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[ "Rolling a pair of fair $6$ -sided dice, the probability of getting a sum of $7$ is $\\frac16:$ Regardless what the first die shows, the second die has exactly one outcome to make the sum $7.$ We consider the complement: The probability of not getting a sum of $7$ is $1-\\frac16=\\frac56.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-\\left(\\frac56\\right)^n.$\nTherefore, we have $1-\\left(\\frac56\\right)^n>\\frac12,$ or \\[\\left(\\frac56\\right)^n<\\frac12.\\] Since $\\left(\\frac56\\right)^4<\\frac12<\\left(\\frac56\\right)^3,$ the least integer $n$ satisfying the inequality is $\\boxed{4}.$", "Let's try the answer choices. We can quickly find that when we roll $3$ dice, either the first and second sum to $7$ , the first and third sum to $7$ , or the second and third sum to $7$ . There are $6$ ways for the first and second dice to sum to $7$ $6$ ways for the first and third to sum to $7$ , and $6$ ways for the second and third dice to sum to $7$ . However, we overcounted (but not by much) so we can assume that the answer is $\\boxed{4}$", "We can start by figuring out what the probability is for each die to add up to $7$ if there is only $1$ roll. We can quickly see that the probability is $\\frac16$ , as there are $6$ ways to make $7$ from $2$ numbers on a die, and there are a total of $36$ ways to add $2$ numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for $n$ rolls would be $\\frac16$ $n$ . The smallest number that satisfies this is $\\boxed{4}$" ]

[ "The total number of combinations when rolling two dice is $6*6 = 36$\nThere are three ways that a sum of 7 can be rolled. $2+5$ $4+3$ , and $6+1$ . There are two 2's on one die and two 5's on the other, so there are a total of 4 ways to roll the combination of 2 and 5. There are two 4's on one die and two 3's on the other, so there are a total of 4 ways to roll the combination of 4 and 3. There are two 6's on one die and two 1's on the other, so there are a total of 4 ways to roll the combination of 6 and 1. Add $4 + 4 + 4 = 12$\nThus, our probability is $\\frac{12}{36} = \\frac{1}{3}$ . The answer is $\\boxed{13}$", "Assume we roll the die with only evens first. For whatever value rolled, there are exactly 2 faces on the odd die that makes the sum 7. The odd die has 6 faces, so our probability is $\\boxed{13}$" ]

[ "Let $M$ and $N$ be the points on $\\overline{AB}$ and $\\overline{AC}$ , respectively, where the paper is folded. Let $D$ be the point on $\\overline{BC}$ where the folded $A$ touches it. We have $AF=6\\sqrt{3}$ and $FD=3$ , so $AD=3\\sqrt{13}$ . Denote $\\angle DAF = \\theta$ ; we get $\\cos\\theta = 2\\sqrt{3}/\\sqrt{13}$\nIn triangle $AXY$ $AY=\\tfrac 12 AD = \\tfrac 32 \\sqrt{13}$ , and $AX=AY\\sec\\theta =\\tfrac{13}{4}\\sqrt{3}$\nIn triangle $AMX$ , we get $\\angle AMX=60^\\circ-\\theta$ and then use sine-law to get $MX=\\tfrac 12 AX\\csc(60^\\circ-\\theta)$ ; similarly, from triangle $ANX$ we get $NX=\\tfrac 12 AX\\csc(60^\\circ+\\theta)$ . Thus \\[MN=\\tfrac 12 AX(\\csc(60^\\circ-\\theta) +\\csc(60^\\circ+\\theta)).\\] Since $\\sin(60^\\circ\\pm \\theta) = \\tfrac 12 (\\sqrt{3}\\cos\\theta \\pm \\sin\\theta)$ , we get \\begin{align*} \\csc(60^\\circ-\\theta) +\\csc(60^\\circ+\\theta) &= \\frac{\\sqrt{3}\\cos\\theta}{\\cos^2\\theta - \\tfrac 14} = \\frac{24 \\cdot \\sqrt{13}}{35} \\end{align*} Then \\[MN = \\frac 12 AX \\cdot \\frac{24 \\cdot \\sqrt{13}}{35} = \\frac{39\\sqrt{39}}{35}\\]\nThe answer is $39 + 39 + 35 = \\boxed{113}$", "Let $P$ and $Q$ be the points on $\\overline{AB}$ and $\\overline{AC}$ , respectively, where the paper is folded.\nLet $D$ be the point on $\\overline{BC}$ where the folded $A$ touches it.\nLet $a$ $b$ , and $x$ be the lengths $AP$ $AQ$ , and $PQ$ , respectively.\nWe have $PD = a$ $QD = b$ $BP = 12 - a$ $CQ = 12 - b$ $BD = 9$ , and $CD = 3$\nUsing the Law of Cosines on $BPD$\n$a^{2} = (12 - a)^{2} + 9^{2} - 2 \\times (12 - a) \\times 9 \\times \\cos{60}$\n$a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a$\n$a = \\frac{39}{5}$\nUsing the Law of Cosines on $CQD$\n$b^{2} = (12 - b)^{2} +3^{2} - 2 \\times (12 - b) \\times 3 \\times \\cos{60}$\n$b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b$\n$b = \\frac{39}{7}$\nUsing the Law of Cosines on $DPQ$\n$x^{2} = a^{2} + b^{2} - 2ab \\cos{60}$\n$x^{2} = (\\frac{39}{5})^2 + (\\frac{39}{7})^2 - (\\frac{39}{5} \\times \\frac{39}{7})$\n$x = \\frac{39 \\sqrt{39}}{35}$\nThe solution is $39 + 39 + 35 = \\boxed{113}$", "Proceed with the same labeling as in Solution 1.\n$\\angle B = \\angle C = \\angle A = \\angle PDQ = 60^\\circ$\n$\\angle PDB + \\angle PDQ + \\angle QDC = \\angle QDC + \\angle CQD + \\angle C = 180^\\circ$\nTherefore, $\\angle PDB = \\angle CQD$\nSimilarly, $\\angle BPD = \\angle QDC$\nNow, $\\bigtriangleup BPD$ and $\\bigtriangleup CDQ$ are similar triangles, so\n$\\frac{3}{12-a} = \\frac{12-b}{9} = \\frac{b}{a}$\nSolving this system of equations yields $a = \\frac{39}{5}$ and $b = \\frac{39}{7}$\nUsing the Law of Cosines on $APQ$\n$x^{2} = a^{2} + b^{2} - 2ab \\cos{60}$\n$x^{2} = (\\frac{39}{5})^2 + (\\frac{39}{7})^2 - (\\frac{39}{5} \\times \\frac{39}{7})$\n$x = \\frac{39 \\sqrt{39}}{35}$\nThe solution is $39 + 39 + 35 = \\boxed{113}$", "We let the original position of $A$ be $A$ , and the position of $A$ after folding be $D$ . Also, we put the triangle on the coordinate plane such that $A=(0,0)$ $B=(-6,-6\\sqrt3)$ $C=(6,-6\\sqrt3)$ , and $D=(3,-6\\sqrt3)$\n\nNote that since $A$ is reflected over the fold line to $D$ , the fold line is the perpendicular bisector of $AD$ . We know $A=(0,0)$ and $D=(3,-6\\sqrt3)$ . The midpoint of $AD$ (which is a point on the fold line) is $(\\tfrac32, -3\\sqrt3)$ . Also, the slope of $AD$ is $\\frac{-6\\sqrt3}{3}=-2\\sqrt3$ , so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of $AD$ , or $\\frac{1}{2\\sqrt3}=\\frac{\\sqrt3}{6}$ . Then, using point slope form, the equation of the fold line is \\[y+3\\sqrt3=\\frac{\\sqrt3}{6}\\left(x-\\frac32\\right)\\] \\[y=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\] Note that the equations of lines $AB$ and $AC$ are $y=\\sqrt3x$ and $y=-\\sqrt3x$ , respectively. We will first find the intersection of $AB$ and the fold line by substituting for $y$ \\[\\sqrt3 x=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\] \\[\\frac{5\\sqrt3}{6}x=-\\frac{13\\sqrt3}{4} \\implies x=-\\frac{39}{10}\\] Therefore, the point of intersection is $\\left(-\\tfrac{39}{10},-\\tfrac{39\\sqrt3}{10}\\right)$ . Now, lets find the intersection with $AC$ . Substituting for $y$ yields \\[-\\sqrt3 x=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\] \\[\\frac{-7\\sqrt3}{6}x=-\\frac{13\\sqrt3}{4} \\implies x=\\frac{39}{14}\\] Therefore, the point of intersection is $\\left(\\tfrac{39}{14},-\\tfrac{39\\sqrt3}{14}\\right)$ . Now, we just need to use the distance formula to find the distance between $\\left(-\\tfrac{39}{10},-\\tfrac{39\\sqrt3}{10}\\right)$ and $\\left(\\tfrac{39}{14},-\\tfrac{39\\sqrt3}{14}\\right)$ \\[\\sqrt{\\left(\\frac{39}{14}+\\frac{39}{10}\\right)^2+\\left(-\\frac{39\\sqrt3}{14}+\\frac{39\\sqrt3}{10}\\right)^2}\\] The number 39 is in all of the terms, so let's factor it out: \\[39\\sqrt{\\left(\\frac{1}{14}+\\frac{1}{10}\\right)^2+\\left(-\\frac{\\sqrt3}{14}+\\frac{\\sqrt3}{10}\\right)^2}=39\\sqrt{\\left(\\frac{6}{35}\\right)^2+\\left(\\frac{\\sqrt3}{35}\\right)^2}\\] \\[\\frac{39}{35}\\sqrt{6^2+\\sqrt3^2}=\\frac{39\\sqrt{39}}{35}\\] Therefore, our answer is $39+39+35=\\boxed{113}$ , and we are done." ]

[ "Consider the quadratic in completed square form: it must be $y=a(x-4)^{2}+2$ . Now substitute $x=2$ and $y=0$ to give $a=-\\frac{1}{2}$ . Now expanding gives $y=-\\frac{1}{2}x^{2}+4x-6$ , so the product is $-\\frac{1}{2} \\cdot 4 \\cdot -6 = 3 \\cdot 4 = 12$ , which is $\\boxed{12}$" ]

[ "We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new $x$ -coordinate will be $1 + 2 + \\frac{1}{2} = \\frac{7}{2}$ because she travels along a distance of $2 \\cdot \\frac{1}{2} = 1$ km because of the side relationships of an equilateral triangle, then $2$ km because the line is parallel to the $x$ -axis, and the remaining distance is $\\frac{1}{2}$ km because she went halfway along and because of the logic for the first part of her route. For her $y$ -coordinate, we can use similar logic to find that the coordinate is $\\sqrt{3} + 0 - \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{2}$ . Therefore, her distance is \\[\\sqrt{\\left(\\frac{7}{2}\\right)^2 + \\left(\\frac{\\sqrt{3}}{2}\\right)^2} = \\sqrt{\\frac{49}{4} + \\frac{3}{4}} = \\sqrt{\\frac{52}{4}} = \\sqrt{13},\\] giving an answer of $\\boxed{13}$" ]

[ "Let the particle's position be represented by a complex number. Recall that multiplying a number by cis $\\left( \\theta \\right)$ rotates the object in the complex plane by $\\theta$ counterclockwise. In this case, we use $a = cis(\\frac{\\pi}{4})$ . Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to\nwhere a is cis $\\left( \\theta \\right)$ . By De-Moivre's theorem, $\\left(cis( \\theta \\right)^n )$ =cis $\\left(n \\theta \\right)$ .\nTherefore,\nFurthermore, $5a^{150} = - 5i$ . Thus, the final answer is\nAs before, consider $z$ as a complex number. Consider the transformation $z \\to (z-\\omega)e^{i\\theta} + \\omega$ . This is a clockwise rotation of $z$ by $\\theta$ radians about the points $\\omega$ . Let $f(z)$ denote one move of $z$ . Then\nTherefore, $z$ rotates along a circle with center $\\omega = 5+(5+5\\sqrt2)i$ . Since $8 \\cdot \\frac{\\pi}{4} = 2\\pi$ $f^9(z) = f(z) \\implies f^{150}(z) = f^6(z) \\implies p+q = \\boxed{019}$ , as desired (the final algebra bash isn't bad).", "Let $T:\\begin{pmatrix}x\\\\y\\end{pmatrix}\\rightarrow R(\\frac{\\pi}{4})\\begin{pmatrix}x\\\\y\\end{pmatrix}+\\begin{pmatrix}10\\\\0\\end{pmatrix}$ . We assume that the rotation matrix $R(\\frac{\\pi}{4}) = R$ here. Then we have\nThis simplifies to\nSince $R+R^{7}=O, R^2+R^6=O, R^3+R^5=O, I+R^4=O$ , so we have $R^6\\begin{pmatrix}5\\\\0\\end{pmatrix}+(-R^6-R^7)\\begin{pmatrix}10\\\\0\\end{pmatrix}$ , giving $p=-5\\sqrt{2}, q=5\\sqrt{2}+5$ . The answer is yet $\\lfloor10\\sqrt{2}+5\\rfloor=\\boxed{019}$" ]

[ "The length of the path (the number of times the particle moves) can range from $l = 5$ to $9$ ; notice that $d = 10-l$ gives the number of diagonals. Let $R$ represent a move to the right, $U$ represent a move upwards, and $D$ to be a move that is diagonal. Casework upon the number of diagonal moves:\nTogether, these add up to $2 + 18 + 42 + 20 + 1 = \\boxed{083}$", "Another possibility is to use block-walking and recursion : for each vertex, the number of ways to reach it is $a + b + c$ , where $a$ is the number of ways to reach the vertex from the left (without having come to that vertex (the one on the left) from below), $b$ is the number of ways to reach the vertex from the vertex diagonally down and left, and $c$ is the number of ways to reach the vertex from below (without having come to that vertex (the one below) from the left).\nAssign to each point $(i,j)$ the triplet $(a_{i,j}, b_{i,j}, c_{i,j})$ . Let $s(i,j) = a_{i,j}+ b_{i,j}+ c_{i,j}$ . Let all lattice points that contain exactly one negative coordinate be assigned to $(0,0,0)$ . This leaves the lattice points of the first quadrant, the positive parts of the $x$ and $y$ axes, and the origin unassigned. As a seed, assign to $(0,1,0)$ . (We will see how this correlates with the problem.) Then define for each lattice point $(i,j)$ its triplet thus: \\begin{align*} a_{i,j} &= s(i-1,j) - c_{i-1,j}\\\\ b_{i,j} &= s(i-1,j-1) \\\\ c_{i,j} &= s(i,j-1) - a_{i,j-1}. \\end{align*} It is evident that $s(i,j)$ is the number of ways to reach $(i,j)$ from $(0,0)$ . Therefore we compute vertex by vertex the triplets $(a_{i,j}, b_{i,j}, c_{i,j})$ with $0 \\leq i, j \\leq 5$ Finally, after simple but tedious calculations, we find that $(a_{5,5}, b_{5,5}, c_{5,5}) = (28,27,28)$ , so $s(i,j)=28+27+28 = \\boxed{083}$" ]

[ "Squares of size $1\\times1,\\ 2\\times2,\\ 3\\times3,\\ ...$ are successively enclosed between the path and the axes.\n\nIt takes $1+1+1$ minutes to enclose the first square, $1+2+2$ minutes to enclose the second, $1+3+3$ minutes to enclose the third, and so on. After odd squares, the particle is on the Y axis; after even squares, the particle is on the X axis.\nFirst we find the highest integer $n$ such that $\\sum_{k=1}^n1+2k\\le1989$ . The sum is equal to \\[\\sum_{k=1}^n1+2\\sum_{k=1}^nk=n+n(n+1)=n(n+2)=(n+1)^2-1\\] so the highest value is $n=43$ for which the sum is $1935$\nAfter $1935$ minutes the 43rd square is enclosed and the particle is at the point $(0,43)$ . During the 1936th minute it moves up to $(0,44)$ . At the end of the 1980th minute it has moved right to $(44,44)$ . After this it moves downward, and at the end of the 1989th minute it is at $(44,35)$ . The answer is $\\boxed{44,35}$" ]

[ "The highest elevation a particle can reach is the vertex of the quadratic. The x-value that can get the maximum is $\\frac{-160}{-2 \\cdot 16} = 5$ , so the highest elevation is $160(5) - 16(5^2) = 400$ feet, which is answer choice $\\boxed{400}$" ]

[ "The clock will display the incorrect time for the entire hours of $1, 10, 11$ and $12$ . So the correct hour is displayed $\\frac 23$ of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a $1$ , so the minutes that will not display correctly are $10, 11, 12, \\dots, 19$ and $01, 21, 31, 41,$ and $51$ . This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is $\\frac 23 \\cdot \\left(1 - \\frac {15}{60}\\right) = \\frac 23 \\cdot \\frac 34 = \\boxed{12}$" ]

[ "The clock will display the incorrect time for the entire hours of $1, 10, 11$ and $12$ . So the correct hour is displayed $\\frac 23$ of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a $1$ , so the minutes that will not display correctly are $10, 11, 12, \\dots, 19$ and $01, 21, 31, 41,$ and $51$ . This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is $\\frac 23 \\cdot \\left(1 - \\frac {15}{60}\\right) = \\frac 23 \\cdot \\frac 34 = \\boxed{12}$" ]

[ "We use case work on the value of $a_3$\nCase 1: $a_3 = 1$ . Since $a_1 + a_2 < a_4 + a_5$ $(a_1, a_2)$ can only be a permutation of $(2, 3)$ or $(2, 4)$ . The values of $a_1$ and $a_2$ , as well as the values of $a_4$ and $a_5$ , are interchangeable, so this case produces a total of $2(2 \\cdot 2) = 8$ solutions.\nCase 2: $a_3 = 2$ . Similarly, we have $(a_1, a_2)$ is a permutation of $(1, 3)$ $(1, 4)$ , or $(1, 5)$ , which gives a total of $3(2 \\cdot 2) = 12$ solutions.\nCase 3: $a_3 = 3$ $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 4)$ , which gives a total of $2(2 \\cdot 2) = 8$ solutions.\nCase 4: $a_3 = 4$ $(a_1, a_2)$ is a permutation of $(1, 2)$ $(1, 3)$ , or $(2, 3)$ , which gives a total of $3(2 \\cdot 2) = 12$ solutions.\nCase 5: $a_3 = 5$ $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 3)$ , which gives a total of $2(2 \\cdot 2) = 8$ solutions.\nTherefore, our answer is $8 + 12 + 8 + 12 + 8 = 48 \\Rightarrow \\boxed{48}$" ]

[ "We claim that the smallest $n$ is $67^2 = \\boxed{4489}$", "This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as \"equivalence relation\":\nIt is possible to write all positive integers $n$ in the form $p\\cdot m^2$ , where $m^2$ is the largest perfect square dividing $n$ , so $p$ is not divisible by the square of any prime. Obviously, one working permutation of $[n]$ is simply $(1, 2, \\ldots, n)$ ; this is acceptable, as $ka_k$ is always $k^2$ in this sequence.\nLemma 1. We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities $p$\nProof. Let $p_k$ and $m_k$ be the values of $p$ and $m$ , respectively, for a given $k$ as defined above, such that $p$ is not divisible by the square of any prime. We can obviously permute two numbers which have the same $p$ , since if $p_j = p_w$ where $j$ and $w$ are 2 values of $k$ , then $j\\cdot w = p_j^2\\cdot m_j^2\\cdot m_w^2$ , which is a perfect square. This proves that we can permute any numbers with the same value of $p$\nEnd Lemma\nLemma 2. We will prove the converse of Lemma 1: Let one number have a $p$ value of $\\phi$ and another, $\\gamma$ $\\phi\\cdot f$ and $\\gamma\\cdot g$ are both perfect squares.\nProof. $\\phi\\cdot f$ and $\\gamma\\cdot g$ are both perfect squares, so for $\\phi\\cdot \\gamma$ to be a perfect square, if $g$ is greater than or equal to $f$ $g/f$ must be a perfect square, too. Thus $g$ is $f$ times a square, but $g$ cannot divide any squares besides $1$ , so $g = 1f$ $g = f$ . Similarly, if $f\\geq g$ , then $f = g$ for our rules to keep working.\nEnd Lemma\nWe can permute $l$ numbers with the same $p$ in $l!$ ways. We must have at least 67 numbers with a certain $p$ so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as $h$ , in general, we need numbers all the way up to $h\\cdot 67^2$ , so obviously, $67^2$ is the smallest such number such that we can get a $67!$ term; here 67 $p$ terms are 1. Thus we need the integers $1, 2, \\ldots, 67^2$ , so $67^2$ , or $\\boxed{4489}$ , is the answer.", "It's well known that there exists $f(n)$ and $g(n)$ such that $n = f(n) \\cdot g(n)$ , no square divides $f(n)$ other than 1, and $g(n)$ is a perfect square.\nWe prove first: If $f(k) = f(a_k)$ $k \\cdot a_k$ is a perfect square.\n$k \\cdot a_k = f(k) \\cdot g(k) \\cdot f(a_k) \\cdot g(a_k) = f(k)^2 \\cdot g(k) \\cdot g(a_k)$ , which is a perfect square.\nWe will now prove: If $k \\cdot a_k$ is a perfect square, $f(k) = f(a_k)$\nWe do proof by contrapositive: If $f(k) \\neq f(a_k)$ $k \\cdot a_k$ is not a perfect square.\n$v_p(k)$ is the p-adic valuation of k. (Basically how many factors of p you can take out of k)\nNote that if $f(k) \\neq f(a_k)$ , By the Fundamental Theorem of Arithmetic, $f(k)$ and $f(a_k)$ 's prime factorization are different, and thus there exists a prime p, such that $v_p(f(k)) \\neq v_p(f(a_k))$ . Also, since $f(k)$ and $f(a_k)$ is squarefree, $v_p(k), v_p(a_k) \\leq 1$ . Thus, $v_p(k \\cdot a_k) = 1$ , making $k \\cdot a_k$ not a square.\nThus, we can only match k with $a_k$ if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the $a_k$ with f value 1, then 2, ... Thus, our answer is:\n$P(n) = \\prod_{1 \\leq i \\leq n, g(i) = 1} \\left\\lfloor \\sqrt{\\frac{n}{i}} \\right \\rfloor !$\nFor all $n < 67^2$ $P(n)$ doesn't have a factor of 67. However, if $n = 67^2$ , the first term will be a multiple of 2010, and thus the answer is $67^2 = \\boxed{4489}$" ]

