles $ABX,BCX,CAX$ does not exceed $a$. So, we have\n$m(ABX)+m(BCX)+m(CAX)\\ge\\frac{2[ABX]}{a}+\\frac{2[BCX]}{a}+\\frac{2[CAX]}{a}$\n$=\\frac{2[ABC]}{a}=m(ABC)$\nas desired.\nNow on to prove the assertion for $X$ outside the triangle $ABC$. We shall assume without loss of generality that out of the points $A,B,C$ point $A$ is that farthest from $X$.\nIf $ABCX$ is concave or a degenerate quadrilateral, assume without loss of generality that $B$ inside or on triangle $ACX$. We shall prove that $m(ACX)\\ge m(ABC)$ to prove the main claim. If the shortest height of $ACX$ was from vertex $A$ then it is clear that the height of $ABC$ from $A$ is smaller than that since ray $CB$ is closer to $A$ than ray $CX$. The case of the height from $C$ being the smallest of triangle $ACX$ is analogous, so we move on to the case of the height from $X$ is the smallest. If this is the case, then it is clear that the height of $ABC$ from $B$ is smaller than $m(ABC)$ since $\\angle XAC\\ge \\angle BAC$. Thus the claim is proved.\nIf, instead, $ABCX$ is convex, we can assign the letter $D$ to represent to point of intersection of $AX$ and $BC$. Before proving the main claim, we shall prove that for triangle $ABX$ we have $m(ABX)\\ge m(ABD)$. We prove this by considering each of the vertices that the shortest height of $ABX$ is on. If the shortest height is that from $X$, then it is obvious that the height from $D$ to side $AB$ is smaller than that from $X$ since $D$ is closer to $A$ than $X$ is and so $m(ABX)\\ge m(ABD)$. If the shortest height of $ABX$ was from $B$, then since the height from $B$ to $XD$ is equal to the height from $B$ to $AD$, we have $m(ABX)\\ge m(ABD)$. If instead the shortest height of $ABX$ was from $A$, the it is clear that $\\angle B<90$. Thus, the projection of $A$ onto $BD$ is on the same side of $B$ as $C$. Now it is obvious that the height of $ABD$ from $A$ is smaller than that of $ABX$ from $A$ since the ray $BD$ is closer to $A$ than the ray $BX$. Thus, we have $m(ABX)\\ge m(ABD)$ in all cases. Notice that the case for triangle $ACX$ is analogous. Therefore, we have $m(ABX)+m(ACX)\\ge m(ABD)+m(ACD)$. But, this is the case of $D$ on the triangle $ABC$, and this case was shown in the first part of this proof.\nSo, we have that $m(ABC) \\leq m(ABX) + m(AXC) + m(XBC)$ is true in all cases, as desired." }, { "Problem": "Find all ordered pairs $(m,n)$ where $m$ and $n$ are positive integers such that $\\frac {n^3 + 1}{mn - 1}$ is an integer.", "Solution": "Suppose $\\frac{n^3+1}{mn-1}=k$ where $k$ is a positive integer. Then $n^3+1=(mn-1)k$ and so it is clear that $k\\equiv -1\\pmod{n}$. So, let $k=jn-1$ where $j$ is a positive integer. Then we have $n^3+1=(mn-1)(jn-1)=mjn^2-(m+j)n+1$ which by cancelling out the $1$s and dividing by $n$ yields $n^2=mjn-(m+j)\\Rightarrow n^2-mjn+m+j=0$. The equation $x^2-mjx+m+j=0$ is a quadratic. We are given that $n$ is one of the roots. Let $p$ be the other root. Notice that since $n+p=mj$ we have that $p$ is an integer, and so from $np=m+j$ we have that $p$ is positive.\nIt is obvious that $j=m=n=p=2$ is a solution. Now, if not, and $j,m,n,p$ are all greater than $1$, we have the inequalities $np>n+p$ and $mj>m+j$ which contradicts the equations $np=m+j, n+p=mj$. \nThus, at least one of $j,m,n,p$ is equal to $1$.\nIf one of $m,j$ is $1$, without loss of generality assume it is $j$. Then we have $np=m+1, n+p=m$. That is, $np-n-p=1\\Rightarrow (n-1)(p-1)=2$ which gives positive solutions $(n,p)=(3,2),(2,3)$. These give $m=5$ and since we assumed $j=1$, we can also have $m=1$ and $j=5$.\nIf one of $n,p$ is $1$, without loss of generality assume it is $p$. Then we have $n=m+j, n+1=mj$. That is, $mj-m-j=1\\Rightarrow (m-1)(j-1)=2$ which gives positive solutions $(m,j)=(3,2),(2,3)$. These give $n=5$ and since we assumed $p=1$, we can also have $n=1$ and $p=5$.\nFrom these, we have all solutions $(m,n)=(2,2),(5,3),(5,2),(1,3),(1,2),(3,5),(2,5),(3,1),(2,1)$." }, { "Problem": "The positive real numbers $x_0, x_1, x_2,.....x_{1994}, x_{1995}$ satisfy the relations\n$x_0=x_{1995}$ and $x_{i-1}+\\frac{2}{x_{i-1}}=2{x_i}+\\frac{1}{x_i}$\nfor $i=1,2,3,....1995$\nFind the maximum value that $x_0$ can have.", "Solution": "First we start by solving for $x_{i}$ in the recursive relation\n$x_{i-1}+\\frac{2}{x_{i-1}}=2x_{i}+\\frac{1}{x_{i}}$\n$\\frac{x_{i-1}^{2}+2}{x_{i-1}}=\\frac{2x_{i}^{2}+1}{x_{i}}$\n$\\left( x_{i} \\right)\\left( x_{i-1}^{2}+2 \\right)=\\left( x_{i-1} \\right)\\left( 2x_{i}^{2}+1 \\right)$\n$x_{i}x_{i-1}^{2}+2x_{i}=2x_{i-1}x_{i}^{2}+x_{i-1}$\n$x_{i}x_{i-1}^{2}-2x_{i-1}x_{i}^{2}+2x_{i}-x_{i-1}=0$\n$x_{i}x_{i-1}\\left( x_{i-1} \\right)-2x_{i}x_{i-1}\\left( x_{i} \\right)+2x_{i}-x_{i-1}=0$\n$x_{i}x_{i-1}\\left( x_{i-1} -2x_{i}\\right)+2x_{i}-x_{i-1}=0$\n$\\left( 2x_{i}-x_{i-1}\\right)\\left( 1-x_{i}x_{i-1} \\right)=0$\n$x_{i}=\\frac{x_{i-1}}{2},\\;$ or $x_{i}=\\frac{1}{x_{i-1}}$\nSo we have two recursive properties to chose from.\nIf we want to maximize $x_{0}$ then we can use $x_{i}=\\frac{x_{i-1}}{2}$ from $i=1$ to $1994$. This will make $x_{0}$ the largest and $x_{1994}$ the smallest.\nThen we can simply use $x_{i}=\\frac{1}{x_{i-1}}$ to get $x_{1995}$ since the reciprocal will make it very large.\nThen we use $x_{0}=x_{1995}$ and solve for $x_{0}$\nThis means that we can write $x_{i}$ as:\n$x_{i}=\\begin{cases} \\frac{x_{0}}{2^{i}} & 1 \\le i \\le 1994 \\\\ \\frac{1}{x_{i-1}} & i=1995\\end{cases}$\nThen $x_{1994}=\\frac{x_0}{2^{1994}}$,\nthus $x_{1995}=\\frac{1}{x_{1994}}=\\frac{2^{1994}}{x_0}={x_0}$\nSolving for ${x_0}$ we get:\n${x_0}^{2}=2^{1994}$\n${x_0}=\\pm\\sqrt{2^{1994}}=\\pm2^{997}$. Since all ${x_i}$ are defined as positive, ${x_0}=2^{997}$.\nTherefore, the maximum value that $x_0$ can have is ${x_0}=2^{997}$\n~ Tomas Diaz. orders@tomasdiaz.com\n\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\n" }, { "Problem": "The positive integers $a$ and $b$ are such that the numbers $15a+16b$ and $16a-15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "An $n \\times n$ matrix whose entries come from the set $S={1,2,...,2n-1}$ is called a $\\textit{silver}$ matrix if, for each $i=1,2,...,n$, the $i$th row and the $i$th column together contain all elements of $S$. Show that\n(a) there is no $\\textit{silver}$ matrix for $n=1997$;\n(b) $\\textit{silver}$ matrices exist for infinitely many values of $n$.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Determine all pairs $(n,p)$ of positive integers such that\n$p$ is a prime, $n$ not exceeded $2p$, and $(p-1)^{n}+1$ is divisible by $n^{p-1}$", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "A magician has one hundred cards numbered $1$ to $100$. He puts them into three boxes,\na red one, a white one and a blue one, so that each box contains at least one card.\nA member of the audience selects two of the three boxes, chooses one card from each\nand announces the sum of the numbers on the chosen cards. Given this sum, the magician\nidentifies the box from which no card has been chosen.\nHow many ways are there to put all the cards into the boxes so that this trick always\nworks? (Two ways are considered different if at least one card is put into a different box.)", "Solution": "Consider $1$, $2$ and $3$. If they are in different boxes, then $4$ must be in the\nsame box as $1$, $5$ in the same box as $2$ and so on. This leads to the solution where all\nnumbers congruent to each other mod $3$ are in the same box.\nSuppose $1$ and $2$ are in box $A$ and $3$ in box $B$. Then $4$ must be in box $A$ or $B$. In\ngeneral, if $k\\ge 4$ is in either box $A$ or $B$, then $k + 1$ also must be in box $A$ or $B$. Thus box\n$C$ is empty which is impossible.\nSimilarly, it is impossible for $1$ and $3$ to be in box $A$ and $2$ in box $B$.\nThus we are left with the case where $1$ is in box $A$ and $2$ and $3$ in box $B$. Suppose box\n$B$ contains $2, . . . k$, where $k \\ge 3$, but does not contain $k + 1$ and $m$ is the smallest number in box $C$. \nThen $m > k$.\nIf $m > k + 1$, then $k + 1$ must be box $A$ and we see that no box can\ncontain $m-1$.\nThus $m = k + 1$. If $k < 99$, we see that no box can contain $k + 2$. Thus\n$k = 99$. It is easy to see that this distribution works. Thus there altogether $12$ ways - $2$ options times permutation of $3$ colors for each of $3$ boxes.\nTaken from: https://sms.math.nus.edu.sg/Simo/IMO_Problems/00.pdf" }, { "Problem": "Let $n_1, n_2, \\dots , n_m$ be integers where $m>1$ is odd. Let $x = (x_1, \\dots , x_m)$ denote a permutation of the integers $1, 2, \\cdots , m$. Let $f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m$. Show that for some distinct permutations $a$, $b$ the difference $f(a) - f(b)$ is a multiple of $m!$.", "Solution": "Notice that if $\\{0,1,2,...,m!-1\\}\\not=\\{f(1),f(2),f(3),...,f(m!)\\}\\pmod{m!}$ then by the pigeon hole princible, there must be some $a,b\\in\\{1,2,...,m!\\}$ such that $f(a)\\equiv f(b)\\pmod{m!}$ as desired. Thus, we must prove that $\\{0,1,2,...,m!-1\\}\\not=\\{f(1),f(2),f(3),...,f(m!)\\}\\pmod{m!}$. Suppose there was such a situation. Then, since the two sets have the same number of elements, for every $t\\in\\{0,1,...,m!-1\\}$, there exists a $v\\in\\{1,2,...,m!\\}$ such that $t\\equiv f(v)\\pmod{m!}$. So, let $t\\equiv f(v_t)\\pmod{m!}$. Consider the sum $f(v_0)+f(v_1)+\\cdots+f(v_{m!-1})\\pmod{m!}$. Using the fact that $f(v_t)\\equiv t\\pmod{m!}$, we find the sum is congruent to $0+1+\\cdots+m!-1=\\frac{m!(m!-1)}{2}$.\nOn the other hand, using the fact that $f(x)=x_1n_1 + x_2n_2 + ... + x_mn_m$, we can combine the terms with the same coefficient ($n_i$). For each $n_1$, since all the $v_j$ are distinct, the coefficient in the sum would be $\\sum x_i$ over all permutations $x$ of $\\{1,2,...,m\\}$. Thus, the coefficient would be $\\sum_{i=1}^m i(m-1)!=\\frac{m!(m+1)}{2}$. Since $m$ is odd, $\\frac{m+1}{2}$ is an integer, so the coefficient is congruent to $0\\pmod{m!}$. Thus, the whole sum is congruent to $0\\pmod{m!}$. Therefore, $\\frac{m!(m!-1)}{2}\\equiv0\\pmod{m!}$. But, since $m>1$, we have $m!$ is even, and thus $m!-1$ is odd. Therefore, the integer $\\frac{m!(m!-1)}{2}$ has one factor of $2$ less than $m!$ does. Contradiction!" }, { "Problem": "Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, and $R$ be the feet of perpendiculars from $D$ to lines $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively. Show that $PQ=QR$ if and only if the bisectors of angles $ABC$ and $ADC$ meet on segment $\\overline{AC}$.", "Solution": "Clearly $PQR$ is the Simson Line and $APDQ$, $BPDR$, $CQDR$ is cyclic. By angle chasing we have $\\triangle DPQ\\sim\\triangle DBC$, $\\triangle DQR\\sim\\triangle DAB$. Then by $PQ=QR$ we have $\\frac{DC}{CB}=\\frac{DQ}{QP}=\\frac{DQ}{QR}=\\frac{DA}{AB}$. Rearranging and using the angle bisector theorem we are done." }, { "Problem": "(Hojoo Lee) Let $n \\geq 3$ be an integer. Let $t_1, t_2, \\dots , t_n$ be positive real numbers such that\n\nShow that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \\leq i < j < k \\leq n$.", "Solution": "For $n=3$, suppose (for sake of contradiction) that $t_3 = t_2 + t_1 + k$ for $k \\ge 0$; then (by Cauchy-Schwarz Inequality)\n\nso it is true for $n=3$. We now claim the result by induction; for $n \\ge 4$, we have\n\nBy AM-GM, $\\frac{t_n}{t_i} + \\frac{t_i}{t_n} \\ge 2$, so $f(n) \\ge f(n-1) + 2(n-1) + 1 = f(n-1) + 2n - 1$. Then the problem is reduced to proving the statement true for $n-1$ numbers, as desired.$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\square$" }, { "Problem": "Determine all positive integers relatively prime to all the terms of the infinite sequence", "Solution": "Let $k$ be a positive integer that satisfies the given condition.\nFor all primes $p>3$, by Fermat's Little Theorem, $n^{p-1} \\equiv 1\\pmod p$ if $n$ and $p$ are relatively prime. This means that $n^{p-3} \\equiv \\frac{1}{n^2} \\pmod p$. Plugging $n = p-3$ back into the equation, we see that the value$\\mod p$ is simply $\\frac{2}{9} + \\frac{3}{4} + \\frac{1}{36} - 1 = 0$. Thus, the expression is divisible by all primes $p>3.$ Since $a_2 = 48 = 2^4 \\cdot 3,$ we can conclude that $k$ cannot have any prime divisors. Therefore, our answer is only $1.$" }, { "Problem": "Determine all pairs $(x, y)$ of integers such that\n", "Solution": "If $(x,y)$ is a solution then obviously $x\\geq 0$ and $(x,-y)$ is a solution too. For $x=0$ we get the two solutions $(0,2)$ and $(0,-2)$.\nNow let $(x,y)$ be a solution with $x > 0$; without loss of generality confine attention to $y > 0$. The equation rewritten as shows that the factors $y-1$ and $y+1$ are even, exactly one of them divisible by $4$. Hence $x\\geq 3$ and one of these factors is divisible by $2^{x-1}$ but not by $2^x$. So\n\nPlugging this into the original equation we obtain\n\nor, equivalently Therefore For $\\varepsilon = 1$ this yields $m^2 - 8 < 0$, i.e. $m=1$, which fails to satisfy $(\\dagger)$. For $\\varepsilon = -1$ equation $(\\dagger)$ gives us implying $2m^2 - m - 17 \\leq 0$. Hence $m\\leq 3$; on the other hand $m$ cannot be $1$ by $(\\dagger)$. Because $m$ is odd, we obtain $m=3$, leading to $x=4$. From $(*)$ we get $y=23$. These values indeed satisfy the given equation. Recall that then $y = -23$ is also good. Thus we have the complete list of solutions $(x,y)$: $(0,2)$, $(0,-2)$, $(4,23)$, $(4,-23)$." }, { "Problem": "In $\\triangle ABC$ the bisector of $\\angle{BCA}$ intersects the circumcircle again at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area.", "Solution_1_.28Efficient.29": "$\\angle{RQL} = 90+\\angle{QCL} = 90+\\dfrac{C}{2}$, and similarly $\\angle{RPK} = 90+\\angle{PCK} = 90+\\dfrac{C}{2}$. Therefore, $\\angle{RQL} = \\angle{RPK}$. Using the triangle area formula $A = \\dfrac{1}{2}bc\\sin{\\angle{A}}$ yields $RQ \\cdot QL = RP \\cdot PK = \\dfrac{PK}{QL} = \\dfrac{RQ}{RP}$ after cancelling the sines and constant. Draw line $QD$ perpendicular to $BC$ that intersects $BC$ at $D$, then $QD=QL$ because the perpendicular bisectors are congruent, (or alternatively $\\triangle QDC\\cong\\triangle QLC$). This presents us $\\dfrac{PK}{QL}=\\dfrac{PK}{QD}=\\dfrac{PC}{QC}$ by similar triangles; now, we have only to prove $\\dfrac{PC}{QC}=\\dfrac{RQ}{RP}$, or $RQ \\cdot QC=RP \\cdot PC$.\nSince $\\angle{OPQ} =1 80-\\angle{RPK} =1 80-\\angle{RQL} = \\angle{OQP}$, we have $\\triangle OPQ$ is isosceles. Draw the perpendicular from $O$ to $RC$, intersecting at $E$. Then $PE = QE = x$ for a real $x$, now because the perpendicular from the center of a circle to a chord bisects that chord, $RE = CE$. Let $y = RE$, and then $RQ \\cdot QC = (y+x) \\cdot (y-x) = PC \\cdot RP$, proving our claim.", "Solution_2": "The area of $\\triangle{RQL}$ is given by $\\dfrac{1}{2}QL*RQ\\sin{\\angle{RQL}}$ and the area of $\\triangle{RPK}$ is $\\dfrac{1}{2}RP*PK\\sin{\\angle{RPK}}$. Let $\\angle{BCA}=C$, $\\angle{BAC}=A$, and $\\angle{ABC}=B$. Now $\\angle{KCP}=\\angle{QCL}=\\dfrac{C}{2}$ and $\\angle{PKC}=\\angle{QLC}=90$, thus $\\angle{RPK}=\\angle{RQL}=90+\\dfrac{C}{2}$. $\\triangle{PKC} \\sim \\triangle{QLC}$, so $\\dfrac{PK}{QL}=\\dfrac{KC}{LC}$, or $\\dfrac{PK}{QL}=\\dfrac{BC}{AB}$. The ratio of the areas is $\\dfrac{[RPK]}{[RQL]}=\\dfrac{BC*RP}{AC*RQ}$. The two areas are only equal when the ratio is 1, therefore it suffices to show $\\dfrac{RP}{RQ}=\\dfrac{AC}{BC}$. Let $O$ be the center of the circle. Then $\\angle{ROK}=A+C$, and $\\angle{ROP}=180-(A+C)=B$. Using law of sines on $\\triangle{RPO}$ we have: $\\dfrac{RP}{\\sin{B}}=\\dfrac{OR}{\\sin{(90+\\dfrac{C}{2})}}$ so $RP*\\sin{(90+\\dfrac{C}{2})}=OR*\\sin{B}$. $OR*\\sin{B}=\\dfrac{1}{2}AC$ by law of sines, and $\\sin{(90+\\dfrac{C}{2})}=\\cos{\\dfrac{C}{2}}$, thus 1) $2RP\\cos{\\dfrac{C}{2}}=AC$. Similarly, law of sines on $\\triangle{ROQ}$ results in $\\dfrac{RQ}{\\sin{(180-A)}}=\\dfrac{OR}{\\sin{(90-\\dfrac{C}{2})}}$ or $\\dfrac{RQ}{\\sin{A}}=\\dfrac{OR}{\\cos{\\dfrac{C}{2}}}$. Cross multiplying we have $RQ\\cos{\\dfrac{C}{2}}=OR*\\sin{A}$ or 2) $2RQ\\cos{\\dfrac{C}{2}}=BC$. Dividing 1) by 2) we have $\\dfrac{RP}{RQ}=\\dfrac{AC}{BC}$ $\\square$\n$(tkhalid)$", "Solution_3": "(Image Link) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi85LzQ5OTJlZGNhYTQ0YjJjODcxMTBmZGNmMTdiZDdkMGRjZGUyOWQ5LnBuZw==&rn=U2NyZWVuIFNob3QgMjAxOS0wOC0wOCBhdCAxMi4yMC4zOCBQTS5wbmc=\nWLOG, let the diameter of $(ACBR)$ be $1.$\nWe see that $PK = \\dfrac{1}{2}a \\tan \\dfrac{1}{2}C$ and $QL = \\dfrac{1}{2}b \\tan \\dfrac{1}{2}C$ from right triangles $\\triangle PKC$ and $\\triangle QLC.$\nWe now look at $AR.$ By the Extended Law of Sines on $\\triangle ACR,$ we get that $AR = \\sin\\frac{1}{2}C.$ Similarly, $BR = \\sin \\frac{1}{2}C.$\nWe now look at $CR.$ By Ptolemy's Theorem, we have which gives us This means that We now seek to relate the lengths computed with the areas.\nTo do this, we consider the altitude from $R$ to $PK.$ This is to find the area of $RPK.$ Finding the area of $\\triangle RQL$ is similar.\nWe claim that $RF = \\dfrac{1}{2}b.$ In order to prove this, we will prove that $\\triangle RFP \\cong \\triangle QLC.$ In other words, we wish to prove that $PR = QC.$ This is equivalent to proving that $PC + QC = CR.$\nNote that $PC = \\dfrac{PK}{\\sin \\frac{1}{2}C}$ and $QC = \\dfrac{QL}{\\sin \\frac{1}{2}C}.$ Therefore, we get that \n\n\n\n\n\n\n\n\n\n\nThus, $RF = \\dfrac{1}{2}b.$ In this way, we get that the altidude from $R$ to $QL$ has length $\\dfrac{1}{2}a.$ Therefore, we see that $[RPK] = \\dfrac{1}{8}ab \\tan \\frac{1}{2}C$ and $[RQL] = \\dfrac{1}{8}ab \\tan \\frac{1}{2}C,$ so the two areas are equal.\nSolution by Ilikeapos", "Solution_4": "$[\\triangle{RPK}]=[\\triangle{RQL}], LQ*RQ*\\sin\\angle{LQR}=KP*PR*\\sin\\angle{RPK}$\nSince $CR$ bisects $\\angle{ACB},\\angle{QCL}=\\angle{PCK}$, $OL,OK$ are perpendicular to sides $AC,BC$ separately, $\\angle{QLC}=\\angle{PKC}=90^{\\circ}, \\angle{CQL}=\\angle{CPK}, \\angle{LQR}=\\angle{RPK}$\nSo now, we only have to prove $RP*PK=LQ*QR$, which is $RP*(CQ+QP)*\\cos\\angle{CPK}=CQ*(QP+PR)*\\cos\\angle{CQL}$, as mentioned above, the two angles are the same, we have to prove that $RP(CQ+QP)=CQ(QP+PR)$, which is equivalent to $RP*QP=QP*CQ$, we have to prove $RP=CQ$\nNow notice that $\\triangle{OQP}$ is isosceles. $\\angle{CQL}=\\angle{OQP}=\\angle{OPQ}$, construct $OJ \\bot CR$, $CJ=JR,JQ=JP,CQ=PR$ as desired\n~bluesoul\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Find all functions $f: (0, \\infty) \\mapsto (0, \\infty)$ (so $f$ is a function from the positive real numbers) such that\nfor all positive real numbers $w,x,y,z,$ satisfying $wx = yz.$\n", "Solution": "Considering $w=1$ and $z=y=\\sqrt{x}$ which satisfy the constraint $wx=yz$ we get the following equation:\n\n\n\n\nAt once considering $x=1$ we get $(f(1))^2 = f(1)$ and knowing that $f : \\mathbb{R}^+ \\rightarrow \\mathbb{R}^+$ the only possible solution is $f(1)=1$ since $f(1)=0$ is impossible.\n\nSo we get the quadratic equation:\n\n\n\nSolving for $f(x)$ as a function of $x$ we get:\n\n\n\nAt once we see that for one value of $x$, $f(x)$ can only take one of 2 possible values:\n\n.\n\nTake into consideration that $f(2) = 2$ but $f(3) = \\frac{1}{3}$ verifies the quadratic equation and thus so far we can't say that $f(x)=x \\, \\forall_{x \\in \\mathbb{R}^+}$ or alternatively $f(x)=\\frac{1}{x} \\, \\forall_{x \\in \\mathbb{R}^+}$. This is indeed the case but we haven't proved it yet.\n\nTo prove the previous assertion consider 2 values $a,b \\in \\mathbb{R}^+$ such that $a\\ne 1 \\wedge b \\ne 1 \\wedge a \\ne b$ while having $f(a) = a \\wedge f(b)=\\frac{1}{b}$\n\nConsider now the original functional equation with $w=a,\\ x=b,\\ y=z=\\sqrt{ab}$ which verifies the constraint. Substituting we have:\n\n\n\nNow either $f(ab)=ab$ or $f(ab)=\\frac{1}{ab}$. (notice that $ab \\ne b \\wedge ab\\ne b$ by hypothesis)\n\n\n\nIf $f(ab)=ab$ then we have $ab = ab\\frac{a^2 + \\frac{1}{b^2}}{a^2 + b^2} \\Leftrightarrow b^4=1$ and since $b>0$ the only solution is $b=1$.\n\nIf $f(ab)=\\frac{1}{ab}$ then we have $\\frac{1}{ab} = ab\\frac{a^2 + \\frac{1}{b^2}}{a^2 + b^2} \\Leftrightarrow a^2+b^2 = a^4b^2 +a^2 \\Leftrightarrow a^4=1$ and since $a>0$ the only solution is $a=1$.\n\nSo the only solutions are $a=1$ or $b=1$ in which case both alternatives imply $f(1)=1$. Thus we conclude that solutions to the functional equation are a subset of $\\left\\{f(x)=x \\ \\forall_{x \\in \\mathbb{R}^+},\\ f(x)=\\frac{1}{x}\\ \\forall_{x \\in \\mathbb{R}^+} \\right\\}$.\n\nFinally, plug each of these 2 functions into the functional equation and verify that they indeed are solutions.\n\nThis is trivial since $f(x)=x$ is an obvious solution and for $f(x)=\\frac{1}{x}$ we have:\n\nprovided that $(wx)^2 = (yz)^2$ which verifies the original constraint.\n\nSo the functional equation has 2 solutions:\n" }, { "Problem": "Let $ABC$ be a triangle with $AB=AC$. The angle bisectors of $\\angle CAB$ and $\\angle ABC$ meet the sides $BC$ and $CA$ at $D$ and $E$, respectively. Let $K$ be the incentre of triangle $ADC$. Suppose that $\\angle BEK=45^\\circ$. Find all possible values of $\\angle CAB$.\nAuthors: Jan Vonk and Peter Vandendriessche, Belgium, and Hojoo Lee, South Korea", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $P$ be a point interior to triangle $ABC$ (with $CA \\neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.", "Solution_1": "Without loss of generality, suppose that $AS > BS$. By Power of a Point, $SP^2 = SC^2 = SB \\cdot SA$, so $\\overline{SP}$ is tangent to the circumcircle of $\\triangle ABP$. Thus, $\\angle KPS = 180 - \\angle SPA = \\widehat{AP}/2 = \\angle ABP$. It follows that after some angle-chasing,\nso $ML = MK$ as desired.", "Solution_2": "Let the tangent at $M$ to $\\Gamma$ intersect $SC$ at $X$. We now have that since $\\triangle{XMC}$ and $\\triangle{SPC}$ are both isosceles, $\\angle{SPC}=\\angle{SCP}=\\angle{XMC}$. This yields that $MX \\| PS$.\nNow consider the power of point $S$ with respect to $\\Gamma$.\n\nHence by AA similarity, we have that $\\triangle{SPA} \\sim \\triangle{SBP}$. Combining this with the arc angle theorem yields that $\\angle{SPA}=\\angle{SBP}=\\angle{PKL}$. Hence $PS \\| LK$.\nThis implies that the tangent at $M$ is parallel to $LK$ and therefore that $M$ is the midpoint of arc $LK$. Hence $MK=ML$.\n", "Solution_3": "Since $SP = SC$, we can construct circle $\\omega$ tangent to $SP$ and $SC$ at $P$ and $C$ respectively. Then, we see that the circumcircle of $\\triangle ABP$ must be tangent to $SP$ at $P$ by the converse of the radical axis theorem, since $\\omega$ is internally tangent to $\\Gamma$ at $C$, and $(ABP)$ meets $\\Gamma$ at $A$ and $B$. Thus, $\\angle SPB = \\angle BAK = \\angle BLK$, so lines $SP$ and $KL$ are parallel.\nIt then follows that $\\angle SCP = \\angle SPC = \\angle KDC = \\angle KLC + \\angle LCD$. Since $SC$ is tangent to $\\Gamma$, we see that $\\angle KLC = \\angle SCK$, and thus $\\angle KCD = \\angle LCD = \\angle SCP - \\angle KLC$. Thus since $CM$ bisects $\\angle KCL$ in cyclic quad $CKML$, we have $MK = ML$. $\\blacksquare$" }, { "Problem": "Let $n > 0$ be an integer. We are given a balance and $n$ weights of weight $2^0,2^1, \\cdots ,2^{n-1}$. We are to place each of\tthe $n$ weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.\nDetermine the number of ways in which this can be done.", "Solution": "Call our answer $W(n)$.\nWe proceed to prove $W(n)=(2n-1)!!$.\nIt is evident $W(1)=1$.\nNow, the key observation is that smaller weights can never add up to the weight of a larger weight, ie which side is heavier is determined completely by the heaviest weight currently placed. It follows, therefore, that the number of ways to place $n$ weights on the balance according to the rule is the same no matter which $n$ distinct powers of two are the weights, as each weight completely overpowers any smaller weight and is completely overpowered by any larger weight. That is, there is the 1st heaviest weight, the 2nd heaviest, the 3rd, ..., the n-th heaviest, and each weight is negligible compared to any heavier weight. Thus, any valid placement of $n$ weights of weight $2^0,2^1, \\cdots ,2^{n-1}$ can changed by replacing $2^i$ with the $(n-i)$-th heaviest weight in the set ${2^{a_k}}$, where $a_k \\in \\mathbb{Z}$, and vice versa, forming a $1:1$ relation. With this in mind, we use recursion upon the last weight placement. There are $2n-1$ choices; namely, you can put any weight on either side except for the heaviest weight on the right. For the first $n-1$ weight placements, the answer reduces to $W(n-1)$. We can reduce $W(n-1)$ in the same way.\n$W(n)=(2n-1)W(n-1)=(2n-1)(2n-3)W(n-2)=...=(2n-1)!!W(1)=(2n-1)!!$\n$\\text{QED}$" }, { "Problem": "Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\\omega_1$ such that $WX$ is a diameter of $\\omega_1$. Analogously, denote by $\\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\\omega_2$. Prove that $X, Y$ and $H$ are collinear.", "Solution_1": "\nLet $T$ be the intersection of $\\omega_1$ and $\\omega_2$ other than $W$.\nLemma 1: $T$ is on $XY$.\nProof: We have $\\angle{XTW} = \\angle{YTW} = 90^\\circ$ because they intercept semicircles. Hence, $\\angle{XTY} = \\angle{XTW} + \\angle{YTW} = 180^\\circ$, so $XTY$ is a straight line.\nLemma 2: $T$ is on $AW$.\nProof: Let the circumcircles of $NBW$ and $MWC$ be $\\omega_1$ and $\\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\\angle{BNC} = \\angle{BMC} = 90^\\circ$), let its circumcircle be $\\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$.\nLemma 3: $YT$ is perpendicular to $AW$.\nProof: This is immediate from $\\angle{YTW} = 90^\\circ$.\nLet $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\\angle{ALB} = 90^\\circ.$\nLemma 4: Quadrilateral $THLW$ is cyclic.\nProof: We know that $NHLB$ is cyclic because $\\angle{BNH}$ and $\\angle{BLH}$, opposite and right angles, sum to $180^\\circ$. Furthermore, we are given that $NTWB$ is cyclic. Hence, by Power of a Point,\n\nThe converse of Power of a Point then proves $THLW$ cyclic.\nHence, $\\angle{WTH} = 180^\\circ - \\angle{WLH} = 90^\\circ$, and so $HT$ is perpendicular to $AW$ as well. Combining this with Lemma 3's statement, we deduce that $T, H, Y$ are collinear. But, as $X$ is on $YT$ (from Lemma 1), $X, Y, H$ are collinear. This completes the proof.\n$\\blacksquare$\n--Suli 13:51, 25 August 2014 (EDT)", "Solution_2": "Probably a simpler solution than above.\nAs above, let $T = \\omega_1 \\cap \\omega_2 \\neq W.$ By Miquel $MTHN$ is cyclic. Then since $\\angle WCY = \\angle WBX = 90^{\\circ}$ we know, because $W,B,X,T \\in \\omega_1$ and $W,C,Y,T \\in \\omega_2,$ that $\\angle WTY = \\angle WTX = 90^{\\circ},$ thus $X,T,Y$ are collinear.\nThere are a few ways to finish.\n(a) so $H,X,Y$ are collinear, as desired $\\square$\n(b) Since $BNMC$ is cyclic we know $AN \\cdot AB = AM \\cdot AC$ which means $p(A,\\omega_1) = p(A, \\omega_2)$ so $A$ is on the radical axis, $TW,$ hence so $H$ lies on this line as well and we may conclude $\\square$\n--mathguy623 03:10, 12 August 2016 (EDT)", "Solution_3_.28Complex_Bash.29": "\nFor any point $P,$ let $p$ denote the complex number for point $P.$ First off, let $C$ be the origin. Now, since $WY$ is the diameter of the circumcircle of triangle $CMW$, we must have $\\angle{YMW}=\\angle{YCH}=90^{\\circ}.$ Since angles inscribed in the same arc are congruent, $\\angle{MYW}=\\angle{MCW}=\\angle{C}.$ This means that $\\frac{|w-m|}{|y-m|}=\\tan{\\angle{C}}.