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[ "If $n \\geq 5$, then $E(n, n-2)=3$. Note that Elizabeth cannot flip every switch in one minute, and after two minutes, some switch (in fact, many switches) must be flipped exactly twice. However, Elizabeth can escape in three minutes using the sequence $\\{1,4,5, \\ldots, n\\},\\{2,4,5, \\ldots, n\\},\\{3,4,5, \\ldots, n\\}$." ]

[ "3" ]

[ "First, we prove that if $n$ is even and $k$ is odd, then $E(n, k)=E(n,n-k)$.\n\n\nBecause $n$ is even, and because each switch must be flipped an odd number of times in order to escape, the total number of flips is even. Because $k$ must be odd, $E(n, k)$ must be even. To show this, consider the case where $E(n, k)$ is odd. If $E(n, k)$ is odd, then an odd number of flips happen an odd number of times, resulting in an odd number of total flips. This is a contradiction because $n$ is even.\n\nCall a switch \"non-flipped\" in any given minute if it is not among the switches flipped in that minute. Because $E(n, k)$ (i.e., the total number of minutes) is even, and each switch is flipped an odd number of times, each switch must also be non-flipped an odd number of times. Therefore any sequence of flips that solves the \" $(n, k)$ puzzle\" can be made into a sequence of flips that solves the \" $(n, n-k)$ \" puzzle by interchanging flips and non-flips. These sequences last for the same number of minutes, and therefore $E(n, k)=E(n, n-k)$.\n\n$E(2020,1993)=76$. By the above prove, conclude that $E(2020,1993)=E(2020,27)$. Compute the latter instead. Because $\\frac{2020}{27}>74$, it will require at least 75 minutes to flip each switch once. Furthermore, $E(2020,27) \\geq 76$ because the prove above implies that $E(2020,27)$ is even.\n\nTo solve the puzzle in exactly 76 minutes, use the following strategy. For the first 33 minutes, flip switch 1, along with the first 26 switches that have not yet been flipped. The end result is that lights 1 through $26 \\cdot 33+1=859$ are on, and the remaining 1161 lights are off. Note that $1161=27 \\cdot 43$, so it takes 43 minutes to flip each remaining switch exactly once, for a total of 76 minutes, as desired." ]

[ "76" ]

[ "$E(2001,501)=5$. First, note that three minutes is not enough time to flip each switch once. In four minutes, Elizabeth can flip each switch once, but has three flips left over. Because there are an odd number of leftover flips to distribute among the 2001 switches, some switch must get an odd number of leftover flips, and thus an even number of total flips. Thus $E(2001,501)>4$.\n\nTo solve the puzzle in five minutes, Elizabeth can flip the following sets of switches:\n\n- in the first minute, $\\{1,2,3, \\ldots, 501\\}$;\n- in the second minute, $\\{1,2,3, \\ldots, 102\\}$ and $\\{502,503,504, \\ldots, 900\\}$;\n- in the third minute, $\\{1,2,3, \\ldots, 102\\}$ and $\\{901,902,903, \\ldots, 1299\\}$;\n- in the fourth minute, $\\{1,2,3, \\ldots, 100\\}$ and $\\{1300,1301,1302, \\ldots, 1700\\}$;\n- in the fifth minute, $\\{1,2,3, \\ldots, 100\\}$ and $\\{1701,1702,1703, \\ldots, 2001\\}$.\n\nThis results in switches $1,2,3, \\ldots, 100$ being flipped five times, switches 101 and 102 being flipped three times, and the remaining switches being flipped exactly once, so that all the lights are on at the end of the fifth minute." ]

[ "5" ]

[ "$I(6,3)=0$. By definition, $I(6,3)=E(6,3)-\\frac{6}{3}$. Because $3 \\mid 6, E(6,3)=\\frac{6}{3}=2$, and so $I(6,3)=2-2=0$." ]

[ "0" ]