[ "The point $(a,b)$ is the foot of the perpendicular from $(12,10)$ to the line $y=-5x+18$ . The perpendicular has slope $\\frac{1}{5}$ , so its equation is $y=10+\\frac{1}{5}(x-12)=\\frac{1}{5}x+\\frac{38}{5}$ . The $x$ -coordinate at the foot of the perpendicular satisfies the equation $\\frac{1}{5}x+\\frac{38}{5}=-5x+18$ , so $x=2$ and $y=-5\\cdot2+18=8$ . Thus $(a,b) = (2,8)$ , and $a+b = \\boxed{10}$", "If the mouse is at $(x, y) = (x, 18 - 5x)$ , then the square of the distance from the mouse to the cheese is $(x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).$ The value of this expression is smallest when $x = 2$ , so the mouse is closest to the cheese at the point $(2, 8)$ , and $a+b=2+8 = \\boxed{10}$", "We are trying to find the point where distance between the mouse and $(12, 10)$ is minimized. This point is where the line that passes through $(12, 10)$ and is perpendicular to $y=-5x+18$ intersects $y=-5x+18$ . By basic knowledge of perpendicular lines, this line is $y=\\frac{x}{5}+\\frac{38}{5}$ . This line intersects $y=-5x+18$ at $(2,8)$ . So $a+b=\\boxed{10}$ . - MegaLucario1001", "If the mouse is at $(x, y) = (x, 18 - 5x)$ , then the square of the distance from the mouse to the cheese is $(x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4)$ .\nThe value of this expression is smallest when $x = 2$ , so the mouse is closest to the cheese at the point $(2, 8)$ , and $a+b=2+8 = \\boxed{10}$ .\n-Paixiao" ]

[ "Note that the fold is the perpendicular bisector of $(0,2)$ and $(4,0)$ . Thus, the fold goes through the midpoint $(2,1)$\nThe fold also has a slope of $-\\frac{1}{m}$ , where the $m$ is the slope of the line connecting these two points. We find $m = \\frac{0 - 2}{4 - 0} = -\\frac{1}{2}$ . Thus, the slope of the fold is $2$ and goes through $(2,1)$ , so the equation of the fold is $y = 2x - 3$\nWe want the line connecting $(7,3)$ and $(m,n)$ to have the same fold as a perpendicular bisector. The slope should be $-\\frac{1}{2}$ , so we get $\\frac{n - 3}{m - 7} = -\\frac{1}{2}$ , which leads to $m = 13 -2n$\nWe also want $y = 2x - 3$ to bisect the segment from $(7,3)$ to $(m,n)$ . Thus, the midpoint $(x,y) = \\left(\\dfrac{7 + m}{2}, \\frac{3 + n}{2}\\right)$ must lie on the line. Plugging into the equation of the line, we find $\\frac{3 + n}{2} = (7 + m) - 3$ , which simplifies to $n = 5 + 2m$\nSolving the system of two equations in two variables $m = 13 - 2n$ and $n = 5 + 2m$ gives $m = \\frac{3}{5}$ and $n = \\frac{31}{5}$ , for a sum of $\\boxed{6.8}$" ]

[ "The line of the fold is the perpendicular bisector of the segment that connects $(0,2)$ and $(4,0)$ . \nThe point $(m,n)$ is the image of the point $(7,3)$ according to this axis.\nThe situation looks as follows.\nNow, we will compute the coordinates of the point $D$ , using the following facts:\nAs the triangles $SCT$ and $SDT$ are congruent, their areas are equal. The area of the triangle $SCT$ is $1/2$ of the size of the vector product $\\overrightarrow{SC}\\times\\overrightarrow{ST}$ , and the area of $SDT$ is $1/2$ of the size of $\\overrightarrow{ST}\\times\\overrightarrow{SD}$\nWe get that $\\overrightarrow{SC}\\times\\overrightarrow{ST} = \\overrightarrow{ST}\\times\\overrightarrow{SD}$\nThe equality remains valid if we multiply the vector $\\overrightarrow{ST}$ by any constant. In other words, instead of $\\overrightarrow{ST}$ we can use any vector with the same direction.\nThe axis of symmetry is perpendicular to $\\overrightarrow{AB}$ . Thus its direction is $(2,4)=(1,2)$\nWe get that $\\overrightarrow{SC}\\times (1,2) = (1,2) \\times\\overrightarrow{SD}$\nSubstituting the coordinates $\\overrightarrow{SC}=(5,2)$ and $\\overrightarrow{SD}=(m-2,n-1)$ we get $5\\cdot 2 - 2\\cdot 1 = 1\\cdot (n-1) - 2\\cdot(m-2)$ . This simplifies to $n=2m+5$\nWe just discovered that the coordinates of $D$ are $(m,2m+5)$ . We will now use the second two facts mentioned above to find $m$\nWe have $|SD| = |SC|$ and therefore $|SD|^2 = |SC|^2$ . We know that $|SC|^2 = 5^2 + 2^2 = 29$ , and $|SD|^2 = (m-2)^2 + (2m+4)^2$ . Simplifying, we get the equation $5m^2 + 12m - 9 = 0$ . This has exactly one positive root $m=0.6$\nIt follows that $D=(0.6,6.2)$ , and that $m+n = 6.2 + 0.6 = \\boxed{6.8}$" ]

[ "In this solution, let $\\boldsymbol{n}$ -line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$ -line points.\nThere are $\\binom{40}{2}=780$ pairs of lines. Among them:\nIt follows that the $2$ -line points account for $780-9-24-50-90=\\boxed{607}$ pairs of lines, where each pair intersect at a single point." ]

[ "The probability of rolling an even number on the first turn is $\\frac{1}{2}$ and the probability of rolling the same number on the next turn is $\\frac{1}{6}$ . The probability of winning is $\\frac{1}{2}\\cdot \\frac{1}{6} =\\frac{1}{12}$ . If the game is to be fair, the amount paid, $5$ dollars, must be $\\frac{1}{12}$ the amount of the prize money, so the answer is $\\boxed{60}.$" ]

[ " The best square's side length is slightly less than $\\sqrt 5$ , yielding an answer of $\\boxed{5.0}$" ]

[ "Any point outside the square with side length $\\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\\frac{3}{5}$ that has the same center and orientation as the unit square has $\\frac{1}{5}\\le d(P)\\le\\frac{1}{3}$\nSince the area of the unit square is $1$ , the probability of a point $P$ with $\\frac{1}{5}\\le d(P)\\le\\frac{1}{3}$ is the area of the shaded region, which is the difference of the area of two squares.\n$\\left(\\frac{3}{5}\\right)^2-\\left(\\frac{1}{3}\\right)^2=\\frac{9}{25}-\\frac{1}{9}=\\frac{56}{225}$\nThus, the answer is $56 + 225 = \\boxed{281}.$", "First, let's figure out $d(P) \\geq \\frac{1}{3}$ which is \\[\\left(\\frac{3}{5}\\right)^2=\\frac{9}{25}.\\] Then, $d(P) \\geq \\frac{1}{5}$ is a square inside $d(P) \\geq \\frac{1}{3}$ , so \\[\\left(\\frac{1}{3}\\right)^2=\\frac{1}{9}.\\] Therefore, the probability that $\\frac{1}{5}\\le d(P)\\le\\frac{1}{3}$ is \\[\\frac{9}{25}-\\frac{1}{9}=\\frac{56}{225}\\] So, the answer is $56+225=\\boxed{281}$", "First, lets assume that point $P$ is closest to a side $S$ of the square. If it is $\\frac{1}{5}$ far from $S$ , then it should be at least $\\frac{1}{5}$ from both the adjacent sides of $S$ in the square. This leaves a segment of $1 - 2 \\cdot \\frac{1}{5} = \\frac{3}{5}$ . If the distance from $P$ to $S$ is $\\frac{1}{3}$ , then notice the length of the side-ways segment for $P$ is $1 - 2 \\cdot \\frac{1}{3} = \\frac{1}{3}$ . Notice that as the distance from $P$ to $S$ increases, the possible points for the side-ways decreases. This produces a trapezoid with parallel sides $\\frac{3}{5}$ and $\\frac{1}{3}$ with height $\\frac{1}{3} - \\frac{1}{5} = \\frac{2}{15}$ . This trapezoid has area (or probability for one side) $\\frac{1}{2} \\cdot \\left(\\frac{1}{3}+\\frac{3}{5}\\right)\\cdot \\frac{2}{15} = \\frac{14}{225}$ . Since the square has $4$ sides, we multiply by $4$ . Hence, the probability is $\\frac{56}{225}$ . The answer is $\\boxed{281}$ . ~Saucepan_man02" ]

[ "Using the Secant-Secant Power Theorem, you can get $9(16)=(13-r)(13+r)$ , where $r$ is the radius of the given circle. Solving the equation, you get a quadratic: $r^2-25$ . A radius cannot be negative so the answer is $\\boxed{5}$" ]

[ "\nThe diagram represents each unit square of the given $2020 \\times 2020$ square.\nWe consider an individual one-by-one block.\nIf we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$ , the area covered by the circles should be $0.5$ . Because of this, and the fact that there are four circles, we write\n\\[4 \\cdot \\frac{1}{4} \\cdot \\pi d^2 = \\frac{1}{2}\\]\nSolving for $d$ , we obtain $d = \\frac{1}{\\sqrt{2\\pi}}$ , where with $\\pi \\approx 3$ , we get $d \\approx \\frac{1}{\\sqrt{6}} \\approx \\dfrac{1}{2.5} = \\dfrac{10}{25} = \\dfrac{2}{5}$ , and from here, we see that $d \\approx 0.4 \\implies \\boxed{0.4}.$", "As in the previous solution, we obtain the equation $4 \\cdot \\frac{1}{4} \\cdot \\pi d^2 = \\frac{1}{2}$ , which simplifies to $\\pi d^2 = \\frac{1}{2} = 0.5$ . Since $\\pi$ is slightly more than $3$ $d^2$ is slightly less than $\\frac{0.5}{3} = 0.1\\bar{6}$ . We notice that $0.1\\bar{6}$ is slightly more than $0.4^2 = 0.16$ , so $d$ is roughly $\\boxed{0.4}.$ emerald_block", "As above, we find that we need to estimate $d = \\frac{1}{\\sqrt{2\\pi}}$\nNote that we can approximate $2\\pi \\approx 6.28318 \\approx 6.25$ and so $\\frac{1}{\\sqrt{2\\pi}}$ $\\approx \\frac{1}{\\sqrt{6.25}}=\\frac{1}{2.5}=0.4$\nAnd so our answer is $\\boxed{0.4}$", "We only need to figure out the probability for a unit square, as it will scale up to the $2020\\times 2020$ square. Since we want to find the probability that a point inside a unit square that is $d$ units away from a lattice point (a corner of the square) is $\\frac{1}{2}$ , we can find which answer will come the closest to covering $\\frac{1}{2}$ of the area.\nSince the closest is $0.4$ which turns out to be $(0.4)^2\\times \\pi = 0.16 \\times \\pi$ which is about $0.502$ , we find that the answer rounded to the nearest tenth is $0.4$ or $\\boxed{0.4}$", "As per the above diagram, realize that $\\pi d^2 = \\frac{1}{2}$ , so $d = \\frac{1}{(\\sqrt{2})(\\sqrt{\\pi})}$\n$\\sqrt{2} \\approx 1.4 = \\frac{7}{5}$\n$\\sqrt{\\pi}$ is between $1.7$ and $1.8$ $((1.7)^2 = 2.89$ and $(1.8)^2 = 3.24)$ , so we can say $\\sqrt{\\pi} \\approx 1.75 = \\frac{7}{4}$\nSo $d \\approx \\frac{1}{(\\frac{7}{5})(\\frac{7}{4})} = \\frac{1}{\\frac{49}{20}} = \\frac{20}{49}$ . This is slightly above $\\boxed{0.4}$", "As above, we have the equation $\\pi d^2 = \\frac{1}{2}$ , and we want to find the most accurate value of $d$ . We resort to the answer choices and can plug those values of $d$ in and see which value of $d$ will lead to the most accurate value of $\\pi$\nStarting off in the middle, we try option C with $d=0.5$ . Plugging this in, we get $\\pi \\left(\\frac{1}{2}\\right)^2 = \\frac{1}{2},$ and after simplifying we get $\\pi = \\frac{1}{2} \\cdot 4 = 2.$ That's not very good. We know $\\pi \\approx 3.14.$\nLet's see if we can do better. Trying option A with $d = 0.3,$ we get $\\pi = \\frac{1}{2} \\cdot \\frac{100}{9} = \\frac{50}{9} = 5 \\frac{5}{9}.$\nHm, let's try option B with $d = 0.4.$ We get $\\pi = \\frac{1}{2} \\cdot \\frac{25}{4} = \\frac{25}{8} = 3 \\frac{1}{8}$ . This is very close to $\\pi$ and is the best estimate for $\\pi$ of the 5 options.\nTherefore, the answer is $\\boxed{0.4}.$ ~ epiconan" ]

[ "\nThe diagram represents each unit square of the given $2020 \\times 2020$ square.\nWe consider an individual one-by-one block.\nIf we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$ , the area covered by the circles should be $0.5$ . Because of this, and the fact that there are four circles, we write\n\\[4 \\cdot \\frac{1}{4} \\cdot \\pi d^2 = \\frac{1}{2}\\]\nSolving for $d$ , we obtain $d = \\frac{1}{\\sqrt{2\\pi}}$ , where with $\\pi \\approx 3$ , we get $d \\approx \\frac{1}{\\sqrt{6}} \\approx \\dfrac{1}{2.5} = \\dfrac{10}{25} = \\dfrac{2}{5}$ , and from here, we see that $d \\approx 0.4 \\implies \\boxed{0.4}.$", "As in the previous solution, we obtain the equation $4 \\cdot \\frac{1}{4} \\cdot \\pi d^2 = \\frac{1}{2}$ , which simplifies to $\\pi d^2 = \\frac{1}{2} = 0.5$ . Since $\\pi$ is slightly more than $3$ $d^2$ is slightly less than $\\frac{0.5}{3} = 0.1\\bar{6}$ . We notice that $0.1\\bar{6}$ is slightly more than $0.4^2 = 0.16$ , so $d$ is roughly $\\boxed{0.4}.$ emerald_block", "As above, we find that we need to estimate $d = \\frac{1}{\\sqrt{2\\pi}}$\nNote that we can approximate $2\\pi \\approx 6.28318 \\approx 6.25$ and so $\\frac{1}{\\sqrt{2\\pi}}$ $\\approx \\frac{1}{\\sqrt{6.25}}=\\frac{1}{2.5}=0.4$\nAnd so our answer is $\\boxed{0.4}$", "We only need to figure out the probability for a unit square, as it will scale up to the $2020\\times 2020$ square. Since we want to find the probability that a point inside a unit square that is $d$ units away from a lattice point (a corner of the square) is $\\frac{1}{2}$ , we can find which answer will come the closest to covering $\\frac{1}{2}$ of the area.\nSince the closest is $0.4$ which turns out to be $(0.4)^2\\times \\pi = 0.16 \\times \\pi$ which is about $0.502$ , we find that the answer rounded to the nearest tenth is $0.4$ or $\\boxed{0.4}$", "As per the above diagram, realize that $\\pi d^2 = \\frac{1}{2}$ , so $d = \\frac{1}{(\\sqrt{2})(\\sqrt{\\pi})}$\n$\\sqrt{2} \\approx 1.4 = \\frac{7}{5}$\n$\\sqrt{\\pi}$ is between $1.7$ and $1.8$ $((1.7)^2 = 2.89$ and $(1.8)^2 = 3.24)$ , so we can say $\\sqrt{\\pi} \\approx 1.75 = \\frac{7}{4}$\nSo $d \\approx \\frac{1}{(\\frac{7}{5})(\\frac{7}{4})} = \\frac{1}{\\frac{49}{20}} = \\frac{20}{49}$ . This is slightly above $\\boxed{0.4}$", "As above, we have the equation $\\pi d^2 = \\frac{1}{2}$ , and we want to find the most accurate value of $d$ . We resort to the answer choices and can plug those values of $d$ in and see which value of $d$ will lead to the most accurate value of $\\pi$\nStarting off in the middle, we try option C with $d=0.5$ . Plugging this in, we get $\\pi \\left(\\frac{1}{2}\\right)^2 = \\frac{1}{2},$ and after simplifying we get $\\pi = \\frac{1}{2} \\cdot 4 = 2.$ That's not very good. We know $\\pi \\approx 3.14.$\nLet's see if we can do better. Trying option A with $d = 0.3,$ we get $\\pi = \\frac{1}{2} \\cdot \\frac{100}{9} = \\frac{50}{9} = 5 \\frac{5}{9}.$\nHm, let's try option B with $d = 0.4.$ We get $\\pi = \\frac{1}{2} \\cdot \\frac{25}{4} = \\frac{25}{8} = 3 \\frac{1}{8}$ . This is very close to $\\pi$ and is the best estimate for $\\pi$ of the 5 options.\nTherefore, the answer is $\\boxed{0.4}.$ ~ epiconan" ]

[ "\\[(x-y)(x+y)=2000^2=2^8 \\cdot 5^6\\]\nNote that $(x-y)$ and $(x+y)$ have the same parities , so both must be even. We first give a factor of $2$ to both $(x-y)$ and $(x+y)$ . We have $2^6 \\cdot 5^6$ left. Since there are $7 \\cdot 7=49$ factors of $2^6 \\cdot 5^6$ , and since both $x$ and $y$ can be negative, this gives us $49\\cdot2=\\boxed{098}$ lattice points." ]

[ "Letting Y stand for a voter who approved of the work, and N stand for a person who didn't approve of the work, the pollster could select responses in $3$ different ways: $\\text{YNN, NYN, and NNY}$ . The probability of each of these is $(0.7)(0.3)^2=0.063$ . Thus, the answer is $3\\cdot0.063=\\boxed{0.189}$", "In more concise terms, this problem is an extension of the binomial distribution. We find the number of ways only 1 person approves of the mayor multiplied by the probability 1 person approves and 2 people disapprove: ${3\\choose 1} \\cdot(0.7)^1\\cdot(1-0.7)^{(3-1)}=3\\cdot0.7\\cdot0.09=\\boxed{0.189}$" ]

[ "From the given information, $121a+11b+c=512+64b+8c+a \\implies 120a=512+53b+7c$ . Since $a$ $b$ , and $c$ have to be positive, $a \\geq 5$ . Since we need to minimize the value of $n$ , we want to minimize $a$ , so we have $a = 5$ . Then we know $88=53b+7c$ , and we can see the only solution is $b=1$ $c=5$ . Finally, $515_{11} = 621_{10}$ , so our answer is $\\boxed{621}$" ]

[ "Solution by e_power_pi_times_i\nNotice that all four-digit palindromes are divisible by $11$ , so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between $1100$ and $2000$ , which also means that the last digit of the three-digit number is $1$ . Checking through the three-digit numbers $101, 111, 121,\\dots, 181$ , we find out that there are $\\boxed{4}$ three-digit prime numbers, which when multiplied by $11$ , result in palindromes." ]

[ "We may let $n = 7^k \\cdot m$ , where $m$ is not divisible by 7. Using the fact that the number of divisors function $d(n)$ is multiplicative, we have $d(n) = d(7^k)d(m) = (k+1)d(m) = 60$ . Also, $d(7n) = d(7^{k+1})d(m) = (k+2)d(m) = 80$ . These numbers are in the ratio 3:4, so $\\frac{k+1}{k+2} = \\frac{3}{4} \\implies k = 2 \\Rightarrow \\boxed{2}$" ]

[ "A positive integer with only four positive divisors has its prime factorization in the form of $a \\cdot b$ , where $a$ and $b$ are both prime positive integers or $c^3$ where $c$ is a prime. One can easily deduce that none of the numbers are even near a cube so the second case is not possible. We now look at the case of $a \\cdot b$ . The four factors of this number would be $1$ $a$ $b$ , and $ab$ . The sum of these would be $ab+a+b+1$ , which can be factored into the form $(a+1)(b+1)$ . Easily we can see that now we can take cases again.\nCase 1: Either $a$ or $b$ is 2.\nIf this is true then we have to have that one of $(a+1)$ or $(b+1)$ is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either $\\frac{2016}{3} - 1$ or $\\frac{2010}{3} - 1$ is a prime. We see that in this case none of them work.\nCase 2: Both $a$ and $b$ are odd primes.\nThis implies that both $(a+1)$ and $(b+1)$ are even which implies that in this case the number must be divisible by $4$ . This leaves only $2012$ and $2016$ $2012={4}\\cdot{503}$ so either $(a+1)$ or $(b+1)$ both have a factor of $2$ or one has a factor of $4$ . If it was the first case, then $a$ or $b$ will equal $1$ . That means that either $(a+1)$ or $(b+1)$ has a factor of $4$ . That means that $a$ or $b$ is $502$ which isn't a prime, so 2012 does not work. $2016 = 4 \\cdot 504$ so we have $(503 + 1)(3 + 1)$ . 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is $\\boxed{1}$", "After deducing that $2012$ and case $1$ is impossible, and since there is no option for $0$ $2016$ is obviously a solution and the answer is $\\boxed{1}$" ]

[ "Let's say that we will have $3$ slips for every number not exceeding $100$ but bigger than $50.$ This is to account for the $3p$ probability part. Let's now say that we will only have one slip for each number below or equal to $50.$ The probability(or $p$ ) will then be $\\frac{1}{200}.$ Now let's have all the squares under $50,$ which are $1,4,9,16,25,36,49.$ The probability for these are $\\frac{7}{200}.$ The numbers above $50$ that are squares are $64,81,100.$ We then need to multiply the probability by $3$ so the probability of these are $\\frac{9}{200}.$ The answer is $\\frac{7}{200}+\\frac{9}{200}=0.08\\implies\\boxed{.08}$" ]