$ Combining this with the fact that $\\angle{YMW}$ is right, we find that Solving, we find that We wish to simplify $(i\\tan{\\angle{C}}-1)m$ first. Note that This means that \n\nThis means that Now, since $WX$ is a diameter of the circumcircle of triangle $NBW,$ we must have $\\angle{WNX}=\\angle{WBX}=90^{\\circ}.$ Since angles inscribed in the same arc are congruent, $\\angle{NXW}=\\angle{NBW}=\\angle{B}.$ This means that $\\frac{|x-n|}{|w-n|}=\\cot{\\angle{B}}.$ Combining this with the fact that $\\angle{WNX}$ is right, we find that Solving, we find that We wish to simplify $n(1-i\\cot{\\angle{B}})$ first. Note that This means that \n\nThis means that This means that the line through complex numbers $x$ and $y$ satisfy the equation If there is a fixed point to the line, then the real and imaginary values of the point must not contin $w$. If the fixed point is $c,$ then we have after comparing the coefficient of $w.$ This means that \n\nFrom the Law of Sines, we find that Substituting this, we find that\n\nThis means that \n\nSince the $w$'s cancel out, we can just discard everything with $w$ in it. Thus,\n\nSince we have \nThus, In conclusion, which is the fixed point $XY$ always passes through. However, by inspection, $c=H=\\text{the orthocenter}.$ Therefore, we conclude that $X,Y,H$ are collinear for all acute triangles $ABC.$\n~pinkpig" }, { "Problem": "Points $P$ and $Q$ lie on side $BC$ of acute-angled $\\triangle{ABC}$ so that $\\angle{PAB}=\\angle{BCA}$ and $\\angle{CAQ}=\\angle{ABC}$. Points $M$ and $N$ lie on lines $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$, and $Q$ is the midpoint of $AN$. Prove that lines $BM$ and $CN$ intersect on the circumcircle of $\\triangle{ABC}$.", "Solution_1": "\nWe are trying to prove that the intersection of $BM$ and $CN$, call it point $D$, is on the circumcircle of triangle $ABC$. In other words, we are trying to prove $\\angle {BDC} + \\angle {BAC} = 180$. \nLet the intersection of $BM$ and $AN$ be point $E$, and the intersection of $AM$ and $CN$ be point $F$. \nLet us assume $\\angle {BDC} + \\angle {BAC} = 180$. Note: This is circular reasoning. If $\\angle {BDC} + \\angle {BAC} = 180$, then $\\angle {BAC}$ should be equal to $\\angle {BDN}$ and $\\angle {CDM}$. We can quickly prove that the triangles $ABC$, $APB$, and $AQC$ are similar, so $\\angle {BAC} = \\angle {AQC} = \\angle {APB}$. We also see that $\\angle {AQC} = \\angle {BQN} = \\angle {APB} = \\angle {CPF}$. Also because angles $BEQ$ and $NED, MFD$ and $CFP$ are equal, the triangles $BEQ$ and $NED$, $MDF$ and $FCP$ must be two pairs of similar triangles. Therefore we must prove angles $CBM$ and $ANC, AMB$ and $BCN$ are equal. \nWe have angles $BQA = APC = NQC = BPM$. We also have $AQ = QN$, $AP = PM$. Because the triangles $ABP$ and $ACQ$ are similar, we have $\\dfrac {EC}{EN} = \\dfrac {BF}{FM}$, so triangles $BFM$ and $NEC$ are similar. So the angles $CBM$ and $ANC, BCN$ and $AMB$ are equal and we are done.", "Solution_2": "Let $L$ be the midpoint of $BC$. Easy angle chasing gives $\\angle{AQP} = \\angle{APQ} = \\angle{BAC}$. Because $P$ is the midpoint of $AM$, the cotangent rule applied on triangle $MBA$ gives us\n\nHence, by the cotangent rule on $ABC$, we have\n\nBecause the period of cotangent is $180^\\circ$, but angles are less than $180^\\circ$, we have $\\angle{BAL} = \\angle{MBC}.$\nSimilarly, we have $\\angle{LAC} = \\angle{NCB}.$ Hence, if $BM$ and $CN$ intersect at $Z$, then $\\angle{BZC} = 180^\\circ - \\angle{BAC}$ by the Angle Sum in a Triangle Theorem. Hence, $BACZ$ is cyclic, which is equivalent to the desired result.\n--Suli 23:27, 7 February 2015 (EST)", "Solution_3": "Let $L$ be the midpoint of $BC$. By AA Similarity, triangles $BAP$ and $BCA$ are similar, so $\\dfrac{BA}{AP} = \\dfrac{BC}{CA}$ and $\\angle{BPA} = \\angle{BAC}$. Similarly, $\\angle{CQA} = \\angle{BAC}$, and so triangle $AQP$ is isosceles. Thus, $AQ = AP$, and so $\\dfrac{BA}{AQ} = \\dfrac{BC}{CA}$. Dividing both sides by 2, we have $\\dfrac{BA}{AN} = \\dfrac{BL}{AC}$, or\n\nBut we also have $\\angle{ABL} = \\angle{CAQ}$, so triangles $ABL$ and $NAC$ are similar by $SAS$ similarity. In particular, $\\angle{ANC} = \\angle{BAL}$. Similarly, $\\angle{BMA} = \\angle{CAL}$, so $\\angle{ANC} + \\angle{BMA} = \\angle{BAC}$. In addition, angle sum in triangle $AQP$ gives $\\angle{QAP} = 180^\\circ - 2\\angle{A}$. Therefore, if we let lines $BM$ and $CN$ intersect at $T$, by Angle Sum in quadrilateral $AMTN$ concave $\\angle{NTM} = 180^\\circ + \\angle{A}$, and so convex $\\angle{BTC} = 180^\\circ - \\angle{A}$, which is enough to prove that $BACT$ is cyclic. This completes the proof.\n\n--Suli 10:38, 8 February 2015 (EST)", "Solution_4": "Let $D_1$ be the second intersection of $NC$ with the circumcircle of $\\triangle ABC,$ and $D_2$ the second intersection of $MB$ with the circumcircle of $\\triangle ABC.$ By inscribed angles, the tangent at $C$ is parallel to $AN.$ Let $P_{\\infty}$ denote the point at infinity along line $AN.$ Note that So, $ABD_1C$ is harmonic. Similarly, we can find $ABD_2C$ is harmonic. Therefore, $D_1=D_2,$ which means that $BM$ and $CN$ intersect on the circumcircle. $\\blacksquare$", "Solution_5": "We use barycentric coordinates. Due to the equal angles, $AC$ is tangent to the circumcircle of $ABQ$ and $AB$ is tangent to the circumcircle of the $APC.$ Therefore, we can use power of a point to solve for side ratios. We have \nTherefore, $D=(-a^2:2b^2:2c^2),$ as $BM$ and $CN$ are cevians. Note that $(x,y,z)$ lies on the circumcircle iff $a^2yz+b^2xz+c^2xy=0.$ Substituting the values in, we have so we are done. $\\blacksquare$", "Solution_6": "Note that the givens immediately imply that $\\triangle{ABC} \\sim \\triangle{QAC} \\sim \\triangle{PBA}$, hence $\\angle{AQP}=\\angle{APQ}=\\angle{A}$. Let $D$ be the midpoint of BC, $E$ be the midpoint of $AC$, and $F$ the midpoint of $AB$. By the similar triangles, we have $\\angle{BAD}=\\angle{AQE}=\\angle{AMC}$. We also have $\\angle{BAD}=\\angle{BPF}=\\angle{MNB}$, so we find $\\angle{AMC}=\\angle{MNB}$. We note that $\\angle{AMC}+\\angle{CMN}=\\angle{AMN}=\\angle{AQP}=\\angle{A}$, so $\\angle{CMN}+\\angle{MNB}=A$, which gives that $\\angle{BKC}=180-\\angle{A}$ and we are done.\nAs an addition, $AK$ is the A-symmedian in $\\triangle{ABC}$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Triangle $ABC$ has circumcircle $\\Omega$ and circumcenter $O$. A circle $\\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\\Gamma$ and $\\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$.\nSuppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$.\nProposed by Silouanos Brazitikos and Evangelos Psychas, Greece", "Solution": "Lemma (On three chords). If two lines pass through different endpoints of two circles' common chord, then the other two chords cut by these lines on the circles are parallel.\nProof The second and the third chords are anti-parallel to the first (common) chord with respect to the given lines, so they are parallel to each other. $\\Box$\nTo solve this problem, it is sufficient to apply the lemma 5 times. Indeed, let the lines $FD, GE, FK, GL$ meet $\\Omega$ second time at $H, I, M, N$ respectively. One of the circles that figure in lemma is always $\\Omega$, while the other is one of three other circles from the problem statement. Applying the lemma to the lines $FDH$ and $GEI$, $FKM$ and $BDC$, $FDH$ and $BKA$, $GLN$ and $CEB$, $GEI$ and $CLA$, we get $DE \\parallel IH$, $KD \\parallel MC$, $KD \\parallel AH$, $LE \\parallel NB$, $LE \\parallel AI$, respectively. From this, $BC \\parallel IH$, $MC \\parallel AH$, $NB \\parallel AI$. Therefore, $AN=IB=HC=AM$. This means that $N$ and $M$ are symmetric wrt $AO$, a diameter of $\\Omega$ through $A$. So are $F$ and $G$, as $AF=AG$. Therefore, the lines $FM$ and $GN$ are symmetric wrt $AO$ and meet on it." }, { "Problem": "A set of positive integers is called fragrant if it contains at least two elements and each of its elements has a prime factor in common with at least one of the other elements. Let $P(n)=n^2+n+1$. What is the least possible positive integer value of $b$ such that there exists a non-negative integer $a$ for which the set $\\{P(a+1),P(a+2),\\ldots,P(a+b)\\}$ is fragrant?", "Solution": "Consider $P(x)$ and $P(x+y)$. We note that in order to $p \\mid P(x)$ and $p \\mid P(x+y)-P(x)$ we must have $p \\mid x^2+x+1$ and $p \\mid y(2x+y+1)$. It is obvious that $p \\equiv 1 \\pmod{6}$ since $p \\mid n^2+n+1 \\mid (2n+1)^2+3$ or $\\left( \\tfrac{-3}{p} \\right)=1$.\nIf $y=1$ then $p \\mid 2x+2$ implies $p \\mid x+1$. WLOG, we can let $x=p-1$ and see that $p \\nmid x^2+x+1$. So there doesn't exists $x$ so that $\\gcd \\left( P(x),P(x+1) \\right)>1$.\nIf $y=2$, we find $p \\mid 2x+3$ and we let $x=\\tfrac{p-3}{2}$. Hence, from $p \\mid x^2+x+1$ we get $p=7$. So there is one prime $p=7$ such that $\\gcd \\left( P(x),P(x+2) \\right)>1$.\nIf $y=3$, it is obvious that $p=3$ satisfy and is the only answer.\nIf $y=4$, we do the similar thing and get $p \\mid 2x+5$ and $p \\mid x^2+x+1$ so $p=19$.\n___________________________\nNow, back to the problem, since there doesn't exists any prime $p$ for $y=1$ so $b \\ge 3$. The only prime for $y=2$ is $p=7$ so if we choose $7 \\mid P(x+1), 7 \\mid P(x+3)$ then there will be a number $7 \\nmid P(x+2)$ remains. This means $b \\ge 5$ since we need a prime $p \\mid P(x+2)$ and $p \\mid P(x+y)$ but $y \\ge 5$ since $p \\ne 7$.\nIf $b=5$, we consider two cases, where there are two numbers that are divisible by $19$ (which means $19 \\mid P(x+1), 19 \\mid P(x+5)$), the middle-three numbers $P(x+2),P(x+3),P(x+4)$ we can't find a way make each two of them have common prime factor. If no two are divisible by $19$ then they can only be divisible by $7,3$, but it can't cover all $5$ \"consecutive\" numbers.\nIf $b=6$ then we can pick $19 \\mid P(x+2),P(x+6), 3 \\mid P(x+1), P(x+4), 7 \\mid P(x+3), 7 \\mid P(x+5)$.\nSo the final answer is $b=6$." }, { "Problem": "Let $R$ and $S$ be different points on a circle $\\Omega$ such that $RS$ is not a diameter. Let $\\ell$ be the tangent line to $\\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\\Omega$ so that the circumcircle $\\Gamma$ of triangle $JST$ intersects $\\ell$ at two distinct points. Let $A$ be the common point of $\\Gamma$ and $\\ell$ that is closer to $R$. Line $AJ$ meets $\\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\\Gamma$.", "Solution": "We construct inversion which maps $KT$ into the circle $\\omega_1$ and $\\Gamma$ into $\\Gamma.$ Than we prove that $\\omega_1$ is tangent to $\\Gamma.$\nQuadrangle $RJSK$ is cyclic $\\implies \\angle RSJ = \\angle RKJ.$\nQuadrangle $AJST$ is cyclic $\\implies \\angle RSJ = \\angle TAJ \\implies AT||RK.$\nWe construct circle $\\omega$ centered at $R$ which maps $\\Gamma$ into $\\Gamma.$\nLet $C = \\omega \\cap RT \\implies RC^2 = RS \\cdot RT.$ Inversion with respect to $\\omega$ swap $T$ and $S \\implies \\Gamma$ maps into $\\Gamma (\\Gamma = \\Gamma').$\nLet $O$ be the center of $\\Gamma.$\nInversion with respect to $\\omega$ maps $K$ into $K'$.\n$K$ belong $KT \\implies$ circle $K'SR = \\omega_1$ is the image of $KT$. Let $Q$ be the center of $\\omega_1.$\n$K'T$ is the image of $\\Omega$ at this inversion, $l = AR$ is tangent line to $\\Omega$ at $R,$ so $K'T||AR.$\n$K'$ is image K at this inversion $\\implies K \\in RK' \\implies RK'||AT \\implies ARK'T$ is parallelogram.\n$S$ is the midpoint of $RT \\implies S$ is the center of symmetry of $ATK'R \\implies$\n$\\triangle RSK'$ is symmetrical to $\\triangle TSA$ with respect to $S \\implies$\n$\\omega_1$ is symmetrical to $\\Gamma$ with respect to $S \\implies$\n$O$ is symmetrycal $Q$ with respect to $S.$\n$S$ lies on $\\Gamma$ and on $\\omega_1 \\implies \\Gamma$ is tangent to $\\omega_1 \\implies$ line $KT$ is tangent to $\\Gamma.$\nvladimir.shelomovskii@gmail.com, vvsss", "Solution_2": "We use the tangent-chord theorem: the angle formed between a chord and a tangent line to a circle is equal to the inscribed angle on the other side of the chord.\nQuadrangle $RJSK$ is cyclic $\\implies \\angle RSJ = \\angle RKJ.$\nQuadrangle $AJST$ is cyclic $\\implies \\angle RSJ = \\angle TAJ$\n(One can use Reim’s theorem – it is shorter way.)\nLet $B$ be symmetric to $A$ with respect to $S \\implies$\n$ATBR$ is parallelogram.\n\n$\\angle RBT = \\angle RAT \\implies \\angle KST + \\angle KBT = 180^\\circ$\n$\\implies SKBT$ is cyclic.\nInscribed angle of $\\Gamma (\\angle SAT)$ is equal to angle between $KT$ and chord $ST \\implies$\n$KT$ is tangent to $\\Gamma$ by the inverse of tangent-chord theorem.\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "A site is any point $(x, y)$ in the plane such that $x$ and $y$ are both positive integers less\nthan or equal to 20.\nInitially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy\ngoing first. On her turn, Amy places a new red stone on an unoccupied site such that the distance\nbetween any two sites occupied by red stones is not equal to $\\sqrt{5}$. On his turn, Ben places a new blue\nstone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from\nany other occupied site.) They stop as soon as a player cannot place a stone.\nFind the greatest $K$ such that Amy can ensure that she places at least $K$ red stones, no matter\nhow Ben places his blue stones.", "Solution": "The maximal K is 100.\nAmy can reach at least 100 by playing only on sites for which x+y is even. There are 200 such sites, none are of distance $\\sqrt{5}$ from each other and Ben can occupy at most half of them.\nOn the other hand Ben can prevent Amy from reaching more than 100 using the following strategy:\nPicture the sites as a 20 by 20 board and divide it into 25 non overlapping 4-by-4 squares. We label each site in the square as follows:\nWhenever Amy plays in a square Ben plays in the same square and in the site with the same label. In each square Amy can place at most 2 stones in sites labeled 1,4,6,7 (no three sites with labels from this set are free from distance $\\sqrt{5}$ and Amy can play one stone on each label since Ben plays the other). Likewise for the sites labeled 2,3,5,8. So in total Amy can place at most 4 stones in each of the 25 squares for a total of 100 stones." }, { "Problem": "Find all pairs $(k,n)$ of positive integers such that\n", "Solution_1": "$LHS$\n$k! = 1$ (when $k = 1$), $2$ (when $k = 2$), $6$ (when $k = 3$)\n$RHS = 1$(when $n = 1$), $6$ (when $n = 2$)\nHence, $(1,1)$, $(3,2)$ satisfy\nFor $k = 2: RHS$ is strictly increasing, and will never satisfy $k$ = 2 for integer n since $RHS = 6$ when $n = 2$.\n\nIn all solutions, for any prime $p$ and positive integer $N$, we will denote by the exponent of the largest power of $p$ that divides $N$. The right-hand side of $(1)$ will be denoted by that is,\n=\nOn the other hand, is expressed by the $Legendre$ $formula$ as\nThus, $k! = L_n$ implies the inequality \nIn order to obtain an opposite estimate, observe that \nWe claim that for all $n \\geq 6$\nFor $n \\geq 6$ the estimate (3) is true because \nand \n~flamewavelight and ~phoenixfire" }, { "Problem": "Let $\\Gamma$ be a circle with centre $I$, and $ABCD$ a convex quadrilateral such that each of\nthe segments $AB, BC, CD$ and $DA$ is tangent to $\\Gamma$. Let $\\Omega$ be the circumcircle of the triangle $AIC$.\nThe extension of $BA$ beyond $A$ meets $\\Omega$ at $X$, and the extension of $BC$ beyond $C$ meets $\\Omega$ at $Z$.\nThe extensions of $AD$ and $CD$ beyond $D$ meet $\\Omega$ at $Y$ and $T$, respectively. Prove that", "Solution_1": "Let $O$ be the centre of $\\Omega$.\nFor $AB=BC$ the result follows simply. By Pitot's Theorem we have so that, $AD = CD.$ The configuration becomes symmetric about $OI$ and the result follows immediately.\nNow assume WLOG $AB < BC$. Then $T$ lies between $A$ and $X$ in the minor arc $AX$ and $Z$ lies between $Y$ and $C$ in the minor arc $YC$.\nConsider the cyclic quadrilateral $ACZX$.\nWe have $\\angle CZX = \\angle CAB$ and $\\angle IAC = \\angle IZC$. So that, Since $I$ is the incenter of quadrilateral $ABCD$, $AI$ is the angular bisector of $\\angle DBA$. This gives us, Hence the chords $IX$ and $IY$ are equal. \nSo $Y$ is the reflection of $X$ about $OI$.\nHence, and now it suffices to prove \nLet $P, Q, N$ and $M$ be the tangency points of $\\Gamma$ with $AB, BC, CD$ and $DA$ respectively. Then by tangents we have, $AD = AM + MD = AP + ND$. So $AD + DT + XA = AP + ND + DT + XA = XP + NT$.\nSimilarly we get, $CD + DY + ZC = ZQ + YM$. So it suffices to prove, \nConsider the tangent $XJ$ to $\\Gamma$ with $J \\ne P$. Since $X$ and $Y$ are reflections about $OI$ and $\\Gamma$ is a circle centred at $I$ the tangents $XJ$ and $YM$ are reflections of each other. Hence By a similar argument on the reflection of $T$ and $Z$ we get $NT = ZQ$ and finally,\n as required.\n$QED$\n~BUMSTAKA", "Solution_2": "Denote $AD$ tangents to the circle $I$ at $N$, $CD$ tangents to the same circle at $M$; $XB$ tangents at $F$ and $ZB$ tangents at $J$. We can get that $AD=AM+MD;CD=DN+CN$.Since $AM=AF,XA=XF-AM;ZC=ZJ-CN$ Same reason, we can get that $DT=TN-DM;DY=YM-DM$\nWe can find that $AD+XA+DT=XF+TN;CD+DY+ZC=ZJ+YM$. Connect $IM,IT,IT,IZ,IX,IN,IJ,IF$ separately, we can create two pairs of congruent triangles. In $\\triangle{XIF},\\triangle{YIM}$, since $\\widehat{AI}=\\widehat{AI},\\angle{FXI}=\\angle{MYI}$ After getting that $\\angle{IFX}=\\angle{IMY};\\angle{FXI}=\\angle{MYI};IF=IM$, we can find that $\\triangle{IFX}\\cong \\triangle{IMY}$. Getting that $YM=XF$, same reason, we can get that $ZJ=TN$.\nNow the only thing left is that we have to prove $TX=YZ$. Since $\\widehat{IX}=\\widehat{IY};\\widehat{IT}=\\widehat{IZ}$ we can subtract and get that $\\widehat{XT}=\\widehat{YZ}$,means $XT=YZ$ and we are done\n~bluesoul", "Solution_3_.28Visual.29": "We use the equality of the tangent segments and symmetry.\nUsing Claim 1 we get $\\overset{\\Large\\frown} {TX}$ symmetric to $\\overset{\\Large\\frown} {ZY}$ with respect $IO.$\nTherefore $\\hspace{10mm} TX = ZY.$\nLet $P, Q, N$ and $M$ be the tangency points of $\\Gamma$ with $AB, BC, CD,$ and $DA,$ respectively.\nUsing Claim 2 we get $TM = QZ, PX = NY.$\nClaim 1\nLet $O$ be the center of $\\Omega.$ Then point $T$ is symmetric to $Z$ with respect $IO,$ point $X$ is symmetric to $Y$ with respect $IO.$\nProof\nLet $\\angle BAD =2\\alpha, \\angle ABC =2\\beta, \\angle BCD =2\\gamma, \\angle ADC =2\\delta.$\nWe find measure of some arcs:\n\n\n\n$\\overset{\\Large\\frown} {IX}= 2\\pi - \\overset{\\Large\\frown} {IY} - \\overset{\\Large\\frown} {XY} = 2\\alpha =\\overset{\\Large\\frown} {IY}\\implies$ symmetry $X$ and $Y.$\n\n$\\overset{\\Large\\frown} {IZ}= 2\\pi - \\overset{\\Large\\frown} {IT} - \\overset{\\Large\\frown} {TZ}= 2\\gamma= \\overset{\\Large\\frown} {IT}\\implies$ symmetry $T$ and $Z.$\nClaim 2\nLet circles $\\omega$ centered at $I$ and $\\Omega$ centered at $O$ be given. Let points $A$ and $A'$ lies on $\\Omega$ and $A$ be symmetric to $A'$ with respect $OI.$ Let $AC$ and $A'B$ be tangents to $\\omega$. Then $AC = A'B.$\nProof\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "Let $ABCDE$ be a convex pentagon such that $BC = DE$. Assume that there is a\npoint $T$ inside $ABCDE$ with $TB = TD$, $TC = TE$ and $\\angle ABT = \\angle TEA$. Let line $AB$ intersect\nlines $CD$ and $CT$ at points $P$ and $Q$, respectively. Assume that the points $P, B, A, Q$ occur on their\nline in that order. Let line $AE$ intersect lines $CD$ and $DT$ at points $R$ and $S$, respectively. Assume\nthat the points $R, E, A, S$ occur on their line in that order. Prove that the points $P, S, Q, R$ lie on\na circle.", "Solution": "$\\hspace{28mm}CDQS$ is cyclic $\\implies \\angle QCD = \\angle QSD.$\n\n\n$\\hspace{43mm}PRQS$ is cyclic.\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "Let $x_1, x_2, \\cdots , x_{2023}$ be pairwise different positive real numbers such that\n\nis an integer for every $n = 1,2,\\cdots,2023$. Prove that $a_{2023} \\ge 3034$.", "Solution": "We solve for $a_{n+2}$ in terms of $a_n$ and $x.$\n$a_{n+2}^2 \\\\ = (\\sum^{n+2}_{k=1}x_k)(\\sum^{n+2}_{k=1}\\frac1{x_k}) \\\\ = (x_{n+1}+x_{n+2}+\\sum^{n}_{k=1}x_k)(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}+\\sum^{n}_{k=1}\\frac1{x_k}) \\\\ = \\frac{x_{n+1}}{x_{n+1}} + \\frac{x_{n+1}}{x_{n+2}} + \\frac{x_{n+2}}{x_{n+1}} + \\frac{x_{n+2}}{x_{n+2}} + \\frac{1}{x_{n+1}}\\sum^{n}_{k=1}x_k + x_{n+1}\\sum^{n}_{k=1}\\frac1{x_k} + \\frac{1}{x_{n+2}}\\sum^{n}_{k=1}x_k + x_{n+2}\\sum^{n}_{k=1}\\frac1{x_k} + (\\sum^{n}_{k=1}x_k)(\\sum^{n}_{k=1}\\frac1{x_k}) \\\\ = \\frac{x_{n+1}}{x_{n+1}} + \\frac{x_{n+1}}{x_{n+2}} + \\frac{x_{n+2}}{x_{n+1}} + \\frac{x_{n+2}}{x_{n+2}} + \\frac{1}{x_{n+1}}\\sum^{n}_{k=1}x_k + x_{n+1}\\sum^{n}_{k=1}\\frac1{x_k} + \\frac{1}{x_{n+2}}\\sum^{n}_{k=1}x_k + x_{n+2}\\sum^{n}_{k=1}\\frac1{x_k} + a_n^2 \\\\ \\text{}$\nAgain, by AM-GM, the above equation becomes \n$a_{n+2}^2 \\ge 4 \\sqrt[4]{(\\frac{x_{n+1}}{x_{n+1}})(\\frac{x_{n+1}}{x_{n+2}})(\\frac{x_{n+2}}{x_{n+1}})(\\frac{x_{n+2}}{x_{n+2}})} + 4\\sqrt[4]{ (\\frac{1}{x_{n+1}}\\sum^{n}_{k=1}x_k)(x_{n+1}\\sum^{n}_{k=1}\\frac1{x_k})(\\frac{1}{x_{n+2}}\\sum^{n}_{k=1}x_k)(x_{n+2}\\sum^{n}_{k=1}\\frac1{x_k}) } + a_n^2 = a_n^2+4a_n+4 = (a_n+2)^2 \\\\ \\text{}$\nHence, $a_{n+2} \\ge a_{n} + 2,$ but equality is achieved only when $\\frac{x_{n+1}}{x_{n+1}},\\frac{x_{n+1}}{x_{n+2}},\\frac{x_{n+2}}{x_{n+1}},$ and $\\frac{x_{n+2}}{x_{n+2}}$ are equal. They can never be equal because there are no two equal $x_k.$So $a_{2023} \\ge a_1 + 3\\times \\frac{2023-1}{2} = 1 + 3033 = 3034$" }, { "Problem": "Consider the cube $ABCDA'B'C'D'$ (with face $ABCD$ directly above face $A'B'C'D'$).\na) Find the locus of the midpoints of the segments $XY$, where $X$ is any point of $AC$ and $Y$ is any point of $B'D'$;\nb) Find the locus of points $Z$ which lie on the segment $XY$ of part a) with $ZY = 2XZ$.", "Solution": "Let $A=(0,0,2)$, $B=(2,0,2)$, $C=(2,0,0)$, $D=(0,0,0)$, $A'=(0,2,2)$, $B'=(2,2,2)$, $C'=(2,2,0)$, and $D'=(0,2,0)$. Then there exist real $x$ and $y$ in the closed interval $[0,2]$ such that $X=(x,0,2-x)$ and $Y=(y,2,y)$.\nThe midpoint of $XY$ has coordinates $((x+y)/2, 1, (2-x+y)/2)$. Let $a$ and $b$ be the $x$- and $z$-coordinates of the midpoint of $XY$, respectively. We then have that $a+b=y+1$ and $a-b=x-1$, so $a+b\\in [1,3]$ and $a-b\\in [-1,1]$. The region of points that satisfy these inequalities is the closed square with vertices at $(1,1,2)$, $(2,1,1)$, $(1,1,0)$, and $(0,1,1)$. For every point $P$ in this region, there exist unique points $X$ and $Y$ such that $P$ is the midpoint of $XY$.\nIf $Z\\in XY$ and $ZY=2XZ$, then $Z$ has coordinates $((2x+y)/3, 2/3, (4-2x+y)/3)$. Let $a$ and $b$ be the $x$- and $z$- coordinates of $Z$. We then have that $a+b=(4/3)+(2/3)y$ and $a-b=(4x-4)/3$, and $a\\in (4/3,8/3)$ and $b\\in (-4/3,4/3)$. The region of points that satisfy these inequalities is the closed rectangle with vertices at $(0,2/3,4/3)$, $(2/3,2/3,1)$, $(1,2/3,2/3)$, and $(4/3,2/3,0)$. For every point $Z$ in this region, there exist unique points $X$ and $Y$ such that $Z\\in XY$ and $ZY=2XZ$." }, { "Problem": "Construct a triangle ABC if the following elements are given: $AC = b, AB = c$, and $\\angle AMB = \\omega \\left(\\omega < 90^{\\circ}\\right)$ where M is the midpoint of BC. Prove that the construction has a solution if and only if\n$b \\tan{\\frac{\\omega}{2}} \\le c < b$\nIn what case does equality hold?\n", "Solution": "Prolong BA to a point D such that $BD = 2AB$. Take circle through B and D such that the minor arc BD is equal to $2*\\omega$ so that for points P on the major arc BD we have $\\angle BPD = \\omega$. Draw a circle with center A and radius AC, and the point of intersection of this circle and the major arc BD will be C. In general there are two possibilities for C.\nLet X be the intersection of the arc BN and the perpendicular to the segment BN through A. For the construction to be possible we require $AX \\geqslant AC > AB$. But $\\frac{AB}{AX} = \\tan{\\frac{\\omega}{2}}$, so we get the condition in the question" }, { "Problem": "On the circle $K$ there are given three distinct points $A,B,C$. Construct (using only straightedge and compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained.", "Solution": "Aviso: La siguiente solución está en español\ngeneralidad ya que es simétrico para todas las combinaciones posibles)\nCircunferencia K en 2 puntos, uno de ello se nombrará con E (va a estar del mismo lado que B con \nrespecto a AC) y el otro se nombrará F\niguales, abrir con el compás la distancia BE y se traza la Circunferencia de Centro en F y Radio BE, \nasí esta nueva circunferencia corta a la circunferencia K en 2 puntos, llámese D al punto que está \ndel mismo lado de A con respecto a la mediatriz de AC, ese es el punto pedido que cumple la" }, { "Problem": "Prove that $\\cos{\\frac{\\pi}{7}}-\\cos{\\frac{2\\pi}{7}}+\\cos{\\frac{3\\pi}{7}}=\\frac{1}{2}$.", "Solution": "Because the sum of the $x$-coordinates of the seventh roots of unity is $0$, we have\n\nNow, we can apply $\\cos{x} = \\cos{\\left(-x\\right)}$ to obtain\n\nFinally, since $\\cos{x} = -\\cos{(\\pi-x)}$,\n\n~mathboy100", "Solution_2": "Let $\\cos{\\frac{\\pi}{7}}-\\cos{\\frac{2\\pi}{7}}+\\cos{\\frac{3\\pi}{7}}=S$. We have\n\nThen, by product-sum formulae, we have\n\nThus $S = 1/2$. $\\blacksquare$", "Solution_3": "Let $a=\\sin{\\frac{\\pi}{7}}$ and $b=\\cos{\\frac{\\pi}{7}}$. From the addition formulae, we have\n\nFrom the Trigonometric Identity, $a^2=1-b^2$, so\n\nWe must prove that $S=1/2$. It suffices to show that $8b^3-4b^2-4b+1=0$.\nNow note that $\\cos{\\frac{4\\pi}{7}}=-\\cos{\\frac{3\\pi}{7}}$. We can find these in terms of $a$ and $b$:\n\n\nTherefore $8b^4-8b^2+1=-(3b-4b^3)\\Rightarrow 8b^4+4b^3-8b^2-3b+1=0$. Note that this can be factored:\n\nClearly $b\\neq -1$, so $8b^3-4b^2-4b+1=0$. This proves the result. $\\blacksquare$", "Solution_4": "Let $\\omega=\\mathrm{cis}\\left(\\frac{\\pi}{14}\\right)$. Thus it suffices to show that $\\omega+\\omega^{-1}-\\omega^2-\\omega^{-2}+\\omega^3+\\omega^{-3}=1$. Now using the fact that $\\omega^k=\\omega^{14+k}$ and $-\\omega^2=\\omega^9$, this is equivalent to\n\n\nBut since $\\omega$ is a $14$th root of unity, $\\omega^{14}=1$. The answer is then $-\\omega^{7}=1$, as desired.\n~yofro\n", "Solution_5": "We let $\\omega = \\mathrm{cis} \\ \\left(\\dfrac{2\\pi}{7} \\right)$. We therefore have $w^i$, where $0 \\leq i \\leq 6$, are the $7^{\\text{th}}$ roots of unity. Since $\\sum_{i = 0}^6 \\omega^i = 0$, then $\\sum_{i = 1}^6 \\omega^i = -1$, so $\\sum_{i = 1}^3 \\mathrm{Re}(\\omega^i) = -\\dfrac{1}{2}$. Therefore, because $\\mathrm{cis} \\ \\alpha = \\cos \\alpha + i \\sin \\alpha, \\mathrm{Re}(\\mathrm{cis} \\ \\alpha) = \\cos \\alpha$, so\nSince $\\cos \\alpha = - \\cos(\\pi - \\alpha)$, we have $\\cos \\left(\\dfrac{\\pi}{7} \\right) - \\cos \\left(\\dfrac{2\\pi}{7} \\right) + \\cos \\left(\\dfrac{3\\pi}{7} \\right) = \\dfrac{1}{2}$ and we are done $\\blacksquare$\n~Yiyj1" }, { "Problem": "Suppose five points in a plane are situated so that no two of the straight lines\njoining them are parallel, perpendicular, or coincident. From each point perpendiculars\nare drawn to all the lines joining the other four points. Determine\nthe maximum number of intersections that these perpendiculars can have.", "Solution": "Suppose, those five points are $(A, B, C, D, E)$. Now, we want to create some special structure. Let, we take the line $BC$ and draw a perpendicular from $A$ on $BC$, andd call it $P_1$. We can do this set up in $\\binom{5}{1}\\binom{4}{2}=30$ ways. There will $30$ such $P_i$ s.