[ "First note that $\\overline{J N}$ is a shared edge of all three pyramids, and that the viewpoint for the figure below is from along the line that is the extension of edge $\\overline{J N}$.\n\n<img_3460>\n\nLet $h$ denote the height of each pyramid. Let $X$ be the center of pyramid JOAN, and consider the plane passing through $H, A$, and $E$. By symmetry, the altitude in pyramid $J O H N$ through $H$ and the altitude in pyramid $J A N E$ through $E$ pass through $X$. Thus points $H, X$, and $A$ are collinear, as are points $E, X$, and $O$. Hence $A H=O E=2 h$. Using the result that the four medians in a tetrahedron are concurrent and divide each other in a $3: 1$ ratio, it follows that $A X=O X=\\frac{3 h}{4}$ and $X E=O E-O X=\\frac{5 h}{4}$. Applying the Law of Cosines to triangle $A X E$ yields $\\cos \\angle X A E=\\cos \\angle H A E=\\frac{2-2 h^{2}}{3 h}$. Suppose, without loss of generality, that the common side length of the pyramids is 1 . Then $h=\\sqrt{\\frac{2}{3}}$ and $\\cos \\angle H A E=\\frac{\\sqrt{6}}{9}$. Hence $\\sin \\angle H A E=\\frac{\\sqrt{75}}{9}$ and therefore $\\tan \\angle H A E=\\frac{5 \\sqrt{2}}{\\mathbf{2}}$." ]

[ "$\\frac{5 \\sqrt{2}}{2}$" ]

[ "Count the digits separately by position, noting that 1 is irrelevant to the product. There are a total of 20 instances of the digit 2 in the thousands place. The digit 0 only occurs in the hundreds place if the thousands digit is 2 , so look at the numbers 1 through 1999. Each non-zero digit contributes an equal number of times, so there are 200 each of $1,2,3,4,6,7,8,9$. The same applies to the tens digit, except there can be the stray digit of 1 among the numbers 2010 through 2019, but again, these do not affect the product. In the units place, there are 202 of each of the digits. Altogether, there are 602 each of $2,3,4,6,7,8$, 9, along with 20 extra instances of the digit 2 . Note that $9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 4 \\cdot 3 \\cdot 2=3024 \\cdot 24=72,576$ leaves a remainder of 576 when divided by 1000 . Also $2^{20}=1024^{2} \\equiv 24^{2}(\\bmod 1000)$, so $2^{20}$ contributes another factor of 576 . The answer is therefore the remainder when $576^{603}$ is divided by 1000 . This computation can be simplified by using the Chinese Remainder Theorem with moduli 8 and 125 , whose product is 1000 . Note $576^{603} \\equiv 0(\\bmod 8)$ because 576 is divisible by 8 . Also $576 \\equiv 76(\\bmod 125)$. By Euler's totient theorem, $576^{100} \\equiv 1(\\bmod 125)$, so $576^{603} \\equiv 76^{3}(\\bmod 125)$. This can quickly be computed by noting that $76^{3}=(75+1)^{3}=75^{3}+3 \\cdot 75^{2}+3 \\cdot 75+1 \\equiv 3 \\cdot 75+1 \\equiv-24(\\bmod 125)$. Observing that $-24 \\equiv 0(\\bmod 8)$, it follows that $576^{603} \\equiv-24(\\bmod 1000)$, hence the desired remainder is 976 ." ]

[ "976" ]