[ "The prime factorization of $12!$ is $2^{10} \\cdot 3^5 \\cdot 5^2 \\cdot 7 \\cdot 11$ . \nThis yields a total of $11 \\cdot 6 \\cdot 3 \\cdot 2 \\cdot 2$ divisors of $12!.$ In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that the divisor can't have any factors of $7$ and $11$ in the prime factorization because there is only one of each in $12!.$ Thus, there are $6 \\cdot 3 \\cdot 2$ perfect squares. (For $2$ , you can have $0$ $2$ $4$ $6$ $8$ , or $10$ $2$ s, etc.)\nThe probability that the divisor chosen is a perfect square is \\[\\frac{6\\cdot 3\\cdot 2}{11\\cdot 6\\cdot 3\\cdot 2\\cdot 2}=\\frac{1}{22} \\implies \\frac{m}{n}=\\frac{1}{22} \\implies m\\ +\\ n = 1\\ +\\ 22 = \\boxed{23}\\]" ]

[ "Since $x\\%$ means $0.01x$ , the statement \" $x\\% \\text{ of } x \\text{ is 4}$ \" can be rewritten as \" $0.01x \\cdot x = 4$ \":\n$0.01x \\cdot x=4 \\Rightarrow x^2 = 400 \\Rightarrow x = \\boxed{20}.$", "Try the answer choices one by one. Upon examination, it is quite obvious that the answer is $\\boxed{20}.$ Very fast." ]

[ "Since $x\\%$ means $0.01x$ , the statement \" $x\\% \\text{ of } x \\text{ is 4}$ \" can be rewritten as \" $0.01x \\cdot x = 4$ \":\n$0.01x \\cdot x=4 \\Rightarrow x^2 = 400 \\Rightarrow x = \\boxed{20}.$", "Try the answer choices one by one. Upon examination, it is quite obvious that the answer is $\\boxed{20}.$ Very fast." ]

[ "Let $p(t)=24t$ be the number of feet the powderman is from the blast at $t$ seconds after the fuse is lit, and let $q(t)=1080t-32400$ be the number of feet the sound has traveled. We want to solve for $p(t)=q(t)$ \\[24t=1080t-32400\\] \\[1056t=32400\\] \\[t=\\frac{32400}{1056}\\] \\[t=\\frac{675}{22}=30.6\\overline{81}\\] The number of yards the powderman is from the blast at time $t$ is $\\frac{24t}3=8t$ , so the answer is $8(30.6\\overline{81})$ , which is about $245$ yards. $\\boxed{245}$" ]

[ "WLOG let the speed of the river be 0. This is allowed because the problem never states that the speed of the current has to have a magnitude greater than 0. In this case, when the powerboat travels from $A$ to $B$ , the raft remains at $A$ . Thus the trip from $A$ to $B$ takes the same time as the trip from $B$ to the raft. Since these times are equal and sum to $9$ hours, the trip from $A$ to $B$ must take half this time, or $4.5$ hours. The answer is thus $\\boxed{4.5}$", "What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Think of the blue arrow as the power boat and the red arrow as the raft in the following three diagrams, which represent different time intervals of the problem.\n\nThinking about the distance covered as their distances with respect to each other, they are $0$ distance apart in the first diagram when they haven't started to move yet, some distance $d$ apart in the second diagram when the power boat reaches $B$ , and again $0$ distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of $d$ on the way there, and again cover a distance of $d$ on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.\nLet $b$ denote the speed of the power boat (only the power boat, not factoring in current) and $r$ denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from $A$ to $B$ , the boat travels at a velocity of $b+r$ , and on the way back, travels at a velocity of $-(b-r)=r-b$ , since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from $A$ to $B$ becomes $(r+b)-r=b$ , and on the way back it becomes $(r-b)-r=-b$ . Since the boat's velocities with respect to the raft are exact opposites, $b$ and $-b$ , we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.\nFrom this, we have that the boat travels a distance $d$ at rate $b$ with respect to the raft both on the way to $B$ and on the way back. Thus, using $\\dfrac{distance}{speed}=time$ , we have $\\dfrac{2d}{b}=9\\text{ hours}$ , and to see how long it took to travel half the distance, we have $\\dfrac{d}{b}=4.5\\text{ hours}\\implies\\boxed{4.5}$", "Let $t$ be the time it takes the power boat to go from $A$ to $B$ in hours, $r$ be the speed of the river current (and thus also the raft), and $p$ to be the speed of the power boat with respect to the river.\nUsing $d = rt$ , the raft covers a distance of $9r$ , the distance from $A$ to $B$ is $(p + r)t$ , and the distance from $B$ to where the raft and power boat met up is $(9 - t)(p - r)$\nThen, $9r + (9 - t)(p - r) = (p + r)t$ . Solving for $t$ , we get $t = 4.5$ , which is $\\boxed{4.5}$" ]

[ "Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$ . The side length of the base is $14$ . The heights of $\\triangle ABE$ and $\\triangle CDE$ are $2\\cdot105\\div14=15$ and $2\\cdot91\\div14=13$ , respectively. Consider a side view of the pyramid from $\\triangle BCE$ . We have a systems of equations through the Pythagorean Theorem:\n$13^2-(14-a)^2=h^2 \\\\ 15^2-a^2=h^2$\nSetting them equal to each other and simplifying gives $-27+28a=225 \\implies a=9$\nTherefore, $h=\\sqrt{15^2-9^2}=12$ , and the volume of the pyramid is $\\frac{bh}{3}=\\frac{12\\cdot 196}{3}=\\boxed{784}$" ]

[ "Let the triangular base be $\\triangle ABC$ , with $\\overline {AB} = 24$ . We find that the altitude to side $\\overline {AB}$ is $16$ , so the area of $\\triangle ABC$ is $(24*16)/2 = 192$\nLet the fourth vertex of the tetrahedron be $P$ , and let the midpoint of $\\overline {AB}$ be $M$ . Since $P$ is equidistant from $A$ $B$ , and $C$ , the line through $P$ perpendicular to the plane of $\\triangle ABC$ will pass through the circumcenter of $\\triangle ABC$ , which we will call $O$ . Note that $O$ is equidistant from each of $A$ $B$ , and $C$ . Then,\n\\[\\overline {OM} + \\overline {OC} = \\overline {CM} = 16\\]\nLet $\\overline {OM} = d$ . Then $OC=OA=\\sqrt{d^2+12^2}.$ Equation $(1)$ \\[d + \\sqrt {d^2 + 144} = 16\\]\nSquaring both sides, we have\n\\[d^2 + 144 + 2d\\sqrt {d^2+144} + d^2 = 256\\]\n\\[2d^2 + 2d\\sqrt {d^2+144} = 112\\]\n\\[2d(d + \\sqrt {d^2+144}) = 112\\]\nSubstituting with equation $(1)$\n\\[2d(16) = 112\\]\n\\[d = 7/2\\]\nWe now find that $\\sqrt{d^2 + 144} = 25/2$\nLet the distance $\\overline {OP} = h$ . Using the Pythagorean Theorem on triangle $AOP$ $BOP$ , or $COP$ (all three are congruent by SSS):\n\\[25^2 = h^2 + (25/2)^2\\]\n\\[625 = h^2 + 625/4\\]\n\\[1875/4 = h^2\\]\n\\[25\\sqrt {3} / 2 = h\\]\nFinally, by the formula for volume of a pyramid,\n\\[V = Bh/3\\]\n\\[V = (192)(25\\sqrt{3}/2)/3\\] This simplifies to $V = 800\\sqrt {3}$ , so $m+n = \\boxed{803}$", "We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length $24$ is at the origin, or $(0, 0, 0)$ . Then, the two other vertices can be $(-12, -16, 0)$ and $(12, -16, 0)$ . Let the fourth vertex have coordinates of $(x, y, z)$ . We have the following $3$ equations from the distance formula.\n\\[x^2+y^2+z^2=625\\]\n\\[(x+12)^2+(y+16)^2+z^2=625\\]\n\\[(x-12)^2+(y+16)^2+z^2=625\\]\nAdding the last two equations and substituting in the first equation, we get that $y=-\\frac{25}{2}$ . If you drew a good diagram, it should be obvious that $x=0$ . Now, solving for $z$ , we get that $z=\\frac{25\\sqrt{3}}{2}$ . So, the height of the pyramid is $\\frac{25\\sqrt{3}}{2}$ . The base is equal to the area of the triangle, which is $\\frac{1}{2} \\cdot 24 \\cdot 16 = 192$ . The volume is $\\frac{1}{3} \\cdot 192 \\cdot \\frac{25\\sqrt{3}}{2} = 800\\sqrt{3}$ . Thus, the answer is $800+3 = \\boxed{803}$", "Label the four vertices of the tetrahedron and the midpoint of $\\overline {AB}$ , and notice that the area of the base of the tetrahedron, $\\triangle ABC$ , equals $192$ , according to Solution 1.\nNotice that the altitude of $\\triangle CPM$ from $\\overline {CM}$ to point $P$ is the height of the tetrahedron. Side $\\overline {PM}$ is can be found using the Pythagorean Theorem on $\\triangle APM$ , giving us $\\overline {PM}=\\sqrt{481}.$\nUsing Heron's Formula, the area of $\\triangle CPM$ can be written as \\[\\sqrt{\\frac{41+\\sqrt{481}}{2}(\\frac{41+\\sqrt{481}}{2}-16)(\\frac{41+\\sqrt{481}}{2}-25)(\\frac{41+\\sqrt{481}}{2}-\\sqrt{481})}\\] \\[=\\frac{\\sqrt{(41+\\sqrt{481})(9+\\sqrt{481})(-9+\\sqrt{481})(41-\\sqrt{481})}}{4}\\]\nNotice that both $(41+\\sqrt{481})(41-\\sqrt{481})$ and $(9+\\sqrt{481})(-9+\\sqrt{481})$ can be rewritten as differences of squares; thus, the expression can be written as \\[\\frac{\\sqrt{(41^2-481)(481-9^2)}}{4}=\\frac{\\sqrt{480000}}{4}=100\\sqrt{3}.\\]\nFrom this, we can determine the height of both $\\triangle CPM$ and tetrahedron $ABCP$ to be $\\frac{100\\sqrt{3}}{8}$ ; therefore, the volume of the tetrahedron equals $\\frac{100\\sqrt{3}}{8} \\cdot 192=800\\sqrt{3}$ ; thus, $m+n=800+3=\\boxed{803}.$", "Notation is shown on diagram. \\[AM = MB = c = 12, AC = BC = b = 20,\\] \\[DA = DB = DC = a = 25.\\] \\[CM = x + y = \\sqrt{b^2-c^2} = 16,\\] \\[x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,\\] \\[x – y = \\frac{x^2 – y^2}{x+y} = \\frac {c^2} {16} = 9,\\] \\[x = \\frac {16 + 9}{2} = \\frac {a}{2},\\] \\[h = \\sqrt{a^2 -\\frac{ a^2}{4}} = a \\frac {\\sqrt{3}}{2},\\] \\[V = \\frac{h\\cdot CM \\cdot c}{3}= \\frac{16\\cdot 25 \\sqrt{3} \\cdot 12}{3} = 800 \\sqrt{3} \\implies \\boxed{803}.\\] vladimir.shelomovskii@gmail.com, vvsss" ]

[ "Let's use the triangle inequality. We know that for a triangle, the sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line.\nSimilarly, for a convex quadrilateral, the sum of the shortest 3 sides must always be longer than the longest side. Thus, the answer is $\\frac{26}{2}-1=13-1=\\boxed{12}$", "Say the chosen side is $a$ and the other sides are $b,c,d$\nBy the Generalised Polygon Inequality, $a<b+c+d$ . We also have $a+b+c+d=26\\Rightarrow b+c+d=26-a$\nCombining these two, we get $a<26-a\\Rightarrow a<13$\nThe largest length that satisfies this is $a=\\boxed{12}$", "The quadrilateral can by cyclic only when it is an isosceles triangle. Without Loss of Generality, lets assume that this quadrilateral is a trapezoid. We can assume this as if we inscribe a trapezoid in a triangle, the base can be the diameter of the circle which is the longest chord in the circle, therefore maximizing the side length. By Brahmagupta's Formula, the area of the quadrilateral is defined by $\\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semi-perimeter. If the perimeter of the quadrilateral is $26$ , then the semi-perimeter will be $13$ . The area of the quadrilateral must be positive so the difference between the semi-perimeter and a side length must be greater than $0$ as otherwise, the area will be $0$ or negative. Therefore, the longest a side can be in this quadrilateral is $\\boxed{12}$", "This is an AMC 10 problem 4, so there is no need for any complex formulas. The largest singular side length from a quadrilateral comes from a trapezoid. So we can set the $2$ sides of the trapezoid equal to $4$ . Next we can split the trapezoid into $5$ triangles, where each base length of the triangle equals $4$ . So the top side equals $8$ , and the bottom side length equals $4+4+4$ $=$ $\\boxed{12}$ ~ kabbybear" ]

[ "Note that the slope of $PQ$ is $\\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\\frac{b+a}{a+b}=1$ . Hence, $PQ\\perp PS$ and we can similarly prove that the other angles are right angles. This means that $PQRS$ is a rectangle. By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$ . Simplifying we get $(a-b)(a+b) = 8$ . Thus $a+b$ and $a-b$ have to be a factor of 8. The only way for them to be factors of $8$ and remain integers is if $a+b = 4$ and $a-b = 2$ . So the answer is $\\boxed{4}$", "Solution by e_power_pi_times_i\nBy the Shoelace Theorem, the area of the quadrilateral is $2a^2 - 2b^2$ , so $a^2 - b^2 = 8$ . Since $a$ and $b$ are integers, $a = 3$ and $b = 1$ , so $a + b = \\boxed{4}$" ]

[ "\nLet $AD$ intersect $OB$ at $E$ and $OC$ at $F.$\n$\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$\n$\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$\nFrom there, $\\triangle{OAB} \\sim \\triangle{ABE}$ , thus:\n$\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$\n$OA = OB$ because they are both radii of $\\odot{O}$ . Since $\\frac{OA}{AB} = \\frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$\n$OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \\boxed{500}$", "To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.\n\nConstruct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Let the intersection of $BD$ and $OC$ be point $E$ . Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.\nWe set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem, \\[\\sqrt{1^2-x^2}+\\sqrt{(\\sqrt{2})^2-x^2}=\\sqrt{2}\\]\nWe solve for $x$ \\[1-x^2+2-x^2+2\\sqrt{(1-x^2)(2-x^2)}=2\\] \\[2\\sqrt{(1-x^2)(2-x^2)}=2x^2-1\\] \\[4(1-x^2)(2-x^2)=(2x^2-1)^2\\] \\[8-12x^2+4x^4=4x^4-4x^2+1\\] \\[8x^2=7\\] \\[x=\\frac{\\sqrt{14}}{4}\\]\nBy Ptolemy's Theorem \\[AB \\cdot CD + BC \\cdot AD = AC \\cdot BD = BD^2 = (2 \\cdot BE)^2\\]\nSubstituting values, \\[1^2+1 \\cdot AD = 4{\\left( \\frac{\\sqrt{14}}{4} \\right)}^2\\] \\[1+AD=\\frac{7}{2}\\] \\[AD=\\frac{5}{2}\\]\nFinally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\\boxed{500}$", "\nLet $s = 200$ . Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\\triangle OBC$ to $\\overline{OB}$ . Since $\\triangle OBC$ is isosceles we can compute its area to be $\\frac{s^2 \\sqrt{7}}{4}$ , hence $CA = 2 \\cdot \\tfrac{2 \\cdot s^2\\sqrt7/4}{s\\sqrt2} = s\\sqrt{\\frac{7}{2}}$\nNow by Ptolemy's Theorem we have $CA^2 = s^2 + AD \\cdot s \\implies AD = \\left(\\frac{7}{2}-1\\right)s.$ This gives us: \\[\\boxed{500.}\\]", "Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\\sqrt{7},200\\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\\frac{100}{200\\sqrt{2}}=\\frac{\\sqrt{2}}{4}$ . Similarly, the cosine is $\\frac{100\\sqrt{7}}{200\\sqrt{2}}=\\frac{\\sqrt{14}}{4}$ .\nSince there are three sides, and since $\\sin\\theta=\\sin\\left(180-\\theta\\right)$ ,we seek to find $2r\\sin 3\\theta$ .\nFirst, $\\sin 2\\theta=2\\sin\\theta\\cos\\theta=2\\cdot\\left(\\frac{\\sqrt{2}}{4}\\right)\\left(\\frac{\\sqrt{14}}{4}\\right)=\\frac{2\\sqrt{2}\\sqrt{14}}{16}=\\frac{\\sqrt{7}}{4}$ and $\\cos 2\\theta=\\frac{3}{4}$ by Pythagorean. \\[\\sin 3\\theta=\\sin(2\\theta+\\theta)=\\sin 2\\theta\\cos\\theta+\\sin \\theta\\cos 2\\theta=\\frac{\\sqrt{7}}{4}\\left(\\frac{\\sqrt{14}}{4}\\right)+\\frac{\\sqrt{2}}{4}\\left(\\frac{3}{4}\\right)=\\frac{7\\sqrt{2}+3\\sqrt{2}}{16}=\\frac{5\\sqrt{2}}{8}\\] \\[2r\\sin 3\\theta=2\\left(200\\sqrt{2}\\right)\\left(\\frac{5\\sqrt{2}}{8}\\right)=400\\sqrt{2}\\left(\\frac{5\\sqrt{2}}{8}\\right)=\\frac{800\\cdot 5}{8}=\\boxed{500}\\]", "For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\\sqrt{OB^2-BM^2}=\\sqrt7$ and $ON=\\sqrt{OA^2-AN^2}=\\sqrt{8-x^2}$ . Thus the height of the trapezoid is $\\sqrt7-\\sqrt{8-x^2}$ , so the area is $\\frac{(2+2x)(\\sqrt7-\\sqrt{8-x^2})}{2}=(x+1)(\\sqrt7-\\sqrt{8-x^2})$ . By Brahmagupta's formula , the area is $\\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Setting these two equal, we get $(x+1)(\\sqrt7-\\sqrt{8-x^2})=\\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Dividing both sides by $x+1$ and then squaring, we get $7-2(\\sqrt7)(\\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$ . Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\\sqrt7)(\\sqrt{8-x^2})=2x+3$ . Rearranging and dividing by two, we get $(\\sqrt7)(\\sqrt{8-x^2})=6-x$ . Squaring both sides, we get $56-7x^2=x^2-12x+36$ . Rearranging, we get $8x^2-12x-20=0$ . Dividing by 4 we get $2x^2-3x-5=0$ . Factoring we get, $(2x-5)(x+1)=0$ , and since $x$ cannot be negative, we get $x=2.5$ . Since $DA=2x$ $DA=5$ . Scaling up by 100, we get $\\boxed{500}$", " Label the points as shown, and let $\\angle{EOF} = \\theta$ . Since $\\overline{OB} = \\overline{OC}$ , and $\\triangle{OFE} \\sim \\triangle{OCB}$ , we get that $\\angle{EFO} = 90-\\frac{\\theta}{2}$ . We assign $\\alpha$ to $90-\\frac{\\theta}{2}$ for simplicity. \nFrom here, by vertical angles $\\angle{CFD} = \\alpha$ . Also, since $\\triangle{OCB} \\cong \\triangle{ODC}$ $\\angle{OCD} = \\alpha$ . This means that $\\angle{CDF} = 180-2\\alpha = \\theta$ , which leads to $\\triangle{OCB} \\sim \\triangle{DCF}$ . \nSince we know that $\\overline{CD} = 200$ $\\overline{DF} = 200$ , and by similar reasoning $\\overline{AE} = 200$ . \nFinally, again using similar triangles, we get that $\\overline{CF} = 100\\sqrt{2}$ , which means that $\\overline{OF} = \\overline{OC} - \\overline{CF} = 200\\sqrt{2} - 100\\sqrt{2} = 100\\sqrt{2}$ . We can again apply similar triangles (or use Power of a Point) to get $\\overline{EF} = 100$ , and finally $\\overline{AD} = \\overline{AE}+\\overline{EF}+\\overline{FD} = 200+100+200=\\boxed{500}$ - ColtsFan10", "We first scale down by a factor of $200\\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresponds to $z^3$ and $B$ corresponds to $z^2$ . We are given that $\\lvert z \\rvert = 1$ and $\\lvert z-1 \\rvert = 1/\\sqrt{2}$ , and we wish to find $\\lvert z^3 - 1 \\rvert=\\lvert z^2+z+1\\rvert \\cdot \\lvert z-1 \\rvert=\\lvert (z^2+z+1)/\\sqrt{2} \\rvert$ . Let $z=a+bi$ , where $a$ and $b$ are real numbers. Then, $a^2+b^2=1$ and $a^2-2a+1+b^2=1/2$ ; solving for $a$ and $b$ yields $a=3/4$ and $b=\\sqrt{7}/4$ . Thus, $AD = \\lvert z^3 - 1 \\rvert = \\lvert (z^2+z+1)/\\sqrt{2} \\rvert = \\lvert (15/8 + 5\\sqrt{7}/8 \\cdot i)/\\sqrt{2} \\rvert = \\frac{5\\sqrt{2}}{4}$ . Scaling back up gives us a final answer of $\\frac{5\\sqrt{2}}{4} \\cdot 200\\sqrt{2} = \\boxed{500}$", "Let angle $C$ be $2a$ . This way $BD$ will be $400sin(a)$ . Now we can trig bash. As the circumradius of triangle $BCD$ is $200\\sqrt{2}$ , we can use the formula \\[R=\\frac{abc}{4A}\\] and \\[A=\\frac{absin(C)}{2}\\] and plug in all the values we got to get \\[200\\sqrt{2}=\\frac{200^2 \\cdot 400sin(a)}{4 \\cdot (\\frac{200^2 sin(2a)}{2})}\\] . This boils down to \\[\\sqrt{2}=\\frac{sin(a)}{sin{2a}}\\] . This expression can further be simplified by the trig identity \\[sin(2a)=2sin(a)cos(a)\\] . This leads to the final simplified form \\[2\\sqrt{2}=\\frac{1}{cos(a)}\\] . Solving this expression gives us \\[cos(a)=\\frac{\\sqrt{2}}{4}\\] . However, as we want $sin(a)$ , we use the identity $sin^2+cos^2=1$ , and substitute to get that $sin(a)=\\frac{\\sqrt{14}}{4}$ , and therefore BD is $100\\sqrt{14}$\nThen, as $ABCD$ is a cyclic quadrilateral, we can use Ptolemy’s Theorem (with $AD=x$ ) to get \\[14 \\cdot 100^2=200x+200^2\\] . Finally, we solve to get $\\boxed{500}$" ]