\nNow, we will find how many other perpendiculars intersect the line. We can do this in total $20$ ways. Why? See, can draw perpendiculars from $B$ and $C$ to other lines( we haven't counted the perpendicular from $B$ to $AC$ and perpendicular from $C$ on $AB$ , as they intersect $P_1$ at the same point) in $5$ ways for each. So, total $10$ ways.\nNow, $5$ perpendiculars from each $D$ and $E$ on the other lines except on $BC$( because in this case teh perpendiculars from $D$ and $E$ will be parallel to $P_1$ , and so shall not intersect). So,total $10$ cases.\nFrom, these two cases we get $P_1$ will be intersected at at most $5*6*(5+5+5+5)=600$ points.\nBut, as we have passed this algorithm over all the five points, we have counted each intersection points twice. So, there are total $\\frac{600}{2}=300$ ways.\nNow, as we had excluded the orthocentres, we have to add now. There are total $\\binom{5}{3}=10$ orthocentres. Also we should add those vertices as these are also point of intersection of silimar perpendiculars, there are $5$ such.\nSo, total ways $300+10+5=315$." }, { "Problem": "Consider $\\triangle OAB$ with acute angle $AOB$. Through a point $M \\neq O$ perpendiculars are drawn to $OA$ and $OB$, the feet of which are $P$ and $Q$ respectively. The point of intersection of the altitudes of $\\triangle OPQ$ is $H$. What is the locus of $H$ if $M$ is permitted to range over (a) the side $AB$, (b) the interior of $\\triangle OAB$?", "Solution": "Let $O(0,0),A(a,0),B(b,c)$.\nEquation of the line $AB: y=\\frac{c}{b-a}(x-a)$.\nPoint $M \\in AB : M(\\lambda,\\frac{c}{b-a}(\\lambda-a))$.\nEasy, point $P(\\lambda,0)$.\nPoint $Q = OB \\cap MQ$, $MQ \\bot OB$.\nEquation of $OB : y=\\frac{c}{b}x$, equation of $MQ : y=-\\frac{b}{c}(x-\\lambda)+\\frac{c}{b-a}(\\lambda-a)$.\nSolving: $x_{Q}=\\frac{1}{b^{2}+c^{2}}\\left[b^{2}\\lambda+\\frac{c^{2}(\\lambda-a)b}{b-a}\\right]$.\nEquation of the first altitude: $x=\\frac{1}{b^{2}+c^{2}}\\left[b^{2}\\lambda+\\frac{c^{2}(\\lambda-a)b}{b-a}\\right] \\quad (1)$.\nEquation of the second altitude: $y=-\\frac{b}{c}(x-\\lambda)\\quad\\quad (2)$.\nEliminating $\\lambda$ from (1) and (2):\n\t\na line segment $MN , M \\in OA , N \\in OB$.\nSecond question: the locus consists in the $\\triangle OMN$." }, { "Problem": "Solve the system of equations\n$|a_1 - a_2| x_2 +|a_1 - a_3| x_3 +|a_1 - a_4| x_4 = 1\\\\ |a_2 - a_1| x_1 +|a_2 - a_3| x_3 +|a_2 - a_4| x_4 = 1\\\\ |a_3 - a_1| x_1 +|a_3 - a_2| x_2 +|a_3-a_4|x_4= 1\\\\ |a_4 - a_1| x_1 +|a_4 - a_2| x_2 +|a_4 - a_3| x_3 = 1$\nwhere $a_1, a_2, a_3, a_4$ are four different real numbers.", "Solution": "Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get:\n\nSubtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:\n\nHence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:\n\nHence $x2 = x3 = 0$, and $x1 = x4 = 1/(a1 - a4)$." }, { "Problem_5": "Let $f$ be a real-valued function defined for all real numbers $x$ such that, for some positive constant $a$, the equation\n\nholds for all $x$.\n(a) Prove that the function $f$ is periodic (i.e., there exists a positive number $b$ such that $f(x + b) = f(x)$ for all $x$).\n(b) For $a = 1$, give an example of a non-constant function with the required properties.", "Solution": "(a) Since \n\nis true for any $x$, and\nWe have:\n\nTherefore $f$ is periodic, with $2a>0$ as a period.\n(b) $f(x) = 1$ when $2n\\le x < 2n+1$ for some integer $n$, and $f(x)=\\frac{1}{2}$ when $2n+1\\le x < 2n+2$ for some integer $n$." }, { "Problem": "Given $n > 4$ points in the plane such that no three are collinear, prove that there are at least $(n-3)/2$ convex quadrilaterals whose vertices are four of the given points.", "Solution": "Orient the points so that one of them corresponds to the origin (A), another lies on the y-axis (B), and all the others are in either quandrant I or IV. (In other words, you are creating a boundary line.) Select one of the points (C) which minimizes the angle BAC. You cannot have two of them because then they would be collinear. Also no points lie inside the area of ABC because if such a point D did exist, then the angle DAC would be less than BAC - contradiction. So there are none of the other n-3 points inside the area of ABC.\nNow we will show that for any two of the remaining n-3 points, you can create a convex quadrilateral containing two of them and two of A,B,C.\nNow select two arbitrary points from the remaining n-3. Call them E and F, and draw a line though them; call this line l. If l does not intersect ABC, then ABCEF is convex and you choose the quadrilateral ABEF.\nThe other case is that it separates ABC such that one of A,B,C is one one side of l and the other two are on the other side. Without loss of generality, A is on one side of l while B and C are on the other side. Consider the quadrilateral BCEF (or CBEF, depending on orientation.) We know that BC are on the same side of EF. Also, since no points are in the third quadrant, we know that EF are on the same side of BC. Now we can choose either BCEF or CBEF to be our convex quadrilateral.\nSo we now have that for the given ABC, all pairs of the other n points give a distinct convex quadrilateral. So we have n-3 points to choose from, and we must choose two of them. So this gives us at least C(n-3,2) = $\\frac{(n-3)(n-4)}{2}$ convex quadrilaterals.\nThe above solution was posted and copyrighted by Philip_Leszczynski. The original thread for this problem can be found here: [1]" }, { "Problem": "In the tetrahedron $ABCD$, angle $BDC$ is a right angle. Suppose that the foot $H$ of the perpendicular from $D$ to the plane $ABC$ in the tetrahedron is the intersection of the altitudes of $\\triangle ABC$. Prove that\nFor what tetrahedra does equality hold?", "Solution": "Let us show first that angles $ADB$ and $ADC$ are also right. Let $H$ be the intersection of the altitudes\nof $ABC$ and let $CH$ meet $AB$ at $X$. Planes $CED$ and $ABC$ are perpendicular and $AB$ is perpendicular to\nthe line of intersection $CE$. Hence $AB$ is perpendicular to the plane $CDE$ and hence to $ED$. So $BD^2 = DE^2 + BE^2.$ Also $CB^2 = CE^2 + BE^2.$ Therefore $CB^2 - BD^2 = CE^2 - DE^2.$ But $CB^2 - BD^2 = CD^2,$ so $CE^2 = CD^2 + DE^2$, so angle $CDE = 90^{\\circ}$. But angle $CDB = 90^{\\circ}$, so $CD$ is perpendicular to\nthe plane $DAB$, and hence angle $CDA$ = $90^{\\circ}$. Similarly, angle $ADB = 90^{\\circ}$.\nHence $AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2)$. But now we are done, because Cauchy's\ninequality gives $(AB + BC + CA)^2 = 3(AB^2 + BC^2 + CA^2).$ We have equality if and only if\nwe have equality in Cauchy's inequality, which means $AB = BC = CA.$\n" }, { "Problem": "Prove that for every natural number $m$; there exists a finite set $S$ of points in a plane with the following property: For every point $A$ in $S$; there are exactly $m$ points in $S$ which are at unit distance from $A$.", "Solution_1": "I shall prove a more general statement about the unit distance graph($V=\\mathbb{R}^2$, adjacency iff the Euclidean distance between the points is $1$) and this will follow as a consequence.\nif $G,H$ occur as unit distance graphs, then so does $G\\times H$( here $G\\times H$ is described as $V(G\\times H)=V(G)\\times V(H), (v_1,w_1)\\leftrightarrow (v_2,w_2) \\Leftrightarrow v_1=v_2,w_1\\leftrightarrow w_2$ or $v_1\\leftrightarrow v_2,w_1=w_2$).\nthis is seen by describing the vertices by complex numbers. suppose there is an embedding of $G$ by the complex numbers $v_1,v_2\\ldots,v_n$ and for $H$ by the numbers $w_1,w_2,\\ldots, w_m$. we claim that for some choice of $0\\leq\\theta<2\\pi$, $v_i+e^{i\\theta}w_j$ will do the job(a suitable rotation).\nwhat we need is that $(v_i+e^{i\\theta}w_j)\\leftrightarrow (v_k+e^{i\\theta}w_l)$ iff either $v_i=v_k,|w_j-w_l|=1$ or $w_j=w_l,|v_i-v_k|=1$. clearly if the condition holds then the adjacency is satisfied. suppose $i\\neq k,j\\neq l$ and that the corresponding complex numbers are at a distance $1$ from one another. Then this gives a quadratic in $e^{i\\theta}$ and hence $\\theta$ can take only finitely many values here.ruling this out for each such set of $i,j,k,l$ hence rules out finitely many values of $\\theta$ and therefore a suitable choice exists.\nfor the given problem we need a unit distance graph which is regular of degree $m$ for any given $m$. since we can form the graph $K_2$, we can form $Q_n=Q_{n-1}\\times K_2$(the unit cube) and that solves the problem.\nThis solution was posted and copyrighted by seshadri. The original thread for this problem can be found here: [1]", "Solution_2": "Suppose $S$ has the formwhere $V_m=\\{\\mathbf v_1,\\mathbf v_2,\\dots,\\mathbf v_m\\}$ is unknown set of distinct unit vectors in $\\mathbb R^2$. We can build $V_m$ inductively, starting from the empty set and adding vectors to it, one by one. We just need to make sure that two thing are respected:\n1. All $2^m$ vectors in $S_m$ are distinct;\n2. Two vector sums are at unit distance from one another if and only if they differ in presence of exactly one summand (i.e. one and only one coefficient $b_i$ in the sum changes from $0$ to $1$).\nIf these two conditions are kept, then each of $2^m$ points at $S_m$ will be at unit distance from exactly $m$ points corresponding to sums at which one and only one of $m$ coefficients differs from coefficients of this point. However, respecting these conditions is not hard because $\\left\\|\\sum_{\\mathbf v \\in U_1} \\mathbf v - \\sum_{\\mathbf v \\in U_2} \\mathbf v\\right\\|=\\left\\|\\sum_{\\mathbf v \\in U_1 \\setminus U_2} \\mathbf v \\,-\\! \\sum_{\\mathbf v \\in U_2 \\setminus U_1} \\mathbf v\\right\\|$ and for each new vector being added to $V_m$ there is at most some finite set of forbidden endpoints given by sums/differences of already determined vectors but the rest of the (infinite) unit circle is permissible.\nThis solution was posted and copyrighted by Bandera. The original thread for this problem can be found here: [2]" }, { "Problem_5": "Determine all possible values of where $a, b, c, d,$ are arbitrary positive numbers.", "Solution": "Note that We will now prove that $S$ can reach any range in between $1$ and $2$.\nChoose any positive number $a$. For some variables such that $k, m, l > 0$ and $k + m + l = 1$, let $b = ak$, $c = am$, and $d = al$. Plugging this back into the original fraction, we get \n\nThe above equation can be further simplified to \n\nNote that $S$ is a continuous function and that $f(m) = m + \\frac{1}{2-m}$ is a strictly increasing function. We can now decrease $k$ and $l$ to make $m$ tend arbitrarily close to $1$. We see $\\lim_{m\\to1} m + \\frac{1}{2-m} = 2$, meaning $S$ can be brought arbitrarily close to $2$. \nNow, set $a = d = x$ and $b = c = y$ for some positive real numbers $x, y$. Then \n\nNotice that if we treat the numerator and denominator each as a quadratic in $y$, we will get $1 + \\frac{g(x)}{2y^2 + 5xy + 2x^2}$, where $g(x)$ has a degree lower than $2$. This means taking $\\lim_{y\\to\\infty} 1 + \\frac{g(x)}{2y^2 + 5xy + 2x^2} = 1$, which means $S$ can be brought arbitrarily close to $1$. Therefore, we are done.\n\n~Imajinary" }, { "Problem": "Determine, with proof, whether or not one can find $1975$ points on the circumference of a circle with unit radius such that the distance between any two of them is a rational number.", "Solution": "Since there are infinitely many primitive Pythagorean triples, there are infinitely many angles $\\theta$ s.t. $\\sin\\theta, \\cos\\theta$ are both rational. Call such angles good. By angle-sum formulas, if $a,b$ are good, then $a+b,a-b$ are also good.\nFor points $A,B$ on the circle $\\omega$, let $\\angle AB$ be the angle subtended by $AB$. Now inductively construct points on $\\omega$ s.t. all angles formed by them are good; for 1,2 take any good angle. If there are $n$ points chosen, pick a good angle $\\theta$ and a marked point $A$ s.t. the point $B$ on $\\omega$ with $\\angle AB=\\theta$ is distinct from the $n$ points. Since there are infinitely many good angles but finitely many marked points, such $\\theta$ exists. For a previously marked point $P$ we have $\\angle BP=\\pm \\angle AP\\pm \\angle AB$ for suitable choices for the two $\\pm$. Since $\\angle AP ,\\angle AB$ are both good, it follows that $\\angle BP$ is good, which finishes induction by adding $B$.\nObserve that these points for $n=1975$ work: since $AB=27sin\\angle AB$ for $A,B$ on the circle, it follows that $AB$ is rational, and so we're done.\nThe above solution was posted and copyrighted by tobash_co. The original thread for this problem can be found here: [1]" }, { "Problem": "We consider the following system\nwith $q = 2p$:\n$\\begin{matrix} a_{11}x_{1} + \\ldots + a_{1q}x_{q} = 0, \\\\ a_{21}x_{1} + \\ldots + a_{2q}x_{q} = 0, \\\\ \\ldots , \\\\ a_{p1}x_{1} + \\ldots + a_{pq}x_{q} = 0, \\\\ \\end{matrix}$\nin which every coefficient is an element from the set $\\{ - 1,0,1\\}$$.$ Prove that there exists a solution $x_{1}, \\ldots,x_{q}$ for the system with the properties:\na.) all $x_{j}, j = 1,\\ldots,q$ are integers$;$\nb.) there exists at least one j for which $x_{j} \\neq 0;$\nc.) $|x_{j}| \\leq q$ for any $j = 1, \\ldots ,q.$", "Solution": "First of all note that we have $(q + 1)^q - 1$ possible nonzero vectors $(x_1,\\cdots,x_q)$ such that $0\\leq x_i\\leq q$ are integers.\nBut $a_{j1}x_1 + \\cdots + a_{jq}x_q$ can only assume $q^2 + 1$ different values, because if it is maximized/minimized by $(M_1,M_2,\\cdots,M_q)/(m_1,m_2,\\cdots,m_q)$, we have that $\\sum_{i = 1}^{q}a_{ji}(M_i - m_i)\\leq q\\times q = q^2$ (if $a_{ji} = 0$, it doesn't affect the sum, if it is $1$, $M_i = q,m_i = 0$, and if it is $- 1$, $M_i = 0,m_i = q$).\nFrom this we conclude that there are at most $(q^2 + 1)^p = (q^2 + 1)^{q/2}$ possible values for the vector $(a_{11}x_{1} + \\ldots + a_{1q}x_{q},\\cdots,a_{p1}x_{1} + \\ldots + a_{pq}x_{q})$.\nBut we have that:\n$(q + 1)^q - 1 = (q^2 + 1 + 2q)^{q/2} - 1 =$\n$= (q^2 + 1)^{q/2} + \\left( - 1 + \\sum_{j = 1}^{q/2}{{q/2}\\choose j}(q^2 + 1)^{q/2 - j}(2q)^j\\right) > (q^2 + 1)^{q/2}$\nWe conclude that by the pigeonhole principle there are two distinct vectors being mapped to the same vector. Taking their difference we have a vector with the desired properties.\nThe above solution was posted and copyrighted by Jorge Miranda. The original thread for this problem can be found here: [1]" }, { "Problem": "Let $a,b$ be two natural numbers. When we divide $a^2+b^2$ by $a+b$, we the the remainder $r$ and the quotient $q.$ Determine all pairs $(a, b)$ for which $q^2 + r = 1977.$", "Solution": "Using $r=1977-q^2$, we have $a^2+b^2=(a+b)q+1977-q^2$, or $q^2-(a+b)q+a^2+b^2-1977=0$, which implies $\\Delta=7908+2ab-2(a^2+b^2)\\ge 0$. If we now assume Wlog that $a\\ge b$, it follows $a+b\\le 88$. If $q\\le 43$, then $r=1977-q^2\\ge 128$, contradicting $r3$). Let $p$ be its perimeter. Prove that:\n\nwhere $[x]$ denotes the greatest integer not exceeding $x$.", "Solution": "Consider all the pairs of non-adjacent sides in the polygon. Each pair uniquely determines two diagonals which intersect in an interior point. The sum of these two diagonals is greater than the sum of the two sides (triangle inequality). Each side will appear in a pair $n-3$ times. And each time the side appears it also includes two diagonals. Thus, $(n-3)p < 2d$.\nNote that for any diagonal, its length is less than the sum of the sides on either side of it (triangle inequality). Consider all the diagonals from a fixed vertex and using the least number of sides in each application of the triangle inequality. We can sum everything up over all the diagonals considering the two cases (even or odd) to get the result.\nThis solution was posted and copyrighted by aznlord1337. The original thread for this problem can be found here: [1]" }, { "Problem": "A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$). Prove that $\\angle OMB = 90^{\\circ}$.", "Solution": "$M$ is the Miquel Point of quadrilateral $ACNK$, so there is a spiral similarity centered at $M$ that takes $KN$ to $AC$. Let $M_1$ be the midpoint of $KA$ and $M_2$ be the midpoint of $NC$. Thus the spiral similarity must also send $M_1$ to $M_2$ and so $BMM_1 M_2$ is cyclic. $OM_1 B M_2$ is also cyclic with diameter $BO$ and thus $M$ must lie on the same circumcircle as $B$, $M_1$, and $M_2$ so $\\angle OMB = 90^{\\circ}$.", "Solution_2": "Let $\\Omega, \\Omega', \\omega$ and $O,O',O''$ be the circumcircles and circumcenters of $AKNC, ABC, BNKM,$ respectively.\nLet $\\angle ACB = \\gamma, AKNC$ is cyclic $\\implies \\angle BKN = \\gamma.$\nThe radius of $\\omega$ is $MO'' = BO'' = \\frac {BN}{2 \\sin \\gamma}.$\nLet $D$ and $E$ be midpoints of $BC$ and $NC$ respectively.\n$OE \\perp BC, OD \\perp BC, OO' \\perp AC, DE = \\frac {BC}{2} - \\frac {NC}{2} = \\frac {BN}{2}$\n$\\implies OO' = \\frac {DE}{\\sin \\gamma} = \\frac {BN}{2 \\sin \\gamma} = MO''.$\n$M$ is the Miquel Point of quadrilateral $ACNK,$ so $MO''O'O$ is cyclic.\n$MO''O'O$ is trapezium $\\implies O''O' || MO.$ $O''O' \\perp BM \\implies MO\\perp BM$ as desired.\nvladimir.shelomovskii@gmail.com, vvsss", "Solution_3_.28No_Miquel.27s_point.29": "Consider $\\triangle MKA$ and $\\triangle MNC$, they are similar because $\\angle MAK$ = $\\angle MCN$, and also $\\angle MKA = \\angle MNC$.\nNow draw $OP \\perp AB$, and intersecting $AB$ at $P$; $OQ \\perp BC$, at $Q$. Naturally $OP$ bisects $AK$, and $OQ$ bisects $CN$. We claim $\\triangle MAP \\sim \\triangle MCQ$, because \n$\\frac {AP}{CQ} = \\frac {AK}{CN} = \\frac {AM}{CB}.$\nThus $\\angle AMP = \\angle CMQ$, this implies $\\angle PMQ = \\angle AMC = \\angle ABC = \\angle PBQ$. Obviously BMPQ is cyclic, and so is BPOQ. Finally, we have $OM \\perp MB$. (by gougutheorem)" }, { "Problem": "Find all (if any) functions $f$ taking the non-negative reals onto the non-negative reals, such that\n(a) $f(xf(y))f(y) = f(x+y)$ for all non-negative $x$, $y$;\n(b) $f(2) = 0$;\n(c) $f(x) \\neq 0$ for every $0 \\leq x < 2$.", "Solution_1": "For all $x+y\\ge2$, there exists a nonnegative real $t$ such that $t+2=x+y$ and $f(x+y)=f(t+2)=f(tf(2))f(2)=0$. Hence, $f(x)=0$ for all $x\\ge 2$.\nFor $x<2$, take $0=f(2)=f((2-x)+(x))=f((2-x)f(x))f(x)$.\nSince $f(x)\\neq0$ for $x<2$, then $f((2-x)f(x))=0$.\nThus, $(2-x)f(x)\\ge2\\Leftrightarrow f(x)\\ge\\frac{2}{2-x}$ for all $x<2$.\nSuppose $f(a)>\\frac{2}{2-a}$ so that $f(a)=\\frac{2}{2-a-\\epsilon}$ for some $\\epsilon>0$ and $0\\le a<2$.\nThen $f\\left((2-a-\\epsilon)f(a)\\right)f(a)=f(2-\\epsilon)$. The left hand side is equivalent to $f\\left((2-a-\\epsilon)\\left(\\frac{2}{2-a-\\epsilon}\\right)\\right)f(a)=f(2)=0$, but the right hand side is nonzero, since $2-\\epsilon<2$ and $f(x)\\neq0$ for all $x<2$.\nHence, $f(x)=\\frac{2}{2-x}$ for $0\\le a<2$.\nFor $x+y<2$,\\begin{eqnarray*}f(xf(y))f(y) &=& f\\left((x)\\left(\\frac{2}{2-y}\\right)\\right)\\left(\\frac{2}{2-y}\\right)\\\\ &=&\\left(\\frac{2}{2-\\frac{2x}{2-y}}\\right)\\left(\\frac{2}{2-y}\\right)\\\\ &=&\\frac{4}{4-2y-2x}\\\\ &=&\\frac{2}{2-y-x}\\\\ f(x+y)&=&\\frac{2}{2-y-x}.\\end{eqnarray*}\nIndeed, $f(xf(y))f(y)=f(x+y)$, so the desired function is\nThis solution was posted and copyrighted by towersfreak2006. The original thread for this problem can be found here: [1]\n", "Solution_2": "Taking $y=2$ gives $f(x+2)=0 \\forall x\\ge 0$, or equivalently $f(x)=0 \\forall x\\ge 2$. Taking $y=2-x$ for $0\\le x<2$ yieldsSince $0<2-x\\le 2$, we must have $f(xf(2-x))=0$, which forcesNow taking $x=2/f(y)$, for $0\\le y<2$ gives,Thus,Hence, this forces $f(x)=2/(2-x)$ for $0\\le x<2$, giving a final answer of\nThis solution was posted and copyrighted by winnertakeover. The original thread for this problem can be found here: [2]", "Solution_3": "Let $P(x,y)$ be the assertion. Note that $P(x,2)\\implies f(x+2)=0.$\nSince $x$ is defined for non-negative reals, we have $x+2\\geq 2.$ So, $\\boxed{f(x)=0\\qquad \\forall x\\geq2}.$\nNow, $P(2-y,y)\\implies f((2-y)f(y))f(y)=0.$\nHere, we are working with $0\\leq y<2.$ Since we may not have $f(y)=0$ due to condition (iii), we must have $f((2-y)f(y))=0.$\nFrom our first solution, it is imperative that $(2-y)f(y)\\geq 2.$ So, $f(y)\\geq \\dfrac{2}{2-y} \\qquad (1).$\nNow consider $P(\\dfrac{2}{f(y),y}$. We get,Similar to the previous step, we have $\\dfrac{2+yf(y)}{f(y)}\\geq 2.$ Re arranging this, we get $f(y)\\leq \\dfrac{2}{2-y}\\qquad (2).$\nSince $f(y)$ satisfies both $(1)$ and $(2),$ we must have $\\boxed{f(x)=\\dfrac{2}{2-x}\\qquad \\forall x\\in [0,2)}.$\nOf course, we plug both solution back to see if they satisfy the three conditions. Sure enough, they do!\nThis solution was posted and copyrighted by proshi. The original thread for this problem can be found here: [3]" }, { "Problem": "Let $n$ be an integer greater than or equal to 3. Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.", "Solution": "Consider the set of points $S = \\{ (x,x^2) \\mid 1 \\le x \\le n , x \\in \\mathbb{N} \\}$ in the $xy$-plane.\nThe distance between any two distinct points $(x_1,x^2_1)$ and $(x_2,x^2_2)$ in $S$ (with $x_1 \\neq x_2$) is:\n$d = \\sqrt{(x_1-x_2)^2+\\left(x^2_1-x^2_2\\right)^2} =$ $\\sqrt{(x_1-x_2)^2+(x_1-x_2)^2(x_1+x_2)^2} = |x_1-x_2|\\sqrt{1+(x_1+x_2)^2}$.\nSince $1+(x_1+x_2)^2$ is an integer and not a perfect square, $\\sqrt{1+(x_1+x_2)^2}$ is irrational. \nSince $|x_1-x_2|$ is a non-zero integer, $d$ is irrational as desired.\nAll the points in $S$ lie on the parabola $y = x^2$. Thus, it is impossible of any set of three points to be collinear, since no line can intersect a parabola at more than two points. Therefore, any triangle with all vertices in $S$ must be non-degenerate as desired.\nSince all the points in $S$ are lattice points, by Pick's Theorem, the area of any triangle with all vertices in $S$ must be in the form $A = I + \\dfrac{B}{2} - 1$ where $I$ and $B$ are integers. Thus, the area of the triangle must be rational as desired.\nThis completes the proof." }, { "Problem": "In a right-angled triangle $ABC$ let $AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ABD, ACD$ intersect the sides $AB, AC$ at the points $K,L$ respectively. If $E$ and $E_1$ dnote the areas of triangles $ABC$ and $AKL$ respectively, show that", "Solution": "Lemma: Through the incenter $I$ of $\\triangle{ABC}$ draw a line that meets the sides $AB$ and $AC$ at $P$ and $Q$, then:\n\nProof of the lemma:\nConsider the general case: $M$ is any point on side $BC$ and $PQ$ is a line cutting AB, AM, AC at P, N, Q. Then:\n$\\frac{AM}{AN}=\\frac{S_{APMQ}}{\\triangle{APQ}}=\\frac{\\triangle{APM}+\\triangle{AQM}}{\\triangle{PQA}}=\\frac{\\frac{AP}{AB}\\triangle{ABM}+\\frac{AQ}{AC}\\triangle{ACM}}{\\frac{AP\\cdot AQ}{AB \\cdot AC}}=$\n$=\\frac{AC}{AQ}\\cdot \\frac{BM}{BC}+\\frac{AB}{AP}\\cdot \\frac{CM}{BC}$\nIf $N$ is the incentre then $\\frac{AM}{AN}=\\frac{AB+BC+CA}{AB+AC}$, $\\frac{BM}{BC}=\\frac{AB}{AB+AC}$ and $\\frac{CM}{BC}=\\frac{AC}{AC+AB}$. Plug them in we get:\nBack to the problem\nLet $I_1$ and $I_2$ be the areas of $\\triangle{ABD}$ and $\\triangle{ACD}$ and $E$ be the intersection of $KL$ and $AD$. Thus apply our formula in the two triangles we get:\n\nand\n\nCancel out the term $\\frac{AD}{AE}$, we get:\n\n\n\n\n\nSo we conclude $AK=AL$.\nHence $\\angle{AKI_1}=45^o=\\angle{ADI_1}$ and $\\angle{ALI_2}=45^o=\\angle{ADI_2}$, thus $\\triangle{AK_1} \\cong \\triangle{ADI_1}$ and $\\triangle{ALI_2} \\cong \\triangle{ADI_2}$. Thus $AK=AD=AL$. So the area ratio is:\nThis solution was posted and copyrighted by shobber. The original thread for this problem can be found here: [1]" }, { "Problem": "Prove that for each positive integer $n$ there exist $n$ consecutive positive integers none of which is an integral power of a prime number.", "Solution_1": "There are at most $1+\\sqrt[2]{n}+\\sqrt[3]{n}+\\sqrt[4]{n}+...+\\sqrt[\\left\\lfloor \\log_2(n)\\right\\rfloor]{n} \\leq 1+ \\sqrt n log_2(n)$ 'true' powers $m^k , k\\geq 2$ in the set $\\{1,2,...,n\\}$. So when $p(n)$ gives the amount of 'true' powers $\\leq n$ we get that $\\lim_{n \\to \\infty} \\frac{p(n)}{n} = 0$.\nSince also $\\lim_{n \\to \\infty} \\frac{\\pi(n)}{n} = 0$, we get that $\\lim_{n \\to \\infty} \\frac{p(n)+\\pi(n)}{n} = 0$. Now assume that there is no 'gap' of lenght at least $k$ into the set of 'true' powers and the primes. Then this would give that $\\frac{p(n)+\\pi(n)}{n} \\geq \\frac{1}{k}$ for all $n$ in contrary to the above (at east this proves a bit more).\nEdit: to elementarize the $\\lim_{n \\to \\infty} \\frac{\\pi(n)}{n} = 0$ part:\nLook $\\mod (k+1)!$. Then all numbers in the residue classes $2,3,4,...,k+1$ are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length $k$, there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each $(k+1)!$ numbers, again contradicting $\\lim_{n \\to \\infty} \\frac{p(n)}{n} = 0$.\nThis solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]", "Solution_2": "By Chinese Remainder theorem, there exists $x$ such that:\n$x \\equiv -1\\;mod\\;p_1 q_1\\\\ x \\equiv -2\\;mod\\;p_2 q_2\\\\ x \\equiv -3\\;mod\\;p_3 q_3\\\\ ...\\newline x \\equiv -n\\;mod\\;p_n q_n$\nWhere $p_1, p_2, ..., p_n, q_1, q_2, ..., q_n$ are distinct primes. \nThe n consecutive numbers $x+1, x+2, ..., x+n$ each have at least two prime factors, so none of them can be expressed as an integral power of a prime." }, { "Problem": "Given an initial integer $n_{0}\\textgreater1$, two players, $\\mathbb{A}$ and $\\mathbb{B}$, choose integers $n_{1}, n_{2},n_{3}$, . . . alternately according to the following rules:\nKnowing $n_{2k}$, $\\mathbb{A}$ chooses any integer $n_{2k+1}$ such that\n.\nKnowing $n_{2k+1}$, $\\mathbb{B}$ chooses any integer $n_{2k+2}$ such that\n\nis a prime raised to a positive integer power.\nPlayer $\\mathbb{A}$ wins the game by choosing the number 1990; player $\\mathbb{B}$ wins by choosing\nthe number 1. For which $n_{0}$ does:\n(a) $\\mathbb{A}$ have a winning strategy?\n(b) $\\mathbb{B}$ have a winning strategy?\n(c) Neither player have a winning strategy?", "Solution": "If $n_0<6$ it is clear that B wins, because A can only choose numbers with at most 2 prime factors (30 is the smallest with three and $30>5^2$), which B can either make smaller (choose the lesser of the two) or the number contains one prime factor in which case B wins. Once $n_{2k}=2$, A can only choose prime powers, so B wins.\nIf $n_0=6,7$ it's a draw, because the only numbers that give A a chance of winning are those with three prime factors (for reasons discussed above). Less than 49, this gives 30 and 42 as the only choices which won't cause A to lose. However, if A chooses 30, B may choose 6, and we have a stalemate. If A chooses 42, B may choose 6, and we still have a stalemate.\nIf $n_0 \\geq 8$, if $n_0$ is less than 45 by choosing (perhaps starting later) 3*4*5, then 3*5*7, then 2*3*5*7, and then 3*4*5*7 A can force B to make $n_{2k} \\geq 60$ for some k.\nIf $45 \\leq n_{2k} \\leq 1990$ for some k, then A wins (he may pick 1990). If $n_{2k}>1990$ pick the least natural value of m such that $2^m 47\\geq n_{2k}$. The best B can do is half it, taking it to $n_{2k+2} = 2^{m-1} 47 < n_{2k}$. However, B cannot make $n_{2k+2}$ drop below 45 because $47*32<1990 \\Rightarrow 2^m \\geq 64$. Thus we will get a descending sequence of integers until we end up with $45 \\leq n_{2k} \\leq 1990$, at which point A wins.