[ "Define a positive integer $n$ to be a semiprime if it is a product of exactly two (not necessarily distinct) primes. Define a lucky trio to be a sequence of three consecutive integers, $n, n+1, n+2$, each of which is a semiprime. Note that a lucky trio must contain exactly one multiple of 3. Also note that the middle number in a lucky trio must be even. To see this, note that if the first and last numbers in a lucky trio were both even, then exactly one of these numbers would be a multiple of 4 . But neither $2,3,4$ nor 4,5,6 is a lucky trio, and if a list of three consecutive integers contains a multiple of 4 that is greater than 4 , this number cannot be a semiprime. Using this conclusion and because $3,4,5$ is not a lucky trio, it follows that the middle number of a lucky trio cannot be a multiple of 4 . Hence it is necessary that a lucky trio has the form $4 k+1,4 k+2,4 k+3$, for some positive integer $k$, with $2 k+1$ being a prime. Note that $k \\neq 1(\\bmod 3)$ because when $k=1$, the sequence $5,6,7$ is not a lucky trio, and when $k>1,4 k+2$ would be a multiple of 6 greater than 6 , hence it cannot be a semiprime. Trying $k=2,3,5,6,8,9, \\ldots$ allows one to eliminate sequences of three consecutive integers that are not lucky trios, and if lucky trios are ordered by their least elements, one finds that the first three lucky trios are 33,34,35; 85,86,87; and 93,94,95. Hence the answer is 93." ]

[ "93" ]

[ "The distance between points $A(1,2,3)$ and $B(3,3,2)$ is $A B=\\sqrt{(3-1)^{2}+(3-2)^{2}+(2-3)^{2}}=\\sqrt{6}$. Denote by $s$ the side length of the cube. Consider three possibilities.\n\n- If $\\overline{A B}$ is an edge of the cube, then $A B=s$, so one possibility is $s_{1}=\\sqrt{6}$.\n- If $\\overline{A B}$ is a face diagonal of the cube, then $A B=s \\sqrt{2}$, so another possibility is $s_{2}=\\sqrt{3}$.\n- If $\\overline{A B}$ is a space diagonal of the cube, then $A B=s \\sqrt{3}$, so the last possibility is $s_{3}=\\sqrt{2}$.\n\nThe answer is then $s_{1}^{3} s_{2}^{3} s_{3}^{3}=\\left(s_{1} s_{2} s_{3}\\right)^{3}=6^{3}=\\mathbf{2 1 6}$." ]

[ "216" ]

[ "The answer is 12 . To give an example in which 12 selfies is possible, consider regular octagon $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6} P_{7} P_{8}$. Each vertex of the octagon represents a student and each of the diagonals and sides of the octagon represents a pair of students. Construct eight triangles $P_{1} P_{2} P_{4}, P_{2} P_{3} P_{5}, P_{3} P_{4} P_{6}, \\ldots, P_{8} P_{1} P_{3}$. Each of the segments in the forms of $\\overline{P_{i} P_{i+1}}, \\overline{P_{i} P_{i+2}}, \\overline{P_{i} P_{i+3}}$ appears exactly once in these eight triangles. Taking 8 three-person selfies (namely $\\left.\\left\\{P_{1}, P_{2}, P_{4}\\right\\},\\left\\{P_{2}, P_{3}, P_{5}\\right\\}, \\ldots,\\left\\{P_{8}, P_{1}, P_{3}\\right\\}\\right)$ and 4 two-person selfies (namely $\\left.\\left\\{P_{1}, P_{5}\\right\\},\\left\\{P_{2}, P_{6}\\right\\},\\left\\{P_{3}, P_{7}\\right\\},\\left\\{P_{4}, P_{8}\\right\\}\\right)$ gives a total of 12 selfies, completing the desired task.\n\nA diagram of this construction is shown below. Each of the eight triangles is a different color, and each of the two-person selfies is represented by a dotted diameter.\n\n<img_3543>\n\nIt remains to show fewer than 12 selfies is impossible. Assume that the students took $x$ three-person selfies and $y$ two-person selfies. Each three-person selfie counts 3 pairs of student appearances (in a selfie), and each two-person selfie counts 1 pair of student appearances (in a selfie). Together, these selfies count $3 x+y$ pairs of student appearances. There are $\\left(\\begin{array}{l}8 \\\\ 2\\end{array}\\right)=28$ pairs of student appearances. Hence $3 x+y=28$. The number of\n\n\n\nselfies is $x+y=28-2 x$, so it is enough to show that $x \\leq 8$.\n\nAssume for contradiction there are $x \\geq 9$ three-person selfies; then there are at least $3 \\cdot 9=27$ (individual) student appearances on these selfies. Because there are 8 students, some student $s_{1}$ had at least $\\lceil 27 / 8\\rceil$ appearances; that is, $s_{1}$ appeared in at least 4 of these three-person selfies. There are $2 \\cdot 4=8$ (individual) student appearances other than $s_{1}$ on these 4 selfies. Because there are only 7 students besides $s_{1}$, some other student $s_{2}$ had at least $[8 / 7\\rceil$ (individual) appearances on these 4 selfies; that is, $s_{2}$ appeared (with $s_{1}$ ) in at least 2 of these 4 three-person selfies, violating the condition that each pair of the students appears in exactly one selfie. Thus the answer is $\\mathbf{1 2}$." ]