[ "\nLet quadrilateral $ABCD$ be inscribed in circle $O$ , where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$\nBy the Pythagorean Theorem, the length of $OH$ is \\begin{align*} \\sqrt{CO^2 - HC^2} &= \\sqrt{(200\\sqrt{2})^2 - \\left(\\frac{200}{2}\\right)^2} \\\\ &= \\sqrt{80000 - 10000} \\\\ &= \\sqrt{70000} \\\\ &= 100\\sqrt{7}. \\end{align*}\nNote that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$ ; then we have that\n$[AOB] + [BOC] + [COD] = \\frac{200 \\times 100\\sqrt{7}}{2} + \\frac{200 \\times 100\\sqrt{7}}{2} + \\frac{200 \\times 100\\sqrt{7}}{2} = \\frac{x \\times h}{2} + \\frac{(100\\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$\nFurthermore, \\begin{align*} h &= \\sqrt{OD^2 - GD^2} \\\\ &= \\sqrt{(200\\sqrt{2})^2 - \\left(\\frac{x}{2}\\right)^2} \\\\ &= \\sqrt{80000 - \\frac{x^2}{4}} \\end{align*}\nSubstituting this value of $h$ into the previous equation and evaluating for $x$ , we get: \\[\\frac{200 \\times 100\\sqrt{7}}{2} + \\frac{200 \\times 100\\sqrt{7}}{2} + \\frac{200 \\times 100\\sqrt{7}}{2} = \\frac{x \\times h}{2} + \\frac{(100\\sqrt{7} - h)(200 + x)}{2}\\] \\[\\frac{3 \\times 200 \\times 100\\sqrt{7}}{2} = \\frac{x\\sqrt{80000 - \\frac{x^2}{4}}}{2} + \\frac{\\left(100\\sqrt{7} - \\sqrt{80000 - \\frac{x^2}{4}}\\right)(200 + x)}{2}\\] \\[60000\\sqrt{7} = \\left(x\\sqrt{80000 - \\frac{x^2}{4}}\\right) + \\left(20000\\sqrt{7}\\right) + \\left(100x\\sqrt{7}\\right) - \\left(200\\sqrt{80000 - \\frac{x^2}{4}}\\right) - \\left(x\\sqrt{80000 - \\frac{x^2}{4}}\\right)\\] \\[40000\\sqrt{7} = 100x\\sqrt{7} - 200\\sqrt{80000 - \\frac{x^2}{4}}\\] \\[400\\sqrt{7} = x\\sqrt{7} - 2\\sqrt{80000 - \\frac{x^2}{4}}\\] \\[(x - 400)\\sqrt{7} = 2\\sqrt{80000 - \\frac{x^2}{4}}\\] \\[7(x-400)^2 = 4\\left(80000 - \\frac{x^2}{4}\\right)\\] \\[7x^2 - 5600x + 1120000 = 320000 - x^2\\] \\[8x^2 - 5600x + 800000 = 0\\] \\[x^2 - 700x + 100000 = 0\\]\nThe roots of this quadratic are found by using the quadratic formula: \\begin{align*} x &= \\frac{-(-700) \\pm \\sqrt{(-700)^2 - 4 \\times 1 \\times 100000}}{2 \\times 1} \\\\ &= \\frac{700 \\pm \\sqrt{490000 - 400000}}{2} \\\\ &= \\frac{700}{2} \\pm \\frac{\\sqrt{90000}}{2} \\\\ &= 350 \\pm \\frac{300}{2} \\\\ &= 200, 500 \\end{align*}\nIf the length of $AD$ is $200$ , then $ABCD$ would be a square. Thus, the radius of the circle would be \\[\\frac{\\sqrt{200^2 + 200^2}}{2} = \\frac{\\sqrt{80000}}{2} = \\frac{200\\sqrt{2}}{2} = 100\\sqrt{2}\\] Which is a contradiction. Therefore, our answer is $\\boxed{500}.$", "\nConstruct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Apply the law of cosines on $\\Delta BOC$ ; let $\\theta = \\angle BOC$ . We get the following equation: \\[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\\cdot OB \\cdot OC\\cdot \\cos\\theta\\] Substituting the values in, we get \\[(200)^{2}=2\\cdot (200)^{2}+ 2\\cdot (200)^{2}- 2\\cdot 2\\cdot (200)^{2}\\cdot \\cos\\theta\\] Canceling out, we get \\[\\cos\\theta=\\frac{3}{4}\\] Because $\\angle AOB$ $\\angle BOC$ , and $\\angle COD$ are congruent, $\\angle AOD = 3\\theta$ . To find the remaining side ( $AD$ ), we simply have to apply the law of cosines to $\\Delta AOD$ . Now, to find $\\cos 3\\theta$ , we can derive a formula that only uses $\\cos\\theta$ \\[\\cos 3\\theta=\\cos (2\\theta+\\theta)= \\cos 2\\theta \\cos\\theta- (2\\sin\\theta \\cos\\theta) \\cdot \\sin \\theta\\] \\[\\cos 3\\theta= \\cos\\theta (\\cos 2\\theta-2\\sin^{2}\\theta)=\\cos\\theta (2\\cos^{2}\\theta-3+2\\cos^{2}\\theta)\\] \\[\\Rightarrow \\cos 3\\theta=4\\cos^{3}\\theta-3\\cos\\theta\\] It is useful to memorize the triple angle formulas ( $\\cos 3\\theta=4\\cos^{3}\\theta-3\\cos\\theta, \\sin 3\\theta=3\\sin\\theta-4\\sin^{3}\\theta$ ). Plugging in $\\cos\\theta=\\frac{3}{4}$ , we get $\\cos 3\\theta= -\\frac{9}{16}$ . Now, applying law of cosines on triangle $OAD$ , we get \\[(AD)^{2}= 2\\cdot (200)^{2}+ 2\\cdot (200)^{2}+2\\cdot 200\\sqrt2 \\cdot 200\\sqrt2 \\cdot \\frac{9}{16}\\] \\[\\Rightarrow 2\\cdot (200)^{2} \\cdot (1+1+ \\frac{9}{8})=(200)^{2}\\cdot \\frac{25}{4}\\] \\[AD=200 \\cdot \\frac{5}{2}=\\boxed{500}\\]", "\nConstruct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\\overline{BC}$ and let these points be $E$ and $F$ , respectively. Also, let $\\theta = \\angle BOC$ . From the Law of Cosines on $\\triangle BOC$ , we have $\\cos \\theta = \\frac{3}{4}$\nNow, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$ , we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$ . In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$ , so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$ , so $\\angle DCF = \\theta$\nThus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$ , so $FC = 150$ . Similarly, $BE = 150$ , and $AD = 150 + 200 + 150 = \\boxed{500}$", " Claim: $[ABCD]$ is an isosceles trapezoid.\nProof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\\angle BAD = 180 - \\angle BCD = 180-(\\angle BCO + \\angle DCO)=180-(\\angle CBO+\\angle ABO) = 180 - \\angle ABC = \\angle CDA$ . Hence, $[ABCD]$ is an isosceles trapezoid.\nLet $\\angle CDA=\\alpha$ . Notice that the length of the altitude from $C$ to $AD$ is $200sin(\\alpha)$ . Furthermore, the length of the altitude from $O$ to $BC$ is $100\\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\\sqrt{7}-200sin\\alpha$ . Let $F$ the feet of the altitude from $O$ to $AD$ . Then, $FD=(200+400cos(\\alpha))/2=100+200cos(\\alpha)$ , because $AOD$ is isosceles.\nTherefore, by the Pythagorean theorem, $(100+200cos(\\alpha))^2+(100\\sqrt{7}-200sin(\\alpha))^2=80000$ . Simplifying, we have $1+cos(\\alpha)=sin(\\alpha) \\cdot sqrt{7} \\implies cos^2(\\alpha)+2cos(\\alpha)+1=sin^2(\\alpha) \\cdot 7 = 7-7cos^2(\\alpha) \\implies 8cos^2(\\alpha)+2cos(\\alpha) - 6 =0$ . Solving this quadratic, we have $cos(\\alpha)=\\frac{3}{4}, -1$ , but $0<\\alpha<180 \\implies cos(\\alpha)=3/4$ . Therefore, $AD=200cos(\\alpha)+200cos(\\alpha)+200=\\boxed{500}$" ]

[ "\nLet $AD$ intersect $OB$ at $E$ and $OC$ at $F.$\n$\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$\n$\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$\nFrom there, $\\triangle{OAB} \\sim \\triangle{ABE}$ , thus:\n$\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$\n$OA = OB$ because they are both radii of $\\odot{O}$ . Since $\\frac{OA}{AB} = \\frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$\n$OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \\boxed{500}$", "To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.\n\nConstruct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Let the intersection of $BD$ and $OC$ be point $E$ . Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.\nWe set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem, \\[\\sqrt{1^2-x^2}+\\sqrt{(\\sqrt{2})^2-x^2}=\\sqrt{2}\\]\nWe solve for $x$ \\[1-x^2+2-x^2+2\\sqrt{(1-x^2)(2-x^2)}=2\\] \\[2\\sqrt{(1-x^2)(2-x^2)}=2x^2-1\\] \\[4(1-x^2)(2-x^2)=(2x^2-1)^2\\] \\[8-12x^2+4x^4=4x^4-4x^2+1\\] \\[8x^2=7\\] \\[x=\\frac{\\sqrt{14}}{4}\\]\nBy Ptolemy's Theorem \\[AB \\cdot CD + BC \\cdot AD = AC \\cdot BD = BD^2 = (2 \\cdot BE)^2\\]\nSubstituting values, \\[1^2+1 \\cdot AD = 4{\\left( \\frac{\\sqrt{14}}{4} \\right)}^2\\] \\[1+AD=\\frac{7}{2}\\] \\[AD=\\frac{5}{2}\\]\nFinally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\\boxed{500}$", "\nLet $s = 200$ . Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\\triangle OBC$ to $\\overline{OB}$ . Since $\\triangle OBC$ is isosceles we can compute its area to be $\\frac{s^2 \\sqrt{7}}{4}$ , hence $CA = 2 \\cdot \\tfrac{2 \\cdot s^2\\sqrt7/4}{s\\sqrt2} = s\\sqrt{\\frac{7}{2}}$\nNow by Ptolemy's Theorem we have $CA^2 = s^2 + AD \\cdot s \\implies AD = \\left(\\frac{7}{2}-1\\right)s.$ This gives us: \\[\\boxed{500.}\\]", "Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\\sqrt{7},200\\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\\frac{100}{200\\sqrt{2}}=\\frac{\\sqrt{2}}{4}$ . Similarly, the cosine is $\\frac{100\\sqrt{7}}{200\\sqrt{2}}=\\frac{\\sqrt{14}}{4}$ .\nSince there are three sides, and since $\\sin\\theta=\\sin\\left(180-\\theta\\right)$ ,we seek to find $2r\\sin 3\\theta$ .\nFirst, $\\sin 2\\theta=2\\sin\\theta\\cos\\theta=2\\cdot\\left(\\frac{\\sqrt{2}}{4}\\right)\\left(\\frac{\\sqrt{14}}{4}\\right)=\\frac{2\\sqrt{2}\\sqrt{14}}{16}=\\frac{\\sqrt{7}}{4}$ and $\\cos 2\\theta=\\frac{3}{4}$ by Pythagorean. \\[\\sin 3\\theta=\\sin(2\\theta+\\theta)=\\sin 2\\theta\\cos\\theta+\\sin \\theta\\cos 2\\theta=\\frac{\\sqrt{7}}{4}\\left(\\frac{\\sqrt{14}}{4}\\right)+\\frac{\\sqrt{2}}{4}\\left(\\frac{3}{4}\\right)=\\frac{7\\sqrt{2}+3\\sqrt{2}}{16}=\\frac{5\\sqrt{2}}{8}\\] \\[2r\\sin 3\\theta=2\\left(200\\sqrt{2}\\right)\\left(\\frac{5\\sqrt{2}}{8}\\right)=400\\sqrt{2}\\left(\\frac{5\\sqrt{2}}{8}\\right)=\\frac{800\\cdot 5}{8}=\\boxed{500}\\]", "For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\\sqrt{OB^2-BM^2}=\\sqrt7$ and $ON=\\sqrt{OA^2-AN^2}=\\sqrt{8-x^2}$ . Thus the height of the trapezoid is $\\sqrt7-\\sqrt{8-x^2}$ , so the area is $\\frac{(2+2x)(\\sqrt7-\\sqrt{8-x^2})}{2}=(x+1)(\\sqrt7-\\sqrt{8-x^2})$ . By Brahmagupta's formula , the area is $\\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Setting these two equal, we get $(x+1)(\\sqrt7-\\sqrt{8-x^2})=\\sqrt{(x+1)(x+1)(x+1)(3-x)}$ . Dividing both sides by $x+1$ and then squaring, we get $7-2(\\sqrt7)(\\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$ . Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\\sqrt7)(\\sqrt{8-x^2})=2x+3$ . Rearranging and dividing by two, we get $(\\sqrt7)(\\sqrt{8-x^2})=6-x$ . Squaring both sides, we get $56-7x^2=x^2-12x+36$ . Rearranging, we get $8x^2-12x-20=0$ . Dividing by 4 we get $2x^2-3x-5=0$ . Factoring we get, $(2x-5)(x+1)=0$ , and since $x$ cannot be negative, we get $x=2.5$ . Since $DA=2x$ $DA=5$ . Scaling up by 100, we get $\\boxed{500}$", " Label the points as shown, and let $\\angle{EOF} = \\theta$ . Since $\\overline{OB} = \\overline{OC}$ , and $\\triangle{OFE} \\sim \\triangle{OCB}$ , we get that $\\angle{EFO} = 90-\\frac{\\theta}{2}$ . We assign $\\alpha$ to $90-\\frac{\\theta}{2}$ for simplicity. \nFrom here, by vertical angles $\\angle{CFD} = \\alpha$ . Also, since $\\triangle{OCB} \\cong \\triangle{ODC}$ $\\angle{OCD} = \\alpha$ . This means that $\\angle{CDF} = 180-2\\alpha = \\theta$ , which leads to $\\triangle{OCB} \\sim \\triangle{DCF}$ . \nSince we know that $\\overline{CD} = 200$ $\\overline{DF} = 200$ , and by similar reasoning $\\overline{AE} = 200$ . \nFinally, again using similar triangles, we get that $\\overline{CF} = 100\\sqrt{2}$ , which means that $\\overline{OF} = \\overline{OC} - \\overline{CF} = 200\\sqrt{2} - 100\\sqrt{2} = 100\\sqrt{2}$ . We can again apply similar triangles (or use Power of a Point) to get $\\overline{EF} = 100$ , and finally $\\overline{AD} = \\overline{AE}+\\overline{EF}+\\overline{FD} = 200+100+200=\\boxed{500}$ - ColtsFan10", "We first scale down by a factor of $200\\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresponds to $z^3$ and $B$ corresponds to $z^2$ . We are given that $\\lvert z \\rvert = 1$ and $\\lvert z-1 \\rvert = 1/\\sqrt{2}$ , and we wish to find $\\lvert z^3 - 1 \\rvert=\\lvert z^2+z+1\\rvert \\cdot \\lvert z-1 \\rvert=\\lvert (z^2+z+1)/\\sqrt{2} \\rvert$ . Let $z=a+bi$ , where $a$ and $b$ are real numbers. Then, $a^2+b^2=1$ and $a^2-2a+1+b^2=1/2$ ; solving for $a$ and $b$ yields $a=3/4$ and $b=\\sqrt{7}/4$ . Thus, $AD = \\lvert z^3 - 1 \\rvert = \\lvert (z^2+z+1)/\\sqrt{2} \\rvert = \\lvert (15/8 + 5\\sqrt{7}/8 \\cdot i)/\\sqrt{2} \\rvert = \\frac{5\\sqrt{2}}{4}$ . Scaling back up gives us a final answer of $\\frac{5\\sqrt{2}}{4} \\cdot 200\\sqrt{2} = \\boxed{500}$", "Let angle $C$ be $2a$ . This way $BD$ will be $400sin(a)$ . Now we can trig bash. As the circumradius of triangle $BCD$ is $200\\sqrt{2}$ , we can use the formula \\[R=\\frac{abc}{4A}\\] and \\[A=\\frac{absin(C)}{2}\\] and plug in all the values we got to get \\[200\\sqrt{2}=\\frac{200^2 \\cdot 400sin(a)}{4 \\cdot (\\frac{200^2 sin(2a)}{2})}\\] . This boils down to \\[\\sqrt{2}=\\frac{sin(a)}{sin{2a}}\\] . This expression can further be simplified by the trig identity \\[sin(2a)=2sin(a)cos(a)\\] . This leads to the final simplified form \\[2\\sqrt{2}=\\frac{1}{cos(a)}\\] . Solving this expression gives us \\[cos(a)=\\frac{\\sqrt{2}}{4}\\] . However, as we want $sin(a)$ , we use the identity $sin^2+cos^2=1$ , and substitute to get that $sin(a)=\\frac{\\sqrt{14}}{4}$ , and therefore BD is $100\\sqrt{14}$\nThen, as $ABCD$ is a cyclic quadrilateral, we can use Ptolemy’s Theorem (with $AD=x$ ) to get \\[14 \\cdot 100^2=200x+200^2\\] . Finally, we solve to get $\\boxed{500}$" ]

[ "\nLet quadrilateral $ABCD$ be inscribed in circle $O$ , where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$\nBy the Pythagorean Theorem, the length of $OH$ is \\begin{align*} \\sqrt{CO^2 - HC^2} &= \\sqrt{(200\\sqrt{2})^2 - \\left(\\frac{200}{2}\\right)^2} \\\\ &= \\sqrt{80000 - 10000} \\\\ &= \\sqrt{70000} \\\\ &= 100\\sqrt{7}. \\end{align*}\nNote that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$ ; then we have that\n$[AOB] + [BOC] + [COD] = \\frac{200 \\times 100\\sqrt{7}}{2} + \\frac{200 \\times 100\\sqrt{7}}{2} + \\frac{200 \\times 100\\sqrt{7}}{2} = \\frac{x \\times h}{2} + \\frac{(100\\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$\nFurthermore, \\begin{align*} h &= \\sqrt{OD^2 - GD^2} \\\\ &= \\sqrt{(200\\sqrt{2})^2 - \\left(\\frac{x}{2}\\right)^2} \\\\ &= \\sqrt{80000 - \\frac{x^2}{4}} \\end{align*}\nSubstituting this value of $h$ into the previous equation and evaluating for $x$ , we get: \\[\\frac{200 \\times 100\\sqrt{7}}{2} + \\frac{200 \\times 100\\sqrt{7}}{2} + \\frac{200 \\times 100\\sqrt{7}}{2} = \\frac{x \\times h}{2} + \\frac{(100\\sqrt{7} - h)(200 + x)}{2}\\] \\[\\frac{3 \\times 200 \\times 100\\sqrt{7}}{2} = \\frac{x\\sqrt{80000 - \\frac{x^2}{4}}}{2} + \\frac{\\left(100\\sqrt{7} - \\sqrt{80000 - \\frac{x^2}{4}}\\right)(200 + x)}{2}\\] \\[60000\\sqrt{7} = \\left(x\\sqrt{80000 - \\frac{x^2}{4}}\\right) + \\left(20000\\sqrt{7}\\right) + \\left(100x\\sqrt{7}\\right) - \\left(200\\sqrt{80000 - \\frac{x^2}{4}}\\right) - \\left(x\\sqrt{80000 - \\frac{x^2}{4}}\\right)\\] \\[40000\\sqrt{7} = 100x\\sqrt{7} - 200\\sqrt{80000 - \\frac{x^2}{4}}\\] \\[400\\sqrt{7} = x\\sqrt{7} - 2\\sqrt{80000 - \\frac{x^2}{4}}\\] \\[(x - 400)\\sqrt{7} = 2\\sqrt{80000 - \\frac{x^2}{4}}\\] \\[7(x-400)^2 = 4\\left(80000 - \\frac{x^2}{4}\\right)\\] \\[7x^2 - 5600x + 1120000 = 320000 - x^2\\] \\[8x^2 - 5600x + 800000 = 0\\] \\[x^2 - 700x + 100000 = 0\\]\nThe roots of this quadratic are found by using the quadratic formula: \\begin{align*} x &= \\frac{-(-700) \\pm \\sqrt{(-700)^2 - 4 \\times 1 \\times 100000}}{2 \\times 1} \\\\ &= \\frac{700 \\pm \\sqrt{490000 - 400000}}{2} \\\\ &= \\frac{700}{2} \\pm \\frac{\\sqrt{90000}}{2} \\\\ &= 350 \\pm \\frac{300}{2} \\\\ &= 200, 500 \\end{align*}\nIf the length of $AD$ is $200$ , then $ABCD$ would be a square. Thus, the radius of the circle would be \\[\\frac{\\sqrt{200^2 + 200^2}}{2} = \\frac{\\sqrt{80000}}{2} = \\frac{200\\sqrt{2}}{2} = 100\\sqrt{2}\\] Which is a contradiction. Therefore, our answer is $\\boxed{500}.$", "\nConstruct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$ . Apply the law of cosines on $\\Delta BOC$ ; let $\\theta = \\angle BOC$ . We get the following equation: \\[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\\cdot OB \\cdot OC\\cdot \\cos\\theta\\] Substituting the values in, we get \\[(200)^{2}=2\\cdot (200)^{2}+ 2\\cdot (200)^{2}- 2\\cdot 2\\cdot (200)^{2}\\cdot \\cos\\theta\\] Canceling out, we get \\[\\cos\\theta=\\frac{3}{4}\\] Because $\\angle AOB$ $\\angle BOC$ , and $\\angle COD$ are congruent, $\\angle AOD = 3\\theta$ . To find the remaining side ( $AD$ ), we simply have to apply the law of cosines to $\\Delta AOD$ . Now, to find $\\cos 3\\theta$ , we can derive a formula that only uses $\\cos\\theta$ \\[\\cos 3\\theta=\\cos (2\\theta+\\theta)= \\cos 2\\theta \\cos\\theta- (2\\sin\\theta \\cos\\theta) \\cdot \\sin \\theta\\] \\[\\cos 3\\theta= \\cos\\theta (\\cos 2\\theta-2\\sin^{2}\\theta)=\\cos\\theta (2\\cos^{2}\\theta-3+2\\cos^{2}\\theta)\\] \\[\\Rightarrow \\cos 3\\theta=4\\cos^{3}\\theta-3\\cos\\theta\\] It is useful to memorize the triple angle formulas ( $\\cos 3\\theta=4\\cos^{3}\\theta-3\\cos\\theta, \\sin 3\\theta=3\\sin\\theta-4\\sin^{3}\\theta$ ). Plugging in $\\cos\\theta=\\frac{3}{4}$ , we get $\\cos 3\\theta= -\\frac{9}{16}$ . Now, applying law of cosines on triangle $OAD$ , we get \\[(AD)^{2}= 2\\cdot (200)^{2}+ 2\\cdot (200)^{2}+2\\cdot 200\\sqrt2 \\cdot 200\\sqrt2 \\cdot \\frac{9}{16}\\] \\[\\Rightarrow 2\\cdot (200)^{2} \\cdot (1+1+ \\frac{9}{8})=(200)^{2}\\cdot \\frac{25}{4}\\] \\[AD=200 \\cdot \\frac{5}{2}=\\boxed{500}\\]", "\nConstruct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\\overline{BC}$ and let these points be $E$ and $F$ , respectively. Also, let $\\theta = \\angle BOC$ . From the Law of Cosines on $\\triangle BOC$ , we have $\\cos \\theta = \\frac{3}{4}$\nNow, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$ , we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$ . In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$ , so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$ , so $\\angle DCF = \\theta$\nThus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$ , so $FC = 150$ . Similarly, $BE = 150$ , and $AD = 150 + 200 + 150 = \\boxed{500}$", " Claim: $[ABCD]$ is an isosceles trapezoid.\nProof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\\angle BAD = 180 - \\angle BCD = 180-(\\angle BCO + \\angle DCO)=180-(\\angle CBO+\\angle ABO) = 180 - \\angle ABC = \\angle CDA$ . Hence, $[ABCD]$ is an isosceles trapezoid.\nLet $\\angle CDA=\\alpha$ . Notice that the length of the altitude from $C$ to $AD$ is $200sin(\\alpha)$ . Furthermore, the length of the altitude from $O$ to $BC$ is $100\\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\\sqrt{7}-200sin\\alpha$ . Let $F$ the feet of the altitude from $O$ to $AD$ . Then, $FD=(200+400cos(\\alpha))/2=100+200cos(\\alpha)$ , because $AOD$ is isosceles.\nTherefore, by the Pythagorean theorem, $(100+200cos(\\alpha))^2+(100\\sqrt{7}-200sin(\\alpha))^2=80000$ . Simplifying, we have $1+cos(\\alpha)=sin(\\alpha) \\cdot sqrt{7} \\implies cos^2(\\alpha)+2cos(\\alpha)+1=sin^2(\\alpha) \\cdot 7 = 7-7cos^2(\\alpha) \\implies 8cos^2(\\alpha)+2cos(\\alpha) - 6 =0$ . Solving this quadratic, we have $cos(\\alpha)=\\frac{3}{4}, -1$ , but $0<\\alpha<180 \\implies cos(\\alpha)=3/4$ . Therefore, $AD=200cos(\\alpha)+200cos(\\alpha)+200=\\boxed{500}$" ]