\nThis solution was posted and copyrighted by Ilthigore. The original thread for this problem can be found here: [1]" }, { "Problem": "Let $\\,ABC\\,$ be a triangle and $\\,P\\,$ an interior point of $\\,ABC\\,$. Show that at least one of the angles $\\,\\angle PAB,\\;\\angle PBC,\\;\\angle PCA\\,$ is less than or equal to $30^{\\circ }$.", "Solution_1": "Let $A_{1}$ , $A_{2}$, and $A_{3}$ be $\\angle CAB$, $\\angle ABC$, $\\angle BCA$, respectively.\nLet $\\alpha_{1}$ , $\\alpha_{2}$, and $\\alpha_{3}$ be $\\angle PAB$, $\\angle PBC$, $\\angle PCA$, respcetively.\nUsing law of sines on $\\Delta PAB$ we get: $\\frac{\\left| PA \\right|}{sin(A_{2}-\\alpha_{2})}=\\frac{\\left| PB \\right|}{sin(\\alpha_{1})}$, therefore, $\\frac{\\left| PA \\right|}{\\left| PB \\right|}=\\frac{sin(A_{2}-\\alpha_{2})}{sin(\\alpha_{1})}$\nUsing law of sines on $\\Delta PBC$ we get: $\\frac{\\left| PB \\right|}{sin(A_{3}-\\alpha_{3})}=\\frac{\\left| PC \\right|}{sin(\\alpha_{2})}$, therefore, $\\frac{\\left| PB \\right|}{\\left| PC \\right|}=\\frac{sin(A_{3}-\\alpha_{3})}{sin(\\alpha_{2})}$\nUsing law of sines on $\\Delta PCA$ we get: $\\frac{\\left| PC \\right|}{sin(A_{1}-\\alpha_{1})}=\\frac{\\left| PA \\right|}{sin(\\alpha_{3})}$, therefore, $\\frac{\\left| PC \\right|}{\\left| PA \\right|}=\\frac{sin(A_{1}-\\alpha_{1})}{sin(\\alpha_{3})}$\nMultiply all three equations we get:\n$\\frac{\\left| PA \\right|}{\\left| PB \\right|}\\frac{\\left| PB \\right|}{\\left| PC \\right|}\\frac{\\left| PC \\right|}{\\left| PA \\right|}=\\frac{sin(A_{2}-\\alpha_{2})}{sin(\\alpha_{1})}\\frac{sin(A_{3}-\\alpha_{3})}{sin(\\alpha_{2})}\\frac{sin(A_{1}-\\alpha_{1})}{sin(\\alpha_{3})}$\n$1=\\frac{sin(A_{2}-\\alpha_{2})}{sin(\\alpha_{1})}\\frac{sin(A_{3}-\\alpha_{3})}{sin(\\alpha_{2})}\\frac{sin(A_{1}-\\alpha_{1})}{sin(\\alpha_{3})}$\n$\\prod_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}=1$\nUsing AM-GM we get:\n$\\frac{1}{3}\\sum_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}\\ge \\sqrt[3]{\\prod_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}}$\n$\\frac{1}{3}\\sum_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}\\ge 1$\n$\\sum_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}\\ge 3$. [Inequality 1]\n$\\sum_{i=1}^{3}\\frac{sin(A_{i})cos(\\alpha_{i})-cos(A_{i})sin(\\alpha_{i})}{sin(\\alpha_{i})}\\ge 3$\n$\\sum_{i=1}^{3}\\left[ sin(A_{i})cot(\\alpha_{i})-cos(A_{i})\\right]\\ge 3$\nNote that for $0<\\alpha_{i}<180^{\\circ}$, $cot(\\alpha_{i})$ decreases with increasing $\\alpha_{i}$ and fixed $A_{i}$\nTherefore, $\\left[ sin(A_{i})cot(\\alpha_{i})-cos(A_{i})\\right]$ decreases with increasing $\\alpha_{i}$ and fixed $A_{i}$\nFrom trigonometric identity:\n$sin(x)+sin(y)=2sin\\left( \\frac{x+y}{2} \\right)cos\\left( \\frac{x-y}{2} \\right)$,\nsince $-1\\le cos\\left( \\frac{x-y}{2} \\right) \\le 1$, then:\n$sin(x)+sin(y) \\le 2sin\\left( \\frac{x+y}{2} \\right)$\nTherefore,\n$sin(A_{1}-30^{\\circ})+sin(A_{2}-30^{\\circ}) \\le 2sin\\left( \\frac{A_{1}+A_{2}-60^{\\circ}}{2} \\right)$\nand also,\n$sin(A_{3}-30^{\\circ})+sin(30^{\\circ}) \\le 2sin\\left( \\frac{A_{3}}{2} \\right)$\nAdding these two inequalities we get:\n$sin(30^{\\circ})+\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 2\\left[ sin\\left( \\frac{A_{1}+A_{2}-60^{\\circ}}{2} \\right)+sin\\left( \\frac{A_{3}}{2} \\right) \\right]$\n$\\frac{1}{2}+\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 2\\left[ 2sin\\left( \\frac{A_{1}+A_{2}+A_{3}-60^{\\circ}}{4} \\right) \\right]$\n$\\frac{1}{2}+\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 2\\left[ 2sin\\left( \\frac{180^{\\circ}-60^{\\circ}}{4} \\right) \\right]$\n$\\frac{1}{2}+\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 4sin\\left( 30^{\\circ} \\right)$\n$\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le \\frac{3}{2}$\n$2\\sum_{i=1}^{3}sin(A_{i}-30^{\\circ})\\le 3$.\n$\\sum_{i=1}^{3}\\frac{sin(A_{i}-30^{\\circ})}{sin(30^{\\circ})}\\le 3$. [Inequality 2]\nCombining [Inequality 1] and [Inequality 2] we see the following:\n$\\sum_{i=1}^{3}\\frac{sin(A_{i}-30^{\\circ})}{sin(30^{\\circ})}\\le \\sum_{i=1}^{3}\\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}$\nThis implies that for at least one of the values of $i=1$,$2$,or $3$, the following is true:\n$\\frac{sin(A_{i}-30^{\\circ})}{sin(30^{\\circ})}\\le \\frac{sin(A_{i}-\\alpha_{i})}{sin(\\alpha_{i})}$\nor\n$\\frac{sin(\\alpha_{i})}{sin(A_{i}-\\alpha_{i})}\\le \\frac{sin(30^{\\circ})}{sin(A_{i}-30^{\\circ})}$\nWhich means that for at least one of the values of $i=1$,$2$,or $3$, the following is true:\n$\\alpha_{i} \\le 30^{\\circ}$\nTherefore, at least one of the angles $\\,\\angle PAB,\\;\\angle PBC,\\;\\angle PCA\\,$ is less than or equal to $30^{\\circ }$.\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.", "Solution_2": "At least one of $\\angle ABC, \\angle BCA, \\angle CAB \\ge 60^\\circ$. Without loss of generality, assume that $\\angle BCA \\ge 60^\\circ$\nIf $\\angle PAB > 30^\\circ$ and $\\angle PBC > 30^\\circ$\nDraw a circle $R$ centered at $O$ and passing through $A, P, B$. Since $P$ is an interior point of $\\triangle ABC$, thus $C$ is outside the circle $R$\nDraw two lines $CD, CE$ passing through $C$ and tangent to $R$. Line $CD$ intersect $R$ at $D$, and line $CE$ intersect $R$ at $E$. Choose $D$ near $A$, and choose $E$ near $B$\nExtends line $BC$, and intersect $R$ at $F$ other than $B$ when $BC$ is not tangent to $R$. If $BC$ is tangent to $R$, we have $B = E$ be the tangent point, and simply let $F = B = E$\nDraw the segment $OE$, and choose a point $G$ on $R$ such that $\\angle GOE = 60^\\circ$. There are two possible points, we choose $G$ near point $P$. Draw segments $OG, GE$, thus $\\triangle GOE$ is an equilateral triangle\nDraw segments $OP, OC, OB, OF, PB, GC$\n$\\angle OCE = \\dfrac{1}{2} \\angle DCE \\ge \\dfrac{1}{2} \\angle BCA \\ge 30^\\circ$. Then we have $\\angle COE = 90^\\circ - \\angle OCE \\le 60^\\circ = \\angle GOE$\n$\\angle POB = 2 \\angle PAB > 60^\\circ, \\angle POF = 2 \\angle PBC > 60^\\circ$, since we have either $\\angle POE \\ge \\angle POB$ or $\\angle POE \\ge \\angle POF$, thus $\\angle POE > 60^\\circ = \\angle GOE$\nThus we have $\\angle COE \\le \\angle GOE < \\angle POE$, then $\\angle OCE \\le \\angle GCE < \\angle PCE$\nBecause $\\angle GCE \\ge \\angle OCE \\ge 30^\\circ = \\angle GEC$, thus $GC \\le GE = OG$, and $\\angle GCO \\ge \\angle GOC$\nFinally, $\\angle PCA = \\angle ACE - \\angle PCE < \\angle ACE - \\angle GCE = \\angle ACO - \\angle GCO$\nSince $\\angle ACO \\le \\angle DCO$, and $\\angle GCO \\ge \\angle GOC$, thus we have $\\angle PCA < \\angle ACO - \\angle GCO \\le \\angle DCO - \\angle GOC = 90^\\circ - \\angle COE - \\angle GOC = 90^\\circ - \\angle GOE = 30^\\circ$\nWe have proved that when $\\angle PAB > 30^\\circ$ and $\\angle PBC > 30^\\circ$, the angle $\\angle PCA$ must be less than $30^\\circ$. Thus at least one of $\\angle PAB, \\angle PBC, \\angle PCA$ should less than or equal to $30^\\circ$\n~Joseph Tsai, mgtsai@gmail.com" }, { "Problem": "Let $S$ be a finite set of points in three-dimensional space. Let $S_{x}$,$S_{y}$,$S_{z}$, be the sets consisting of the orthogonal projections of the points of $S$ onto the $yz$-plane, $zx$-plane, $xy$-plane, respectively. Prove that\n\nwhere $|A|$ denotes the number of elements in the finite set $|A|$. (Note: The orthogonal projection of a point onto a plane is the foot of the perpendicular from that point to the plane)", "Solution": "Let $Z_{i}$ be planes with index $i$ such that $1 \\le i \\le n$ that are parallel to the $xy$-plane that contain multiple points of $S$ on those planes such that all points of $S$ are distributed throughout all planes $Z_{i}$ according to their $z$-coordinates in common.\nLet $a_{i}$ be the number of unique projected points from each $Z_{i}$ to the $yz$-plane\nLet $b_{i}$ be the number of unique projected points from each $Z_{i}$ to the $xz$-plane\nThis provides the following: $|Z_{i}| \\le a_{i}b_{i}\\;$ [Equation 1]\nWe also know that $|S|=\\sum_{i=1}^{n}|Z_{i}|\\;$ [Equation 2]\nSince $a_{i}$ be the number of unique projected points from each $Z_{i}$ to the $yz$-plane,\nif we add them together it will give us the total points projected onto the $yz$-plane.\nTherefore, $|S_{x}|=\\sum_{i=1}^{n}a_{i}\\;$ [Equation 3]\nlikewise, $|S_{y}|=\\sum_{i=1}^{n}b_{i}\\;$ [Equation 4]\nWe also know that the total number of elements of each $Z_{i}$ is less or equal to the total number of elements in $S_{z}$\nThat is, $|Z_{i}| \\le |S_{z}|\\;$ [Equation 5]\nMultiplying [Equation 1] by [Equation 5] we get:\n$|Z_{i}|^{2} \\le a_{i}b_{i}|S_{z}|$\nTherefore, $|Z_{i}| \\le \\sqrt{a_{i}b_{i}}\\sqrt{|S_{z}|}$\nAdding all $|Z_{i}|$ we get:\n$\\sum_{i=1}^{n}|Z_{i}| \\le \\sqrt{|S_{z}|}\\sum_{i=1}^{n}\\sqrt{a_{i}b_{i}}\\;$[Equation 6]\nSubstituting [Equation 2] into [Equation 6] we get:\n$|S| \\le \\sqrt{|S_{z}|}\\sum_{i=1}^{n}\\sqrt{a_{i}b_{i}}$\n$|S|^{2} \\le |S_{z}| \\left( \\sum_{i=1}^{n}\\sqrt{a_{i}b_{i}} \\right)^{2}$\nSince, $\\sum_{i=1}^{n}\\sqrt{a_{i}b_{i}} \\le \\sqrt{\\sum_{i=1}^{n}a_{i}}\\sqrt{\\sum_{i=1}^{n}b_{i}}$,\nThen, $|S|^{2} \\le |S_{z}| \\left( \\sqrt{\\sum_{i=1}^{n}a_{i}}\\sqrt{\\sum_{i=1}^{n}b_{i}} \\right)^{2}$\n$|S|^{2} \\le |S_{z}| \\left( \\sum_{i=1}^{n}a_{i}\\right)\\left( \\sum_{i=1}^{n}b_{i}\\right)\\;$ [Equation 7]\nSubstituting [Equation 3] and [Equation 4] into [Equation 7] we get:\n$|S|^{2} \\le |S_{x}| \\cdot |S_{y}| \\cdot |S_{z}|$\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": "Let $\\mathbb{N} = \\{1,2,3, \\ldots\\}$. Determine if there exists a strictly increasing function $f: \\mathbb{N} \\mapsto \\mathbb{N}$ with the following properties:\n(i) $f(1) = 2$;\n(ii) $f(f(n)) = f(n) + n, (n \\in \\mathbb{N})$.", "Solution": "Here is my Solution https://artofproblemsolving.com/community/q2h62193p16226748\nFind as ≈ Ftheftics" }, { "Problem": "Let $S$ be the set of real numbers strictly greater than $-1$. Find all functions $f:S \\to S$ satisfying the two conditions:\n1. $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$ for all $x$ and $y$ in $S$;\n2. $\\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1 0.$ We get $t = 2s + s^2 > s.$ Since $\\frac{f(x)}{x}$ is strictly increasing for $x>0$ and $t > s$ in this domain, we must have $\\frac{f(t)}{t} > \\frac{f(s)}{s}$ but since $f(s) = s$ and $f(t) = t,$ we also have that $\\frac{f(s)}{s} = 1 = \\frac{f(t)}{t}$ which is a contradiction. Therefore $s \\leq 0$\nConsider $-1 < s < 0.$ Using a similar argument, we will get that $t = 2s + s^2 < s$ but also $\\frac{f(s)}{s} = 1 = \\frac{f(t)}{t},$ which is a contradiction.\nHence, $s$ must be $0.$ Since $s=0,$ we can conclude that $x+f(x)+xf(x) = 0$ and therefore, $f(x) = \\frac{-x}{x+1}$ for all $x$.\n" }, { "Problem": "Let $ABCDEF$ be a convex hexagon with $AB=BC=CD$ and $DE=EF=FA$, such that $\\angle BCD=\\angle EFA=\\frac{\\pi}{3}$. Suppose $G$ and $H$ are points in the interior of the hexagon such that $\\angle AGB=\\angle DHE=\\frac{2\\pi}{3}$. Prove that $AG+GB+GH+DH+HE\\ge CF$.", "Solution": "Draw $AE$ and $BD$ to make equilateral $\\triangle EFA$ and $\\triangle BCD$, and draw points $I$ and $J$ such that $IA=IB$, $JD=JE$, directed angle $\\measuredangle IAB=-\\measuredangle CDB$, and directed angle $\\measuredangle JDE=-\\measuredangle FAE$ to make equilateral $\\triangle AIB$ and $\\triangle DJE$. Notice that $G$ is on the circumcircle of $\\triangle AIB$ and $H$ is on the circumcircle of $\\triangle DJE$. By Ptolemy, $GA+GB=GI$ and $HD+HE=HJ$. So, Notice that octagon $AIBCDJEF$ is symmetric about $\\overline{BE}$. So, $IG+GH+HJ\\ge IJ=CF$. --tigerzhang" }, { "Problem": "Let $ABCDEF$ be a convex hexagon such that $AB$ is parallel to $DE$, $BD$ is parallel to $EF$, and $CD$ is parallel to $FA$. Let $R_{A}$, $R_{C}$, $R_{E}$ denote the circumradii of triangles $FAB$, $BCD$, $DEF$, respectively, and let $P$ denote the perimeter of the hexagon. Prove that\n$R_{A}+R_{C}+R_{E} \\ge \\frac{P}{2}$", "Solution": "Let $s_{1}=\\left| AB \\right|,\\;s_{2}=\\left| BC \\right|,\\;s_{3}=\\left| CD \\right|,\\;s_{4}=\\left| DE \\right|,\\;s_{5}=\\left| EF \\right|,\\;s_{6}=\\left| FA \\right|$\nLet $d_{1}=\\left| FB \\right|,\\;d_{2}=\\left| BD \\right|,\\;d_{1}=\\left| DF \\right|$\nLet $\\alpha_{1}=\\angle FAB,\\;\\alpha_{2}=\\angle ABC,\\;\\alpha_{3}=\\angle BCD,\\;\\alpha_{4}=\\angle CDE,\\;\\alpha_{5}=\\angle DEF,\\;\\alpha_{6}=\\angle EFA\\;$\nFrom the parallel lines on the hexagon we get:\n$\\alpha_{1}=\\alpha_{4},\\;\\alpha_{2}=\\alpha_{5},\\;\\alpha_{3}=\\alpha_{6}$ [Equations 1]\nSo now we look at $\\Delta FAB$. We construct a perpendicular from $A$ to $FE$ and a perpendicular from $A$ to $BC$.\nWe find out the length of these two perpendiculars and add them to get the distance between parallel lines $FE$ and $BC$ and because of the triangle inequality the distance $\\left| FB \\right|$ is greater or equal to tha the distance between parallel lines $FE$ and $BC$:\nThis provides the following inequality:\n$d_{1} \\ge s_{6}sin(\\alpha{6})+s_{1}sin(\\alpha{2})$\nUsing the [Equations 1] we simplify to:\n$d_{1} \\ge s_{6}sin(\\alpha{3})+s_{1}sin(\\alpha{2})$ [Equation 2]\nWe now construct a perpendicular from $D$ to $FE$ and a perpendicular from $D$ to $BC$. Then we find out the length of these two perpendiculars and add them to get the distance between parallel lines $FE$ and $BC$ and get:\n$d_{1} \\ge s_{3}sin(\\alpha{3})+s_{4}sin(\\alpha{5})$\nUsing the [Equations 1] we simplify to:\n$d_{1} \\ge s_{3}sin(\\alpha{3})+s_{4}sin(\\alpha{2})$ [Equation 3]\nWe now add [Equation 2] and [Equation 3] to get:\n$2d_{1} \\ge (s_{3}+s_{6})sin(\\alpha_{3})+(s_{1}+s_{4})sin(\\alpha_{2})$ [Equation 4]\nWe now use the Extended law of sines on $\\Delta FAB$ with $R_{A}$ to get:\n$\\frac{d_{1}}{sin(\\alpha_{1})}=2R_{A}$\n$d_{1}=2R_{A}sin(\\alpha_{1})$ [Equation 5]\nSubstitute [Equation 5] into [Equation 4]:\n$4R_{A}sin(\\alpha_{1}) \\ge (s_{3}+s_{6})sin(\\alpha_{3})+(s_{1}+s_{4})sin(\\alpha_{2})$\n$4R_{A} \\ge (s_{3}+s_{6})\\frac{sin(\\alpha_{3})}{sin(\\alpha_{1})}+(s_{1}+s_{4})\\frac{sin(\\alpha_{2})}{sin(\\alpha_{1})}$ [Equation 6]\nTo find the equivalent inequality for $4R_{C}$ and $4R_{E}$ we just need shift the indexes by two. That is to add $2$ to each of the indexes of $s_{i}$ and $\\alpha_{i}$ and adjust the indexes so that for $s$ the indexes are 1 through 6, and for $\\alpha$ the indexes are 1 through 3.\n$4R_{C} \\ge (s_{2}+s_{5})\\frac{sin(\\alpha_{2})}{sin(\\alpha_{3})}+(s_{3}+s_{6})\\frac{sin(\\alpha_{1})}{sin(\\alpha_{3})}$ [Equation 7]\n$4R_{E} \\ge (s_{1}+s_{4})\\frac{sin(\\alpha_{1})}{sin(\\alpha_{2})}+(s_{2}+s_{5})\\frac{sin(\\alpha_{3})}{sin(\\alpha_{2})}$ [Equation 8]\nAdding [Equation 6], [Equation 7], and [Equation 8] we get:\n$4\\left( R_{A}+R_{C}+R_{E} \\right) \\ge (s_{1}+s_{4})\\left( \\frac{sin(\\alpha_{2})}{sin(\\alpha_{1})}+ \\frac{sin(\\alpha_{1})}{sin(\\alpha_{2})}\\right)+(s_{2}+s_{5})\\left( \\frac{sin(\\alpha_{2})}{sin(\\alpha_{3})}+ \\frac{sin(\\alpha_{3})}{sin(\\alpha_{2})}\\right)+(s_{3}+s_{6})\\left( \\frac{sin(\\alpha_{3})}{sin(\\alpha_{1})}+ \\frac{sin(\\alpha_{1})}{sin(\\alpha_{3})}\\right)$\nFrom AM-GM inequality we get:\n$x+\\frac{1}{x} \\ge 2\\;$ Therefore, $\\left( \\frac{sin(\\alpha_{a})}{sin(\\alpha_{b})}+ \\frac{sin(\\alpha_{b})}{sin(\\alpha_{a})}\\right) \\ge 2\\;$ for any index $a$ and $b$\nTherefore,\n$4\\left( R_{A}+R_{C}+R_{E} \\right) \\ge 2(s_{1}+s_{4})+2(s_{2}+s_{5})+2(s_{3}+s_{6})$\n$4\\left( R_{A}+R_{C}+R_{E} \\right) \\ge 2\\sum_{i=1}^{6}s_{i}$\n$R_{A}+R_{C}+R_{E} \\ge \\frac{1}{2}\\sum_{i=1}^{6}s_{i}$\nSince $P=\\sum_{i=1}^{6}s_{i}$, then $R_{A}+R_{C}+R_{E} \\ge \\frac{P}{2}$\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": "Find all pairs $(a,b)$ of integers $a,b \\ge 1$ that satisfy the equation\n$a^{b^{2}}=b^{a}$", "Solution": "Case 1: $(1 \\le a \\le b)$\n$(a^{b})^{b}=b^{a}$\nLooking at this expression since $b \\ge a$ then $a^{b} \\le b$.\nHere we look at subcase $a>1$ which gives $a^{b}>b$ for all $(1 < a \\le b)$. This contradicts condition $a^{b} \\le b$, and thus $a$ can't be more than one giving the solution of $a=1$ with $b \\ge 1$. So we substitute the value of $a=1$ into the original equation to get $1^{b^2}=b^{1}$ which solves to $b=1$ and our first pair $(a,b)=(1,1)$\nCase 2: $(1 \\le b < a)$\n$a^{b^2}=b^{a}$\nsince $a>b$, then $b^{2}1$ and $b^{a-2b^{2}}>0$. This gives $a-2b^{2}>1$\nThis implies that $a>2b^{2}$ for $b>1$\nLet $a=kb^{2}$ with $k \\in \\mathbb{Z} ^{+}$. Since $a>2b^{2}$, then $k \\ge 3$\n$(kb^{2}b^{-2})^{b^{2}}=b^{kb^{2}-2b^{2}}$\n$k^{b^{2}}=b^{(k-2)b^{2}}$\n$k=b^{k-2}$, which gives $b \\ge 2$\nsubcase $k=3$:\n$3=b^{3-2}=b$ and $a=kb^{2}=(3)(3)^{2}=27$. which provides 2nd pair $(a,b)=(27,3)$\nsubcase $k=4$:\n$4=b^{4-2}=b^{2}$, thus $b=2$ and $a=kb^{2}=(4)(2)^{2}=16$. which provides 3rd pair $(a,b)=(16,2)$\nsubcase $k \\ge 5$:\n$k=b^{k-2}$, thus $b=k^{1/(k-2)}$ which decreases with $k$ and $b \\to 1$ as $k \\to \\infty$ . From subcase $k=4$, we know that $b=2$, thus for subcase $k \\ge 5$, $11$ with integer coefficients, and let $k$ be a positive integer. Consider the polynomial $Q(x) = P( P ( \\ldots P(P(x)) \\ldots ))$, where $P$ occurs $k$ times. Prove that there are at most $n$ integers $t$ such that $Q(t)=t$.", "Solution": "We use the notation $P^k(x)$ for $Q(x)$.\nLemma 1. The problem statement holds for $k=2$.\nProof. Suppose that $a_1, \\dotsc, a_k$, $b_1, \\dotsc, b_k$ are integers such that $P(a_j) = b_j$ and $P(b_j) = a_j$ for all indices $j$. Let the set $\\{ a_1, \\dotsc, a_k, b_1, \\dotsc, b_k \\}$ have $m$ distinct elements. It suffices to show that $\\deg (P) \\ge m$.\nIf $a_j = b_j$ for all indices $j$, then the polynomial $P(x)-x$ has at least $m$ roots; since $P$ is not linear, it follows that $\\deg P \\ge m$ by the division algorithm.\nSuppose on the other hand that $a_i \\neq b_i$, for some index $i$. In this case, we claim that $a_j + b_j$ is constant for every index $j$. Indeed, we note that\n\nso $\\lvert a_j - a_i \\rvert = \\lvert b_j - b_i \\rvert$. Similarly,\n\nso $\\lvert a_j - b_i \\rvert = \\lvert b_j - a_i \\rvert$. It follows that $a_j + b_j = a_i + b_i$.\nThis proves our claim. It follows that the polynomial\n\nhas at least $m$ roots. Since $P$ is not linear it follows again that $\\deg P \\ge m$, as desired. Thus the lemma is proven. $\\blacksquare$\nLemma 2. If $a$ is a positive integer such that $P^r(a)$ for some positive integer $r$, then $P^2(a) = a$.\nProof. Let us denote $a_0 = a$, and $a_j = P^j(a)$, for positive integers $j$. Then $a_0 = a_r$, and\n\nIt follows that $\\lvert a_{j+1} - a_j \\rvert$ is constant for all indices $j$; let us abbreviate this quantity $d$. Now, since\n\nit follows that for some index $j$,\n\nor $a_j = a_{j+2} = P^2(a_j)$. Since $a = a_r = P^{r-j}(a_j)$, it then follows that $P^2(a) = a$, as desired. $\\blacksquare$\nNow, if there are more than $n$ integers $t$ for which $Q(t) = t$, then by Lemma 2, there are more than $n$ integers $t$ such that $P^2(t) = t$, which is a contradiction by Lemma 1. Thus the problem is solved. $\\blacksquare$\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "(Kevin Buzzard and Edward Crane, United Kingdom)\nLet $a$ and $b$ be positive integers. Show that if $4ab-1$ divides $(4a^2-1)^2$, then $a=b$.", "Solution": "Lemma. If there is a counterexample for some value of $a$, then there is a counterexample $(a,b)$ for this value of $a$ such that $ba$. Note that $4ab -1 \\equiv -1 \\pmod{4a}$, but $(4a^2-1)^2 \\equiv 1 \\pmod{4a}$. It follows that $(4a^2-1)^2/(4ab-1) \\equiv -1 \\pmod{4a}$. Since\n\nit follows that $(4a^2-1)^2/(4ab-1)$ can be written as $4ab'-1$, with $01$. Then $f$ is a periodic function. But then $f$ is bounded by a constant $M$, i.e. $f(x)2M$. We get that $a,f(b)$ and $f(b+f(a)-1)$ are sides of the triangle, but the first number is greater than $2M$ and other two are less than $M$, which is imposible. We get the contradiction, so $u$ could not be greater than 1.\nSo $u=f(1)=1$.\nProperty 1. For any $x$\nProof. Consider the given condition for $a=x$, $b=1$ and use Lemma.\nProperty 2. For any $x$ and $y$\nProof. Consider the given condition for $a=f(x+1)$, $b=y$ and use triangle inequality and Property 1.\nLet $g(x)=f(x+1)-1$.\nProperty 3. For any $x$ and $y$\nProof. Follows from Property 2.\nProperty 4. For any $x$, $y$, $m$, $n$\nProof. Follows from Property 3.\nProperty 5. For any $n$, there is $k>n$, s.t. $g(k) \\ge k$.\nProof. Because of the Property 1.\nProperty 6. $g(x)=x$.\nProof. Suppose, for some $x$, $g(x)=y\\ne x$. Without lost of generality we can assume that $xn$ could be represented as $k=ax+by$, where $a>b$. Then $g(k)=g(ax+by)\\ge ag(x)+bg(y)=ay+bx>ax+by=k$. That contradicts to the Property 5.\nTherefore by definion of $g$, $f(x)=x$." }, { "Problem": "Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed\nType 1) Choose a non-empty box $B_j$, $1\\leq j \\leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$;\nType 2) Choose a non-empty box $B_k$, $1\\leq k \\leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$.\nDetermine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.\nAuthor: Hans Zantema", "Solution": "Let the notation $[a_1,a_2,a_3,a_4,a_5,a_6]$ be the configuration in which the $x$-th box has $a_x$ coin,\nLet $T=2010^{2010^{2010}}$.\n\nOur starting configuration is $[1,1,1,1,1,1]$\n\nCompound move 1: $[a,0]\\rightarrow[0,2a]$, this is just repeated type 1 move on all $a$ coins.\nCompound move 2: $[a,0,0]\\rightarrow[0,2^a,0]$, apply type 1 move on 1 of the coin to get $[a-1,2,0]$, then apply compound move 1 to the 2 coins to get $[a-1,0,2^2]$, apply type 2 move and get $[a-2,2^2,0]$, and repeat compound move 1 and type 2 move until $[0,2^a,0]$ is achieve.\nCompound move 3: $[a,b,0,0]\\rightarrow[a-1,2^b,0,0]$, apply compound move 2 to obtain $[a,0,2^b,0]$ and use type 2 move to get $[a-1,2^b,0,0]$\nCompound move 4: $[a,0,0,0]\\rightarrow[0,2^{2^{.^{.^{.^2}}}},0,0]$ with $a$ $2$'s. Apply compound move 3 $a$ times.\nCompound move 5: $[a,0,0]\\rightarrow[b,0,0]$ with $a>b$, use type 2 move $(a-b)$ times.\nLet's follow this move:\nUsing Compound move 1, 4 and 5, We can obtain:\n$[0,1,19,0,0,0]\\rightarrow[0,1,0,X,0,0]\\rightarrow[0,0,X,0,0,0]\\rightarrow[0,0,0,Y,0,0]\\rightarrow[0,0,0,T/4,0,0]\\rightarrow[0,0,0,0,T/2,0]\\rightarrow[0,0,0,0,0,T]$\n, where $X, Y = 2^{2^{.^{.^{.^2}}}}$ where X has $19$ 2's and Y has $X$ 2's, and $Y$ is clearly bigger then $T/4$\n" }, { "Problem": "Let $ABC$ be a triangle with $\\angle BCA=90^{\\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M=\\overline{AL}\\cap \\overline{BK}$. Prove that $MK=ML$.", "Solution": "Let $\\Gamma$, $\\Gamma'$, $\\Gamma''$ be the circumcircle of triangle $ABC$, the circle with its center as $A$ and radius as $AC$, and the circle with its center as $B$ and radius as $BC$, Respectively.\nSince the center of $\\Gamma$ lies on $BC$, the three circles above are coaxial to line $CD$.\nLet Line $AX$ and Line $BX$ collide with $\\Gamma$ on $P$ ($\\neq A$) and $Q$ ($\\neq B$), Respectively. Also let $R = AQ \\cap BP$.\nThen, since $\\angle AYB = \\angle AZB = 90^{\\circ}$, by ceva's theorem, $R$ lies on $CD$.\nSince triangles $ABC$ and $ACD$ are similar, $AL^2 = AC^2 = AD \\cdot AB$, Thus $\\angle ALD = \\angle ABL$.\nIn the same way, $\\angle BKD = \\angle BAK$.\nTherefore, $\\angle ARD = \\angle ABQ = \\angle ALD$ making $(A, R, L, D)$ concyclic.\nIn the same way, $(B, R, K, D)$ is concyclic.\nSo $\\angle ADR = \\angle ALR = 90^{\\circ}$, and in the same way $\\angle BKR = 90^{\\circ}$. Therefore, the lines $RK$ and $RL$ are tangent to $\\Gamma'$ and $\\Gamma''$, respectively.\nSince $R$ is on $CD$, and $CD$ is the concentric line of $\\Gamma'$ and $\\Gamma''$, $RK^2 = RL^2$, Thus $RK = RL$. Since $RM$ is in the middle and $\\angle ADR = \\angle BKR = 90^{\\circ}$,\nwe can say triangles $RKM$ and $RLM$ are congruent. Therefore, $KM = LM$.\nEdit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here.\n~ Latex edit by Kscv\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Let $\\mathbb Q_{>0}$ be the set of all positive rational numbers. Let $f:\\mathbb Q_{>0}\\to\\mathbb R$ be a function satisfying the following three conditions:\n(i) for all $x,y\\in\\mathbb Q_{>0}$, we have $f(x)f(y)\\geq f(xy)$;\n(ii) for all $x,y\\in\\mathbb Q_{>0}$, we have $f(x+y)\\geq f(x)+f(y)$;\n(iii) there exists a rational number $a>1$ such that $f(a)=a$.\nProve that $f(x)=x$ for all $x\\in\\mathbb Q_{>0}$.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "For each positive integer $n$, the Bank of Cape Town issues coins of denomination $\\tfrac{1}{n}$. Given a finite collection of such coins (of not necessarily different denominations) with total value at most $99+\\tfrac{1}{2}$, prove that it is possible to split this collection into $100$ or fewer groups, such that each group has total value at most $1$.", "Solution": "The bound is not tight. We'll prove the result for at most $k - \\frac{k}{2k+1}$ with $k$ groups.\nFirst, perform the following optimizations.\n- If any coin of size $\\frac{1}{2m}$ appears twice, then replace it with a single coin of size $\\frac{1}{m}$.\n- If any coin of size $\\frac{1}{2m+1}$ appears $2m+1$ times, group it into a single group and induct downwards.\nApply this operation repeatedly until it cannot be done anymore.\nNow construct boxes $B_0$, $B_1$, ...., $B_{k-1}$. In box $B_0$ put any coins of size $\\tfrac 12$ (clearly there is at most one).\nIn the other boxes $B_m$, put coins of size $\\frac{1}{2m+1}$ and $\\frac{1}{2m+2}$ (at most $2m$ of the former and at most one of the latter).\nNote that the total weight in the box is less than $1$.\nFinally, place the remaining ``light coins of size at most $\\frac{1}{2k+1}$ in a pile.\nThen just toss coins from the pile into the boxes arbitrarily, other than the proviso that no box should have its weight exceed $1$.\nWe claim this uses up all coins in the pile. Assume not, and that some coin remains in the pile when all the boxes are saturated.\nThen all the boxes must have at least $1 -\\frac{1}{2k+1}$, meaning the total amount in the boxes is strictly greater than\n\nwhich is a contradiction. (The inequality is strict since the pile has a coin leftover.)\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 $f$:$\\mathbb{R}\\rightarrow\\mathbb{R}$ satisfying the equation\n$f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)$\nfor all real numbers $x$ and $y$.\nProposed by Dorlir Ahmeti, Albania", "Solution": "$f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)$ for all real numbers $x$ and $y$.\n(1) Put $x=y=0$ in the equation,\nWe get$f(0 + f(0)) + f(0) = 0 + f(0) + 0$\nor $f(f(0)) = 0$\nLet $f(0) = k$, then $f(k) = 0$\n(2) Put $x=0, y=k$ in the equation,\nWe get $f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)$\nBut $f(k) = 0$ and $f(0) = k$\nso, $f(0) + f(0) = f(0)^2$\nor $f(0)[f(0) - 2] = 0$\nHence $f(0) = 0, 2$\nCase $1$ : $f(0) = 0$\nPut $x=0, y=x$ in the equation,\nWe get $f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)$\nor, $f(f(x)) = f(x)$\nSay $f(x) = z$, we get $f(z) = z$\nSo, $f(x) = x$ is a solution -- fallacy\nCase $2$ : $f(0) = 2$\nAgain put $x=0, y=x$ in the equation,\nWe get $f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)$\nor, $f(f(x)) + 2 = f(x) + 2x$\nWe observe that $f(x)$ must be a polynomial of power $1$ as any other power (for that matter, any other function) will make the $LHS$ and $RHS$ of different powers and will not have any non-trivial solutions. -- fallacy\nAlso, if we put $x=0$ in the above equation we get $f(2) = 0$\n$f(x) = 2-x$ satisfies both the above.\nHence, the solutions are $\\boxed{\\color{red}{f(x) = x}}$ and $\\boxed{\\color{red}{f(x) = 2-x}}$." }, { "Problem": "The equation\nis written on the board, with $2016$ linear factors on each side. What is the least possible value of $k$ for which it is possible to erase exactly $k$ of these $4032$ linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "An integer $N \\ge 2$ is given. A collection of $N(N + 1)$ soccer players, no two of whom are of the same height, stand in a row. Sir Alex wants to remove $N(N - 1)$ players from this row leaving a new row of $2N$ players in which the following $N$ conditions hold:\n($1$) no one stands between the two tallest players,\n($2$) no one stands between the third and fourth tallest players,\n$\\;\\;\\vdots$\n($N$) no one stands between the two shortest players.\nShow that this is always possible.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $a_1, a_2, \\dots$ be an infinite sequence of positive integers. Suppose that there is an integer$N > 1$ such that, for each $n \\geq N$, the number $\\frac{a_1}{a_2}+\\frac{a_2}{a_3}+\\dots +\\frac{a_{n-1}}{a_n}+\\frac{a_n}{a_1}$ is an integer. Prove that there is a positive integer $M$ such that $a_m = a_{m+1}$ for all $m \\geq M.$", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "The Bank of Bath issues coins with an $H$ on one side and a $T$ on the other. Harry has $n$ of these coins arranged in a line from left to right. He repeatedly performs the following operation:\nIf there are exactly $k > 0$ coins showing $H$, then he turns over the $k^{th}$ coin from the left; otherwise, all coins show $T$ and he stops. For example, if $n = 3$ the process starting with the configuration $THT$ would be $THT \\rightarrow HHT \\rightarrow HTT \\rightarrow TTT$, which stops after three operations.\n(a) Show that, for each initial configuration, Harry stops after a finite number of operations.\n(b) For each initial configuration $C$, let $L(C)$ be the number of operations before Harry stops. For\nexample, $L(THT) = 3$ and $L(TTT) = 0$. Determine the average value of $L(C)$ over all $2^n$\npossible initial configurations $C$.", "Solution_1_.28Induction_and_Case_Division.29": "Don't worry, this is way simpler than it's length may suggest :)\nClaim: The expected value of $L(C)$ is $\\frac{n(n+1)}{4}$.\nWe prove parts A and B simultaneously using induction.\nBase case: $n=1$\n$L(T)=0$ and $L(H)=1$, and these are the only possibilities, so clearly the process terminates and $E[L(C)] = \\frac{1}{2} = \\frac{1*2}{4}$.\nInductive step: Assume that for $n-1$ coins, the process always terminates, and $E[L(C)] = \\frac{n(n-1)}{4}$. Define $E_{n-1} = E[L(C)]$ here. Now, for $n$ coins, define $E_{n} = E[L{C}]$ across all $2^n$ configurations. Consider cases on the last coin:\nCase 1: Final coin is $T$\nIf the last coin is $T$, then we simply can forget about it, as we can flip it to heads only if we have n heads, which is impossible. This occurs with probablity $P_{T} = 1/2$, and will be identical to the general $n-1$ case; thus, this process terminates, and we will have an expected number of moves here of $E_{n-1}$.\nCase 2: Final coin is $H$\nThe probability that this occurs is $P_{H} = 1/2$. The only way we have for the process to terminate now is for all coins to show heads, after which we can progressively flip each coin from right to left (this will take $n$ moves). Thus, we now have a new goal for the first $n-1$ coins (given the head at the end); get all heads.\nWe claim that this process is symmetrical to the original process on $n-1$ coins, and thus it terminates and has the expected number of moves $E_{n-1}$.\nProof: Let us have $k \\ge 1$ heads in our group of n coins, of which $k-1$ are in the first n-1. We are supposed to flip the $k$'th coin from the left in the first n-1 coins, which is equivalent to flipping the $(n-k)$'th coin from the right (in the first $n-1$ coins, NOT all $n$ coins), but there are $n-k$ tails in the first $n-1$ coins! We now have a new process defined as follows:\nConsider a configuration of $n-1$ coins with $x \\ge 1$ tails, and flip the x'th coin from the right, and keep flipping until all coins are heads.\nThis is clearly symmetric to the original process, and will terminate if and only if the original process on $n-1$ coins does, and has the same expected number of moves as the original. Thus, the claim is proven.\nBack to the case: We now have a process to get to all heads that will terminate with an expected number of steps of $E_{n-1}$, and then an n-step algorithm to reach the finish. Thus, here also, the process terminates, and we have a total expected number of steps as $E_{n-1}+n$.\nThus, the overall process on n coins will terminate, since the process terminates in both the representative cases on n-1 coins. $\\boxed{Q.E.D}$ Also, the expected value of steps can be easily computed now; $E_{n} = (P_{T})(E_{n-1})+(P_{H})(E_{n-1}+n) = E_{n-1}+n/2 = \\frac{n(n-1}{4}+\\frac{n}{2} = \\boxed{\\frac{n(n+1)}{4}}$ as desired. We are now done.\n-Solution by thanosaops", "Solution_2_.28Strong_Induction_and_Alternative_Case_Division.29": "We define a sequence $\\{a_i\\}_{i=1}^n$: $a_i=1$ if the $i$'th coin is $H$ and $a_i=0$ if the $i$'th coin is $T$. Then, $k$ will be the number of $1$'s in $\\{a_i\\}$ and after each operation, $a_k$ will be changed to $1-a_k$. Let $f(n)=\\sum L\\{a_i\\}=$ sum of $L\\{a_i\\}$ over all possible $\\{a_i\\}$ (of length $n$).\nWe shall prove that the process terminates and find the value of $f(n)$ using strong induction.\n\nBase case: Clearly, for $n=1$, the process terminates.\nInduction hypothesis: For $n=1, 2, 3, \\cdots, k$, the process terminates.\nInductive step: We claim that for $n=k+1$, the process terminates as well.\n\nCase 1: $a_1=1$\nThis means that $\\{a_i\\}=\\{1, a_2, a_3, \\cdots, a_n\\}$. With the proof of the claim in Solution 1, we can similarly prove that operating on $\\{a_i\\}$ is equivalent to operating on $\\{a_2, a_3, \\cdots, a_n\\}$.\nBy our induction hypothesis, this process will terminate, and hence $\\{a_i\\}$ will be changed to $\\{1, 0, 0, \\cdots, 0\\}$ at the end of this process. Now, we operate one more time to change $a_1$ from $1$ to $0$. Hence, the process terminates.\nWe can also see that $L\\{a_i\\}=L\\{a_2, a_3, \\cdots, a_n\\}+1$. Hence, sum of $L\\{a_i\\}$ over all possible $\\{a_i\\}$ in this case\n\nCase 2: $a_1=0$ and $a_n=1$\nThis means that $\\{a_i\\}=\\{0, a_2, a_3, \\cdots, a_{n-1}, 1\\}$. Operating on $\\{a_i\\}$ is now equivalent to operating on $\\{a_2, a_3, \\cdots, a_{n-1}\\}$.\nBy our induction hypothesis, this process will terminate, and hence $\\{a_i\\}$ will be changed to $\\{0, 0, 0, \\cdots, 0, 1\\}$ at the end of this process. Now, since $k=1$, $a_1$ will become $1$. Now, $k=2$, so $a_2$ will become $1$ as well. This process will repeat until $k=n$ and $\\{a_i\\}$ has been changed to $\\{1, 1, 1, \\cdots, 1\\}$. Now, since $k=n$, $a_n$ will become $0$. Now, $k=n-1$, so $a_{n-1}$ will become $0$ as well. This process will repeat again until $k=0$ and $\\{a_i\\}$ has been changed to $\\{0, 0, 0, \\cdots, 0\\}$. Hence, the process terminates.\nWe can also see that $L\\{a_i\\}=L\\{a_2, a_3, \\cdots, a_{n-1}\\}+(2n-1)$. Hence, sum of $L\\{a_i\\}$ over all possible $\\{a_i\\}$ in this case\n\nCase 3: $a_n=0$\nThis means that $\\{a_i\\}=\\{a_1, a_2, a_3, \\cdots, a_{n-1}, 0\\}$. Operating on $\\{a_i\\}$ is now equivalent to operating on $\\{a_1, a_2, \\cdots, a_{n-1}\\}$.\nBy our induction hypothesis, this process will terminate, and hence $\\{a_i\\}$ will be changed to $\\{0, 0, 0, \\cdots, 0, 0\\}$ at the end of this process. Hence, the process terminates.\nWe can also see that $L\\{a_i\\}=L\\{a_1, a_2, \\cdots, a_{n-1}\\}$. Hence, sum of $L\\{a_i\\}$ over all possible $\\{a_i\\}$ in this case\n\nThis is already sufficient to prove part A, but we want to find $f(n)$ too, and we have overcounted!\nCase 4: $a_1=1$ and $a_n=0$ (This is what we have overcounted)\nThis means that $\\{a_i\\}=\\{1, a_2, a_3, \\cdots, a_{n-1}, 0\\}$. Operating on $\\{a_i\\}$ is now equivalent to operating on $\\{a_2, a_3, \\cdots, a_{n-1}\\}$.\nBy our induction hypothesis, this process will terminate, and hence $\\{a_i\\}$ will be changed to $\\{1, 0, 0, \\cdots, 0, 0\\}$ at the end of this process. Now, we operate one more time to change $a_1$ from $1$ to $0$. Hence, the process terminates.\nWe can also see that $L\\{a_i\\}=L\\{a_2, a_3, \\cdots, a_{n-1}\\}+1$. Hence, sum of $L\\{a_i\\}$ over all possible $\\{a_i\\}$ in this case\n\nHence, we have proven that for all configurations, the process terminates, $\\boxed{Q.E.D.}$ Now, all that remains is to calculate $f(n)$.\n\nThis can be rearranged to give us\n\nWhen $n=1$, $f(n)-n(n+1)\\cdot2^{n-2}=0$. This is good, because now we know that $\\forall n\\in\\mathbb{Z}^{+}, f(n)-n(n+1)\\cdot2^{n-2}=0$ i.e. $f(n)=n(n+1)\\cdot2^{n-2}$.\nNow the average will be $\\frac{n(n+1)\\cdot2^{n-2}}{2^n}=\\boxed{\\frac{1}{4}n(n+1)}$\n~IraeVid13" }, { "Problem": "Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$.\nProve that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a < k < b$.", "Solution": "We will start by introducing some notation.\nThe proof works by counting the parity of upcomming bad nuts and hinges on the fact that 2021 is odd.\nWe start by proving that at any point in time there are an odd number of bad\nnuts. Let $B$ be the number of bad nuts and $G$ the number of good nuts. A bad\nnut is either part of two increasing or two decreasing pairs. Let the number\nof increasing bad nuts be $B_1$ and decreasing bad nuts be $B_2$. A good nut\nis part of one increasing and one decreasing pair. Let $I$ and $D$ be the\nnumber of increasing and decreasing pairs respectively. Then\n$2I = G + 2B_1$\n$2D = G + 2B_2$\nsince we are double-counting each pair. Therefore $G$ must be even, and since $2021 = G + B$ is odd, $B$ must be odd.\nIf we can prove the existence of a bad current nut, we are done. We show that when a current nut is good, the number of bad upcomming nuts does not change parity after the move. Since there are and odd number of upcomming bad nuts before the first move (every nut is upcomming) and 0 upcomming bad nuts after the last move (there are 0 upcomming nuts), this will show not all current nuts can be good which completes the proof.\nWe now show that when the current nut is good, the number of bad upcomming nots does not change parity. Consider the $k$'th move and assume the current $n$ is good and lies in hole $H_i$. After the move, the current nut is no longer upcomming, but since it was asummed to be good, this does not contrubute to the number of bad upcomming nuts. The only nuts whose status can possibly change is the nuts in holes $H_{i-2}, H_{i-1}, H_{i+1}, H_{i+2}$. Since the current nut is good, its two neighbors $a,b$ in $H_{i-1}, H_{i+1}$ respectively are either both smaller or larger than $n$. We tackle the two cases seperately.\nCase $a,b < n$\nIn this case neither $a$ nor $b$ are upcomming nuts at the time of the move and their status after the move is \nirrelevant for the parity of bad upcomming nuts. Consider the nut $c$ in hole $H_{i-2}$. If $c$ is not upcomming, it is irrelevant. Assume $c$ is upcomming. Then $c > n$ and thus $c > a,b$. Therefore the nut pair $H_{i-2}, H_{i-1}$ is decreasing both before and after the move, so $c$ cannot change status. One can make an analogous argument for the nut on $H_{i+2}$. This completes the proof the the parity of upcomming bad nuts are unchanged in this case.\nCase $a,b > n$\nIn this case the nuts $a,b$ are upcomming, so their status matters. Let $c$ be the nut in $H_{i-2}$. If the status of either of the holes $H_{i-2}, H_{i-1}$ change, it changes for both of them. This is because the pair $H_{i-1}, H_i$ is decreasing both before and after the move, so for a status change to occur it must be the case that it is the pair $H_{i-2}, H_{i-1}$ which changes direction (which in turn changes the status for both holes). If swapping $a$ and $b$ changes the direction of $H_{i-2}, H_{i-1}$, then either $a < c < b$ or $b < c < a$ is true. In both cases, $c > n$, so $c$ is an upcomming nut. Since the nuts in both $H_{i-1}$ and $H_{i+1}$ are upcomming both before and after the move, changing the status of either $H_{i-2}$ or $H_{i-1}$ yields a status change for two upcomming nuts. A completely analogous argument can be made for $H_{i+1}, H_{i+2}$.\nThis shows that if some upcomming nut changes status after the move, then an even number of upcomming nuts change status. This preserves the parity of the number of upcomming bad nuts and completes the proof." }, { "Problem": "Find all triples $(a,b,p)$ of positive integers with $p$ prime and", "Solution": "Case 1: $b < p$\nCase 2: $p \\le b < 2p$\n\nCase 3: $2p \\le b$\nJust like in case 2, $b!$ is divisible by $p$ so $a$ must also be divisible by $p$. However, $b!$ and $a^p$ are also both divisible by $p^2$, so remainders modulo $p^2$ tell us that no solutions exist.\nConclusion:\nThe only solutions are $(a,b,p) = (2,2,2), (3,4,3)$.", "Solution_2": "I considered the cases:\n1) If \\(a\\) is even, then it must:\n2) If \\(a\\) is odd, then it must:\nExamining 1.a), we end up with an equation of the form \\(a^p = p+1\\), which has no integer solutions.\nIn case 1.b), \\(b!\\) can take values: 2, 6, 24, 120, 720, ..., so \\(b!+2\\) takes values: 4, 8, 26, 122, 722, .... We observe that the only perfect square is 4 among the possible cases, as for \\(b \\geq 5\\), the result ends in 2, which is not a perfect square. Therefore, we have the triple \\((2,2,2)\\).\nCase 2.a) yields \\(a^2 = 3\\), which is rejected.\nExamining the last case 2.b), we have for \\(b!\\) the values: 2, 6, 24, 120, 720, 5040,\nThe analysis of the last case is incomplete, which is why I wasn't initially sure about the number of triples. Therefore, with this approach (which is not strictly documented), we find the triples: \\((2,2,2), (3,4,3)\\)." }, { "Problem": "Let $n$ be a positive integer. A Japanese triangle consists of $1 + 2 + \\dots + n$ circles arranged in an equilateral triangular shape such that for each $i = 1$, $2$, $\\dots$, $n$, the $i^{th}$ row contains exactly $i$ circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with $n = 6$, along with a ninja path in that triangle containing two red circles.\n[Image to be inserted; also available in solution video]", "Solution": "https://www.youtube.com/watch?v=jZNIpapyGJQ [Video contains solutions to all day 2 problems]" }, { "Problem": "Two planes, $P$ and $Q$, intersect along the line $p$. The point $A$ is in the plane $P$, and the point ${C}$ is in the plane $Q$; neither of these points lies on the straight line $p$. Construct an isosceles trapezoid $ABCD$ (with $AB$ parallel to $DC$) in which a circle can be inscribed, and with vertices $B$ and $D$ lying in the planes $P$ and $Q$, respectively.", "Solution": "We first observe that we must have both lines $AB$ (which we shall denote $a$) and $DC$ (which we shall denote $c$) parallel to $p$, since if one of them is not, then neither can be and they must both intersect $p$ (since they are both coplanar with $p$), making them skew.\nNow we note since a circle can be inscribed in the trapezoid, we must have $AB + DC = AD + BC$, and since the trapezoid is isosceles, this implies that each of the trapezoid's legs has length equal to the average of the lengths of the bases.\nWe can find this average by dropping perpendicular $AA'$ to $c$ such that $A'$ is on $c$. The average will be $A'C$, which is one of the sides of the rectangle with sides on $a$ and $c$ with vertices at $A$ and ${C}$.\nWe now draw a circle with center ${C}$ that contains $A'$. The intersections of this circle with $a$ are the two possible values of $B$, from either of which it is trivial to determine the corresponding location for $D$. It is worth noting that the intersection points may concur (in which case there is only one distinct possibility (a square)), or they may not occur at all. Q.E.D.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\n" }, { "Problem": "Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ be the volume of the cylinder.\na) Prove that $V_1 \\neq V_2$;\nb) Find the smallest number $k$ for which $V_1 = kV_2$; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.", "Solution_1": "Part (a):\nLet $R$ denote the radius of the cone, and let $r$ denote the radius of the cylinder and sphere. Let $l$ denote the slant height of the cone, and let $h$ denote the height of the cone.\nConsider a plane that contains the axis of the cone. This plane will slice the cone and sphere into a circle $\\omega$ inscribed in an isosceles triangle $T$.\nThe area of $T$ may be computed in two different ways:\n\n\nFrom this, we deduce that $r = \\frac{Rh}{l+R}$.\nNow, we calculate our volumes:\n\n\nNow, we will compute the quantity $\\frac{3(l+R)^3}{\\pi R^5 h} (V_1 - V_2)$ and prove that it is always greater than $0$.\nLet $x = \\frac{h}{R}$. Clearly, $x$ can be any positive real number. Define $W_1 = \\frac{3(l+R)^3}{\\pi R^5 h} V_1$ and $W_2 = \\frac{3(l+R)^3}{\\pi R^5 h} V_2$. We will calculate $W_1$ and $W_2$ in terms of $x$ and then compute the desired quantity $W_1 - W_2$.\nIt is easy to see that:\nNow, let $u = \\sqrt{x^2+1}$. Since $x > 0$, it follows that $u > 1$. We now have:\nDefine $f(u) = W_1 - W_2$. It follows that:\nWe see that $f(u) > 8$ for all allowed values of $u$. Thus, $V_1 - V_2 > \\frac{8\\pi R^5 h}{3(l+R)^3}$, meaning that $V_1 > V_2$. We have thus proved that $V_1 \\ne V_2$, as desired.\nPart (b):\nFrom our earlier work in calculating the volumes $V_1$ and $V_2$, we easily see that:\n\nRe-expressing and simplifying, we have:\n\nBy the AM-GM Inequality, $(u-1) + \\frac{4}{u-1} \\ge 4$, meaning that $\\frac{V_1}{V_2} \\ge \\frac{4}{3}$.\nEquality holds if and only if $u-1 = \\frac{4}{u-1}$, meaning that $u=3$ and $x = 2\\sqrt{2}$.\nIf we check the case $x = \\frac{h}{R} = 2\\sqrt{2}$, we may calculate $V_1$ and $V_2$:\n\n\nIndeed, we have $\\frac{V_1}{V_2} = \\frac{4}{3}$, meaning that our minimum of $k = \\frac{4}{3}$ can be achieved.\nThus, we have proved that the minimum value of $k$ such that $V_1 = kV_2$ is $\\frac{4}{3}$.\nNow, let $\\theta$ be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following:\n\nFrom the double-angle formula for tangent,\n\nThis angle is easy to construct. Simply take any segment and treat it as a unit segment. Create a right triangle with legs of lengths $4\\sqrt{2}$ and $7$. This is straightforward, and the angle opposite the leg of length $4\\sqrt{2}$ will be the desired angle $\\theta$.\nIt follows that we have successfully constructed the desired angle $\\theta$.", "Solution_2": "Part a):\nLet $r$ and $a$ denote the radius of the sphere and the radius of the base of the cone, respectively.\nConsider a plane that contains a cross section of the cone. We get a circle (the sphere) inscribed in both an isosceles triangle (the cone) and a rectangle (the cylinder). Let $A$ and $A_1$ be the vertices at the base of the triangle and let $H_1 H_2 = h$ be the altitude through the third vertex $H_1$. Let $\\theta$ be the angle with arms $AO$ and $AA_1$, where $O$ is the center of the circle; $0 < \\theta < \\frac{\\pi}{4}$ .\nWe have $\\tan \\theta = \\frac{r}{a}$ ; $0 < \\tan \\theta < 1$. For the two volumes we have $V_2 = \\pi r^2 \\times 2r = 2\\pi r^3$ and $V_1 = \\frac{\\pi a^2 \\times H_1 H_2}{3} = \\frac{\\pi a^3 \\tan (\\angle A_1 A H_1) }{3}$ .\nKnowing $\\angle A_1 A H_1 = 2 \\times \\angle A_1 A O = 2 \\theta$, and using the formula for a double-angle tangent, we can simplify the result:\n$V_1 = \\frac{\\pi a^3 \\tan (\\angle A_1 A H_1)}{3}$\n$V_1 = \\frac{2 \\pi r^3 \\tan \\theta}{3 \\tan^3 \\theta (1 - \\tan^2 \\theta)}$\n$V_1 = k V_2$, where $k = \\frac{1}{3\\tan^2 \\theta (1 - \\tan^2 \\theta)}$\nWe see that the biquadratic equation $3 \\tan^4 \\theta - 3 \\tan^2 \\theta + 1 = 0$ has no real roots for $\\tan \\theta$, which means that $k \\neq 1$. Therefore $V_1 \\neq V_2$ .\nPart b):\nLet $\\tan^2 \\theta = x; 0 < x < 1$. Consider the function $f(x) = 3x - 3x^2$. The coefficient in front of $x^2$ is negative, which means the function has an upper bound, i.e. k has a lower bound (since it is the reciprocal of $f(x)$). Using the vertex formula for a quadratic function, we get $x = -\\frac{b}{2a} = \\frac{1}{2}$ , which gives us $f(x) = y_{max} = \\frac{3}{4}$ . From there $k_{min} = \\frac{4}{3}$ .\nFor $\\tan^2 \\theta$ we have $\\tan^2 \\theta = \\frac{1}{2}; \\tan \\theta = +/- \\frac{\\sqrt{2}}{4}$ . We defined $\\tan \\theta$ to be in the interval (0;1), so $\\tan \\theta = \\frac{\\sqrt{2}}{4}$ . That gives us $\\tan \\angle A_1 A H_1 = \\frac{4 \\sqrt{2}}{7}$ . To find $2\\theta$ we construct a right triangle with legs equal to $7$ and $4\\sqrt{2}$ using a unit length reference." }, { "Problem": "Consider a plane $\\epsilon$ and three non-collinear points $A,B,C$ on the same side of $\\epsilon$; suppose the plane determined by these three points is not parallel to $\\epsilon$. In plane $\\epsilon$ take three arbitrary points $A',B',C'$. Let $L,M,N$ be the midpoints of segments $AA', BB', CC'$; Let $G$ be the centroid of the triangle $LMN$. (We will not consider positions of the points $A', B', C'$ such that the points $L,M,N$ do not form a triangle.) What is the locus of point $G$ as $A', B', C'$ range independently over the plane $\\epsilon$?", "Solution": "We will consider the various points in terms of their coordinates in space. We have $L=\\frac{A+A^\\prime}{2},M=\\frac{B+B^\\prime}{2},N=\\frac{C+C^\\prime}{2}$. Since the centroid of a triangle is the average of the triangle's vertices, we have $G=\\frac{1}{3}\\left(L+M+N\\right)=\\frac{1}{6}\\left(A+B+C+A^\\prime+B^\\prime+C^\\prime\\right)$. It is clear now that $G$ is midpoint of the line segment connecting the centroid of $ABC$ and the centroid of $A^\\prime B^\\prime C^\\prime$. It is obvious that the centroid of $A^\\prime B^\\prime C^\\prime$ can be any point on plane $\\epsilon$. Thus, the locus of $G$ is the plane parallel to $\\epsilon$ and halfway between the centroid of $ABC$ and $\\epsilon$." }, { "Problem": "Consider an isosceles triangle. Let $r$ be the radius of its circumscribed circle and $\\rho$ the radius of its inscribed circle. Prove that the distance $d$ between the centers of these two circles is", "Solution": "d1f93636341cbe0bc2f98c788171d8a55d94f8c8\nInstead of an isosceles triangle, let us consider an arbitrary triangle $ABC$. Let $ABC$ have circumcenter $O$ and incenter $I$. Extend $AI$ to meet the circumcircle again at $L$. Then extend $LO$ so it meets the circumcircle again at $M$.\nConsider the point where the incircle meets $AB$, and let this be point $D$. We have $\\angle ADI = \\angle MBL = 90^{\\circ}, \\angle IAD = \\angle LMB$; thus, $\\triangle ADI \\sim \\triangle MBL$, or $\\frac {ID}{BL} = \\frac {AI} {ML} \\iff ID \\cdot ML = 2r\\rho = AI \\cdot BL$.\nNow, drawing line $BI$, we see that $\\angle BIL = \\frac {1}{2}\\angle A + \\frac {1}{2}\\angle ABC, \\angle IBL = \\frac {1}{2}\\angle ABC + \\angle CBL = \\frac {1}{2}\\angle ABC + \\frac {1}{2}\\angle A$. Therefore, $BIL$ is isosceles, and $IL = BL$.\nSubstituting this back in, we have $2r\\rho = AI\\cdot IL$. Extending $OI$ to meet the circumcircle at $P,Q$, we see that $AI\\cdot IL = PI\\cdot QI$ by Power of a Point. Therefore, $2r\\rho = PI \\cdot QI = (PO + OI)(QO - OI) = (r + d)(r - d)$, and we have $2r\\rho = r^2 - d^2 \\iff d = \\sqrt {r(r - 2\\rho)}$, and we are done." }, { "Problem": "Five students, $A,B,C,D,E$, took part in a contest. One prediction was that the contestants would finish in the order $ABCDE$. This prediction was very poor. In fact no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order $DAECB$. This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished.", "Solution": "We are given that no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so in order $ABCDE$. None of them finished in that order. Also only two of them had their actual positions in $DAECB$. After imposing these two conditions the list of possible outcomes is: \n(1)$CAEBD$,\n(2)$DCAEB$,\n(3)$DCEBA$,\n(4)$EDACB$. \nOne more condition is that two disjoint pairs of students predicted to finish consecutively actually did so. Out of the above four in the list, (1) and (2) have $AE$ as the correctly predicted consecutive finishers(but only 1 pair), (3) has no correctly predicted consecutive finishers. But (4) has 2 disjoint correctly predicted consecutive finishers who are $DA$ and $CB$.\nHence, order is $EDACB$." }, { "Problem": "In tetrahedron $ABCD$, vertex $D$ is connected with $D_0$, the centroid of $\\triangle ABC$. Lines parallel to $DD_0$ are drawn through $A,B$ and $C$. These lines intersect the planes $BCD, CAD$ and $ABD$ in points $A_1, B_1,$ and $C_1$, respectively. Prove that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$. Is the result true if point $D_o$ is selected anywhere within $\\triangle ABC$?", "Solution": "\nLet $A_{2}$ be the point where line $AD_{0}$ intersects line $BC$\nLet $B_{2}$ be the point where line $BD_{0}$ intersects line $AC$\nLet $C_{2}$ be the point where line $CD_{0}$ intersects line $AB$\nFrom centroid properties we have:\n$|AA_{2}|=3|D_{0}A_{2}|$\n$|BB_{2}|=3|D_{0}B_{2}|$\n$|CC_{2}|=3|D_{0}C_{2}|$\nTherefore,\n$\\frac{|AA_{2}|}{|D_{0}A_{2}|}=3$\n$\\frac{|BB_{2}|}{|D_{0}B_{2}|}=3$\n$\\frac{|CC_{2}|}{|D_{0}C_{2}|}=3$\nSince $\\Delta D_{0}A_{2}A_{1}\\sim \\Delta AA_{2}A_{1}$, then $|AA_{1}|=|DD_{0}| \\frac{|AA_{2}|}{|D_{0}A_{2}|}=3|DD_{0}|$\nSince $\\Delta D_{0}B_{2}B_{1}\\sim \\Delta BB_{2}B_{1}$, then $|BB_{1}|=|DD_{0}| \\frac{|BB_{2}|}{|D_{0}B_{2}|}=3|DD_{0}|$\nSince $\\Delta D_{0}C_{2}C_{1}\\sim \\Delta CC_{2}C_{1}$, then $|CC_{1}|=|DD_{0}| \\frac{|CC_{2}|}{|D_{0}C_{2}|}=3|DD_{0}|$\nSince $|AA_{2}|=|BB_{2}|=|CC_{2}|$ and $AA_{1} \\parallel BB_{1} \\parallel CC_{1} \\parallel DD_{0}$,\nthen $\\Delta A_{1}B_{1}C_{1}\\parallel \\Delta ABC$, and $Area_{\\Delta A_{1}B_{1}C_{1}}=Area_{\\Delta ABC}$\nLet $h_{D}$ be the perpendicular distance from $D$ to $\\Delta ABC$\nLet $h_{\\Delta A_{1}B_{1}C_{1}}$ be the perpendicular distance from $\\Delta A_{1}B_{1}C_{1}$ to $\\Delta ABC$\n$\\frac{h_{\\Delta A_{1}B_{1}C_{1}}}{h_{D}}=\\frac{|AA_{2}|}{|D_{0}A_{2}|}=\\frac{|BB_{2}|}{|D_{0}B_{2}|}=\\frac{|CC_{2}|}{|D_{0}C_{2}|}=3$\n$h_{\\Delta A_{1}B_{1}C_{1}}=3h_{D}$\n$\\frac{1}{3}h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}=3\\frac{1}{3}h_{D}Area_{\\Delta ABC}$\n$\\frac{h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}}{3}=3\\frac{h_{D}Area_{\\Delta ABC}}{3}$\nSince $Volume_{ABCD}=\\frac{h_{D}Area_{\\Delta ABC}}{3}$ and $Volume_{A_{1}B_{1}C_{1}D_{0}}=\\frac{h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}}{3}$\nthen, $Volume_{A_{1}B_{1}C_{1}D_{0}}=3Volume_{ABCD}$\nthus, $Volume_{ABCD}=\\frac{1}{3}Volume_{A_{1}B_{1}C_{1}D_{0}}$\nthis proves that the volume of $ABCD$ is one third the volume of $A_1B_1C_1D_0$\nThe result is NOT true if point $D_o$ is selected anywhere within $\\triangle ABC$ as ratios of $\\frac{|AA_{2}|}{|D_{0}A_{2}|}$, $\\frac{|BB_{2}|}{|D_{0}B_{2}|}$, and $\\frac{|CC_{2}|}{|D_{0}C_{2}|}$ will have values other than 3 as the point is no longer a centroid. Also, it will make the ratios $\\frac{|AA_{2}|}{|D_{0}A_{2}|} \\ne \\frac{|BB_{2}|}{|D_{0}B_{2}|}\\ne \\frac{|CC_{2}|}{|D_{0}C_{2}|}$ which means that $\\Delta A_{1}B_{1}C_{1} \\nparallel \\Delta ABC$ and the volume relationship will no longer hold true.