[ "First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.\n$11_{12}=15_8$\n$22_{12}=32_8$\n$33_{12}=47_8$\n$44_{12}=64_8$\n$55_{12}=101_8$\nWe stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number.\nCompare the ones places to check if they are equal. We find that they are equal if $b=2$ or $b=4$ .\nEvaluating the places to the right side of the decimal point gives us $22.23_{12}$ or $44.46_{12}$ .\nWhen the numbers are converted into base 8, we get $32.14_8$ and $64.30_8$ . Since $d\\neq0$ , the first value is correct. Compiling the necessary digits leaves us a final answer of $\\boxed{321}$", "The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$ . Simplifying the first equation gives $a = (3/2)b$ . Plugging this into the second equation gives $3b/32 = c/8 + d/64$ . Multiplying both sides by 64 gives $6b = 8c + d$ $a$ and $b$ are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using $a = 3/2b$ $(a,b) = (3,2)$ or $(6,4)$ . Testing these gives that $(6,4)$ doesn't work, and $(3,2)$ gives $a = 3, b = 2, c = 1$ , and $d = 4$ . Therefore $abc = \\boxed{321}$", "Converting to base $10$ we get\n$4604a+72c+9d=6960b$\nSince $72c$ and $9d$ are much smaller than the other two terms, dividing by $100$ and approximating we get\n$46a=70b$\nWriting out the first few values of $a$ and $b$ , the first possible tuple is\n$a=3, b=2, c=1, d=4$\nand the second possible tuple is\n$a=6, b=4, c=3, d=0$\nNote that $d$ can not be $0$ , therefore the answer is $\\boxed{321}$", "In the problem, we are given that \\[8a+b+\\dfrac c8+\\dfrac d{64}=12b+b+\\dfrac b{12}+\\dfrac a{144}.\\] We multiply by the LCM of the denominators, which is $576$ to get \\[4608a+576b+72c+9d=6912b+576b+48b+4a.\\] We then group like terms and factor to get \\[4604a+72c+9d=48b(144+1)=145\\cdot48b=24\\cdot290b.\\]\nObserve that the coefficients of $a$ $c$ , and $b$ are all divisible by $4$ . Therefore, we know that $d$ must also be divisible by $4$ to compensate. (To observe this, one could rearrange the terms to see $9d=24\\cdot290b-4604a-72c=4(6\\cdot290b-1151a-18c)$ . At this point, it is obvious that $9d$ must be divisible by $4$ , so $d$ must be divisible by $4$ .) Thus, we let $d=4y$ to see $4604a+72c+9\\cdot4y=24\\cdot290b$ . We divide by $4$ in the equation to get \\[1151a+18c+9y=6\\cdot290b.\\]\nObserve that the coefficients of $c$ $y$ , and $b$ are all divisible by $3$ . By similar reasoning, $a$ must be divisible by $3$ , so we let $a=3x$ . Substituting and dividing by $3$ , we get \\[1151x+6c+3y=2\\cdot290b.\\] We observe that this can no longer be reduced by similar means.\nWe know that $0<a,b,c<8\\implies0<3x,b,4y<8$ . We examine $0<4y<8$ ; this becomes $0<y<2$ . It is apparent that $y=1\\implies d=4$ . However, the problem does not even ask for $d$ , so it may appear that this find is meaningless. However, we substitute our value of $y$ in to get \\[1151x+6c+3=580b.\\]\nWe know that $x$ is either $1$ or $2$ , since $0<3x<8$ . We take our equation modulo $6$ to find \\[5x+3\\equiv4b\\pmod{6}\\implies3\\equiv4b+x\\pmod{6}.\\] If $x=2$ , then $1\\equiv4b\\pmod{6}$ - this is equivalent to saying that $1+6n=4b\\implies 1=4b-6n$ . However, $4b-6n$ is always even, and $1$ is odd. Thus, this case is not possible, so $x=1\\implies a=3$ . We know that $x=1$ ; thus, $3\\equiv4b+1\\pmod{6}\\implies2\\equiv4b\\pmod{6}\\implies1\\equiv2b\\pmod{3}$ . Obviously, $b=2$ and $b=5$ work; however, if $b=5$ , then the RHS of our original equation ( $2\\cdot290b$ ) is much too large to be equal to the LHS (the maximum possible value of the LHS is $1151\\cdot1+6\\cdot7+3\\cdot1$ , which is less than $1200$ while the RHS would become $2\\cdot290\\cdot5=2900$ ), so we have $b=2$\nWe recall our equation of $1151x+6c+3=580b$ . Plugging in what we know, we have $1151\\cdot1+6c+3=580\\cdot2\\implies6c=6\\implies c=1$\nTherefore, $\\overline{abc}=\\boxed{321}$" ]

[ "Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be $0$ , but that would be impossible. Therefore we start by working backwards. Since angle $CDA$ is $8$ , the reflection would give us a triangle with angles $16, 90$ , and $74$ . Then, when we reflect again, we will have $180 - 74 - 74$ $32$ . Since the other side of the reflection when we had the $82$ degrees had carried over to the other side, we have a $32-82-66$ triangle.\nNotice that we keep decreasing by increments of $8$ . This is because the starting angle was $8$ and since we always have to decrease $8$ every time and that every triangle has every increasing angles of $8$ , we must decrease by $8$ every time. This is the most optimal path of the light beam.\nThe pattern of light will be $82-74-66-58-50-42-34-26-18-10$ . When we get to the angle of $2$ degrees, we have reached angle $A$ . Therefore, we don't count the $2$ , so our total number of reflections between $CD$ and $AD$ is $\\boxed{10}$" ]

[ "The polynomial we are given is rather complicated, so we could use Rational Root Theorem to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, $x = 1, -1, 2, -2$ are all possible rational roots. Upon plugging these roots into the polynomial, $x = -2$ and $x = 1$ make the polynomial equal 0 and thus, they are roots that we can factor out.\nThe polynomial becomes:\n$(x - 1)(x + 2)(x^2 + (2a - 1)x + 1)$\nSince we know $1$ and $-2$ are real numbers, we only need to focus on the quadratic.\nWe should set the discriminant of the quadratic greater than or equal to 0.\n$(2a - 1)^2 - 4 \\geq 0$\nThis simplifies to:\n$a \\geq \\dfrac{3}{2}$\nor\n$a \\leq -\\dfrac{1}{2}$\nThis means that the interval $\\left(-\\dfrac{1}{2}, \\dfrac{3}{2}\\right)$ is the \"bad\" interval. The length of the interval where $a$ can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the \"good\" interval is 36 units long.\n$\\dfrac{36}{38} = \\dfrac{18}{19}$\n$18 + 19 = \\boxed{037}$" ]

[ "Substituting $a = b$ we get \\[f(2a) + f(0) = 2f(a)^2\\] Substituting $a= 0$ we find \\[2f(0) = 2f(0)^2 \\implies f(0) \\in \\{0, 1\\}.\\] This gives \\[f(2a) = 2f(a)^2 - f(0) \\geq 0-1\\] Plugging in $a = \\frac{1}{2}$ implies $f(1) \\geq -1$ , so answer choice $\\boxed{2}$ is impossible.", "First, we set $a \\leftarrow 0$ and $b \\leftarrow 0$ .\nThus, the equation given in the problem becomes $[ f(0) + f(0) = 2 f(0) \\times f(0) . ]$\nThus, $f(0) = 0$ or 1.\nCase 1: $f(0) = 0$\nWe set $b \\leftarrow 0$ .\nThus, the equation given in the problem becomes $[ 2 f(a) = 0 . ]$\nThus, $f(a) = 0$ for all $a$\nCase 2: $f(0) = 1$\nWe set $b \\leftarrow a$ .\nThus, the equation given in the problem becomes \\[[ f(2a) + 1 = 2 \\left( f(a) \\right)^2. ]\\]\nThus, for any $a$ \\begin{align*} f(2a) & = -1 + 2 \\left( f(a) \\right)^2 \\\\ & \\geq -1 . \\end{align*}\nTherefore, an infeasible value of $f(1)$ is $\\boxed{2}.$", "The relationship looks suspiciously like a product-to-sum identity. In fact, \\[\\cos(\\alpha)\\cos(\\beta) = \\frac{1}{2}(\\cos(\\alpha-\\beta)+\\cos(\\alpha+\\beta))\\] which is basically the relation. So we know that $f(x) = \\cos(x)$ is a valid solution to the function. However, if we define $x=ay,$ where $a$ is arbitrary, the above relation should still hold for $f(x) = \\cos(ay) = \\cos(a(1))$ so any value in $[-1,1]$ can be reached, so choices $A,B,$ and $C$ are incorrect.\nIn addition, using the similar formula for hyperbolic cosine, we know \\[\\cosh(\\alpha)\\cosh(\\beta) = \\frac{1}{2}(\\cosh(\\alpha-\\beta)+\\cosh(\\alpha+\\beta))\\] The range of $\\cosh(ay)$ is $[1,\\infty)$ so choice $D$ is incorrect.\nTherefore, the remaining answer is choice $\\boxed{2}.$" ]

[ "We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.\nLet's say that $500\\text{ sheets}=5\\text{ cm}\\Rightarrow \\frac{500 \\text{ sheets}}{5 \\text{ cm}} = 1$ . So by multiplying $7.5 \\text{ cm}$ by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets\n\\begin{align*} \\frac{7.5 \\times 500}{5} &= 7.5 \\times 100 \\\\ &= 750 \\text{ sheets} \\\\ \\end{align*}\n$750$ is $\\boxed{750}$", "We can set up a direct proportion relating the amount of sheets to the thickness because according to the problem, all the papers have the same thickness.\nOur proportion is \\[\\frac{5}{500}=\\frac{7.5}{x}\\] where $x$ is the number we are looking for.\nNext, we cross-multiply to get $5x=500 \\times 7.5$ so $x=750$ which is $\\boxed{750}$" ]

[ "When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.\n\nBy Heron's Formula, the area of each isosceles triangle is $\\sqrt{(30)(12)(12)(6)}=\\sqrt{180\\times 12^2}=72\\sqrt{5}$ . So the area of both is $144\\sqrt{5}$ . From the rectangle, our original area is $36a$ . The area of the rectangle in the hexagon is $24a$ . So we have \\[24a+144\\sqrt{5}=36a\\implies 12a=144\\sqrt{5}\\implies a=12\\sqrt{5}\\implies a^2=\\boxed{720}.\\]", "Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let $a/6=p$ . Then, the dimensions of the central rectangle in the hexagon is p x 4, and the original rectangle is 6 x p. By Pythagorean theorem and splitting the end triangles of the hexagon into two right triangles, the altitude of the end triangles is $\\sqrt{3^2-2^2}=\\sqrt{5}$ given 2 as the base of the constituent right triangles. The two end triangles form a large rectangle of area $\\sqrt{5}$ $4$ . Then, the area of the hexagon is $4p+4\\sqrt{5}$ , and the area of the rectangle is $6p$ . Equating them, $p=2\\sqrt{5}$ . Multiply by scale factor of 6 and square it to get $36(20)= 720 \\implies a^2=\\boxed{720}$" ]

[ "\nDraw $BE$ and $CF$ , forming a trapezoid . Since it's cyclic, this trapezoid must be isosceles . Also, drop altitudes from $E$ to $AC$ $B$ to $DF$ , and $C$ to $DF$ , and let the feet of these altitudes be $G$ $H$ , and $I$ respectively. $AGED$ is a rectangle since it has $4$ right angles . Therefore, $AG=DE=3$ , and $GB=4-3=1$ . By the same logic, $GBHE$ is also a rectangle, and $EH=GB=1$ $BH=CI$ since they're both altitudes to a trapezoid, and $BE=CF$ since the trapezoid is isosceles. Therefore, $\\triangle BHE \\cong \\triangle CIF$ by HL congruence , so $IF=EH=1$ . Also, $BCIH$ is a rectangle from $4$ right angles, and $HI=BC=5$ . Therefore, $EF=EH+HI+IF=1+5+1=\\boxed{7}$" ]

[ "The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of $5\\cdot4\\cdot3\\cdot3\\cdot3=\\boxed{540}$", "Case 1: All the rectangles are different colors. It would be $5! = 120$ choices.\nCase 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us $5\\cdot4\\cdot3\\cdot2 = 120$ . But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left and bottom right, the top right and bottom left, or the bottom right and bottom left, which gives us a grand total of $360$\nCase 3: We have two sets of rectangles chosen from these choices (top right & bottom left, top left & bottom right) that have the same color. However, the choice of the bottom left and bottom right does not work for this case, as the second pair would be chosen from two touching rectangles. Again, grouping the same-colored rectangles gives us $5\\cdot4\\cdot3 = 60$\nTherefore, we have $120 + 360 + 60 = \\boxed{540}$" ]

[ "The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of $5\\cdot4\\cdot3\\cdot3\\cdot3=\\boxed{540}$", "Case 1: All the rectangles are different colors. It would be $5! = 120$ choices.\nCase 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us $5\\cdot4\\cdot3\\cdot2 = 120$ . But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left and bottom right, the top right and bottom left, or the bottom right and bottom left, which gives us a grand total of $360$\nCase 3: We have two sets of rectangles chosen from these choices (top right & bottom left, top left & bottom right) that have the same color. However, the choice of the bottom left and bottom right does not work for this case, as the second pair would be chosen from two touching rectangles. Again, grouping the same-colored rectangles gives us $5\\cdot4\\cdot3 = 60$\nTherefore, we have $120 + 360 + 60 = \\boxed{540}$" ]

[ "Put the rectangle on the coordinate plane so its vertices are at $(\\pm4,\\pm3)$ , for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$\nNote that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be $A(4,y)$ $B(-x,3)$ $C(-4,-y)$ and $D(x,-3)$ for nonnegative $x,y$ . Then this is a rectangle, so $OA=OB$ , or $16+y^2=9+x^2$ , so $x^2=y^2+7$\nReflect $D$ across the side of the rectangle containing $C$ to $D'(-8-x,-3)$ . Then $BD'=\\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10$ is constant, and the perimeter of the rectangle is equal to $2(BC+CD')$ . The midpoint of $\\overline{BD'}$ is $(-4-x,0)$ , and since $-4>-4-x$ and $-y\\le0$ $C$ always lies below $\\overline{BD'}$\nIf $y$ is positive, it can be decreased to $y'<y$ . This causes $x$ to decrease as well, to $x'$ , where $x'^2=y'^2+7$ and $x'$ is still positive. If $B$ and $D'$ are held in place as everything else moves, then $C$ moves $(y-y')$ units up and $(x-x')$ units left to $C'$ , which must lie within $\\triangle BCD'$ . Then we must have $BC'+C'D'<BC+CD'$ , and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with $y=0$ , so $x=\\sqrt7$\nBy the distance formula, this minimum perimeter is \\[2\\left(\\sqrt{(4-\\sqrt7)^2+3^2}+\\sqrt{(4+\\sqrt7)^2+3^2}\\right)=4\\left(\\sqrt{8-2\\sqrt7}+\\sqrt{8+2\\sqrt7}\\right)\\] \\[=4(\\sqrt7-1+\\sqrt7+1)=8\\sqrt7=\\sqrt{448}.\\] Therefore $N$ would equal $\\boxed{448}.$", "Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let $a$ and $b$ be the sides of the rectangle. Then $ab = 3(8) = 24$ since both are twice the area of the same right triangle, and $a^2+b^2 = 64$ . So $(a+b)^2 = 64+2(24) = 112$ , so $2(a+b) = \\sqrt{\\boxed{448}$" ]

[ "Let $l$ represent the length of one of the smaller rectangles, and let $w$ represent the width of one of the smaller rectangles, with $w < l$\nFrom the large rectangle, we see that the top has length $3w$ , the right has length $l + w$ , the bottom has length $2l$ , and the left has length $l + 2$\nSince the perimeter of the large rectangle is $176$ , we know that $172 = 3w + l + w + 2l + l + w$ , or $172 = 5w + 4l$\nFrom the top and bottom of the large rectangle, we know that $3w = 2l$ , or $l = 1.5w$\nPlugging that into the first equation, we get $176 = 5w + 4(1.5)w$\n$176 = 11w$\n$w = 16$\n$l = 1.5w = 24$\n$P = 2l + 2w = 2(16 + 24) = 80$ , and the answer is $\\boxed{80}$" ]

[ "Let the rectangle's length be $a$ and its width be $b$ . Its area is $ab$ and the perimeter is $2a+2b$\nThen $A + P = ab + 2a + 2b$ . Factoring, we have $(a + 2)(b + 2) - 4$\nThe only one of the answer choices that cannot be expressed in this form is $102$ , as $102 + 4$ is twice a prime. There would then be no way to express $106$ as $(a + 2)(b + 2)$ , keeping $a$ and $b$ as positive integers.\nOur answer is then $\\boxed{102}$" ]