\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": "In a plane a set of $n$ points ($n\\geq 3$) is given. Each pair of points is connected by a segment. Let $d$ be the length of the longest of these segments. We define a diameter of the set to be any connecting segment of length $d$. Prove that the number of diameters of the given set is at most $n$.", "Solution": "Image of problem Solution. Credits to user awe-sum.\n" }, { "Problem": "In the interior of sides $BC, CA, AB$ of triangle $ABC$, any points $K, L,M$, respectively, are selected. Prove that the area of at least one of the triangles $AML, BKM, CLK$ is less than or equal to one quarter of the area of triangle $ABC$.", "Solution": "Let the lengths of sides $BC$, $CA$, and $AB$ be $a$, $b$, and $c$, respectively. Let $BK=d$, $CL=e$, and $AM=f$.\nNow assume for the sake of contradiction that the areas of $\\Delta AML$, $\\Delta BKM$, and $\\Delta CLK$ are all at greater than one fourth of that of $\\Delta ABC$. Therefore\n\nIn other words, $AM\\cdot AL>\\frac{1}{4}AB\\cdot AC$, or $f(b-e)>\\frac{bc}{4}$. Similarly, $d(c-f)>\\frac{ac}{4}$ and $e(a-d)>\\frac{ab}{4}$. Multiplying these three inequalities together yields\n\nWe also have that $d(a-d)\\leq \\frac{a^2}{4}$, $e(b-e)\\leq \\frac{b^2}{4}$, and $f(c-f)\\leq \\frac{c^2}{4}$ from the Arithmetic Mean-Geometric Mean Inequality. Multiplying these three inequalities together yields\n\nThis is a contradiction, which shows that our assumption must have been false in the first place. This proves the desired result.", "Solution_2": "Let $AR : AB = x, BP : BC = y, CQ : CA = z$. Then it is clear that the ratio of areas of $AQR, BPR, CPQ$ to that of $ABC$ equals $x(1-y), y(1-z), z(1-x)$, respectively. Suppose all three quantities exceed $\\frac{1}{4}$. Then their product also exceeds $\\frac{1}{64}$. However, it is clear by AM-GM that $x(1-x) \\le \\frac{1}{4}$, and so the product of all three quantities cannot exceed $\\frac{1}{64}$ (by the associative property of multiplication), a contradiction. Hence, at least one area is less than or equal to $\\frac{1}{4} [ABC]$." }, { "Problem": "For every natural number $n$, evaluate the sum\n\n(The symbol $[x]$ denotes the greatest integer not exceeding $x$.)", "Solution": "I shall prove that the summation is equal to $n$.\nLet the binary representation of $n$ be $\\overline{d_xd_{x-1}\\dots d_1d_0}_2$, where $d_i\\in\\{0,1\\}$ for all $i$, and $x=\\lfloor \\log_2 n\\rfloor$. Note that if $d_i=0$, then $\\Big[\\frac{n + 2^i}{2^{i+1}}\\Big]=\\Big[\\frac{n}{2^{i+1}}\\Big]$; and if $d_i=1$, then $\\Big[\\frac{n + 2^i}{2^{i+1}}\\Big]=\\Big[\\frac{n}{2^{i+1}}\\Big]+1$. Also note that $\\Big[\\frac{n + 2^i}{2^{i+1}}\\Big]=0$ for all $i\\geq x+1$. Therefore the given sum is equal to\n\nwhere $S$ is the number of 1's in the binary representation of $n$. Legendre's Formula states that $n-S=\\sum_{k=0}^{x} \\bigg[\\frac{n}{2^{k + 1}}\\bigg]$, which proves the assertion. $\\blacksquare$\n", "Solution_2": "We observe\nBut\nso the result is just $n$. $\\square$\n~ilovepi3.14\n", "Solution_3": "By Hermite's identity, for real numbers $x,$\nHence our sum telescopes:\n~Maximilian113" }, { "Problem": "Prove that for all real numbers $x_1, x_2, y_1, y_2, z_1, z_2$, with $x_1 > 0, x_2 > 0, y_1 > 0, y_2 > 0, z_1 > 0, z_2 > 0, x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0$, the inequalityis satisfied. Give necessary and sufficient conditions for equality.", "Solution": "Let $A=x_1y_1 - z_1^2>0$ and $B=x_2y_2 - z_2^2>0$\nFrom AM-GM:\n$\\sqrt{AB} \\le \\frac{A+B}{2}$ with equality at $A=B$\n$4AB \\le (A+B)^2$\n$\\frac{4}{A+B} \\le \\frac{A+B}{AB}$\n$\\frac{8}{2(A+B)} \\le \\frac{A+B}{AB}$\n$\\frac{8}{2(A+B)} \\le \\frac{1}{A}+\\frac{1}{B}$ [Equation 1]\n$(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2$\n$(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=(A+B)+x_1y_2+x_2y_1-2z_1z_2$\nsince $x_1y_1>z_1^2$ and $x_2y_2>z_2^2$, and using the Rearrangement inequality\nthen $x_1y_1+x_2y_2-z_1z_1-z_2z_2 \\le x_1y_2+x_2y_1-z_1z_2-z_1z_2$\n$(A+B) \\le x_1y_2+x_2y_1-2z_1z_2$\n$2(A+B) \\le x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2$\n$2(A+B) \\le (x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2$ [Equation 2]\nTherefore, we can can use [Equation 2] into [Equation 1] to get:\n$\\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \\le \\frac{1}{A}+\\frac{1}{B}$\nThen, from the values of $A$ and $B$ we get:\n$\\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \\leq \\frac{1}{x_1y_1 - z_1^2} + \\frac{1}{x_2y_2 - z_2^2}$\nWith equality at $x_1y_1 - z_1^2=x_2y_2 - z_2^2>0$ and $x_1=x_2, y_1=y_2, z_1=z_2$\n~Tomas Diaz. orders@tomasdiaz.com", "Solution_2": "This solution is actually more difficult but I added it here for fun to see the generalized case as follows:\nProve that for all real numbers $a_i, b_i$, for $i=1,2,...,n$ with $a_i > 0, b_i > 0$\nand $\\prod_{i=1}^{n-1}a_i-a_n^{n-1}>0, \\prod_{i=1}^{n-1}b_i-b_n^{n-1}>0$ the inequality\nis satisfied.\nLet $A=\\prod_{i=1}^{n-1}a_i-a_n^{n-1}$ and $\\prod_{i=1}^{n-1}b_i-b_n^{n-1}>0$\nFrom AM-GM:\n$\\sqrt{AB} \\le \\frac{A+B}{2}$ with equality at $A=B$\n$4AB \\le (A+B)^2$\n$\\frac{4}{A+B} \\le \\frac{A+B}{AB}$\n$\\frac{2^n}{2^{n-2}(A+B)} \\le \\frac{A+B}{AB}$\n$\\frac{2^n}{2^{n-2}(A+B)} \\le \\frac{1}{A}+\\frac{1}{B}$ [Equation 3]\nHere's the difficult part where I'm skipping steps:\nwe prove that $2^{n-2}(A+B) \\le \\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}$\nand replace in [Equation 3] to get:\n$\\frac{2^n}{\\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \\le \\frac{1}{A}+\\frac{1}{B}$\nand replace the values of $A$ and $B$ to get:\n$\\frac{2^n}{\\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \\leq \\frac{1}{\\prod_{i=1}^{n-1}a_i-a_n^{n-1}} + \\frac{1}{\\prod_{i=1}^{n-1}b_i-b_n^{n-1}}$\nwith equality at $a_i=b_i$ for all $i=1,2,...,n$\nThen set $n=3, a_1=x_1, a_2=y_1, a_3=z_1, b_1=x_2, b_2=y_2, b_3=z_3$ and substitute in the generalized inequality to get:\n$\\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \\leq \\frac{1}{x_1y_1 - z_1^2} + \\frac{1}{x_2y_2 - z_2^2}$\nwith equality at $x_1=x_2, y_1=y_2, z_1=z_2$\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": "In a plane there are $100$ points, no three of which are collinear. Consider all possible triangles having these point as vertices. Prove that no more than $70 \\%$ of these triangles are acute-angled.", "Solution": "At most $3$ of the triangles formed by $4$ points can be acute. It follows that at most $7$ out of the $10$ triangles formed by any $5$ points can be acute. For given $10$ points, the maximum number of acute triangles is: the number of subsets of $4$ points times $\\frac{3}{\\text{the number of subsets of 4 points containing 3 given points}}$. The total number of triangles is the same expression with the first $3$ replaced by $4$. Hence at most $\\frac{3}{4}$ of the $10$, or $7.5$, can be acute, and hence at most $7$ can be acute.\nThe same argument now extends the result to $100$ points. The maximum number of acute triangles formed by $100$ points is: the number of subsets of $5$ points times $\\frac{7}{\\text{the number of subsets of 5 points containing 3 given points}}$. The total number of triangles is the same expression with $7$ replaced by $10$. Hence at most $\\frac{7}{10}$ of the triangles are acute." }, { "Problem": "Let $A = (a_{ij})(i, j = 1, 2, \\cdots, n)$ be a square matrix whose elements are non-negative integers. Suppose that whenever an element $a_{ij} = 0$, the sum of the elements in the $i$th row and the $j$th column is $\\geq n$. Prove that the sum of all the elements of the matrix is $\\geq n^2 / 2$.", "Solution_1": "Take the the row or column (without loss of generality, a row) with the minimum possible sum $S$ of its elements. Let $z$ be the number of zeros in this row. We will assume $S < \\frac{n}{2}$ once the other case is obvious. Clearly $S \\ge n - z \\Rightarrow z \\ge n - S$. The total sum $T$ of all elements of the matrix is at least the number of zeros in this row multiplicated by $n - S$ (because the sum of the elements of a row and a column meeting in a zero is $\\ge n$) plus the number of nonzero elements times $S$ (that is the minimum possible sum), it is,Note that, being $z \\ge n - S$ and $n - S \\ge S$, we can put $z - 1$ instead of $z$ and $n - z + 1$ instead of $n - z$ until we get $z = n - S$, what makes the sum become smaller. So we haveby AM - QM inequality.\nThe above solution was posted and copyrighted by Renan. The original thread for this problem can be found here: [1]", "Solution_2": "Denote $\\ell_i,c_i$ the $i$th line and column and $s(\\ell_i), s(c_i)$ the sum of the elements on that row. Let $M=\\max_{\\sigma \\in S_n} | \\{ 1\\le i\\le n| a_{i\\sigma(i)}=0\\} |$. For each permutation $\\sigma\\in S_n$, assign to it the numberNow pick a permutation $\\sigma$ such that it generates $M$ and its assigned number is minimal. Suppose wlog that $\\{ 1\\le i\\le n| a_{i\\sigma(i)}=0\\}=\\{1,2,..,k\\}$. If $k=n$, we are done as twice the sum of all elements is\nSo suppose $k 0 a solution for kai + k'ai', q is kbi + k'bi'. Also a \"near\" solution for ah = 1, other ai = 0 is b1 = qh-1, b2 = qh-2, ... , bh-1 = q, bh = 1, bh+1 = q, ... , bn = qn-h. \"Near\" because the inequalities in (a) and (b) are not strict.\nHowever, we might reasonably hope that the inequalities would become strict in the linear combination, and indeed that is true. Define br = qr-1a1 + qr-2a2 + ... + qar-1 + ar + qar+1 + ... + qn-ran. Then we may easily verify that (a) - (c) hold." }, { "Problem": "Let $P$ be a non-constant polynomial with integer coefficients. If $n(P)$ is the number of distinct integers $k$ such that $(P(k))^2=1,$ prove that $n(P)-\\deg(P)\\le2,$ where $\\deg(P)$ denotes the degree of the polynomial $P.$", "Solution": "Lemma: Let $P(x)$ be a polynomial with integer coefficients which is not constant. Then if $P(x)$ obtains $1$ (or $-1$) as its values for at least four times then $P(x)\\neq -1$ ( or $P(x)\\neq 1$) for all $x$.\nProof. Assume that $P(a)=P(b)=P(c)=P(d)=1$ for $a,b,c,d$ distince. Then if there's $u$ which $P(u)=-1$ then $2=P(a)-P(u)=...$ so $P(x)-P(u)-2=(x-a)(x-b)(x-c)(x-d)Q(x)$ where $Q(x)$ is a polynomial with the integer coefficients! So\n$-2=(u-a)(u-b)(u-c)(u-d)Q(u)$ which is impossible cause $-2$ can not presents as product of more than three distince numbers! This proved the lemma!\nBack to our problem: For convinet put $m=n(P)$ and $n=\\deg P$. Firstly if $n\\leq 2$ then $m-n\\leq2$. Assume $n\\geq3$. If equation $P(x)=1$ with more than three integer points (ie.. at least $4$) then equation $(P(x))^2=1$ implies $P(x)=1$ so $m\\leq n$, ie... $m-n\\leq 0\\leq2$. The same case for equation $P(x)=-1$. So $m\\leq 6$. If $n\\geq4$ then $m-n\\leq 6-n\\leq 2$. Now assume that $n=3$. In this case if $m\\leq 5$ then $m-n\\leq 2$.\nSo let us show that $m<6$. In fact if $m=6$ then $P(x)-1=0$ has three integers distince roots, and the same for $P(x)+1=0$. So\n$P(x)-1=k_1(x-a_1)(x-a_2)(x-a_3)$ and $P(x)+1=k_2(x-b_1)(x-b_2)(x-b_3)$ where $a_i$ distince and $b_j$ distince and all with $k_1,k_2$ are integers! Then\n$k_2(x-b_1)(x-b_2)(x-b_3)-k_1(x-a_1)(x-a_2)(x-a_3)=2$ for all $x$. So $k_1=k_2=k$.\nFinally, we have\n$2=k(a_i-b_1)(a_i-b_2)(a_i-b_3)$ for $i=1,2,3$ and because that $\\pm1$ can not presents as products of three distince numbers so $k=\\pm1$, we may assume $k=1$. Because $2=(-2)\\cdot 1\\cdot -1$ so\n$\\{-2,1,-1\\}=\\{a_i-b_1,a_i-b_2,a_i-b_3\\}$\nThis means $3a_i-(b_1+b_2+b_3)=-2+1-1=-2$. So we must have $3a_1=3a_2=3a_3$ which follows $a_1=a_2=a_3$, which contracts!. So $m\\leq 5$ and we're done.\nThe above solution was posted and copyrighted by pluricomplex. The original thread for this problem can be found here: [1]" }, { "Problem": "Find all polynomials $P$, in two variables, with the following properties:\n(i) for a positive integer $n$ and all real $t, x, y$ (that is, $P$ is homogeneous of degree $n$),\n(ii) for all real $a, b, c$,\n(iii)", "Solution_1": "(i) If $n = 0$ : Clearly no solution\n(ii) If $n = 1$ : $P(x, y) = ax+by \\implies$ the identification yields directly $P(x,y) = x-2y$\n(iii) If $n > 1$, $a = 1, \\ b = 1, \\ c =-2 \\implies P(2,-2)+P(-1, 1)+P(-1, 1) = 0$ $\\implies ((-2)^{n}+2) P(-1, 1) = 0 \\implies P(-1, 1) = 0 \\implies P(-y, y) = 0 \\implies$ $P(x,y)$ is divisible by $(x+y)$\nIt is then easy to see that $\\frac{P(x, y)}{(x+y)}$, of degree $n-1$ verifies all the equations.\nThe only solutions are thus $P(x, y) = (x-2y)(x+y)^{n-1}$\nThe above solution was posted and copyrighted by mathmanman. The original thread for this problem can be found here: [1]", "Solution_2": "$F(a,a,a) \\implies P(2a,a)=0 \\implies (x-2y)$ is a factor of $P(x,y)$.\nWe may write $P(x,y)=(x-2y)Q(x,y)$\n$F(a,b,b) \\implies 2P(a+b,b)+P(2b,a)=0 \\implies 2(a+b-2b)Q(a+b,b)+2(b-a)Q(2b,a) \\implies (a-b)(Q(a+b,b)-Q(2b,a))=0$\nThus $Q(a+b,b)=Q(2b,a) \\forall a \\neq b$\nWe may rewrite it as $Q(x,y)=Q(2y,x-y)=Q(2x-2y,3y-x=\\cdots$\n$Q(x+d,y-d)-Q(x,y)$ is a polynomial in $d$ of degree $n-1$ for any two fixed $x,y$,which has infinitely many zeroes,i.e,$0,2y-x,x-2y,6y-3x,\\cdots$.Thus $Q(x+d,y-d)=Q(x,y)$ holds for all $d$.In particular it holds for $d=y$,i.e, $Q(x+y,0)=Q(x,y)$.Now consider the polynomial $R(x,y)=Q(x+y,0)-Q(x,y)$.Suppose that its not the zero polynomial.Then its degree $d$ is defined.With $t=\\frac{x}{y}$ it can be wriiten as $y^dS(t)=y^d(A(t)-B(t))$.But $S(t)$ has infinitely many zeroes and this forces $A(t)=B(t)$,forcing $R(x,y)$ to be a zero polynomial.Contradiction!.Thus $Q(x+y,0)$ and $Q(x,y)$ are identical polynomials.This forces $Q(x,y)=c(x+y)^{n-1}$.With $Q(1,0)=1$ we get $c=1$.Thus\nThe above solution was posted and copyrighted by JackXD. The original thread for this problem can be found here: [2]" }, { "Problem": "A sequence $(u_{n})$ is defined by\n\nProve that for any positive integer $n$ we have\n\n(where $\\lfloor x\\rfloor$ denotes the smallest integer $\\leq$ $x$)$.$", "Solution": "Let the sequence $(x_n)_{n \\geq 0}$ be defined as \n\\[\nx_{0}=0,x_{1}=1, x_{n}=x_{n-1}+2x_{n-2} \n\\]\nWe notice\n\nBecause the roots of the characteristic polynomial $x_{n}=x_{n-1}+2x_{n-2}$ are $-1$ and $2$. \\\\newline We also see $\\frac{2^1-(-1)^1}{3}=1=x_1$, $\\frac{2^2-(-1)^2}{3}=1=x_2$\nWe want to prove\n\nThis is done by induction\nBase Case:\nFor $n=1$ ses det $2^{1-0}+2^{0-1}=2^{1}+2^{-1}$\nInductive step:\nAssume $2^{x_{n-1}-2x_{n-2}}+2^{-x_{n-1}+2x_{n-2}}=2^{x_1}+2^{-x_1}$\nWe notice\n \nWe then want to show\n\nThis can be done using induction\nBase Case\nFor $n=1$, it is clear that and Therefore, the base case is proved.\nInductive Step\nAssume for all natural $k0$\nFinally we conclude" }, { "Problem": "Let $f(n)$ be a function $f: \\mathbb{N}^{+}\\to\\mathbb{N}^{+}$. Prove that if for each positive integer $n$, then $f(n)=n$.", "Solution": "We will prove this via induction. First we will prove there is a $t$ such that $f(t)=1$ and then that $t=1$ is the only such solution.\nDefine the sequence $a_n$ with $a_0>1$ for $a_0\\in \\mathbb{N}$ and $a_k=f(a_{k-1}-1)$. By the given inequality we have that $f(a_n)>f(a_{n+1})$, this can be used to form a inequality chain of decreasing positive integers: \nBy Infinite Descent, this sequence must terminate, and the only way it can terminate is if we input something into $f$ that is outside of its range. This can only happen if $a_n=1$ since the range and domain of $f$ are the positive integers. Since $a_0\\neq 1$, there is a integer $t$ ($a_{n-1}-1$) such that $f(t)=1$.\nNow if $t\\neq 1$, then $f(t)=1>f(f(t-1))$, which is impossible since $f(f(t-1))\\ge 1$ by the range of $f$, so we have $t=1$ is the only time when $f(t)=1$.\nNow for the inductive step.\nAssume that $f(n)=n$ for all $nk$, by the reasoning above, there is a $a_m$ such that $f(a_m)=1$, by the inductive assumption, this means that $a_m=f(a_{m-1}-1)=1$, we can repeat the inductive assumption to get that $a_{m-k+1}=k$. This implies that $f(a_{m-k}-1)=k$. Thus, there is a $t$ such that $f(t)=k$.\nNow for that $t$, we have $k>f(f(t-1))$, which means that $k+1>t$ by the inductive assumption which implies $t=k$ since we must have $t>k-1$, otherwise $f(t) 0$. After some manipulation, the inequality becomes:\n$xy^3 + yz^3 + zx^3 \\geq xyz(x+y+z)$.\nBy Cauchy, we have:\n$(xy^3 + yz^3 + zx^3)(z+x+y) \\geq xyz(y+z+x)^2$ with equality if and only if $\\frac{xy^3}{z} = \\frac{yz^3}{x} =\\frac{zx^3}{y}$. So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral.", "Solution_2": "Without loss of generality, let $a \\geq b \\geq c > 0$. By Muirhead or by AM-GM, we see that $a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \\geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$.\nIf we can show that $a^3 b + b^3 c+ c^3 a \\geq a^3 c + b^3 a + c^3 b$, we are done, since then $2(a^3 b + b^3 c+ c^3 a ) \\geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \\geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$, and we can divide by $2$.\nWe first see that, $(a^2 + ac + c^2) \\geq (b^2 + bc + c^2)$, so $(a-c)(b-c)(a^2 + ac + c^2) \\geq (a-c)(b-c)(b^2 + bc + c^2)$.\nFactoring, this becomes $(a^3 - c^3)(b-c) \\geq (a-c)(b^3 - c^3)$. This is the same as:\n$(a^3 - c^3)(b-c) + (b^3 - c^3)(c-a) \\geq 0$.\nExpanding and refactoring, this is equal to $a^3 (b-c) + b^3(c-a) + c^3 (a-b) \\geq 0$. (This step makes more sense going backwards.)\nExpanding this out, we have\n$a^3b + b^3 c + c^3 a \\geq a^3 c + b^3 a + c^3 b$,\nwhich is the desired result.", "Solution_3": "Let $s$ be the semiperimeter, $\\frac{a+b+c}{2}$, of the triangle. Then, $a=s-\\frac{-a+b+c}{2}$, $b=s-\\frac{a-b+c}{2}$, and $c=s-\\frac{a+b-c}{2}$. We let $x=\\frac{-a+b+c}{2},$ $y=\\frac{a-b+c}{2}$, and $z=\\frac{a+b-c}{2}.$ (Note that $x,y,z$ are all positive, since all sides must be shorter than the semiperimeter.) Then, we have $a=s-x$, $b=s-y$, and $c=s-z$. Note that $x+y+z=s$, so Plugging this into and doing some expanding and cancellation, we get The fact that each term on the left hand side has at least two variables multiplied motivates us to divide the inequality by $2xyz$, which we know is positive from earlier so we can maintain the sign of the inequality. This gives We move the negative terms to the right, giving We rewrite this as where $r$ is any real number. (This works because if we evaulate the cyclic sum, then as long as the coefficients of $x$ and $y$ on the right sum to 1 the right side will be $x+y+z$.\nThus, we need to show that there exists a real number $r$ such that $\\frac{x^2}{y}\\geq rx+(1-r)y$ for all positive $x,y$. We claim that $r=2$ works. This becomes $\\frac{x^2}{y}\\geq2x-y$, and since $y$ is positive we can multiply by $y$ to yield $x^2\\geq2xy-y^2$, or $(x-y)^2\\geq0$, which is obviously true by the Trivial Inequality. Thus, we are done with part (a).\n(To clarify how this works, we have $\\frac{x^2}{y}\\geq 2x-y$, $\\frac{y^2}{z}\\geq 2y-z$, and $\\frac{z^2}{x}\\geq 2z-x$, we add these inequalities to get $\\frac{x^2}{y}+\\frac{y^2}{z}+\\frac{z^2}{x}\\geq x+y+z.$)\nEquality occurs if and only if $\\frac{x^2}{y}=2x-y$, $\\frac{y^2}{z}=2y-z$, and $\\frac{z^2}{x}=2z-x$ at the same time, since if they are not all equal then the left side will be greater than the right side. From there, it is easy to see that the equality case is $x=y=z$, which is $a=b=c$.\n~john0512", "Solution4": "\"Solution from 111 problems in Algebra and Number Theory\"\nWLOG, we can assume that $a\\geq b\\geq c$. getting that $\\frac{1}{a}\\leq \\frac{1}{b} \\leq \\frac{1}{c}; a(b+c-a)\\leq b(a+c-b)\\leq c(a+b-c)$\nApplying rearrangement inequality to it, getting that $\\frac{a(b+c-a)}{c}+\\frac{b(a+c-b)}{a}+\\frac{c(a+b-c)}{b}\\leq \\frac{a(b+c-a)}{a}+\\frac{b(a+c-b)}{b}+\\frac{c(a+b-c)}{c}=a+b+c$, which equivalent to $\\frac{a(b-a)}{c}+\\frac{b(c-b)}{a}+\\frac{c(a-c)}{b} \\leq 0$ Time $abc$ at both side and get the desired inequality ~bluesoul" }, { "Problem": "Let $a,b,c,d$ be odd integers such that $02^{m-1}$, which is in contradiction with the fact that $b<\\dfrac{b+c}{2}=2^{m-1}$. Thereby, $v_2(a+b)\\ge m-1,\\ v_2(b-a)=1$.\nWrite $a+b=2^{m-1}\\alpha$. If $\\alpha \\ge 2\\Rightarrow2^m\\le a+b2$ otherwise $b+c=2\\Leftrightarrow b=c=1$ or $b+c=4\\Leftrightarrow c=3,b=1\\Rightarrow a=0$ , contradiction )\nSubstituting back, $(b-a)(a+b)=2^m(b-2^{k-m}a)\\Leftrightarrow 2^m\\beta=2^m(2^{m-2}+\\beta-2^{k-m}a)\\Leftrightarrow 2^{k-m}a=2^{m-2}$\nAs $m>2$ and $a$ is odd, we get that $a=1$. Furthermore, $k=2m-2$, hence $d=2^{2m-2}-1$. Now $b(2^m-b)=2^{2m-2}-1\\Leftrightarrow \\left ( b-(2^{m-1}-1) \\right ) \\left ( b-(2^{m-1}+1)\\right )=0$ which together with the fact that $bm$ must hold.\nNext, observe that, $b+c =2^m$ and $b1-\\frac{1}{n} \\Leftrightarrow P_n^{-1}(x_n)>P_n^{-1}(1-\\frac{1}{n}) \\Leftrightarrow x_1>a_n$;\n(b) $\\displaystyle x_n<1 \\Leftrightarrow P_n^{-1}(x_n)1=P_{n+1}(b_{n+1})$, and $x_1>b_{n+1}$. It follows that $b_n>b_{n+1},\\forall n$; and\nThus, $a_nx_n^2\\ge 1$ and so $x_{n+m}>(x_{m+1})^{2^{m-1}}$ which then gives $\\lim_{n\\to\\infty}x_n=+\\infty$.\n(d) If $x_1=0$, then for each $n$, $x_n=0$. If $x_1\\to\\infty$ then for each $n$ (fixed), $x_n\\to\\infty$. Thus, we can denote a mapping $f_n:\\mathbb{R}^{\\ge 0}\\to \\mathbb{R}^{\\ge 0}$ that maps $x_1$ to $x_n$, which is continuous and monotonically increasing, with $\\lim_{x\\to +\\infty} f_n(x)=+\\infty$ so $f_n$ is bijective.\nLet's first show uniqueness. Suppose that $\\{x_n\\}$ and $\\{y_n\\}$ are both such sequences. We have $\\lim_{n\\to\\infty}x_n=\\lim_{n\\to\\infty}y_n=1$ and suppose that $y_1>x_1$. Then for all $n$,\n so inductively, $\\frac{y_{n+1}}{x_{n+1}}>(\\frac{y_{1}}{x_{1}})^n$ with $\\lim_{n\\to\\infty}\\frac{y_{n+1}}{x_{n+1}}=+\\infty$.\nHowever, $\\lim_{n\\to\\infty}x_n=\\lim_{n\\to\\infty}y_n=1$ gives $\\lim_{n\\to\\infty}\\frac{y_{n+1}}{x_{n+1}}=\\frac{1}{1}=1$, contradiction.\nHence, the $x_1$ that satisfies this must be unique.\nNext, let's show existence. We have seen from above that, $y_1>x_1$ implies $y_n>x_n$, and that if $x_n>1$ for some $n$ then $\\{x_n\\}$ is monotonically increasing after some point. Suppose no such $x_1$ exists. Let\n\\[\nA=\\{x_1: \\exists n: x_{n+1}1\\}\n\\]then if $x\\in A$, $y\\in A$ for all $yB$. Notice also an wasy fact that $1\\in B$, so $A$ is bounded. Define, now, $c=glb(A)$. As we assumed $A\\cup B=\\mathbb{R}^+$, this $c$ implies that $x_1c\\to x_1\\in B$. It remains to ask whether $c\\in A$ or $c\\in B$.\nIf $c\\in A$, then for this $x_1:=c$, $x_n\\le 1-\\frac{1}{n}$ and so $x_{n+1}<1-\\frac{1}{n+1}$. Let $y_{n+1}$ be such that $x_{n+1}x_1=c$ by the monotonicity of $f_{n+1}$. This means that there exists $y_1>c\\in A$, contradicting the definition of glb. A similar contradiction (but opposite direction) can also be established for the case $c\\in B$.\nHence $c$ is neither in $A$ or $B$, means that $x_1=c$ should satisfy the problem condition. Q.E.D.\nThis solution was posted and copyrighted by Anzoteh. The original thread for this problem can be found here: [3]" }, { "Problem": "Given a finite set of points in the plane, each with integer coordinates, is it always possible to color the points red or white so that for any straight line $L$ parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on $L$ is not greater than $1$?", "Solution_1": "I'll use a well-known result: if a connected graph has $2k>0$ vertices of odd degree, then its edge set can be partitioned into $k$ paths, and if all its vertices have even degree, then it has an eulerian circuit.\nWe have a bipartite graph with bipartition $A,B$ here, constructed as follows: the horizontal lines which contain points from our set represent the vertices in $A$, the vertical lines represent the vertices in $B$, and the point represent the edges between the vertices corresponding to the rows and columns which contain them. What we must prove is that we can color the edge set of a bipartite graph in two colors s.t. the difference between the number of red edges and white edges adjacent to each vertex is $\\le 1$ in absolute value. It suffices to prove this for connected bipartite graphs, since then we can apply the result to each connected component of the graph.\nIf all the vertices have even degree, then we can find an eulerian circuit. The graph is bipartite, so this circuit has an even number of edges, and we can thus color the edges in the circuit alternatively red and white so that two edges which are consecutive in the circuit have different colors. It's clear that this coloration satisfies the requirements.\nIf, on the other hand, the graph has $2k>0$ vertices with odd degre, then we partition the edge set into $k$ paths, and in each path we color the edges alternatively red and white. Again, it's easy to verify the required properties of the coloration.\nThis solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]", "Solution_2": "We induct on number of points. The small cases are easily checked. Let there exist such a function for $n$ points. We will show there is a function for $n+1$ points.\nIf there exists a line $\\ell$, parallel to any of the coordinate axes (from the next time, any line will be parallel to either of the coordinate axes, unless otherwise mentioned ), containing odd number of points, then choose a point $P_x \\in \\ell$, and consider $S \\setminus P_x$. By inductive hypothesis there exists such a function $f : P \\to \\left \\{ -1, 1 \\right \\}$ for $S \\setminus P_x$. Since $P \\in \\ell, P \\ne P_x$ contain even number of points, we have $\\sum_{\\substack{P \\in \\ell \\\\ P \\ne P_x}} f(P) = 0$. Let $\\ell '$ be the line $\\perp$ to $\\ell$, passing through $P_x$. Let $\\sum _{\\substack{P \\in \\ell ' \\\\ P \\ne P_x}} f(P) = t$, where $t \\in \\left \\{ -1,0 , 1 \\right \\}$. Now for $S$ (with $n+1$ points) define $g$ as $g(P) = f(P)$ for $P \\in S \\setminus P_x$, and $g(P_x) = -t$, if $t \\ne 0$, or $1$ or $-1$ if $t=0$. It indeed works as such a function for $n+1$ points.\nIf the above is not the case, i.e. if all the lines contain even number (greater than zero) of points, pick up an arbitrary point $P_y \\in S$. Let $\\ell$ and $\\ell '$ be the two lines containing $P_y$. Also $\\ell \\perp \\ell '$. Consider $S \\setminus P_y$, for this set , by the inductive hypothesis, there exists such a function $f$. In $S \\setminus P_y$ , $\\ell$ and $\\ell '$ contains odd number of points and if $L$ is a line different from $\\ell$ and $\\ell '$, then it contains even number of points. So $\\sum_{\\substack{P \\in L \\\\ L \\ne \\ell, \\ell '}} f(P) = 0$. Therefore it is seen that\n$\\sum_{\\substack{P \\in \\ell \\\\ P \\ne P_y}} f(P)$ $= - \\sum _{\\substack{P \\in S \\\\ P \\notin \\ell , \\ell ' }} f(P)$ $= \\sum_{\\substack{P \\in \\ell ' \\\\ P \\ne P_y}} f(P) = t$.\nSince $\\ell \\setminus P_y$ contain odd number of points, $t$ is either $-1$ or $1$. Wlog $t = 1$. Now for $S$, (containing $n+1$ points) define $g$ as : $g(P) = f(P)$, for $P \\in S \\setminus P_y$ and $g(P_y) = -t$.\nSo the induction is complete, and the statement is established.\nThis solution was posted and copyrighted by Learner94. The original thread for this problem can be found here: [2]" }, { "Problem": "Let $n$ be an integer greater than or equal to 2. Prove that if $k^2 + k + n$ is prime for all integers $k$ such that $0 \\leq k \\leq \\sqrt{n/3}$, then $k^2 + k + n$ is prime for all integers $k$ such that $0 \\leq k \\leq n - 2$.", "Solution": "First observe that if $m$ is relatively prime to $b+1$, $b+2$, $\\cdots$, $2b$, then $m$ is relatively prime to any number less than $2b$. Since if $c\\leq b$, then we can choose some $i$ to make $2^ic$ lies in range $b+1,b+2,\\cdots,2b$, so $2^ic$ is relatively prime to $m$. Hence $c$ is also. If we also have $(2b+1)^2>m$, then we can conclude that $m$ is a prime. Since there must be a factor of $m$ less than $\\sqrt{m}$.\nLet $n=3r^2+h$ where $0\\leq h<6r+3$, so $r$ is the greatest integer less than or equal to $\\sqrt{n/3}$.(to see this, just let $r=\\lfloor\\sqrt{n/3}\\rfloor$, then we can write $n=3(r+\\epsilon)^2(0\\leq\\epsilon< 1)$, so $h=6r\\epsilon+3\\epsilon^2\\leq 6r+3$).\nAssume that $n+k(k+1)$ is prime for $k=1,2,3\\cdots,r$. We show that $N=n+(r+s)(r+s+1)$ is prime for $s=0,1,2,\\cdots,n-r-2$. By our observation above, it is sufficient to show that $(2s+2r+t)^2>N$ and $N$ is relatively prime to all of $r+s+1,r+s+2,\\cdots,2r+2s$. We have $(2r+2s+1)^2=4r^2+4s^2+8rs+4r+4s+1$. Since $s,t\\ge1$, we have $4s+1>s+2$, $4s^2>s^2$ and $6rs>3r$. Hence \nNow if $N$ has a factor which divides $2r-i$ in the range $-2s$ to $r-s-1$, then so does $N-(i+2s+1)(2r-i)=n+(r-i-s-1)(r-i-s)$ which have the form $n+s'(s'+1)$ with $s'$ in range $0$ to $r$." }, { "Problem": "Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^{2} + b^{2}$. Show that\n$\\frac {a^{2} + b^{2}}{ab + 1}$\nis the square of an integer.", "Solution_1": "Choose integers $a,b,k$ such that $a^2+b^2=k(ab+1)$\nNow, for fixed $k$, out of all pairs $(a,b)$ choose the one with the lowest value of $\\min(a,b)$. Label $b'=\\min(a,b), a'=\\max(a,b)$. Thus, $a'^2-kb'a'+b'^2-k=0$ is a quadratic in $a'$. Should there be another root, $c'$, the root would satisfy: $b'c'\\leq a'c'=b'^2-k-1$. We conclude that $c'=0$ so that $b'^2=k$.\nThis construction works whenever there exists a solution $(a,b)$ for a fixed $k$, hence $k$ is always a perfect square.", "Solution_2_.28Sort_of_Root_Jumping.29": "We proceed by way of contradiction.\nWLOG, let $a\\geq{b}$ and fix $c$ to be the nonsquare positive integer such that such that $\\frac{a^2+b^2}{ab+1}=c,$ or $a^2+b^2=c(ab+1).$ Choose a pair $(a, b)$ out of all valid pairs such that $a+b$ is minimized. Expanding and rearranging, \n\nThis quadratic has two roots, $r_1$ and $r_2$, such that\n\nWLOG, let $r_1=a$. By Vieta's, \n$\\textbf{(1) } r_2=bc-a,$ and\n$\\textbf{(2) } r_2=\\frac{b^2-c}{a}.$\nFrom $\\textbf{(1)}$, $r_2$ is an integer, because both $b$ and $c$ are integers.\nFrom $\\textbf{(2)},$ $r_2$ is nonzero since $c$ is not square, from our assumption.\nWe can plug in $r_2$ for $a$ in the original expression, because $P(r_2)=P(a)=0,$ yielding $c=\\frac{r^2_2+b^2}{r_2b+1}$. If $c>0,$ then $r_2b+1>0,$ and $r_2b+1\\neq{0},$ and because $b>0, r_2$ is a positive integer.\nWe construct the following inequalities: $r_2=\\frac{b^2-c}{a} Q_n$\n$\\implies (2n)! > 2Q_n$\nFirst let $t_i$ represent $|(-1)^{i}2^i\\binom{n}{i}(2n-i)!|$\n$\\therefore Q_n = t_2-t_3+t_4-t_5+ \\cdots -t_{2n-1} + t_{2n}$\nWe see that $2t_2 = (-1)^{2}2^3\\binom{n}{2}(2n-2)! = (2^2)\\left(\\frac{n!}{(n-2)!2!}\\right)(2n-2)! = (2^3)\\left(\\frac{n(n-1)}{2}\\right)(2n-2)! = 2n(2n-2)(2n-2)!$\nBut $(2n)! = (2n)(2n-1)(2n-2)!$\n$(2n)(2n-1)(2n-2)! > 2n(2n-2)(2n-2)!$\n$\\implies (2n)! > 2t_2$\nNext take $t_3$ and $t_4$ for example.\n$t_3$ means at least 3 pairs satisfy T and $t_4$ means at least 4 pairs satisfy $T$.\nBut at least 4 pairs is a subset of at least 3 pairs which means $t_4 \\subset t_3 \\implies |t_3| \\ge |t_4|$\nGeneralising this leads to $t_i \\ge t_{i+1}$\nSo $2Q_n = 2t_2 + 2(t_4-t_3) + \\cdots 2(t_{2n} - t_{2n-1})$\n$\\{(t_4-t_3), \\cdots, (t_{2n}-t_{2n-1})\\} \\le 0$\n$\\implies 2t_2 + 2(t_4-t_3) + \\cdots 2(t_{2n} - t_{2n-1}) \\le 2t_2 < (2n)!$\n$\\therefore 2Q_n < (2n)!$\n$\\mbox{Q.E.D}$", "Solution_2": "Let $F$ be the number of permutations with property $T$, and $F_i$ be the set of permutations $\\{ x_1, x_2, \\ldots, x_{2n} \\}$ such that $|x_i - x_{i+1}| = n$. By the inclusion-exclusion principle,\n\nfor some $E \\geq 0$. Let’s calculate the first two sums on the far right.\nFor any $i$, $|F_i| = 2n \\cdot (2n - 2)!$, since there are are $2n$ choices for $x_i$, which fixes $x_{i+1}$, and $(2n-2)!$ choices for the remaining elements. Thus\n\nNow let’s compute $|F_i \\cap F_j|$, where $1 \\leq i < j \\leq 2n - 1$. It’s not possible that $|x_i - x_{i+1}| = |x_{i+1}- x_{i+2}| = n$, since that would imply $x_i = x_{i+2}$. So $|F_i \\cap F_j| = 0$ if $j = i + 1$. If $j > i + 1$, then $|F_i \\cap F_j| = 2n \\cdot (2n - 2) \\cdot (2n - 4)!$ (there are $2n$ choices for $x_i, x_{i+1}$, $2n - 2$ for $x_j, x_{j+1}$, and $(2n - 4)!$ for everything else.)\nHow often is $|F_i \\cap F_j| \\neq 0$? We know $i$ is at least $1$ and at most $2n - 3$, and for any value of $i$, there are $2n - 2 - i$ possible values for $j$. So the number of non-empty intersections $|F_i \\cap F_j|$ is\n\nNow we can compute\nLet’s plug this into $(1)$:\n\nSo more than half of the $(2n)!$ permutations of $\\{1, \\ldots, 2n\\}$ have property $T$.", "Solution_3": "Let $A_n$ be the set of permutations of $1,2, \\ldots,2n$ not having property $T$ and let $B_n$ be the set of permutations of $1,2, \\ldots,2n$ having exactly one value of $k\\geq2$ such that $|x_k-x_{k+1}|=n$. We will prove a 1-1 correspondence between $A_n$ and $B_n$.\nConsider any permutation in $A_n$. Now take $x_1$ and for the unique value $k$ such that $|x_1-x_k|=n$, put $x_1$ in the spot right before $x_k$. This will give us a permutation that belongs in the set $B_n$. Now consider a permutation in $B_n$. If you take $x_k$ and put it as the first term, you'll get a permutation that belongs in set $A_n$.Therefore, we've proved a 1-1 correspondence between the two.\nBecause $|A_n|$ is clearly less than the number of permutations with property $T$, we finally have that the number of permutations with property $T$ is greater than the number of permutations without $T$.\n~BennettHuang" }, { "Problem": "Prove that there exists a convex $1990\\text{-gon}$ with the following two properties:\n(a) All angles are equal.\n(b) The lengths of the $1990$ sides are the numbers $1^2, 2^2,3^2,\\dots, 1990^2$ in some order.", "Solution_1": "Let $\\{a_1,a_2,\\cdots,a_{1990}\\}=\\{1,2,\\cdots,1990\\}$ and $\\theta=\\frac{\\pi}{995}$.\nThen the problem is equivalent to that there exists a way to assign $a_1,a_2,\\cdots,a_{1990}$ such that $\\sum_{i=1}^{1990}a_i^2e^{i\\theta}=0$.\nNote that $e^{i\\theta}+e^{i\\theta+\\pi}=0$, then if we can find a sequence $\\{b_n\\}$ such that $b_i=a_p^2-a_q^2, (p\\not=q)$ and $\\sum_{i=1}^{995}b_ie^{i\\theta}=0$, the problem will be solved.\nNote that $2^2-1^2=3,4^2-3^2=7,\\cdots,1990^2-1989^2=3979$, then $3,7,\\cdots,3979$ can be written as a sum from two elements in sets $A=\\{3,3+4\\times199,3+4\\times398,3+4\\times597,3+4\\times796\\}$ and $B=\\{0,4\\times1,4\\times2,\\cdots,4\\times 198\\}$.\nIf we assign the elements in $A$ in the way that $l_{a1}=3,l_{a2}=3+4\\times199,l_{a3}=3+4\\times398,l_{a4}=3+4\\times597,l_{a5}=3+4\\times796$, then clearly $\\sum_{i=1}^{199}l_{aj}e^{i\\frac{2\\pi}{199}}=0, (j\\in\\{1,2,3,4,5\\})$\nSimilarly, we could assign elements in $B$ in that way ($l_{b1}=0,l_{b2}=4,\\cdots$) to $e^{i\\frac{2\\pi}{5}}$.\nThen we make $b_i$ according to the previous steps. Let $i\\equiv j$ (mod 5), $i\\equiv k$ (mod 199), and $b_i=l_{aj}+l_{bk}$, then each $b_i$ will be some $a_ p^2-a_q^2$ and $\\sum_{i=1}^{995}b_ie^{i\\theta}=\\sum_{j=1}^5\\sum_{i=1}^{199}l_{aj}e^{i\\frac{2\\pi}{199}}+\\sum_{k=1}^{199}\\sum_{i=1}^{5}l_{bk}e^{i\\frac{2\\pi}{5}}=0$\nAnd we are done.\nThis solution was posted and copyrighted by YCHU. The original thread for this problem can be found here: [1]", "Solution_2": "Throughout this solution, $\\omega$ denotes a primitive $995$th root of unity.\nWe first commit to placing $1^2$ and $2^2$ on opposite sides, $3^2$ and $4^2$ on opposite sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent tobeing true for some permutation $(c_0, \\dots, c_{994})$ of $(3, 7, 11, \\dots, 3981)$.\nDefine $z = 3 \\omega^0 + 7 \\omega^{199} + 11\\omega^{398} + 15 \\omega^{597} + 19 \\omega^{796}$. Then notice that and so summing yields the desired conclusion, as the left-hand side becomesand the right-hand side is the desired expression.\nThis solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]" }, { "Problem": "An infinite sequence $x_0, x_1, x_2,\\ldots$ of real numbers is said to be bounded if there is a constant $C$ such that $|x_i| \\leq C$ for every $i \\geq 0$.\nGiven any real number $a > 1$, construct a bounded infinite sequence $x_0,x_1,x_2,\\ldots$ such that $|x_i-x_j|\\cdot |i-j|^a\\geq 1$ for every pair of distinct nonnegative integers $i,j$.", "Solution_1": "Since $a>1$, the series $\\sum_{k=1}^\\infty\\frac{1}{k^a}$ is convergent; let $L$ be the sum of this convergent series. Let $I\\subset \\mathbb{R}$ be the interval $[-L,L]$ (or any bounded subset of measure $\\geq 2L$).\nSuppose that we have chosen points $x_0,x_1,\\ldots,x_{m-1}$ satisfying\n$\\qquad(\\ast)\\quad |x_i-x_j|\\cdot |i-j|^a\\geq 1$\nfor all distinct $i,j1$. But Osmun Nal argues in this video that $x_k=C\\cdot\\left(k\\sqrt{2}-\\lfloor k\\sqrt{2}\\rfloor\\right), C=2\\sqrt2+1$ satisfies the stronger inequality $|x_i-x_j|\\cdot |i-j|\\geq 1$ for all distinct $i,j$; in other words, this sequence simultaneously solves the problem for all $a\\geq 1$ simultaneously.", "Solution_3": "Let $n = \\max \\left( \\left\\lceil \\dfrac{1}{a - 1} \\right\\rceil, 3 \\right)$, then we have $\\dfrac{\\ln \\left( n + 1 \\right)}{\\ln n} = \\dfrac{\\ln n + \\ln \\left( 1 + 1/n \\right)}{\\ln n} < \\dfrac{\\ln n + 1/n}{\\ln n} = 1 + \\dfrac{1}{n \\ln n} < 1 + \\dfrac{1}{n} \\le a$\nThus $n^a > n + 1$\nFor any non-negative integer $i$, represent it as base-$n$ positional notation:\n$i = i_m n^m + i_{m-1} n^{m-1} + \\cdots + i_2 n^2 + i_1 n + i_0$, where $i_m, i_{m-1}, \\cdots, i_2, i_1, i_0 \\in \\left\\{ 0, 1, 2, \\cdots, n - 1 \\right\\}$\nWe directly construct $x_i = n \\left( i_0 + i_1 \\left( n + 1 \\right)^{-1} + i_2 \\left( n + 1 \\right)^{-2} + \\cdots + i_m \\left( n + 1 \\right)^{-m} \\right)$\nWe allow leading zeros for representing $i$ due to the leading zeros do not effect the value of $x_i$\nThen $0 \\le x_i \\le n \\left( \\left( n - 1 \\right) + \\left( n - 1 \\right) \\left( n + 1 \\right)^{-1} + \\cdots + \\left( n - 1 \\right) \\left( n + 1 \\right)^{-m} \\right) < n \\left( n - 1 \\right) \\dfrac{1}{1 - \\left( n + 1 \\right)^{-1}} = n^2 - 1$\nThus $\\left| x_i \\right| < n^2 - 1$, which is bounded\nFor any two distinct non-negative integers $i \\neq j$, represent them as base-$n$ positional notation:\n$i = i_m n^m + i_{m-1} n^{m-1} + \\cdots + i_2 n^2 + i_1 n + i_0$\n$j = j_m n^m + j_{m-1} n^{m-1} + \\cdots + j_2 n^2 + j_1 n + j_0$\nwhere $i_m, i_{m-1}, \\cdots, i_2, i_1, i_0, j_m, j_{m-1}, \\cdots, j_2, j_1, j_0 \\in \\left\\{ 0, 1, 2, \\cdots, n - 1 \\right\\}$\nLet $\\ell$ be the minimum number such that $i_\\ell \\neq j_\\ell$, that is, $i_0 = j_0, i_1 = j_1, \\cdots, i_{\\ell - 1} = j_{\\ell - 1}, i_\\ell \\neq j_\\ell$\nThen we have $\\left| i - j \\right| \\ge n^\\ell$\nWithout loss of generality, assume that $i_\\ell > j_\\ell$, then we have $i_\\ell - j_\\ell \\ge 1, j_{\\ell + 1} - i_{\\ell + 1}, j_{\\ell + 2} - i_{\\ell + 2}, \\cdots, j_m - i_m \\le n - 1$\n$\\left| x_i - x_j \\right| = n \\left( \\left( i_\\ell - j_\\ell \\right) \\left( n + 1 \\right)^{-\\ell} - \\left( j_{\\ell+1} - i_{\\ell+1} \\right) \\left( n + 1 \\right)^{-\\ell-1} - \\left( j_{\\ell+2} - i_{\\ell+2} \\right) \\left( n + 1 \\right)^{-\\ell-2} - \\cdots - \\left( j_m - i_m \\right) \\left( n + 1 \\right)^{-m} \\right) \\\\~\\\\ \\ge n \\left( n + 1 \\right)^{-\\ell} - n \\left( n - 1 \\right) \\left( \\left( n + 1 \\right)^{-\\ell-1} + \\left( n + 1 \\right)^{-\\ell-2} + \\cdots + \\left( n + 1 \\right)^{-m} \\right) \\\\~\\\\ > n \\left( n + 1 \\right)^{-\\ell} - n \\left( n - 1 \\right) \\left( n + 1 \\right)^{-\\ell-1} \\dfrac{1}{1 - \\left( n + 1 \\right)^{-1}} = \\left( n + 1 \\right)^{-\\ell}$\nThus we have $\\left| x_i - x_j \\right| \\left| i - j \\right|^a > \\left( n + 1 \\right)^{-\\ell} n^{a \\ell} > \\left( n + 1 \\right)^{-\\ell} \\left( n + 1 \\right)^\\ell = 1$\n~Joseph Tsai, mgtsai@gmail.com" }, { "Problem": "For each positive integer $n$, $S(n)$ is defined to be the greatest integer such that, for every positive integer $k \\le S(n)$, $n^{2}$ can be written as the sum of $k$ positive squares.\n(a) Prove that $S(n) \\le n^{2}-14$ for each $n \\ge 4$.\n(b) Find an integer $n$ such that $S(n)=n^{2}-14$.\n(c) Prove that there are infinitely many integers $n$ such that $S(n)=n^{2}-14$.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "There are $n$ lamps $L_0, \\ldots , L_{n-1}$ in a circle ($n > 1$), where we denote $L_{n+k} = L_k$. (A lamp at all times is either on or off.) Perform steps $s_0, s_1, \\ldots$ as follows: at step $s_i$, if $L_{i-1}$ is lit, switch $L_i$ from on to off or vice versa, otherwise do nothing. Initially all lamps are on. Show that:\n(a) There is a positive integer $M(n)$ such that after $M(n)$ steps all the lamps are on again;\n(b) If $n = 2^k$, we can take $M(n) = n^2 - 1$;\n(c) If $n = 2^k + 1$, we can take $M(n) = n^2 - n + 1.$", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Show that there exists a set $A$ of positive integers with the following property: For\nany infinite set $S$ of primes there exist two positive integers $m \\in A$ and $n \\not\\in A$ each of which is a product of $k$ distinct elements of $S$ for some $k \\ge 2$.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $p, q, n$ be three positive integers with $p+q L > M > N$ are positive integers such that $KM + LN = (K + L - M + N)(-K + L + M + N)$. Prove that $KL + MN$ is not prime.", "Solution": "First, $(KL+MN)-(KM+LN)=(K-N)(L-M)>0$ as $K>N$ and $L>M$. Thus, $KL+MN>KM+LN$.\nSimilarly, $(KM+LN)-(KN+LM)=(K-L)(M-N)>0$ since $K>L$ and $M>N$. Thus, $KM+LN>KN+LM$.\nPutting the two together, we have\nNow, we have:\n\n\n\nSo, we have:\n\n\n\n\nThus, it follows that \nNow, since $KL+MN>KM+LN$ if $KL+MN$ is prime, then there are no common factors between the two. So, in order to have we would have to have This is impossible as $KM+LN>KN+LM$. Thus, $KL+MN$ must be composite." }, { "Problem": "Let $n \\ge 3$ be a positive integer. Let $C_1,C_2,...,C_n$ be unit circles in the plane, with centers $O_1,O_2,...,O_n$ respectively. If no line meets more than two of the circles, prove that\n", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "p is a prime number. Prove that for every p there exists a q for every positive integer n, so that $n^p-p$ can't be divided by q.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it.\nLet N be $1 + p + p^2 + ... + p^{p-1}$ which equals $\\frac{p^p-1}{p-1}$\n$N\\equiv{p+1}\\pmod{p^2}$\nWhich means there exists q which is a prime factor of n that doesn't satisfy $q\\equiv{1}\\pmod{p^2}$.\n\\\\unfinished" }, { "Problem": "We call a positive integer alternating if every two consecutive digits in its decimal\nrepresentation are of different parity. Find all positive integers $n$ which have an\nalternating multiple.", "Solution": "We claim that all positive integers $n$ except multiples of $20$ have a multiple that is alternating. If $n$ is a multiple of $20$, then the units digit is $0$ and the tens digit is a multiple of $2$, so both digits are even. Now, we will prove that if $20\\nmid n$, then there is a multiple of $n$ that is alternating.\nWe claim that there exists a $2k$ digit alternating integer that is a multiple of $2^{2k+1}$ and there exists an alternating integer with at most $k$ digits that is a multiple of $2\\cdot5^k$. We will prove this by induction.\nBase Case: $k=1$\nWe can let the number be $16$.\nInductive Step:\nLet $a$ be a $2k-2$ digit alternating number such that $2^{2k-1}\\mid a$. Since $a$ is even, this means that the first digit of $a$ is odd. Let $b\\cdot2^{2k-1}\\equiv a\\pmod{2^{2k+1}}$, where $b\\in\\{0,1,2,3\\}$. Let $x=16-2b$. We see that $x$ is alternating, so the number $x\\cdot10^{2k-2}+a$ is also alternating. We also have $x\\equiv-2b\\pmod8$. Since $a\\equiv b\\cdot2^{2k-1}\\pmod{2^{2k+1}}$, this means that we have\n\\begin{align*}\nx\\cdot10^{2k-2}+a&\\equiv x\\cdot10^{2k-2}+b\\cdot2^{2k-1}\\pmod{2^{2k+1}}\\\\\n&\\equiv2^{2k-2}\\left(x\\cdot5^{2k-2}+2b\\right)\\pmod{2^{2k+1}}\\\\\n&\\equiv2^{2k-2}\\left(-2b\\cdot5^{2k-2}+2b\\right)\\pmod{2^{2k+1}}\\\\\n&\\equiv -b\\cdot2^{2k-1}\\left(5^{2k-2}-1\\right)\\pmod{2^{2k+1}}.\\end{align*}\nSince $4\\mid5^{2k-2}-1$, this means that $x\\cdot10^{2k-2}+a\\equiv0\\pmod{2^{2k+1}}$, so $x\\cdot10^{2k-2}+a$ is a $2k$ digit alternating number that is a multiple of $2^{2k+1}$.\nNow, we will prove that there exists an alternating integer with at most $k$ digits that is a multiple of $2\\cdot5^k$.\nBase Case: $k=1$\nWe can let the number be $0$.\nInductive Step:\nLet $a$ be a $k-1$ digit alternating number such that $2\\cdot5^{k-1}\\mid a$. Let $S$ be the set of digits such that $x\\cdot10^{k-1}+a$ is alternating. We see that either $S=\\{0,2,4,6,8\\}$ or $S=\\{1,3,5,7,9\\}$. In each possible set, each $\\mod5$ residue appears exactly once. Let $a=2\\cdot5^{k-1}\\cdot b$. Then, $x\\cdot10^{k-1}+a=2\\cdot5^{k-1}\\left(x\\cdot2^{k-2}+b\\right)$. Therefore, there exists an $x$ such that $2\\cdot5^k\\mid x\\cdot10^{k-1}+a$, so there exists an alternating integer with at most $k$ digits that is a multiple of $2\\cdot5^k$. If $k$ is even, then $0\\not\\in S$, so it has exactly $k$ digits.\nNow, let $n=2^a\\cdot5^b\\cdot m$. Then, if $20\\nmid n$, then $a<2$ or $b=0$. If $b=0$, then there exists a $2a$ digit alternating integer $x$ that is a multiple of $2^{2a+1}$. Consider the sequence $a_i=x\\cdot\\frac{10^{2ai}-1}{10^{2a}-1}$. Then, the decimal representation of $a_i$ is the result when $x$ is written $i$ times. For any prime $p$ that divides $n$, if $p\\mid10^{2a}-1$, then $\\nu_p\\left(10^{2ai}-1\\right)=\\nu_p\\left(10^{2a}-1\\right)+\\nu_p(i)$, so $\\nu_p(a_i)=\\nu_p(x)+\\nu_p(i)\\geq\\nu_p(i)$. Therefore, there exists an $i$ such that $p^{\\nu_p(n)}\\mid a_{ci}$ for all $c$. If $p\\nmid10^{2a}-1$, then if $i=\\phi(n)$, then $p^{\\nu_p(n)}\\mid10^{2ai}-1$ if $p\\neq2,5$. Since $5\\nmid n$, this means that we only need to make sure $2^a\\mid x\\cdot\\frac{10^{2ai}-1}{10^{2a}-1}$. Since $2^{2a+1}\\mid x$, this means that this is true, so if $5\\nmid n$, then there exists a multiple of $n$ that is alternating.\nIf $a<2$, then there exists an alternating integer $x$ with exactly $2b$ digits that is a multiple of $2\\cdot5^{2b}$. Then, the first digit of $x$ is odd. Let $a_i=x\\cdot\\frac{10^{2bi}-1}{10^{2b}-1}$. Then, if $p\\mid10^{2b}-1$, then $\\nu_p(a_i)\\geq\\nu_p(i)$, so there exists an $i$ such that $p^{\\nu_p(n)}\\mid a_i$. If $p\\nmid10^{2a}-1$, then we can let $i=\\phi(n)$, so $p^{\\nu_p(n)}\\mid a_i$. If $p=5$, then $5^b\\mid5^{2b}\\mid x$, and if $p=2$, then $a=0,1$, so $p^a\\mid x$. Therefore, $n\\mid a_i$ for some $i$.\nTherefore, all positive integers except multiples of $20$ have an alternating multiple." }, { "Problem": "In a mathematical competition, in which 6 problems were posed to the participants, every two of these problems were solved by more than 2/5 of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 problems each.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $n$ be a positive integer. Consider\n\nas a set of $(n+1)^3-1$ points in three-dimensional space. \nDetermine the smallest possible number of planes, the union of which contain $S$ but does not include $(0,0,0)$.", "Solution": "We will prove the result using the following Lemma, which has an easy proof by induction.\nLemma Let $S_1 = \\{0, 1, \\ldots, n_1\\}$, $S_2 = \\{0, 1, \\ldots, n_2\\}$ and $S_3 = \\{0, 1, \\ldots, n_3\\}$. If $P$ is a polynomial in $\\mathbb{R}[x, y, z]$ that vanishes on all points of the grid $S = S_1 \\times S_2 \\times S_3$ except at the origin, then\nProof. We will prove this by induction on $n = n_1 + n_2 + n_3$. If $n = 0$, then the result follows trivially. Say $n > 0$. \nWLOG, we can assume that $n_1 > 0$. \nBy polynomial division over $\\mathbb{R}[y, z][x]$ we can write \nSince $x-n_1$ is a monomial, the remainder $R$ must be a constant in $\\mathbb{R}[y, z]$, i.e., $R$ is a polynomial in two variables $y$, $z$. \nPick an element of $S$ of the form $(n_1, y, z)$ and substitute it in the equation. Since $P$ vanishes on all such points, we get that $R(y, z) = 0$ for all $(y, z) \\in S_2 \\times S_3$. \nLet $S_1' = \\{0, 1, \\ldots, n_1 - 1\\}$ and $S' = S_1' \\times S_2 \\times S_3$. \nFor every point $(x, y, z)$ in $S'$ we have where $x - n_1 \\neq 0$. \nTherefore, the polynomial $Q$ vanishes on all points of $S'$ except the origin. By induction hypothesis, we must have $\\deg Q \\geq n_1 - 1 + n_2 + n_3$. But, $\\deg P \\geq \\deg Q + 1$ and hence we have $\\deg P \\geq n_1 + n_2 + n_3$.\nNow, to solve the problem let $H_1, \\ldots, H_m$ be $m$ planes that cover all points of $S$ except the origin. Since these planes don't pass through origin, each $H_i$ can be written as $a_i x + b_i y + c_i z = 1$. Define $P$ to be the polynomial $\\prod (a_ix + b_iy + c_iz - 1)$. Then $P$ vanishes at all points of $S$ except at the origin, and hence $\\deg P = m \\geq 3n$.\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page." }, { "Problem": "Let $ABCD$ be a convex quadrilateral with $BA \\ne BC$. Denote the incircles of triangles $ABC$ and $ADC$ by $\\omega _1$ and $\\omega _2$ respectively. Suppose that there exists a circle $\\omega$ tangent to ray $BA$ beyond $A$ and to the ray $BC$ beyond $C$, which is also tangent to the lines $AD$ and $CD$. Prove that the common external tangents to $\\omega _1$ and $\\omega _2$ intersect on $\\omega$", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it.\nHere are some hints:\nLet B be the top vertex of triangle ABC, O and K are the centers of the incircles of triangles ABC and ADC with radii R and r, respectively. S is the center of the circumcircle tangential to the extensions of AB, AC and DC. And let\nE be the foot of the projection of O to AB.\nU be the foot of the projection of O to AC.\nV be the foot of the projection of K to AC.\nM be the foot of the projection of S to DC.\nL be the foot of the projection of K to DC.\nL be the intercept of DC and AB.\nWe have\n/_EOB = /_LSC\n/_AOU = /_SOC\n/_ASO = /_KSC\n/_ASK = /_OSC\n/_LSB = /_KCO\nUV = BC – AB\nAU = VC\nOA. AK. cos(/_OAK) = OC. KC. cos(/_OCK)\nOK**2 = (R + r)**2 + UV**2\nSK**2 = (R' + r)**2 + ML**2\nUse sin (90-x) = cos x and cos(90-x) = sinx\nand characteristic of triangle a**2 = b**2 + c**2 - 2.b.c.cosine(angle) to solve.\nVo Duc Dien" }, { "Problem": "Let $a_1,a_2,\\ldots,a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s=a_1+a_2+\\ldots+a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1,a_2,\\ldots,a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M$.\nAuthor: Dmitry Khramtsov, Russia", "Solution": "We will use strong induction on $n$. When $n = 1$, there are no elements in $M$, so the one jump can be made without landing on a point in $M$. When $n = 2$, we consider two cases. If $a_1$ is not in $M$, then the order $a_1, a_2$ will work. If $a_1$ is in $M$, then $a_2$ is not in $M$ and the order $a_2, a_1$ will work. We will assume that the order can be chosen in such a way for all integers $n < k$ for $k \\ge 3$.\nWhen $n = k$, we can assume WLOG that $a_1 a_x$, all jumps before $a_n$ will land on a integer less than $m_1$. Since $m_1$ is the smallest element in $M$, none of the jumps before $a_n$ will land on a point in $M$. Since no jumps after $a_n$ will land on an element in $M$ by the construction, the grasshopper must then land on an element $m_y > m_1$ in $M$ after making the jump $a_n$. However, this would imply that the original construction would land on another element $m_y$ different from $m_1$, but this is a contradiction. So, a valid sequence must exist.\nSince we have an induction on $n$, the statement must be true for all $n$, and we are done.\n" }, { "Problem": "Let $a_1, a_2, a_3, \\ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that\n\nProve there exist positive integers $\\ell \\leq s$ and $N$, such that", "Solution": "So for solving This Problem, we need to take a assumption that," }, { "Problem": "Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, \\ldots, a_n$ such that", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "Let $n \\ge 3$ be an integer, and consider a circle with $n + 1$ equally spaced points marked on it. Consider all labellings of these points with the numbers $0, 1, ... , n$ such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels $a < b < c < d$ with $a + d = b + c$, the chord joining the points labelled $a$ and $d$ does not intersect the chord joining the points labelled $b$ and $c$.\nLet $M$ be the number of beautiful labelings, and let N be the number of ordered pairs $(x, y)$ of positive integers such that $x + y \\le n$ and $\\gcd(x, y) = 1$. Prove that", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "A set of lines in the plane is in $\\textit{general position}$ if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its $\\textit{finite regions}$. Prove that for all sufficiently large $n$, in any set of $n$ lines in general position it is possible to colour at least $\\sqrt{n}$ of the lines blue in such a way that none of its finite regions has a completely blue boundary.", "Solution": "Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\nCall the $\\binom{n}{2}$ intersections, well, points. Then each line will have $n-1$ points. We call 2 points (on a line) neighbors if there are no other points on the line segment joining those 2. Then each finite region has to be a convex polygon whose any pair of neighboring vertices are, well, neighbors. Now start by coloring any line blue, and all points are uncolored (neither red nor blue).\nSuppose there is an uncolored line which does not pass through any red points. Color that line blue. Now consider each of the intersections that line makes with the other blue lines. For each intersection point $X$ of the 2 blue lines $l_1,l_2$, color it blue (it was originally uncolored). Now consider the neighbors of $X$ on $l_1,l_2$, call them $A_1,A_2$ on $l_1$ and $B_1,B_2$ on $l_2$ (if there is only 1 neighbor, let the other point be an uncolored dummy). If neither $A_1$ nor $A_2$ is blue, we colour them both red. Else if neither $B_1$ nor $B_2$ is blue, we colour them both red. Otherwise, we colour the uncolored ones (out of the 4 points) red. Anyway, we would have colored at most 2 points red. Once we have done this for all the intersection points, proceed back to the start of this paragraph until no such line exists.\nNow to show it works. Suppose there is a finite region polygon $P_1...P_m$ with all blue boundaries (with $P_{m+1}=P_1$). Then all the points must be blue. Consider the first point colored blue, WLOG let it be $P_2$. Suppose the previous line to be colored blue is $P_1P_2$. Then $P_2P_3$ has to be colored blue before that, and $P_3P_4$ has to be uncolored (otherwise $P_3$ will be colored blue first). So $P_3$ is uncolored. Then if $P_1$ is blue, $P_3$ will be colored red by our coloring algorithm. $P_1$ has to be uncolored and by our coloring algorithm at least one of $P_1,P_3$ will be colored red. Note that no blue line will pass through a red point. Thus there are no finite region with blue boundaries.