[ "We use Image:2021_AMC_10B_(Nov)_Problem_25,_sol.png to facilitate our analysis.\nDenote $\\angle AFE = \\theta$ . Thus, $\\angle FIB = \\angle CEF = \\angle EKG = \\angle KLC = \\theta$\nHence, \\begin{align*} AB & = AF + FB \\\\ & = EF \\cos \\angle EFA + IF \\sin \\angle FIB \\\\ & = 3 \\cos \\theta + \\sin \\theta , \\end{align*} and \\begin{align*} AC & = AE + EK + KC \\\\ & = EF \\sin \\angle EFA + EG \\cos \\angle CEG + KG \\cos \\angle EKG + KL \\sin \\angle CLK \\\\ & = 3 \\sin \\theta + \\cos \\theta + \\cos \\theta + \\sin \\theta \\\\ & = 2 \\cos \\theta + 4 \\sin \\theta . \\end{align*}\nBecause $AB = AC$ $3 \\cos \\theta + \\sin \\theta = 2 \\cos \\theta + 4 \\sin \\theta$ .\nHence, $\\tan \\theta = \\frac{1}{3}$ .\nHence, $\\sin \\theta = \\frac{1}{\\sqrt{10}}$ and $\\cos \\theta = \\frac{3}{\\sqrt{10}}$\nHence, $AB = AC = BD = CD = \\sqrt{10}$\nNow, we put the graph to a coordinate plane by setting point $A$ as the origin, putting $AB$ in the $x$ -axis and $AC$ on the $y$ -axis.\nHence, $A = \\left( 0 , 0 \\right)$ $B = \\left( \\sqrt{10} , 0 \\right)$ $C = \\left( 0 , \\sqrt{10} \\right)$ $D = \\left( \\sqrt{10} , \\sqrt{10} \\right)$ $E = \\left( 0 , \\frac{3}{\\sqrt{10}} \\right)$ $F = \\left( \\frac{9}{\\sqrt{10}} , 0 \\right)$ $G = \\left( \\frac{1}{\\sqrt{10}} , \\frac{6}{\\sqrt{10}} \\right)$ $H = \\left( \\frac{4}{\\sqrt{10}} , \\frac{7}{\\sqrt{10}} \\right)$ $I = \\left( \\sqrt{10} , \\frac{3}{\\sqrt{10}} \\right)$\nDenote $P = \\left( \\frac{10 - u}{\\sqrt{10}} , \\sqrt{10} \\right)$ $Q = \\left( \\sqrt{10} , \\frac{10 - v}{\\sqrt{10}} \\right)$\nBecause $HPQJ$ is a rectangle, $HP \\perp PQ$ . Hence, $m_{HP} m_{PQ} = -1$ .\nWe have $m_{HP} = \\frac{3}{6 - u}$ and $m_{PQ} = - \\frac{v}{u}$ .\nHence, \\[ \\frac{3}{6 - u} \\cdot \\left( - \\frac{v}{u} \\right) = - 1 . \\hspace{1cm} (1) \\]\nBecause $HPQJ$ is a rectangle, $x_J + x_P = x_H + x_Q$ and $y_J + y_P = y_H + y_Q$ .\nHence, $J = \\left( \\frac{4 + u}{\\sqrt{10}} , \\frac{7 - v}{\\sqrt{10}} \\right)$\nThe equation of line $GI$ is \\begin{align*} y & = \\frac{\\frac{3}{\\sqrt{10}} - \\frac{6}{\\sqrt{10}}}{\\sqrt{10} - \\frac{1}{\\sqrt{10}}} \\left( x - \\frac{1}{\\sqrt{10}} \\right) + \\frac{6}{\\sqrt{10}} \\\\ & = - \\frac{x}{3} + \\frac{19}{3 \\sqrt{10}} . \\end{align*} Because point $J$ is on line $GI$ , plugging the coordinates of $J$ into the equation of line $GI$ , we get \\[ \\frac{7 - v}{\\sqrt{10}} = - \\frac{\\frac{4 + u}{\\sqrt{10}}}{3} + \\frac{19}{3 \\sqrt{10}} . \\hspace{1cm} (2) \\]\nBy solving Equations (1) and (2), we get $\\left( u , v \\right) = \\left( 2 , \\frac{8}{3} \\right)$ or $\\left( 3 , 3 \\right)$\nCase 1: $\\left( u , v \\right) = \\left( 2 , \\frac{8}{3} \\right)$\nWe have $P = \\left( \\frac{8}{\\sqrt{10}} , \\sqrt{10} \\right)$ and $Q = \\left( \\sqrt{10} , \\frac{22}{3 \\sqrt{10}} \\right)$ .\nThus, $HP = \\frac{5}{\\sqrt{10}}$ and $PQ = \\frac{10}{3\\sqrt{10}}$\nTherefore, ${\\rm Area} \\ R = HP \\cdot PQ = \\frac{5}{3}$\nCase 2: $\\left( u , v \\right) = \\left( 3 , 3 \\right)$\nWe have $P = \\left( \\frac{7}{\\sqrt{10}} , \\sqrt{10} \\right)$ and $Q = \\left( \\sqrt{10} , \\frac{7}{ \\sqrt{10}} \\right)$ .\nThus, $HP = \\frac{3 \\sqrt{2}}{\\sqrt{10}}$ and $PQ = \\frac{3 \\sqrt{2}}{\\sqrt{10}}$\nTherefore, ${\\rm Area} \\ R = HP \\cdot PQ = \\frac{9}{5}$\nPutting these two cases together, the sum of all possible values of the area of $R$ is $\\frac{5}{3} + \\frac{9}{5} = \\frac{52}{15}$\nTherefore, the answer is $\\boxed{67}$", " We will scale every number up by a factor of $\\sqrt{10}.$ This implies our final area will be $\\frac{1}{10}$ of the answer we receive.\nWe have \\[FAE\\sim EOH \\sim IOH\\sim JDI\\sim KCJ \\sim NQK\\sim GPF.\\] Let $AE=a$ and $FA=b.$ We have \\[FP=AE=OH=JC=\\frac13ID=a\\] and \\[PG=AF=EO=OI=KC=\\frac13 DJ=b\\] As $ABCD$ is a square, we have $AD=DC$ or \\[a+2b+3a=3b+a \\Rightarrow 3a=b.\\] Since $a^2+b^2=10,$ we have \\[a=1,b=3.\\] We have $\\triangle GPL\\cong \\triangle MQN$ which implies \\[MQ=GP=3.\\] Denote $QK=x.$ As $\\triangle NQK\\sim \\triangle JDI,$ we have $NQ=3x.$\nWe have \\begin{align*}BM&=BC-(CK+QK+MQ)\\\\ &=4-x.\\end{align*}\nIn addition, \\begin{align*}LB&=AB-(AF+FP+PL)\\\\&=6-3x.\\end{align*}\nSince $\\triangle LBM\\sim \\triangle MQN,$ we have \\[\\frac{LB}{BM}=\\frac{MQ}{QN}\\Rightarrow \\frac{6-3x}{4-x}=\\frac{3}{3x}=\\frac{1}{x}.\\] Simplifying we have \\[3x^2-7x+4=0 \\Rightarrow x=\\frac43, 1.\\] We have \\begin{align*}[GLMN]&=MN\\cdot LM\\\\ &= 3\\sqrt{x^2+1}\\cdot \\sqrt{10x^2-44x+52}.\\end{align*}\nPlugging in $x=1,$ we have $[GLMN]=18.$\nPlugging in $x=\\frac43,$ we have $[GLMN]=\\frac{50}{3}.$\nThe sum of the two possible $R$ s is \\[\\frac{1}{10}\\cdot\\frac{104}{3}=\\frac{52}{15}.\\] Hence, $52+15=\\boxed{67}.$" ]

[ "If we extend the lines, we have the following diagram: Therefore, we see that the answer is $\\boxed{1}.$", "Note that the $y$ -intercepts of line $AB$ and line $CD$ are $0$ and $10$ . If the analytic expression for line $AB$ is $y=k_{1}x$ , and the analytic expression for line $CD$ is $y=k_{2}x+10$ , we have equations: $3k_{1} = 1$ and $2k_{2} + 10 = 9$ . Solving these equations, we can find out that $k_{1} = \\frac{1}{3}$ and $k_{2} = -\\frac{1}{2}$ . Therefore, we can determine that the expression for line $AB$ is $y=\\frac{1}{3}x$ and the expression for line $CD$ is $y=-\\frac{1}{2}x + 10$ . When $x=15$ , the coordinates that line $AB$ and line $CD$ pass through are $(15, 5)$ and $\\left(15, \\frac{5}{2}\\right)$ , and $(15, 5)$ lies perfectly on one vertex of the rectangle while the $y$ coordinate of $\\left(15, \\frac{5}{2}\\right)$ is out of the range $3 \\leq y \\leq 5$ (lower than the bottom left corner of the rectangle $(15, 3)$ ). Considering that the $y$ value of the line $CD$ will only decrease, and the $y$ value of the line $AB$ will only increase, there will not be another point on the rectangle that lies on either of the two lines. Thus, we can conclude that the answer is $\\boxed{1}.$" ]

[ "The numbers that are shaded are the triangular numbers, which are numbers in the form $\\frac{(n)(n+1)}{2}$ for positive integers. They can also be generated by starting with $1$ , and adding $1, 2, 3, 4...$ as in the description of the problem.\nSquares that have the same remainder after being divided by $8$ will be in the same column. Thus, we want to find when the last remainder, from $0$ to $7$ , is found.\nSo, instead of adding $1, 2, 3, 4, 5, 6, 7, 8, 9, 10...$ , we can effectively either add $1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3...$ or subtract $7, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5...$ if we are only concerned about remainders when divided by $8$ . We will pick the number that keeps the terms on the list between $1$ and $8$ . We get:\n$1$\n$1 + 2 = 3$\n$3 + 3 = 6$\n$6 - 4 = 2$\n$2 + 5 = 7$\n$7 - 2 = 5$\n$5 - 1 = 4$\n$4 + 0 = 4$\n$4 + 1 = 5$\n$5 + 2 = 7$\n$7 - 5 = 2$\n$2 + 4 = 6$\n$6 - 3 = 3$\n$3 - 2 = 1$\n$1 - 1 = 0$\nFinally, a term with $0$ is found, and checking, all numbers $1$ through $7$ are also on the right side of the list. This means the last term in our sequence is the first time that column $8$ is shaded. There are $15$ terms in the sequence, leading to an answer of $\\frac{15\\cdot 16}{2} = 120$ , which is choice $\\boxed{120}$", "Note that the triangular numbers up to $120$ are $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120$ . When you divide each of those numbers by $8$ , all remainders must be present. We first search for number(s) that are evenly divisible by $8$ ; if two such numbers exist, we search for numbers that leave a remainder of $1$ , etc.\nQuickly scanning the list, only $6, 10, 28, 36, 66, 78$ and $120$ are even. That smaller list doesn't have any multiples of $8$ until it hits $120$ . So $\\boxed{120,}$ must be the answer.", "The numbers shaded are triangular numbers of the form $\\frac{(n)(n+1)}{2}$ . For this number to be divisible by $8$ , the numerator must be divisible by $16$ . Since only one of $n$ and $n+1$ can be even, only one of them can have factors of $2$ . Therefore, the first time the whole expression is divisible by $8$ is when either $n+1=16$ or when $n=16$ . This gives $n=15$ as the first time $\\frac{n(n+1)}{2}$ is divisible by $8$ , which gives $120$ . No other triangular number lower than that is divisible by $8$ , and thus the $8^{th}$ column on the checkerboard won't be filled until then. That gives $\\boxed{120}$ as the right answer." ]

[ "Box P has dimensions $l$ $w$ , and $h$ . \nIts surface area is \\[2lw+2lh+2wh=384,\\] and the sum of all its edges is \\[l + w + h = \\dfrac{4l+4w+4h}{4} = \\dfrac{112}{4} = 28.\\]\nThe diameter of the sphere is the space diagonal of the prism, which is \\[\\sqrt{l^2 + w^2 +h^2}.\\] Notice that \\[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,\\] so the diameter is \\[\\sqrt{l^2 + w^2 +h^2} = \\sqrt{400} = 20.\\] The radius is half of the diameter, so \\[r=\\frac{20}{2} = \\boxed{10}.\\]", "As in the previous solution, we have that $2lw+2lh+2wh=384$ and $l + w + h = \\dfrac{4l+4w+4h}{4} = \\dfrac{112}{4} = 28$ , and the diameter of the sphere is the space diagonal of the prism, $\\sqrt{l^2 + w^2 + h^2}$\nNow, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that $h=0$ . (This essentially means that we have an infinitesimally thin box.) We now have that $2lw = 384$ and $l + w = 28$ , and we are solving for $\\sqrt{l^2 + w^2}$ . Because \\[(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,\\] this means that \\[l^2 + w^2 = 28^2 - 384 = 400,\\] so the space diagonal is $\\sqrt{400} = 20$ . Since the diameter of the sphere is $20$ , the radius is $\\boxed{10}$ . ~ emerald_block" ]

[ "Let the smallest side length be $x$ . Then the volume is $x \\cdot 3x \\cdot 4x =12x^3$ . If $x=2$ , then $12x^3 = 96 \\implies \\boxed{96.}$" ]

[ "Let the height of the box be $x$\nAfter using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\\sqrt{\\left(\\frac{x}{2}\\right)^2 + 64}$ , and $\\sqrt{\\left(\\frac{x}{2}\\right)^2 + 36}$ . Since the area of the triangle is $30$ , the altitude of the triangle from the base with length $10$ is $6$\nConsidering the two triangles created by the altitude, we use the Pythagorean theorem twice to find the lengths of the two line segments that make up the base of $10$\nWe find: \\[10 = \\sqrt{\\left(28+x^2/4\\right)}+x/2\\]\nSolving for $x$ gives us $x=\\frac{36}{5}$ . Since this fraction is simplified: \\[m+n=\\boxed{041}\\]", "We may use vectors. Let the height of the box be $2h$ . Without loss of generality, let the front bottom left corner of the box be $(0,0,0)$ . Let the center point of the bottom face be $P_1$ , the center of the left face be $P_2$ and the center of the front face be $P_3$\nWe are given that the area of the triangle $\\triangle P_1 P_2 P_3$ is $30$ . Thus, by a well known formula, we note that $\\frac{1}{2}|\\vec{P_1P_2} \\text{x} \\vec{P_1P_3}|=30$ We quickly attain that $\\vec{P_1P_2}=<-6,0,h>$ and $\\vec{P_1P_3}=<0,-8,h>$ (We can arbitrarily assign the long and short ends due to symmetry)\nComputing the cross product, we find: \\[\\vec{P_1P_2} x \\vec{P_1P_3}=-<6h,8h,48>\\]\nThus: \\[\\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60\\] \\[h=3.6\\] \\[2h=7.2\\]\n\\[2h=36/5\\]\n\\[m+n=\\boxed{041}\\]", "Let the height of the box be $x$\nAfter using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\\sqrt{(x/2)^2 + 64}$ , and $\\sqrt{(x/2)^2 + 36}$ . Therefore, we can use Heron's formula to set up an equation for the area of the triangle.\nThe semiperimeter is $\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2$ . Therefore, when we square Heron's formula, we find\n\\begin{align*}900 &= \\frac{1}{2}\\left(\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2\\right)\\times\\left(\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2 - 10\\right)\\\\&\\qquad\\times\\left(\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2 - \\sqrt{(x/2)^2 + 64}\\right)\\\\&\\qquad\\times\\left(\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2 - \\sqrt{(x/2)^2 + 36}\\right).\\end{align*}\nSolving, we get $\\boxed{041}$", "Let half the height be $a$ (we want to find $2a$ ), then we see that the three sides of the triangle are (by Pyth Theorem) $10, \\sqrt{a^2+36}, \\sqrt{a^2+64}$ . Using the Law of Sines with the angle as the one included between the square roots, we see that this angle's cosine is $\\frac{a^2}{\\sqrt{(a^2+36)(a^2+64)}}$ by the Law of Cosines, meaning that its sine is $\\frac{\\sqrt{100a^2+2304}}{\\sqrt{(a^2+36)(a^2+64)}}$ . Finally, multiply the two square-rooted sides by this sine and one-half, and equate to 30.\nYou get $\\sqrt{25a^2+576} = 30$ , giving $a=\\frac{18}{5}$ , so our answer is $\\boxed{041}$", "Let $x$ be $\\frac12$ the height of the box. We will solve for $x$ and then multiply by $2$ at the end. The three side lengths of the triangle, by the Pythagorean Theorem, are $\\sqrt{x^2+6^2},\\sqrt{x^2+8^2},$ and $10$\nHeron's formula states that the area of the triangle is $\\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semiperimeter. Applying difference of squares to make the formula less computational, we get $\\frac{\\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{4}=\\frac{\\sqrt{(a^2+b^2+2ab-c^2)(c^2-b^2-a^2+2ab)}}4=\\frac{\\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}4$\nNow, we plug in $\\sqrt{x^2+6^2}$ for $a$ $\\sqrt{x^2+8^2}$ for $b$ , and $10$ for $c$ . This gives us \\[\\frac{\\sqrt{4(x^2+6^2)(x^2+8^2)-(x^2+6^2+x^2+8^2-10^2)^2}}4=30\\] \\[\\sqrt{4x^4+400x^2+4\\cdot48^2-4x^4}=120\\] \\[400x^2+4\\cdot48^2=120^2\\] \\[25x^2+24^2=30^2\\] \\[25x^2=6^2(5^2-4^2)\\] \\[25x^2=18^2\\] \\[5x=18\\] \\[x=\\frac{18}5\\]\nNow multiplying by $2$ , we get that the height is $\\frac{36}5$ , and $m+n=36+5=\\boxed{041}$" ]

[ "We need \\[abc = 2(ab+bc+ac) \\quad \\text{ or } \\quad (a-2)bc = 2a(b+c).\\] Since $a\\le b, ac \\le bc$ , from the first equation we get $abc \\le 6bc$ . Thus $a\\le 6$ . From the second equation we see that $a > 2$ . Thus $a\\in \\{3, 4, 5, 6\\}$\nThus, there are $5+3+1+1 = \\boxed{10}$ solutions.", "The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields\n\\[2(ab+bc+ca)=abc.\\]\nDivide both sides by $2abc$ to obtain \\[\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{2}.\\]\nFirst consider the bound of the variable $a$ . Since $\\frac{1}{a}<\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{2},$ we have $a>2$ , or $a\\geqslant3$\nAlso note that $c \\geq b \\geq a > 0$ , hence $\\frac{1}{a} \\geq \\frac{1}{b} \\geq \\frac{1}{c}$ .\nThus, $\\frac{1}{2}=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\leq \\frac{3}{a}$ , so $a \\leq 6$\nSo we have $a=3, 4, 5$ or $6$\nBefore the casework, let's consider the possible range for $b$ if $\\frac{1}{b}+\\frac{1}{c}=k>0$ . From $\\frac{1}{b}<k$ , we have $b>\\frac{1}{k}$ . From $\\frac{2}{b} \\geq \\frac{1}{b}+\\frac{1}{c}=k$ , we have $b \\leq \\frac{2}{k}$ . Thus $\\frac{1}{k}<b \\leq \\frac{2}{k}$\nWhen $a=3$ , we get $\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ $(3, 8, 24)$ $(3, 9, 18)$ $(3, 10, 15)$ $(3, 12, 12)$ , for a total of $5$ solutions.\nWhen $a=4$ , we get $\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{4}$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ $(4, 6, 12)$ $(4, 8, 8)$ , for a total of $3$ solutions.\nWhen $a=5$ , we get $\\frac{1}{b}+\\frac{1}{c}=\\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$\nWhen $a=6$ $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$\nThus, there are $5+3+1+1 = \\boxed{10}$ solutions." ]

[ "We need \\[abc = 2(ab+bc+ac) \\quad \\text{ or } \\quad (a-2)bc = 2a(b+c).\\] Since $a\\le b, ac \\le bc$ , from the first equation we get $abc \\le 6bc$ . Thus $a\\le 6$ . From the second equation we see that $a > 2$ . Thus $a\\in \\{3, 4, 5, 6\\}$\nThus, there are $5+3+1+1 = \\boxed{10}$ solutions.", "The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields\n\\[2(ab+bc+ca)=abc.\\]\nDivide both sides by $2abc$ to obtain \\[\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{2}.\\]\nFirst consider the bound of the variable $a$ . Since $\\frac{1}{a}<\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{2},$ we have $a>2$ , or $a\\geqslant3$\nAlso note that $c \\geq b \\geq a > 0$ , hence $\\frac{1}{a} \\geq \\frac{1}{b} \\geq \\frac{1}{c}$ .\nThus, $\\frac{1}{2}=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\leq \\frac{3}{a}$ , so $a \\leq 6$\nSo we have $a=3, 4, 5$ or $6$\nBefore the casework, let's consider the possible range for $b$ if $\\frac{1}{b}+\\frac{1}{c}=k>0$ . From $\\frac{1}{b}<k$ , we have $b>\\frac{1}{k}$ . From $\\frac{2}{b} \\geq \\frac{1}{b}+\\frac{1}{c}=k$ , we have $b \\leq \\frac{2}{k}$ . Thus $\\frac{1}{k}<b \\leq \\frac{2}{k}$\nWhen $a=3$ , we get $\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ $(3, 8, 24)$ $(3, 9, 18)$ $(3, 10, 15)$ $(3, 12, 12)$ , for a total of $5$ solutions.\nWhen $a=4$ , we get $\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{4}$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ $(4, 6, 12)$ $(4, 8, 8)$ , for a total of $3$ solutions.\nWhen $a=5$ , we get $\\frac{1}{b}+\\frac{1}{c}=\\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$\nWhen $a=6$ $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$\nThus, there are $5+3+1+1 = \\boxed{10}$ solutions." ]

[ "Because the unpainted part of the floor covers half the area, then the painted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$ . With this information we can make the equation:\n\\begin{eqnarray*} ab &=& 2\\left((a-2)(b-2)\\right) \\\\ ab &=& 2ab - 4a - 4b + 8 \\\\ ab - 4a - 4b + 8 &=& 0 \\end{eqnarray*} Applying Simon's Favorite Factoring Trick , we get\n\\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\\\ (a-4)(b-4) &=& 8 \\end{eqnarray*}\nSince $b > a$ , then we have the possibilities $(a-4) = 1$ and $(b-4) = 8$ , or $(a-4) = 2$ and $(b-4) = 4$ . This allows for 2 possibilities: $(5,12)$ or $(6,8)$ which gives us $\\boxed{2}$" ]

[ "The number of tiles the bug visits is equal to $1$ plus the number of times it crosses a horizontal or vertical line. As it must cross $16$ horizontal lines and $9$ vertical lines, it must be that the bug visits a total of $16+9+1 = \\boxed{26}$ squares.", "We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into $4$ sections, and just draw one of these $5$ feet by $8.5$ feet sections.\n\nThough it may appear that the line we drew comes very close to several points, we know that since $10$ and $17$ are relatively prime (numbers where the only positive integer that divides both of them is 1, a.k.a. numbers with a gcd of 1), the line will not actually pass through any of these points, so the total number of squares crossed will be the same regardless of which side we count. If we count the number of squares the line passes through using the diagram, we get $13$ squares. We can then multiply this by 2 to find out the total number of squares the bug passes through on the rectangular floor giving us a total of $2 \\cdot 13 = \\boxed{26}$" ]

[ "We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$ . To get the greatest area while keeping a perimeter of $160$ , the sides should all be $40$ . that means an area of $1600$ . Right now, the area is $20 \\times 60$ which is $1200$ $1600-1200=400$ which is $\\boxed{400}$" ]

[ "Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a $36$ m by $60$ m rectangle, so the $60$ m side should be parallel to the wall. That means the two fences perpendicular to the wall are $36$ m. We can start by counting $60\\div12+1$ on the $60$ m fence (since we also count the $0$ m post). Next, we have the two $36$ m fences. There are a total of $36\\div12+1-1$ fences on that side since the $0$ m and $60$ m fence posts are also part of the $36$ m fences. So we have $6+2\\cdot3=12$ minimum fence posts needed to box a $36$ m by $60$ m grazing area, or answer $\\boxed{12}$" ]