\nNow suppose $k$ lines are blue. By our algorithm, each blue point intersection will introduce at most 2 red points. Since we can't color any more lines blue, those red points will cover the remaining lines. Thus $2\\binom{k}{2}\\geq n-k\\implies k\\geq \\sqrt{n}$ exact.\nNote: this works for all ${n}\\geq{2}$, provided there is no mistake." }, { "Problem": "The sequence $a_1,a_2,\\dots$ of integers satisfies the conditions:\n(i) $1\\le a_j\\le2015$ for all $j\\ge1$,\n(ii) $k+a_k\\neq \\ell+a_\\ell$ for all $1\\le k<\\ell$.\nProve that there exist two positive integers $b$ and $N$ for whichfor all integers $m$ and $n$ such that $n>m\\ge N$.\nProposed by Ivan Guo and Ross Atkins, Australia.", "Solution": "We can prove the more general statement.\nTheorem Let $T$ be a non-negative integer parameter. If given a sequence $a_1,a_2,\\dots$ that satisfies the conditions:\n(i) $1 \\le a_j \\le T+1$ for all $j \\le 1;$\n(ii) $k+a_k \\ne \\ell+a_\\ell$ for all $1 \\le k<\\ell,$\nthen there exist two integers $v$ and $N$, $0 \\le v \\le T$ and $N>0$, for which for all integers $m$ and $n$ such that $n>m\\ge N$.\nProof Consider the set of points on a plane: $\\{(x,y) \\mid x,y \\in \\mathbb N,\\,y \\le T+1\\}$ (let's call it the base strip). The sequence is represented on it by the subset $\\{(i,a_i) \\mid i \\in \\mathbb N\\}$ which is painted red. Each line $L_n$ of kind $x+y=n$, where $n \\ge 2$ is an integer, will be called a carrier line. Note that the condition (ii) states that each carrier line is occupied by at most one red point. Every such line with exactly one red point on it will be called occupied while the rest carriers will be called free.\nFirst we prove that the set of free carriers is finite. Indeed, all the carriers $L_n$ with $2 \\le n \\le T+m+1$, where $m \\in \\mathbb N$, cover the part of the base strip having $x \\le m$, so there are at least $m$ red points lying on them. Thus, the number of free carriers among them is at most $T+m-m=T$. Since $m$ is arbitrary, the total number of free cariers doesn't exceed $T$. Take $N$ such that all the carriers $L_n$ with $n \\ge N+2$ are occupied.\nNext, take any red point and consider the points on the base strip which are lying on the point's carrier below it (i.e. having strictly greater $x$). Call it the point's trace and paint black. Finally, the set $P_x=\\{\\,y \\mid (x,y) \\text { is black}\\}$ will be called a pattern, the number elements $|P_x|$ in it will be called its volume and the sum $w_x$ of all elements in $P_x$ will be called its weight.\nNow let $j>N$. Lets track how $P_j$, its volume and its weight change when $j$ increases by 1. Obviously $P_{j+1}$ is $P_j$ plus added $a_j$ (red point) and shifted down by $1$; a point that goes to $0$ vanishes. This means that the volume doesn't change at all. Indeed, if $1 \\notin P_j$, then $a_j=1$ or the carrier $L_{j+1}$ will remain unoccupied, so one point always vanishes going to $0$, and exactly one point is added before the shift (it may be the same point). Let $v$ be this constant volume. As a pattern never contains $T+1$, $v$ ranges from $0$ to $T$. Now it is clearly that $w_{j+1}$ is $w_j$ plus $a_j$ (red point added) and minus $v+1$ (all elements are shifted by $1$), i.e. $a_j-(v+1)=w_{j+1}-w_j$. Thus, when $n>m\\ge N$: To finish, it remains to note that the weight ranges from $1+2+\\dots+v$ to $(T-v+1)+(T-v+2)+\\dots+T$, so its swing is:" }, { "Problem": "There are $n\\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.\n(a) Prove that Geoff can always fulfill his wish if $n$ is odd.\n(b) Prove that Geoff can never fulfill his wish if $n$ is even.", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "An ordered pair $(x, y)$ of integers is a primitive point if the greatest common divisor of $x$ and $y$ is $1$. Given a finite set $S$ of primitive points, prove that there exist a positive integer $n$ and integers $a_0, a_1, \\ldots , a_n$ such that, for each $(x, y)$ in $S$, we have:", "Solution": "This problem needs a solution. If you have a solution for it, please help us out by adding it." }, { "Problem": "A convex quadrilateral $ABCD$ satisfies $AB\\cdot CD=BC \\cdot DA.$ Point $X$ lies inside\n$ABCD$ so that\n$\\angle XAB = \\angle XCD$ and $\\angle XBC = \\angle XDA.$\nProve that $\\angle BXA + \\angle DXC = 180^{\\circ}$", "Solution": "We want to find the point $X.$ Let $E$ and $F$ be the intersection points of $AB$ and $CD,$ and $BC$ and $DA,$ respectively.\nThe poinx $X$ is inside $ABCD,$ so points $E,A,X,C$ follow in this order.\n$\\angle XAB = \\angle XCD \\implies \\angle XAE + \\angle XCE = 180^\\circ$\n$\\implies AXCE$ is cyclic $\\implies X$ lie on circle $ACE.$\nSimilarly, $X$ lie on circle $BDF.$\nPoint $X$ is the point of intersection of circles $ACE$ and $\\Omega = BDF.$\nSpecial case\nLet $AD = CD$ and $AB = BC \\implies AB \\cdot CD = BC \\cdot DA.$\nThe points $B$ and $D$ are symmetric with respect to the circle $\\theta = ACEF$ (Claim 1).\nThe circle $BDF$ is orthogonal to the circle $\\theta$ (Claim 2).\n$\\hspace{10mm} \\angle FCX = \\angle BCX = \\frac {\\overset{\\Large\\frown} {XAF}}{2}$ of $\\theta.$ $\\hspace{10mm} \\angle CBX = \\angle XDA = \\frac {\\overset{\\Large\\frown} {XBF}}{2}$ of $\\Omega.$\n$\\overset{\\Large\\frown} {XAF} + \\overset{\\Large\\frown} {XBF} = 180^\\circ$ (Claim 3) $\\implies$\n$\\angle XCB + \\angle XBC = 90^\\circ \\implies \\angle CXB = 90^\\circ.$\nSimilarly, $\\angle AXD = 90^\\circ \\implies$\nCommon case\nDenote by $O$ the intersection point of $BD$ and the perpendicular bisector of $AC.$ Let $\\omega$ be a circle (red) with center $O$ and radius $OA = R.$\nWe will prove $\\sin\\angle BXA =\\sin \\angle DXC$ using point $Y$ symmetric to $X$ with respect to $\\omega.$\nThe points $B$ and $D$ are symmetric with respect to $\\omega$ (Claim 1).\nThe circles $BDF$ and $BDE$ are orthogonal to the circle $\\omega$ (Claim 2).\nCircles $ACF$ and $ACE$ are symmetric with respect to the circle $\\omega$ (Lemma).\nDenote by $Y$ the point of intersection of circles $BDF$ and $ACF.$\nQuadrangle $BYDF$ is cyclic $\\implies \\angle CBY = \\angle ADY.$\nQuadrangle $AYCF$ is cyclic $\\implies \\angle YAD = \\angle YCB.$\nThe triangles $\\triangle YAD \\sim \\triangle YCB$ by two angles, so\nThe points $X$ and $Y$ are symmetric with respect to the circle $\\omega$, since they lie on the intersection of the circles $ACF$ and $ACE$ symmetric with respect to $\\omega$ and the circle $BDF$ orthogonal to $\\omega.$\nThe point $B$ is symmetric to $D$ with respect to $\\omega \\implies$\n\n\nThe point $B$ is symmetric to $D$ and the point $X$ is symmetric to $Y$ with respect to $\\omega,$ hence\nDenote $\\angle XAB = \\angle XCD = \\alpha, \\angle BXA = \\varphi, \\angle DXC = \\psi.$\nBy the law of sines for $\\triangle ABX,$ we obtain $\\frac {AB}{\\sin \\varphi} = \\frac{BX}{\\sin \\alpha}.$\nBy the law of sines for $\\triangle CDX,$ we obtain $\\frac {CD}{\\sin \\psi} = \\frac {DX}{\\sin \\alpha}.$\nHence we get $\\frac{\\sin \\psi} {\\sin \\varphi}= \\frac {CD}{DX} \\cdot \\frac{BX}{AB} = 1.$\nIf $\\varphi = \\psi,$ then $\\triangle XAB \\sim \\triangle XCD \\implies \\frac {CD}{AB} = \\frac {BX}{DX} = \\frac{AX}{CX} = \\frac {AD}{BC}.$\n$CD \\cdot BC = AB \\cdot AD \\implies AD = CD, AB = BC.$ This is a special case.\nIn all other cases, the equality of the sines follows $\\varphi + \\psi = 180^\\circ .$\nClaim 1 Let $A, C,$ and $E$ be arbitrary points on a circle $\\omega, l$ be the perpendicular bisector to the segment $AC.$ Then the straight lines $AE$ and $CE$ intersect $l$ at the points $B$ and $D,$ symmetric with respect to $\\omega.$\nClaim 2 Let points $B$ and $D$ be symmetric with respect to the circle $\\omega.$ Then any circle $\\Omega$ passing through these points is orthogonal to $\\omega.$\nClaim 3 The sum of the arcs between the points of intersection of two perpendicular circles is $180^\\circ.$\nIn the figure they are a blue and red arcs $\\overset{\\Large\\frown} {CD}, \\alpha + \\beta = 180^\\circ.$\nLemma The opposite sides of the quadrilateral $ABCD$ intersect at points $E$ and $F$ ($E$ lies on $AB$). The circle $\\omega$ centered at the point $O$ contains the ends of the diagonal $AC.$ The points $B$ and $D$ are symmetric with respect to the circle $\\omega$ (in other words, the inversion with respect to $\\omega$ maps $B$ into $D).$ Then the circles $ACE$ and $ACF$ are symmetric with respect to $\\omega.$\nProof We will prove that the point $G,$ symmetric to the point $E$ with respect to $\\omega,$ belongs to the circle $ACF$ becouse $\\angle AGC = \\angle AFC.$\nA circle $BDE$ containing points $B$ and $D$ symmetric with respect to $\\omega,$ is orthogonal to $\\omega$ (Claim 2) and maps into itself under inversion with respect to the circle $\\omega.$ Hence, the point $E$ under this inversion passes to some point $G,$ of the same circle $BDE.$\nA straight line $ABE$ containing the point $A$ of the circle $\\omega,$ under inversion with respect to $\\omega,$ maps into the circle $OADG.$ Hence, the inscribed angles of this circle are equal $\\angle ADB = \\angle AGE.$\n$\\angle OCE = \\angle CGE (CE$ maps into $CG)$ and $\\angle OCE = \\angle BCD (BC$ maps into $DC).$ \nConsequently, the angles $\\angle AFC = \\angle ADB – \\angle FBD = \\angle AGE - \\angle CGE = \\angle AGC.$\nThese angles subtend the $\\overset{\\Large\\frown} {AC}$ of the $ACF$ circle, that is, the point $G,$ symmetric to the point $E$ with respect to $\\omega,$ belongs to the circle $ACF.$\nvladimir.shelomovskii@gmail.com, vvsss\n" }, { "Problem": "Let $I$ be the incenter of acute triangle $ABC$ with $AB \\neq AC$. The incircle $\\omega$ of $ABC$ is tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\\omega$ again at $R$. Line $AR$ meets ω again at $P$. The circumcircles of triangles $PCE$ and $PBF$ meet again at $Q$.\nProve that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$.", "Solution": "Step 1\nWe find an auxiliary point $S.$\nLet $G$ be the antipode of $D$ on $\\omega, GD = 2R,$ where $R$ is radius $\\omega.$\nWe define $A' = PG \\cap AI.$\n$RD||AI, PRGD$ is cyclic $\\implies \\angle IAP = \\angle DRP = \\angle DGP.$\n$RD||AI, RD \\perp RG, RI=GI \\implies \\angle AIR = \\angle AIG \\implies$\n\nAn inversion with respect $\\omega$ swap $A$ and $A' \\implies A'$ is the midpoint $EF.$\nLet $DA'$ meets $\\omega$ again at $S.$ We define $T = PS \\cap DI.$\nOpposite sides of any quadrilateral inscribed in the circle $\\omega$ meet on the polar line of the intersection of the diagonals with respect to $\\omega \\implies DI$ and $PS$ meet on the line through $A$ perpendicular to $AI.$\nThe problem is reduced to proving that $Q \\in PST.$\nStep 2\nWe find a simplified way to define the point $Q.$\nWe define $\\angle BAC = 2 \\alpha \\implies \\angle AFE = \\angle AEF = 90^\\circ – \\alpha \\implies$\n$\\angle BFE = \\angle CEF = 180^\\circ – (90^\\circ – \\alpha) = 90^\\circ + \\alpha = \\angle BIC$\n$(AI, BI,$ and $CI$ are bisectrices).\nWe use the Tangent-Chord Theorem and get\n$\\angle BQC = \\angle BQP + \\angle PQC = \\angle BFP + \\angle CEP =$\n$=\\angle BFE – \\angle EFP + \\angle CEF – \\angle FEP =$\n$= 90^\\circ + \\alpha + 90^\\circ + \\alpha – (90^\\circ + \\alpha) =$\n$90^\\circ + \\alpha = \\angle BIC \\implies$\nPoints $Q, B, I,$ and $C$ are concyclic.\nStep 3\nWe perform inversion around $\\omega.$ The straight line $PST$ maps onto circle $PITS.$ We denote this circle $\\Omega.$ We prove that the midpoint of $AD$ lies on the circle $\\Omega.$\nIn the diagram, the configuration under study is transformed using inversion with respect to $\\omega.$ The images of the points are labeled in the same way as the points themselves. Points $D,E,F,P,S,$ and $G$ have saved their position. Vertices $A, B,$ and $C$ have moved to the midpoints of the segments $EF, FD,$ and $DE,$ respectively.\nLet $M$ be the midpoint $AD.$\nWe define $\\angle MID = \\beta, \\angle MDI = \\gamma \\implies$\n$\\angle IMA = \\angle MID + \\angle MDI = \\beta + \\gamma = \\varphi.$\n$DI = IS \\implies \\angle ISD = \\gamma.$\n$MI$ is triangle $DAG$ midline $\\implies MI || AG \\implies$\n\n\n\n$PI = IS \\implies \\angle PIS = 180^\\circ – 2 \\varphi =\\angle DPA \\implies$ point $M$ lies on $\\Omega.$\n$ABDC$ is parallelogram $\\implies M$ is midpoint $BC.$\nStep 4\nWe prove that image of $Q$ lies on $\\Omega.$\nIn the inversion plane the image of point $Q$ lies on straight line $BC$ (It is image of circle $BIC)$ and on circle $PCE.$\npoint $Q$ lies on $\\Omega$.\nvladimir.shelomovskii@gmail.com, vvsss" }, { "Problem": "Prove that there exists a positive constant $c$ such that the following statement is true:\nConsider an integer $n > 1$, and a set $S$ of n points in the plane such that the distance between any two different points in $S$ is at least $1$. It follows that there is a line $\\ell$ separating $S$ such that the distance from any point of $S$ to $\\ell$ is at least $cn^{- \\frac{1}{3}}$.\n(A line $\\ell$ separates a set of points $S$ if some segment joining two points in $S$ crosses $\\ell$.)\nNote. Weaker results with $cn^{- \\frac{1}{3}}$ replaced by $cn^{- \\alpha}$ may be awarded points depending on the value\nof the constant $\\alpha > \\frac{1}{3}$.", "Solution": "For any unit vector $v$, let $a_v=\\min_{p\\in S} p \\cdot v$ and $b_v = \\max_{p\\in S} p\\cdot v$. If $b_v - a_v\\geq n^{2/3}$ then we can find a line $\\ell$ perpendicular to $v$ such that $\\ell$ separates $S$, and any point in $S$ is at least $\\Omega(n^{2/3}/n) = \\Omega(n^{-1/3})$ away from $\\ell$.\nSuppose there is no such direction $v$, then $S$ is contained in a box with side length $n^{2/3}$ by considering the direction of $(1, 0)$ and $(0, 1)$, respectively. Hence, $S$ is contained in a disk with radius $n^{2/3}$. Now suppose that $D$ is the disk with the minimum radius, say $r$, which contains $S$. Then, $r=O(n^{2/3})$. Since the distance between any two points in $S$ is at least $1$, $r=\\Omega(\\sqrt{n})$ too.\nLet $p$ be any point in $S$ on the boundary of $D$. Let $\\ell_1$ be the line tangent to $D$ at $p$, and $\\ell_2$ the line obtained by translating $\\ell_1$ by distance $1$ towards the inside of $D$. Let $H$ be the region sandwiched by $\\ell_1$ and $\\ell_2$. It is easy to show that both the area and the perimeter of $H\\cap D$ is bounded by $O(\\sqrt{r})$ (since $r=\\Omega(\\sqrt{n})$). Hence, there can only be $O(\\sqrt{r})=O(n^{1/3})$ points in $H\\cap S$, by that any two points in $S$ are distance $1$ apart. Since the width of $H$ is $1$, there must exist a line $\\ell$ parallel to $\\ell_1$ such that $\\ell$ separates $S$, and any point in $S$ is at least $1/O(n^{1/3}) = \\Omega(n^{-1/3})$ away from $\\ell$. Q.E.D.\nNote. One can also show that $\\Omega(n^{-1/3})$ is best possible." }, { "Problem": "Let $n$ be a positive integer. A Nordic square is an $n \\times n$ board containing all the integers from $1$ to $n^2$ so that each cell contains exactly one number. Two different cells are considered adjacent if they share an edge. Every cell that is adjacent only to cells containing larger numbers is called a valley. An uphill path is a sequence of one or more cells such that:\n(i) the first cell in the sequence is a valley,\n(ii) each subsequent cell in the sequence is adjacent to the previous cell, and\n(iii) the numbers written in the cells in the sequence are in increasing order.\nFind, as a function of $n$, the smallest possible total number of uphill paths in a Nordic square.", "Solution": "The minimum total number of uphill paths in a Nordic square is $2n(n-1)+1$\nProof:\nFor every pair of adjacent cells in the grid, there is at least one pathway that begins at the higher valued cell, steps to the lower valued cell, and then continues stepping to lower adjacent cells until (by finite descent) a valley is reached. When reversed, this forms an uphill pathway that terminates at that exactly that pair of adjacent cells.\nThere are $2n(n-1)$ such adjacent pairs (each with at least one unique uphill path). In addition, the cell containing the value 1 will always be a valley which, by itself, forms an additional (singleton) uphill path. Therefore the minimum total number of uphill paths is at least $2n(n-1)+1$.\nThe diagram below shows a general family of constructions of Nordic squares that achieve this lower bound, thus proving the claim:\nIn these constructions, numbers are progressively inserted into the white spaces in ascending order emanating outwards from the \"1\" in the corner in the directions of the red arrows, and the remaining j largest numbers are then inserted into the cells labelled as \"mountains\". In this manner, the pathway of finite descent from each pair of adjacent cells is unique, and all terminate at the \"1\" valley, therefore the above-derived minimum is achieved. $\\blacksquare$\n- Griphomaniac (talk) 13:16, 21 July 2022 (EDT)" }, { "Problem": "Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and\nLet $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$\nProve that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$ and $CC_1C_2$ all pass through two common points.\n(Note: a scalene triangle is one where no two sides have equal length.)", "Solution": "https://www.youtube.com/watch?v=jZNIpapyGJQ [Video contains solutions to all day 2 problems]" }, { "Problem": "An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.\na) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;\nb) Calculate the distance of $P$ from either base;\nc) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.", "Solution": "(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.\n(b) Let $x$ be the distance from $P$ to one of the bases; then $h - x$ must be the distance from $P$ to the other base. Similar triangles give $\\frac{x}{\\frac{a}{2}} = \\frac{\\frac{c}{2}}{h - x}$, so $x^2 - hx + \\frac{ac}{4} = 0$ and so $x = \\frac{h \\pm \\sqrt{h^2 - ac}}{2}.$\n(c) When $h^2 \\ge ac$.\nIn our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0).\nLet our point P be on the axis of symmetry at z distance from the origin O.\nThe coordinates of the points A,B,C,D,E,F and P are given in the figure.\nNow,\nSlope of the line $PC= (z-0)/(0-c/2) = -2z/c$\nSlope of the line $PB= (z-h)/(0-a/2) = -2(z-h)/a$\nSince the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1.\ni.e\n$4z(z-h)=-ac$\nor $z^2 - zh + ac/4= 0$\nNow, solving for z, we get, $z= [(h + ( h^2 - ac ) ^1/2 ]/2$ and $[(h - ( h^2 - ac ) ^1/2 ]/2$\nSo, z is the distance of the points from the base CD..\nAlso the points are possible only when , $h^2 - ac >= 0$.. and doesn't exist for $h^2 -ac <0$" }, { "Problem": "The tetrahedron $SABC$ has the following property: there exist five spheres, each tangent to the edges $SA, SB, SC, BC, CA, AB$, or to their extensions.\n(a) Prove that the tetrahedron $SABC$ is regular.\n(b) Prove conversely that for every regular tetrahedron five such spheres exist.", "Solution": "\nPart (a)\nLet points $P_{SA}, P_{SB}, P_{SC}, P_{BC}, P_{CA}, P_{AB}$ be the points where the smallest sphere is tangent to the edges $SA, SB, SC, BC, CA, AB$ respectively.\nFor each face of the tetrahedron there is a circular cross section of the smallest sphere. Since that circle also needs to be tangent to the edges, then that circle is the incircle of each triangular face.\nFrom the properties of an incircle we know that:\n$|AP_{AB}|=|AP_{AC}|$, $|BP_{BC}|=|BP_{BA}|$, and $|CP_{CA}|=|CP_{CB}|$ in $\\Delta ABC$\nThere is a larger circle that is tangent to $AB$ and the extensions of $CA$ and $CB$.\nThis larger circle is a cross section of the sphere that's also tangent to the extension of $CS$ away from $S$ and the edges of $\\Delta ABC$\nTherefore, this larger circle and the incircle of $\\Delta ABC$ are part of that same sphere.\nIn order for these two circles to be part of the same sphere and also tangent to line $AB$, then the point of the tangent of this larger circle needs to be the same as point $P_{AB}$\nThe only way these to circles can share the same tangent point on edge $AB$ is if $|AP_{AB}|=|BP_{AB}|$\nUsing the same argument with the larder circle of edge $BC$ then $|BP_{BC}|=|CP_{BC}|$\nand with the larger circle of edge $AC$ then $|CP_{AC}|=|AP_{AC}|$\nThis results in:\n$|AP_{AB}|=|AP_{AC}|=|BP_{BC}|=|BP_{BA}|=|CP_{CA}|=|CP_{CB}|$ in $\\Delta ABC$\nwhich means that\n$|AB|=|BC|=|AC|$ in $\\Delta ABC$, thus $\\Delta ABC$ is an equilteral triangle.\nLikewise,\n$|AB|=|BS|=|AS|$ in $\\Delta ABS$, thus $\\Delta ABS$ is an equilteral triangle.\n$|AS|=|SC|=|AC|$ in $\\Delta ASC$, thus $\\Delta ASC$ is an equilteral triangle.\n$|SB|=|BC|=|SC|$ in $\\Delta SBC$, thus $\\Delta SBC$ is an equilteral triangle.\nSince all four faces are equilateral triangles, then tetrahedron $ABCS$ is a regular tetrahedron.\n\nPart (b)\nLet's consider a regular tetrahedron in the cartesian space with center at $(0,0,0)$ with $r_c$ as the circumradius.\nWe can write the coordinates for the four vertices as:\n$V_1=\\left( \\frac{2\\sqrt{2}}{3}r_c ,\\;0,\\;-\\frac{r_c}{3}\\right)$\n$V_2=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;\\frac{\\sqrt{6}}{3}r_c,\\;\\frac{-r_c}{3}\\right)$\n$V_3=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;-\\frac{\\sqrt{6}}{3}r_c,\\;\\frac{-r_c}{3}\\right)$\n$V_4=\\left(0 ,\\;0,\\;r_c\\right)$\nAnd the midpoints of all edges as:\n$M_{12}=\\frac{V_1-V_2}{2}=\\left( \\frac{\\sqrt{2}}{6}r_c ,\\;\\frac{\\sqrt{6}}{6}r_c,\\;-\\frac{r_c}{3}\\right)$\n$M_{13}=\\frac{V_1-V_3}{2}=\\left( \\frac{\\sqrt{2}}{6}r_c ,\\;-\\frac{\\sqrt{6}}{6}r_c,\\;-\\frac{r_c}{3}\\right)$\n$M_{23}=\\frac{V_2-V_3}{2}=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;0,\\;-\\frac{r_c}{3}\\right)$\n$M_{14}=\\frac{V_1-V_4}{2}=\\left( \\frac{\\sqrt{2}}{3}r_c ,\\;0,\\;\\frac{r_c}{3}\\right)$\n$M_{24}=\\frac{V_2-V_4}{2}=\\left( -\\frac{\\sqrt{2}}{6}r_c ,\\;\\frac{\\sqrt{6}}{6}r_c,\\;\\frac{r_c}{3}\\right)$\n$M_{34}=\\frac{V_3-V_4}{2}=\\left( -\\frac{\\sqrt{2}}{6}r_c ,\\;-\\frac{\\sqrt{6}}{6}r_c,\\;\\frac{r_c}{3}\\right)$\nNow we calculate the following six dot products:\n$(V_1-V_2) \\bullet M_{12}=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left( \\frac{\\sqrt{6}}{3}r_c\\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0$\n$(V_1-V_3) \\bullet M_{13}=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left(- \\frac{\\sqrt{6}}{3}r_c\\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0$\n$(V_2-V_3) \\bullet M_{23}=0+\\left(- \\frac{\\sqrt{6}}{6}r_c\\right)\\left( 0\\right)+ 0=0$\n$(V_1-V_4) \\bullet M_{13}=\\left( 2\\frac{\\sqrt{2}}{3}r_c \\right)\\left( \\frac{\\sqrt{2}}{3}r_c \\right)+0+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{4}{9}r_c^2-\\frac{4}{9}r_c^2=0$\n$(V_2-V_4) \\bullet M_{13}=\\left( -\\frac{\\sqrt{2}}{3}r_c \\right)\\left( -\\frac{\\sqrt{2}}{6}r_c \\right)+\\left( \\frac{\\sqrt{6}}{3}r_c \\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{1}{9}r_c^2+\\frac{1}{3}r_c^2-\\frac{4}{9}r_c^2=0$\n$(V_3-V_4) \\bullet M_{13}=\\left( -\\frac{\\sqrt{2}}{3}r_c \\right)\\left( -\\frac{\\sqrt{2}}{6}r_c \\right)+\\left( -\\frac{\\sqrt{6}}{3}r_c \\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{1}{9}r_c^2+\\frac{1}{3}r_c^2-\\frac{4}{9}r_c^2=0$\nSince all these dot producs equal to $0$ that means that the lines from the center to each of the midpoints are all perpendicular.\nNow we calculate the distances from $(0,0,0)$ to all the midpoints:\n$|M_{12}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c$\n$|M_{13}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c$\n$|M_{23}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c$\n$|M_{14}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c$\n$|M_{24}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c$\n$|M_{34}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c$\nSince all distances are all the same and all dot products are $0$, then we have our first sphere at $(0,0,0)$ with a radius of $\\frac{\\sqrt{3}}{3}r_c$\nNow we will look at the other 4 spheres.\nLet $E_{ij}$ be the point of tangent of the larger sphere on the extension of line $V{i}V{j}$ in the direction of $V{i}$ to $V{j}$ and beyond $V{j}$ for $i=1,2,3,4$; $j=1,2,3,4$; and $i \\ne j$\nSince $|V_j E_{ij}|=|V_j M_{ij}|=\\frac{1}{2} |V_j V_i|$, then $\\frac{1}{2}(V_j-V_i)=E_{ij}-V_j$, thus $E_{ij}=\\frac{3V_j-V_i}{2}$\nUsing this formula we calculate the following:\n$E_{41}=\\left( \\sqrt{2} r_c ,\\;0,\\;-r_c \\right)$\n$E_{42}=\\left( -\\frac{\\sqrt{2}}{2}r_c ,\\;\\frac{\\sqrt{6}}{2}r_c,\\;-r_c \\right)$\n$E_{43}=\\left( -\\frac{\\sqrt{2}}{2}r_c ,\\;-\\frac{\\sqrt{6}}{2}r_c,\\;-r_c \\right)$\nWe will start with the sphere below the base of the tetrahedron opposite of vertex $V_{4}$ below $\\Delta V{1}V{2}V{3}$\nThe center of this larger sphere is at $C_1=(0,0,-2r_c)$ and it is tangent at points $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$\nWe calculate the following dot products:\n$(V_1-V_2) \\bullet (M_{12}-C_1)=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left( \\frac{\\sqrt{6}}{3}r_c\\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0$\n$(V_1-V_3) \\bullet (M_{13}-C_1)=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left(- \\frac{\\sqrt{6}}{3}r_c\\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0$\n$(V_2-V_3) \\bullet (M_{23}-C_1)=0+\\left(- \\frac{\\sqrt{6}}{6}r_c\\right)\\left( 0\\right)+ 0=0$\n$(E_{41}-V_4) \\bullet (E_{41}-C_1) = \\left( \\sqrt{2}r_c \\right)^2+0+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0$\n$(E_{42}-V_4) \\bullet (E_{42}-C_1) = \\left( -\\frac{\\sqrt{2}}{2}r_c \\right)^2+\\left( \\frac{\\sqrt{6}}{2}r_c \\right)^2+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0$\n$(E_{43}-V_4) \\bullet (E_{43}-C_1) = \\left( -\\frac{\\sqrt{2}}{2}r_c \\right)^2+\\left( -\\frac{\\sqrt{6}}{2}r_c \\right)^2+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0$\nSince all these dot producs equal to $0$ that means that the lines from the center $C_1$ to each of the point $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$ are all perpendicular.\nNow we calculate the distances from $C_1=(0,0,-2r_c)$ to points $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$\n$|C_1M_{12}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c$\n$|C_1M_{13}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c$\n$|C_1M_{23}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c$\n$|C_1E_{41}|=\\sqrt{\\left( \\sqrt{2}r_c\\right)^2+\\left(0\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c$\n$|C_1E_{42}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{2}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{2}r_c\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c$\n$|C_1E_{43}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{2}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{2}r_c\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c$\nSince all distances are all the same and all dot products are $0$, then we have one of the larger spheres at $(0,0,-2r_c)$ with a radius of $\\sqrt{3}r_c$\nThen, the other three larger spheres which are the same size as the sphere with center at $C_1$ are congruent and tangent to their respective sides near the other faces of the tetrahedron.\nand this proves that a tetrahedron with any circumradius $r_c$ will have these 5 spheres, one with radius $\\frac{\\sqrt{3}}{3}r_c$, at the center of the tetrahedron and the other 4 with radius $\\sqrt{3}r_c$ at centers that are at a distance of $3r_c$ away from any of the vertices of the tetrahedron in the direction from that vertex to the center of its opposite face.\nThus these five spheres exist for any regular tetrahedron.\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." } ]