[ "Let the sides of the rectangular parking lot be $a$ and $b$ . Then $a^2 + b^2 = 625$ and $ab = 168$ . Add the two equations together, then factor. \\begin{align*} a^2 + 2ab + b^2 &= 625 + 168 \\times 2\\\\ (a + b)^2 &= 961\\\\ a + b &= 31 \\end{align*} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \\boxed{62}$", "We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy $a^2+b^2=25^2$ . In other words, $(a,b,25)$ is a set of Pythagorean triples. Guessing and checking, we have $(7,24,25)$ as the triplet, as the area is $7 \\cdot 24 = 168$ as requested. Therefore, the perimeter is $2(7+24)=\\boxed{62}$" ]

[ "In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is $8 \\times 10 = 80$ square inches. The height of the whole frame (including the photograph) would be $8+2+2 = 12$ , and the width of the whole frame, $10+2+2 = 14$ . Therefore, the area of the whole figure would be $12 \\times 14 = 168$ square inches. Subtracting the area of the photograph from the area of both the frame and photograph, we find the answer to be $168-80 = \\boxed{88}$" ]

[ "Each triangle has leg length $\\frac 12 \\cdot (25 - 15) = 5$ meters and area $\\frac 12 \\cdot 5^2 = \\frac {25}{2}$ square meters. Thus the flower beds have a total area of $25$ square meters. The entire yard has length $25$ m and width $5$ m, so its area is $125$ square meters. The fraction of the yard occupied by the flower beds is $\\frac {25}{125} = \\boxed{15}$" ]

[ "Each triangle has leg length $\\frac 12 \\cdot (25 - 15) = 5$ meters and area $\\frac 12 \\cdot 5^2 = \\frac {25}{2}$ square meters. Thus the flower beds have a total area of $25$ square meters. The entire yard has length $25$ m and width $5$ m, so its area is $125$ square meters. The fraction of the yard occupied by the flower beds is $\\frac {25}{125} = \\boxed{15}$" ]

[ "We multiply $z$ and $(\\frac{3}{4}+\\frac{3}{4}i)$ to get \\[(\\frac{3}{4}x-\\frac{3}{4}y)+(\\frac{3}{4}xi+\\frac{3}{4}yi).\\] Since we want to find the probability that this number is in $S$ , we need the real and complex coefficients of this number to be less than or equal to $1$ or greater than or equal to $-1.$ This gives us the equations \\[-1\\le \\frac{3}{4}x-\\frac{3}{4}y \\le 1\\] and \\[-1\\le \\frac{3}{4}x+\\frac{3}{4}y\\le 1.\\] Now, we see that we can solve this by graphing. We can graph our barriers $-1\\le x\\le 1$ and $-1\\le y\\le 1$ to form a $2$ by $2$ square centered at the origin. Graphing our two equations gives us the four lines \\[x-y=\\frac{4}{3},\\] \\[x-y=-\\frac{4}{3},\\] \\[x+y=\\frac{4}{3},\\] \\[x+y=-\\frac{4}{3}.\\] The square that is formed is the region that satisfies these four equations. Now, the barriers and this square gives us an octagon as the desired region. The area of this octagon is the total area of the square minus the 4 small triangles on each corner, each with $\\frac{2}{9}$ area. Therefore, the octagon has area of $\\frac{28}{9}.$ Finally, to find the probability of it working, we find the area of the octagon divided by the area of the entire square which is $\\frac{\\frac{28}{9}}{4}=\\frac{7}{9}$ or $\\boxed{79}.$" ]

[ "Since the side of the square is the diameter of the semicircle, the radius of the semicircle is $\\frac{1}{2}\\cdot\\frac{2}{\\pi}=\\frac{1}{\\pi}$\nSince the length of one of the semicircular arcs is half the circumference of the corresponding circle, the length of one arc is $\\frac{1}{2}\\cdot2\\cdot\\pi\\cdot\\frac{1}{\\pi}=1$\nSince the desired perimeter is made up of four of these arcs, the perimeter is $4\\cdot1=\\boxed{4}$" ]

[ "From consideration of a smaller regular polygon with an odd number of sides (e.g. a pentagon), we see that the lines of symmetry go through a vertex of the polygon and bisect the opposite side. Hence $L=15$ , the number of sides / vertices. The smallest angle for a rotational symmetry transforms one side into an adjacent side, hence $R = 360^\\circ / 15 = 24^\\circ$ , the number of degrees between adjacent sides. Therefore the answer is $L + R = 15 + 24 = \\boxed{39}$" ]

[ "$A_{\\triangle} = \\frac{s_t^2\\sqrt{3}}{4}$\n$A_{hex} = \\frac{6s_h^2\\sqrt{3}}{4}$ since a regular hexagon is just six equilateral triangles.\nSetting the areas equal, we get:\n$s_t^2 = 6s_h^2$\n$\\left(\\frac{s_t}{s_h}\\right)^2 = 6$\n$\\frac{s_t}{s_h} = \\sqrt{6}$ , and the answer is $\\boxed{6}$" ]

[ "To find the number of sides on the regular polygons that surround the decagon, we can find the interior angles and work from there. Knowing that the measure of the interior angle of any regular polygon is $\\frac{(n-2)*180}{n}$ , the measure of the decagon's interior angle is $\\frac{8*180}{10} = 144$ degrees.\nThe regular polygons meet at every vertex such that the angle outside of the decagon is divided evenly in two. With this, we know that the angle of the regular polygon is $216/2=108$ degrees. Using the previous formula, $n=5$ $\\boxed{5}$" ]

[ "Note that the distance from the center of the circle to each of the vertices of the inscribed regular polygon equals the radius $R$ . Since each side of a regular polygon is the same length, all the angles between the two lines from the center to the two vertices of a side is the same.\nThat means each of these angles between the two lines from the center to the two vertices of a side equals $\\frac{360}{n}$ degrees. Thus, the area of the polygon is \\[n \\cdot \\frac{1}{2}R^2\\sin\\left(\\frac{360}{n}^{\\circ}\\right) = 3R^2\\] Dividing both sides by $R^2$ yields \\[\\frac{n}{2}\\sin\\left(\\frac{360}{n}^{\\circ}\\right) = 3\\] Multiply both sides by $\\frac{2}{n}$ to get \\[\\sin\\left(\\frac{360}{n}^{\\circ}\\right) = \\frac{6}{n}\\] At this point, use trial-and-error for each of the answer choices. When checking $n = 12$ , the equation results in $\\sin(30^{\\circ}) = \\frac{1}{2}$ , which is correct. Thus, the answer is $\\boxed{12}$" ]

[ "Let $m$ be the number of main courses the restaurant serves, so $2m$ is the number of appetizers. Then the number of dinner combinations is $2m\\times m\\times3=6m^2$ . Since the customer wants to eat a different dinner in all $365$ days of $2003$ , we must have\n\\begin{align*} 6m^2 &\\geq 365\\\\ m^2 &\\geq 60.83\\ldots.\\end{align*} Also, year 2003 is not a leap year, because 2003 divided by 4 does not equal an integer.\nThe smallest integer value that satisfies this is $\\boxed{8}$", "Let $m$ denote the number of main courses needed to meet the requirement. Then the number of dinners available is $3\\cdot m \\cdot 2m = 6m^2$ . Thus $m^2$ must be at least $365/6 \\approx 61$ . Since $7^2 = 49<61<64 = 8^2$ $\\boxed{8}$", "Let $x$ be the number of main courses and let $2x$ be the number of appetizers.\nSince there are 3 desserts, the number of possible dinner choices would be $2x \\cdot x \\cdot 3 = 6x^2$ for any number $x$ . Since a year has $365$ days, we can assume that: \\begin{align*} 6x^2 \\ge 365 \\end{align*} \\begin{align*} x^2 \\ge 61 \\end{align*} \\begin{align*} x \\ge 7.8 \\end{align*} The least option that is greater than $7.8$ is $8$ , so the answer is $\\boxed{8}$" ]

[ "Note Euler's formula where $\\text{Vertices}+\\text{Faces}-\\text{Edges}=2$ . There are $12$ faces and the number of edges is $24$ because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are $14$ vertices on the figure. Now note that the sum of the degrees of all the points is twice the number of edges. Let $x=$ the amount of vertices with $3$ edges. Now we know $\\frac{3x+4(14-x)}{2}=24$ . Solving this system of equations gives $x = 8$ so the answer is $\\boxed{8}$ .\n~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)", "Let $x$ be the number of vertices with three edges, and $y$ be the number of vertices with four edges. Since there are $\\frac{4*12}{2}=24$ edges on the polyhedron, we can see that $\\frac{3x+4y}{2}=24$ . Then, $3x+4y=48$ . Notice that by testing the answer choices, (D) is the only one that yields an integer solution for $y$ . Thus, the answer is $\\boxed{8}$", "With $12$ rhombi, there are $4\\cdot12=48$ total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have $\\dfrac{48}2=24$ total edges.\nLet $A$ be the number of vertices with $3$ edges (this is what the problem asks for) and $B$ be the number of vertices with $4$ edges. We have $3A + 4B = 48$\nEuler's formula states that, for all convex polyhedra, $V-E+F=2$ . In our case, $V-24+12=2\\implies V=14.$ We know that $A+B$ is the total number of vertices as we are given that all vertices are connected to either $3$ or $4$ edges. Therefore, $A+B=14.$\nWe now have a system of two equations. There are many ways to solve for $A$ ; choosing one yields $A=\\boxed{8}$", "Note that Euler's formula is $V+F-E=2$ . We know $F=12$ from the question. We also know $E = \\frac{12 \\cdot 4}{2} = 24$ because every face has $4$ edges and every edge is shared by $2$ faces. We can solve for the vertices based on this information.\nUsing the formula we can find: \\[V + 12 - 24 = 2\\] \\[V = 14\\] Let $t$ be the number of vertices with $3$ edges and $f$ be the number of vertices with $4$ edges. We know $t+f = 14$ from the question and $3t + 4f = 48$ . The second equation is because the total number of points is $48$ because there are 12 rhombuses of $4$ vertices.\nNow, we just have to solve a system of equations. \\[3t + 4f = 48\\] \\[3t + 3f = 42\\] \\[f = 6\\] \\[t = 8\\] Our answer is simply just $t$ , which is $\\boxed{8}$ ~musicalpenguin", "Each of the twelve rhombi has two pairs of angles across from each other that must be congruent. If both pairs of angles occur at $4$ -point intersections, we have a grid of squares. If both occur at $3$ -point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a $3$ -point intersection and two at a $4$ -point intersection.\nSince each $3$ -point intersection has $3$ adjacent rhombuses, we know the number of $3$ -point intersections must equal the number of $3$ -point intersections per rhombus times the number of rhombuses over $3$ . Since there are $12$ rhombuses and two $3$ -point intersections per rhombus, this works out to be:\n$\\frac{2\\cdot12}{3}$\nHence: $\\boxed{8}$ ~hollph27\n~Minor edits by FutureSphinx", "Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen here ), thus there are 6 4-vertices (corresponding to each face of the cube) and 8 3-vertices (corresponding to each corner of the cube). Thus the answer is $\\boxed{8}$", "Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has $8$ triangular faces, which correspond to $\\boxed{8}$ vertices on a rhombic dodecahedron that have $3$ edges." ]

[ "Let $m$ be the number of $4$ -edge vertices, and $n$ be the number of $3$ -edge vertices. The total number of vertices is $m+n$ . Now, we know that there are $4 \\cdot 12 = 48$ vertices, but we have overcounted. We have overcounted $m$ vertices $3$ times and overcounted $n$ vertices $2$ times. Therefore, we subtract $3m$ and $2n$ from $48$ and set it equal to our original number of vertices. \\[48 - 3m - 2n = m+n\\] \\[4m + 3n = 48\\] From here, we reduce both sides modulo $4$ . The $4m$ disappears, and the left hand side becomes $3n$ . The right hand side is $0$ , meaning that $3n$ must be divisible by $4$ . Looking at the answer choices, this is only possible for $n = \\boxed{8}$" ]

[ "The path is a circle with radius equal to the slant height of the cone, which is $\\sqrt {r^{2} + h^{2}}$ . Thus, the length of the path is $2\\pi\\sqrt {r^{2} + h^{2}}$\nAlso, the length of the path is 17 times the circumference of the base, which is $34r\\pi$ . Setting these equal gives $\\sqrt {r^{2} + h^{2}} = 17r$ , or $h^{2} = 288r^{2}$ . Thus, $\\dfrac{h^{2}}{r^{2}} = 288$ , and $\\dfrac{h}{r} = 12\\sqrt {2}$ , giving an answer of $12 + 2 = \\boxed{014}$" ]

[ "We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases.\nCase 1: vertices are on one base. Then we can call one of the vertices $A$ for distinction. Either the triangle can have sides $1, 1, \\sqrt{3}$ with 6 cases or $\\sqrt{3}, \\sqrt{3}, \\sqrt{3}$ with 2 cases. This can be repeated on the other base for $16$ cases.\nCase 2: The vertices span two bases. WLOG call the only vertex on one of the bases $X$ . Call the closest vertex on the other base $B$ , and label clockwise $C, D, E, F, G$ . We will multiply the following scenarios by $12$ , because the top vertex can have $6$ positions and the top vertex can be on the other base. We can have $XCG, XDF$ , but we are not done! Don't forget that the problem statement implies that the longest diagonal in a base is $2$ and the height is $2$ , so $XBE$ is also correct! Those are the only three cases, so there are $12*3=36$ cases for this case.\nIn total there's $\\boxed{052}$ cases.", "If there are two edges on a single diameter, there would be six diameters. There are four ways to put the third number, and four equilateral triangles. There are $4+ 2 \\cdot 2\\cdot 6 = 28$ ways. Then, if one length was $\\sqrt{3}$ but no side on the diameter, there would be twelve was to put the $\\sqrt{3}$ side, and two ways to put the other point. $2 \\cdot 12 = 24$ for four ways to put the third point. Adding the number up, the final answer is $24+28 = \\boxed{052}.$", "To start, let's find the distances from any vertex (call it A, doesn't matter since the prism is symmetrical) to all the other 11 vertices. Using Pythagorean relations, we find that the distances are $1$ $\\sqrt{3}$ $2$ $\\sqrt{3}$ $1$ $\\sqrt{7}$ $\\sqrt{5}$ $2$ $\\sqrt{5}$ $\\sqrt{7}$ , and $\\sqrt{8}$\nWe can clearly form an isosceles triangle using any vertex and any two neighboring edges that are equal. There are 12 total vertices, all of which are symmetrical to the vertex A. Thus, for each vertex, we can form 5 isosceles triangles (with neighboring edges $1$ $1$ $2$ $2$ $\\sqrt{3}$ $\\sqrt{3}$ $\\sqrt{5}$ $\\sqrt{5}$ , and $\\sqrt{7}$ $\\sqrt{7}$ ) , so have so far $12 \\times 5 = 60$ isosceles triangles.\nTo check for overcounting, note that isosceles triangles can be split into two major categories: equilateral isosceles, and non-equilateral isosceles. \nWe only counted non-equilateral isosceles triangles once since there is only one vertex whose two neighboring edges are equal. \nBut equilateral triangles were counted three times (namely once for each vertex).\nBy observation, there are only 4 equilateral triangles (1-1-1 side lengths), two on each hexagonal face. Since we counted each two more times than we should of (three instead of once), we will subtract ( $4$ eq. triangles) $\\times$ $2$ overcounts per eq. triangle) = $8$ overcounts.\nFinally, we have $60 - 8$ overcounts = $\\boxed{052}$ isoceles triangles.", "Start by drawing the hexagonal prism. Start from one of the points, let's call it $A$ . From simple inspection, we can see that each point we choose creates $4$ isosceles triangles; $A$ and the two adjacent points, $A$ and the two points adjacent to those, $A$ and the adjacent point on the other face, and $A$ and the two points adjacent to those. Here, we have $12\\cdot4\\Rightarrow48$ isosceles triangles.\nNext, notice that you can create an isosceles triangle with the bases of the hexagonal prism. Let's say that our base, starting from the very left-most point of the hexagon, has points $A, B, C, D, E, F$ , rotating counter-clockwise. We can create a triangle with either points $\\bigtriangleup{BCE}$ or $\\bigtriangleup{BCF}$ . WLOG, assuming we use $\\bigtriangleup{BCE}$ , we can use the Law of Cosines and find that $\\overline{CE}$ $\\sqrt{3}$ . Thus, this would mean $\\overline{CF}$ is $2$ , which is the exactly the same length as the height of the hexagonal prism. There are $4$ possibilities of this case, so our final solution is $48+4=\\boxed{052}$" ]

[ "Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid. Then $\\angle CAB = 120^\\circ$ . Let $X$ be the foot of the altitude from $A$ to $\\overline{BC}$ . Then since $\\triangle ABX$ is a $30-60-90$ triangle, $AX = 6$ . Since the dihedral angle between $\\triangle ABC$ and $\\triangle BCD$ is $60^\\circ$ $\\triangle AXD$ is a $30-60-90$ triangle and $AD = 6\\sqrt{3} = h$ . Thus $h^2 = \\boxed{108}$", "Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid. Notice that we can already find some lengths. We have $AB=AC=12$ (given) and $BC=BD=\\sqrt{144+h^2}$ by the Pythagorean Theorem. Let $M$ be the midpoint of $BC$ . Then, we have $AM=6$ $30-60-90$ ) triangles and $DM=\\sqrt{36+h^2}$ by the Pythagorean Theorem. Applying the Law of Cosines, since $\\angle AMD=60^{\\circ}$ , we get \\[h^2=36+h^2+36-\\frac12 \\cdot 12 \\sqrt{36+h^2} \\implies h^2=\\boxed{108},\\] as desired." ]

[ "Let $P'$ be the prism similar to $P$ , and let the sides of $P'$ be of length $x,y,z$ , such that $x \\le y \\le z$ . Then\n\\[\\frac{x}{a} = \\frac{y}{b} = \\frac zc < 1.\\]\nNote that if the ratio of similarity was equal to $1$ , we would have a prism with zero volume. As one face of $P'$ is a face of $P$ , it follows that $P$ and $P'$ share at least two side lengths in common. Since $x < a, y < b, z < c$ , it follows that the only possibility is $y=a,z=b=1995$ . Then,\n\\[\\frac{x}{a} = \\frac{a}{1995} = \\frac{1995}{c} \\Longrightarrow ac = 1995^2 = 3^25^27^219^2.\\]\nThe number of factors of $3^25^27^219^2$ is $(2+1)(2+1)(2+1)(2+1) = 81$ . Only in $\\left\\lfloor \\frac {81}2 \\right\\rfloor = 40$ of these cases is $a < c$ (for $a=c$ , we end with a prism of zero volume). We can easily verify that these will yield nondegenerate prisms, so the answer is $\\boxed{040}$" ]

[ "The surface area of this right rectangular prism is $2(\\log_{2}x\\log_{3}x+\\log_{2}x\\log_{4}x+\\log_{3}x\\log_{4}x).$\nThe volume of this right rectangular prism is $\\log_{2}x\\log_{3}x\\log_{4}x.$\nEquating the numerical values of the surface area and the volume, we have \\[2(\\log_{2}x\\log_{3}x+\\log_{2}x\\log_{4}x+\\log_{3}x\\log_{4}x)=\\log_{2}x\\log_{3}x\\log_{4}x.\\] Dividing both sides by $\\log_{2}x\\log_{3}x\\log_{4}x,$ we get \\[2\\left(\\frac{1}{\\log_{4}x}+\\frac{1}{\\log_{3}x}+\\frac{1}{\\log_{2}x}\\right)=1. \\hspace{15mm} (\\bigstar)\\] Recall that $\\log_{b}a=\\frac{1}{\\log_{a}b}$ and $\\log_{b}\\left(a^n\\right)=n\\log_{b}a,$ so we rewrite $(\\bigstar)$ as \\begin{align*} 2(\\log_{x}4+\\log_{x}3+\\log_{x}2)&=1 \\\\ 2\\log_{x}24&=1 \\\\ \\log_{x}576&=1 \\\\ x&=\\boxed{576} ~MRENTHUSIASM" ]

[ "Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\\frac{6}{\\sqrt{2}} = 3\\sqrt{2}$ ). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$ . Because of the symmetrical property of the pyramid,\nwe can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.\nSince the volume is $54 = \\frac{1}{3} \\cdot S \\cdot h = \\frac{1}{3} \\cdot 6^2 \\cdot h$ , where $h=\\frac{9}{2}$ is the height of this pyramid, we have: $l^2=\\left(\\frac{9}{2}-l\\right)^2+\\left(3\\sqrt{2}\\right)^2$ according to the Pythagorean theorem.\nSolve this equation will give us $l = \\frac{17}{4},$ therefore $m+n=\\boxed{021}.$", "To start, we find the height of the pyramid. By the volume of a pyramid formula, we have \\[\\frac13 \\cdot 6^2 \\cdot h=54 \\implies h=\\frac92.\\] Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is $\\frac12 \\cdot 6\\sqrt2=3\\sqrt2$ , we have \\[x=\\sqrt{\\left(\\frac92\\right)^2+(3\\sqrt2)^2}=\\sqrt{\\frac{153}4}=\\frac{3\\sqrt{17}}2.\\] Taking the cross section of the pyramid and transforming the problem into $2$ -d, it suffices to find the radius of the circumcircle of a triangle of side lengths $\\frac{3\\sqrt{17}}2$ $\\frac{3\\sqrt{17}}2$ $6\\sqrt2$ . This turns out to be easy by the formula $R=\\frac{abc}{4A}$ , and through computing this value (the work has been left out) we find that $R=\\frac{17}4$ , so our answer is $\\boxed{021}$", "By the volume of a pyramid formula, we have that the height of the pyramid is $\\frac{9}{2}$ . Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the $xy$ plane. Then, the vertices of the base would be $(3,3,0), (3,-3,0), (-3,3,0), (-3,-3,0)$ in some order. Also, let the vertex be $(0,0,\\frac{9}{2})$ . Recall that the formula for a sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ where the center is $(a,b,c)$ and the radius is $r$ . Symmetry gives that $a=b=0$ . Plug in $(3,3,0)$ and $(0,0,\\frac{9}{2})$ and you get the system of equation\n$18+c^2=r^2$\n$(\\frac{9}{2}-c)^2=r^2$\nSolving gives $c=1/4$ and $r=17/4$ , so our answer is $17+4=\\boxed{021}$ .~ Ddk001" ]

[ "The list starts with $12345$ . There are $\\binom{8}{4} = 70$ four-digit rising numbers that do not begin with $1$ , and thus also $70$ five digit rising numbers that do begin with $1$ that are formed by simply putting a $1$ before the four digit number.\nThus, the $71^{\\text{st}}$ number is $23456$ . There are $\\binom{6}{3} = 20$ three-digit rising numbers that do not begin with a $1,2$ or $3$ , and thus $20$ five digit rising numbers that begin with a $23$\nThus, the $91^{\\text{st}}$ number is $24567$ . Counting up, $24568, 24569, 24578, 24579, 24589, 24678$ is the $97^{\\text{th}}$ number, which does not contain the digit $\\boxed{5}$" ]

[ "The roll of tape is $1/0.015=$ 66 layers thick. In order to find the total length, we have to find the average of each concentric circle and multiply it by $66$ . Since the diameter of the small circle is $2$ inches and the diameter of the large one is $4$ inches, the \"middle value\" is $3$ . Therefore, the average circumference is $3\\pi$ . Multiplying $3\\pi \\cdot 66$ gives $(B) \\boxed{600}$", "There are about $\\dfrac{1}{0.015}=\\dfrac{200}{3}$ \"full circles\" of tape, and with average circumference of $\\dfrac{4+2}{2}\\pi=3\\pi.$ $\\dfrac{200}{3} \\cdot 3\\pi=200\\pi,$ which means the answer is $\\boxed{600}$", "The volume of the tape is always the same, but we can either calculate it when the tape is unrolled as a really long, thin rectangular prism, or we can calculate it as a cylinder with a hole cut out of it. When we calculate it as a long rectangular prism, we can say that the length is $X$ (this is what the problem wants!) and the width is $Y$ . Then, the volume is, of course, $0.015 \\cdot X \\cdot Y.$ Now, notice that the \"width\" of our rectangular prism is also the \"height\" of our cylinder with a hole cut out of it. Then, we can calculate the volume as base times height, or in this case, $3\\pi \\cdot Y.$ Now, since the volume always stays the same, we know that $3\\pi \\cdot Y = 0.015 \\cdot X \\cdot Y.$ Cancelling the $Y$ 's give us an equation for $X$ , and if we approximate $\\pi$ as $3$ , then $X = \\boxed{600}$ . Yay!", "If you cannot notice that the average diameter is $3$ , you can still solve this problem by the following method.\nThe same with solution 1, we have $\\frac{1000}{0.015}$ layers of tape. If we consider every layers with the diameter $2$ , the length should be $\\frac{1000}{0.015}2\\pi\\approx 400$ . If the diameter is seem as $4$ , the length should be $800$ . So, the length is between $400$ and $800$ , the only possible answer is $\\boxed{600}$" ]

[ "Let the length of the inner rectangle be $x$\nThen the area of that rectangle is $x\\cdot1 = x$\nThe second largest rectangle has dimensions of $x+2$ and $3$ , making its area $3x+6$ . The area of the second shaded area, therefore, is $3x+6-x = 2x+6$\nThe largest rectangle has dimensions of $x+4$ and $5$ , making its area $5x + 20$ . The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is $(5x+20) - (3x+6) = 2x + 14$\nThe problem states that $x, 2x+6, 2x+14$ is an arithmetic progression, meaning that the terms in the sequence increase by the same amount each term.\nTherefore, $(2x+6) - (x) = (2x+14) - (2x+6)\\implies x+6 = 8\\implies x =2\\implies \\boxed{2}$" ]

[ "The minimum range occurs in the set $\\{7,7,12,12,12\\}$ , so the answer is $\\boxed{5}$" ]

[ "Let the mode be $x$ , which we let appear $n > 1$ times. We let the arithmetic mean be $M$ , and the sum of the numbers $\\neq x$ be $S$ . Then \\begin{align*} D &= \\left|M-x\\right| = \\left|\\frac{S+xn}{121}-x\\right| = \\left|\\frac{S}{121}-\\left(\\frac{121-n}{121}\\right)x\\right| \\end{align*} As $S$ is essentially independent of $x$ , it follows that we wish to minimize or maximize $x$ (in other words, $x \\in [1,1000]$ ). Indeed, $D(x)$ is symmetric about $x = 500.5$ ; consider replacing all of numbers $x_i$ in the sample with $1001-x_i$ , and the value of $D$ remains the same. So, without loss of generality , let $x=1$ . Now, we would like to maximize the quantity\n$S$ contains $121-n$ numbers that may appear at most $n-1$ times. Therefore, to maximize $S$ , we would have $1000$ appear $n-1$ times, $999$ appear $n-1$ times, and so forth. We can thereby represent $S$ as the sum of $n-1$ arithmetic series of $1000, 999, \\ldots, 1001 - \\left\\lfloor \\frac{121-n}{n-1} \\right\\rfloor$ . We let $k = \\left\\lfloor \\frac{121-n}{n-1} \\right\\rfloor$ , so\nwhere $R(n)$ denotes the sum of the remaining $121-(n-1)k$ numbers, namely $R(n) = (121-(n-1)k)(1000-k)$\nAt this point, we introduce the crude estimate that $k=\\frac{121-n}{n-1}$ , so $R(n) = 0$ and \\begin{align*}2S+2n &= (121-n)\\left(2001-\\frac{121-n}{n-1}\\right)+2n = (120-(n-1))\\left(2002-\\frac{120}{n-1}\\right)\\end{align*} Expanding (ignoring the constants, as these do not affect which $n$ yields a maximum) and scaling, we wish to minimize the expression $5(n-1) + \\frac{36}{n-1}$ . By AM-GM , we have $5(n-1) + \\frac{36}{n-1} \\ge 2\\sqrt{5(n-1) \\cdot \\frac{36}{n-1}}$ , with equality coming when $5(n-1) = \\frac{36}{n-1}$ , so $n-1 \\approx 3$ . Substituting this result and some arithmetic gives an answer of $\\boxed{947}$", "With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of numbers as $\\underbrace{1,1,1...1,}_\\text{n times}\\underbrace{2,2,2...2,}_\\text{n times}\\underbrace{3,3,3...3,}_\\text{n times}........\\underbrace{1000,1000,1000....}_\\text{n+1 times}$ And, we need to find the value of n that makes the sum as low as possible. And, we can create a formula to make it easier. It isn't hard to find the sum. The numbers which are not 1000, average to $\\frac{120}{2n}$ or $\\frac{60}{n}$ , and there are $120-n$ of them. So, they sum to $\\frac{60}{n}(120-n)$ . And, the sum of the numbers that are 1000 is $1000(n+1)$ so, our total sum gets us $1000(n+1)+120/2n(120-n)$ We want to minimize it, since the mode will always be 1000. And, testing the values n = 1, n = 2, n = 3, n = 4, we get these results.\n$n = 1: 2000+60*119 = 9140$\n$n = 2: 3000+30*118 = 6540$\n$n = 3: 4000+20*117 = 6340$\n$n = 4: 5000+15*116 = 6740$\nAnd, as n grows larger and larger from 4, the values will start increasing. Thus, the lowest possible sum is 6340. Dividing by 121, the lowest possible mean is 52.396...., and thus, the highest possible value of $D$ is 947.604, and the floor of that is $\\boxed{947}$" ]

[ "\nImagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\\boxed{1022.}$", "\\[\\begin{tabular}{|c|c|c|c|c|c|c|} \\hline T & T & T & X & T & T & T \\\\ \\hline T & T & T & Y& T & T & T \\\\ \\hline T & T & T & Z & T & T & T \\\\ \\hline X & Y & Z & W & Z & Y & X \\\\ \\hline T & T & T & Z & T & T & T \\\\ \\hline T & T & T & Y & T & T & T \\\\ \\hline T & T & T & X & T & T & T \\\\ \\hline \\end{tabular}\\]\nThere are $3 \\times 3$ squares in the corners of this $7 \\times 7$ square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)\n\\[\\begin{tabular}{|c|c|c|} \\hline A & B & C \\\\ \\hline B & D & E \\\\ \\hline C & E & F \\\\ \\hline \\end{tabular}\\] (Note that coloring one $3 \\times 3$ square will also determine the colorings of the other $3$ because the large $7 \\times 7$ square must look the same when it is rotated by $90^\\circ$\nThere are $6$ different letters and $2$ choices of color (black and white) for each letter, so there are $2^6=64$ colorings of a proper $3 \\times 3$ square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are $4$ different letters, so there are $2^4=16$ different colorings. Multiplying them together gives $16 \\times 64 = 1024$ . Going back to the question, we see that \"there must be at least one square of each color in this grid of $49$ squares.\" We must then eliminate $2$ options: an all-black grid and an all-white grid. $1024-2= \\boxed{1022}$ -hansenhe" ]

[ "\nImagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\\boxed{1022.}$", "\\[\\begin{tabular}{|c|c|c|c|c|c|c|} \\hline T & T & T & X & T & T & T \\\\ \\hline T & T & T & Y& T & T & T \\\\ \\hline T & T & T & Z & T & T & T \\\\ \\hline X & Y & Z & W & Z & Y & X \\\\ \\hline T & T & T & Z & T & T & T \\\\ \\hline T & T & T & Y & T & T & T \\\\ \\hline T & T & T & X & T & T & T \\\\ \\hline \\end{tabular}\\]\nThere are $3 \\times 3$ squares in the corners of this $7 \\times 7$ square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)\n\\[\\begin{tabular}{|c|c|c|} \\hline A & B & C \\\\ \\hline B & D & E \\\\ \\hline C & E & F \\\\ \\hline \\end{tabular}\\] (Note that coloring one $3 \\times 3$ square will also determine the colorings of the other $3$ because the large $7 \\times 7$ square must look the same when it is rotated by $90^\\circ$\nThere are $6$ different letters and $2$ choices of color (black and white) for each letter, so there are $2^6=64$ colorings of a proper $3 \\times 3$ square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are $4$ different letters, so there are $2^4=16$ different colorings. Multiplying them together gives $16 \\times 64 = 1024$ . Going back to the question, we see that \"there must be at least one square of each color in this grid of $49$ squares.\" We must then eliminate $2$ options: an all-black grid and an all-white grid. $1024-2= \\boxed{1022}$ -hansenhe" ]

[ "The formula for expected values is \\[\\text{Expected Value}=\\sum(\\text{Outcome}\\cdot\\text{Probability}).\\] We have \\begin{align*} t &= 50\\cdot\\frac15 + 20\\cdot\\frac15 + 20\\cdot\\frac15 + 5\\cdot\\frac15 + 5\\cdot\\frac15 \\\\ &= (50+20+20+5+5)\\cdot\\frac15 \\\\ &= 100\\cdot\\frac15 \\\\ &= 20, \\\\ s &= 50\\cdot\\frac{50}{100} + 20\\cdot\\frac{20}{100} + 20\\cdot\\frac{20}{100} + 5\\cdot\\frac{5}{100} + 5\\cdot\\frac{5}{100} \\\\ &= 25 + 4 + 4 + 0.25 + 0.25 \\\\ &= 33.5. \\end{align*} Therefore, the answer is $t-s=\\boxed{13.5}.$" ]

[ "The formula for expected values is \\[\\text{Expected Value}=\\sum(\\text{Outcome}\\cdot\\text{Probability}).\\] We have \\begin{align*} t &= 50\\cdot\\frac15 + 20\\cdot\\frac15 + 20\\cdot\\frac15 + 5\\cdot\\frac15 + 5\\cdot\\frac15 \\\\ &= (50+20+20+5+5)\\cdot\\frac15 \\\\ &= 100\\cdot\\frac15 \\\\ &= 20, \\\\ s &= 50\\cdot\\frac{50}{100} + 20\\cdot\\frac{20}{100} + 20\\cdot\\frac{20}{100} + 5\\cdot\\frac{5}{100} + 5\\cdot\\frac{5}{100} \\\\ &= 25 + 4 + 4 + 0.25 + 0.25 \\\\ &= 33.5. \\end{align*} Therefore, the answer is $t-s=\\boxed{13.5}.$" ]

[ "We will show that $22$ $11$ $22$ $44$ , and $22$ meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height $11$ meters, we can deduce that Trees 1 and 3 both have a height of $22$ meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of $11$ and $22$ $44$ and $88$ , or $44$ and $22$ . Checking each of these, in the first case, the average is $17.6$ meters, which doesn't end in $.2$ as the problem requires. Therefore, we consider the other cases. With $44$ and $88$ , the average is $37.4$ meters, which again does not end in $.2$ , but with $44$ and $22$ , the average is $24.2$ meters, which does. Consequently, the answer is $\\boxed{24.2}$", "Notice the average height of the trees ends with $0.2$ ; therefore, the sum of all five heights of the trees must end with $1$ or $6$ . ( $0.2$ $\\cdot$ $5$ $1$ ) \nWe already know Tree $2$ is $11$ meters tall. Both Tree $1$ and Tree $3$ must $22$ meters tall - since neither can be $5.5$ . \nOnce again, apply our observation for solving for the Tree $4$ 's height. Tree $4$ can't be $11$ meters for the sum of the five tree heights to still end with $1$ . Therefore, the Tree $4$ is $44$ meters tall. \nNow, Tree $5$ can either be $22$ or $88$ . Find the average height for both cases of Tree $5$ . Doing this, we realize the Tree $5$ must be $22$ for the average height to end with $0.2$ and that the average height is $\\boxed{24.2}$", "As in Solution 1, we shall show that the heights of the trees are $22$ $11$ $22$ $44$ , and $22$ meters. Let $S$ be the sum of the heights, so that the average height will be $\\frac{S}{5}$ meters. We note that $0.2 = \\frac{1}{5}$ , so in order for $\\frac{S}{5}$ to end in $.2$ $S$ must be one more than a multiple of $5$ . Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both $22$ meters. At this point, our table looks as follows: \\[\\begingroup \\setlength{\\tabcolsep}{10pt} \\renewcommand{\\arraystretch}{1.5} \\begin{tabular}{|c|c|} \\hline Tree 1 & 22 meters \\\\ Tree 2 & 11 meters \\\\ Tree 3 & 22 meters \\\\ Tree 4 & \\rule{0.5cm}{0.15mm} meters \\\\ Tree 5 & \\rule{0.5cm}{0.15mm} meters \\\\ \\hline Average height & \\rule{0.5cm}{0.15mm}\\text{ .}2 meters \\\\ \\hline \\end{tabular} \\endgroup\\]\nIf Tree 4 now has a height of $11$ , then Tree 5 would need to have height $22$ , but in that case $S$ would equal $88$ , which is not $1$ more than a multiple of $5$ . So we instead take Tree 4 to have height $44$ . Then the sum of the heights of the first 4 trees is $22+11+22+44 = 99$ , so using a height of $22$ for Tree 5 gives $S=121$ , which is $1$ more than a multiple of $5$ (whereas $88$ gives $S = 187$ , which is not). Thus the average height of the trees is $\\frac{121}{5} = \\boxed{24.2}$ meters.", "Since we know that the tree heights have to be integers, then it is immediate that Tree 1 and 3 have a height of $22$ . Now using the information given by the last column (that the average of the heights of the trees ends in $.2$ ), we can tell that the sum of all the heights of the trees ends in either $1$ or $6$ , because those are the only numbers from $0$ to $9$ that are congruent to $1$ after taking modulo $5$ . The two multiples of eleven (eleven because all of the tree heights have to be a multiple of eleven if they are integers) that come to mind are $66$ and $121$ . Since the sum of the heights of Tree 1, 2, and 3 is already $55$ , we know that $66$ is impossible to obtain. Then, we can decide with relative confidence that the answer should be $\\frac{121}{5} = \\boxed{24.2}$", "Since her average ends with 0.2, the sum of her tree heights must be $\\equiv 1 \\pmod {5}$ .\nIf Tree 1's height is $a$ , Tree 3's is $b$ , Tree 4's is $c$ , and Tree 5's is $d$ , then we $a$ and $b$ are both 22, since all of the tree heights are integers.\nNow we have $22+11+22+c+d \\equiv 1 \\pmod {5}$ . Simplifying, we get $c+d \\equiv 1 \\pmod {5}$ .\nThe only possible combination of $c$ and $d$ that abides the condition are 44 and 22, respectively.\nAdding these up, we get 121, and $\\frac{121}{5}$ is $\\boxed{24.2}$" ]

[ "\\begin{align*} \\mbox{Expenses} &= 1000 \\cdot \\frac25 = 400 \\\\ \\mbox{Revenue} &= 1000 \\cdot \\frac12 = 500 \\\\ \\mbox{Profit} &= \\mbox{Revenue} - \\mbox{Expenses} = 500-400 = \\boxed{100} Note: Revenue is a gain.", "Note that the troop buys $10$ candy bars at a price of $4$ dollars and sells $10$ bars at a price of $5$ dollars. So the troop gains $1$ dollar for every $10$ bars. So therefore we divide $1000 \\div 10 = 100$ . So our answer is $\\boxed{100}$ .\n~HyperVoid" ]

[ "\\begin{align*} \\mbox{Expenses} &= 1000 \\cdot \\frac25 = 400 \\\\ \\mbox{Revenue} &= 1000 \\cdot \\frac12 = 500 \\\\ \\mbox{Profit} &= \\mbox{Revenue} - \\mbox{Expenses} = 500-400 = \\boxed{100} Note: Revenue is a gain.", "Note that the troop buys $10$ candy bars at a price of $4$ dollars and sells $10$ bars at a price of $5$ dollars. So the troop gains $1$ dollar for every $10$ bars. So therefore we divide $1000 \\div 10 = 100$ . So our answer is $\\boxed{100}$ .\n~HyperVoid" ]

[ "The maximum salary for a single player occurs when the other 20 players receive the minimum salary. The total of all players' salaries is 700000. The answer is $700000-15000*20=400000\\Rightarrow \\boxed{400,000}$" ]

[ "We say the sequence $(a_n)$ completes at $i$ if $i$ is the minimal positive integer such that $a_i = a_{i + 1} = 1$ . Otherwise, we say $(a_n)$ does not complete.\nNote that if $d = \\gcd(999, a_2) \\neq 1$ , then $d|a_n$ for all $n \\geq 1$ , and $d$ does not divide $1$ , so if $\\gcd(999, a_2) \\neq 1$ , then $(a_n)$ does not complete. (Also, $a_{2006}$ cannot be 1 in this case since $d$ does not divide $1$ , so we do not care about these $a_2$ at all.)\nFrom now on, suppose $\\gcd(999, a_2) = 1$\nWe will now show that $(a_n)$ completes at $i$ for some $i \\leq 2006$ . We will do this with 3 lemmas.\nLemma: If $a_j \\neq a_{j + 1}$ , and neither value is $0$ , then $\\max(a_j, a_{j + 1}) > \\max(a_{j + 2}, a_{j + 3})$\nProof: There are 2 cases to consider.\nIf $a_j > a_{j + 1}$ , then $a_{j + 2} = a_j - a_{j + 1}$ , and $a_{j + 3} = |a_j - 2a_{j + 1}|$ . So $a_j > a_{j + 2}$ and $a_j > a_{j + 3}$\nIf $a_j < a_{j + 1}$ , then $a_{j + 2} = a_{j + 1} - a_j$ , and $a_{j + 3} = a_j$ . So $a_{j + 1} > a_{j + 2}$ and $a_{j + 1} > a_{j + 3}$\nIn both cases, $\\max(a_j, a_{j + 1}) > \\max(a_{j + 2}, a_{j + 3})$ , as desired.\nLemma: If $a_i = a_{i + 1}$ , then $a_i = 1$ . Moreover, if instead we have $a_i = 0$ for some $i > 2$ , then $a_{i - 1} = a_{i - 2} = 1$\nProof: By the way $(a_n)$ is constructed in the problem statement, having two equal consecutive terms $a_i = a_{i + 1}$ implies that $a_i$ divides every term in the sequence. So $a_i | 999$ and $a_i | a_2$ , so $a_i | \\gcd(999, a_2) = 1$ , so $a_i = 1$ . For the proof of the second result, note that if $a_i = 0$ , then $a_{i - 1} = a_{i - 2}$ , so by the first result we just proved, $a_{i - 2} = a_{i - 1} = 1$\nLemma: $(a_n)$ completes at $i$ for some $i \\leq 2000$\nProof: Suppose $(a_n)$ completed at some $i > 2000$ or not at all. Then by the second lemma and the fact that neither $999$ nor $a_2$ are $0$ , none of the pairs $(a_1, a_2), ..., (a_{1999}, a_{2000})$ can have a $0$ or be equal to $(1, 1)$ . So the first lemma implies \\[\\max(a_1, a_2) > \\max(a_3, a_4) > \\cdots > \\max(a_{1999}, a_{2000}) > 0,\\] so $999 = \\max(a_1, a_2) \\geq 1000$ , a contradiction. Hence $(a_n)$ completes at $i$ for some $i \\leq 2000$\nNow we're ready to find exactly which values of $a_2$ we want to count.\nLet's keep in mind that $2006 \\equiv 2 \\pmod 3$ and that $a_1 = 999$ is odd. We have two cases to consider.\nCase 1: If $a_2$ is odd, then $a_3$ is even, so $a_4$ is odd, so $a_5$ is odd, so $a_6$ is even, and this pattern must repeat every three terms because of the recursive definition of $(a_n)$ , so the terms of $(a_n)$ reduced modulo 2 are \\[1, 1, 0, 1, 1, 0, ...,\\] so $a_{2006}$ is odd and hence $1$ (since if $(a_n)$ completes at $i$ , then $a_k$ must be $0$ or $1$ for all $k \\geq i$ ).\nCase 2: If $a_2$ is even, then $a_3$ is odd, so $a_4$ is odd, so $a_5$ is even, so $a_6$ is odd, and this pattern must repeat every three terms, so the terms of $(a_n)$ reduced modulo 2 are \\[1, 0, 1, 1, 0, 1, ...,\\] so $a_{2006}$ is even, and hence $0$\nWe have found that $a_{2006} = 1$ is true precisely when $\\gcd(999, a_2) = 1$ and $a_2$ is odd. This tells us what we need to count.\nThere are $\\phi(999) = 648$ numbers less than $999$ and relatively prime to it ( $\\phi$ is the Euler totient function). We want to count how many of these are odd. Note that \\[t \\mapsto 999 - t\\] is a 1-1 correspondence between the odd and even numbers less than and relatively prime to $999$ . So our final answer is $648/2 = 324$ , or $\\boxed{324}$" ]