[ "12" ]

[ "For $0 \\leq t<\\pi / 2$, all three values are $t$, so the desired $t$ does not lie in this interval.\n\nFor $\\pi / 2<t<\\pi$,\n\n$$\n\\begin{aligned}\n\\operatorname{Arcsin}(\\sin (t)) & =\\pi-t \\in(0, \\pi / 2) \\\\\n\\operatorname{Arccos}(\\cos (t)) & =t \\quad \\in(\\pi / 2, \\pi) \\\\\n\\operatorname{Arctan}(\\tan (t)) & =t-\\pi \\in(-\\pi / 2,0)\n\\end{aligned}\n$$\n\nA graph of all three functions is shown below.\n\n<img_3325>\n\nThus if the three numbers are to form an arithmetic progression, they should satisfy\n\n$$\nt-\\pi<\\pi-t<t\n$$\n\nThe three numbers will be in arithmetic progression if and only if $t+(t-\\pi)=2(\\pi-t)$, which implies $t=\\frac{\\mathbf{3 \\pi}}{\\mathbf{4}}$. Note that if $t=\\frac{3 \\pi}{4}$, the arithmetic progression is $-\\frac{\\pi}{4}, \\frac{\\pi}{4}, \\frac{3 \\pi}{4}$, as required." ]

[ "$\\frac{3 \\pi}{4}$" ]

[ "Let $\\theta=\\frac{1}{5} \\mathrm{~m} \\angle A$. Because $\\mathrm{m} \\angle P A Q=\\theta<5 \\theta=\\mathrm{m} \\angle A$, it follows that either $\\mathrm{m} \\angle B=\\theta$ or $\\mathrm{m} \\angle C=\\theta$. Thus there are two cases to consider.\n\nIf $\\mathrm{m} \\angle C=\\theta$, then it follows that $\\mathrm{m} \\angle A Q P=\\mathrm{m} \\angle Q A C+\\mathrm{m} \\angle A C B=4 \\theta$, and hence $\\mathrm{m} \\angle B=4 \\theta$. So $\\triangle A B C$ has angles of measures $5 \\theta, 4 \\theta, \\theta$, and thus $\\theta=18^{\\circ}$. However, this implies $\\mathrm{m} \\angle A=5 \\theta=90^{\\circ}$, which is not the case.\n\n<img_3197>\n\nIf instead $\\mathrm{m} \\angle B=\\theta$, then it follows that $\\mathrm{m} \\angle A P Q=\\mathrm{m} \\angle B A P+\\mathrm{m} \\angle A B P=2 \\theta$, and hence $\\mathrm{m} \\angle C=2 \\theta$. So $\\triangle A B C$ has angles of measures $5 \\theta, 2 \\theta, \\theta$, and thus $\\theta=22.5^{\\circ}$. Hence $\\mathrm{m} \\angle B=\\theta=\\mathbf{2 2 . 5}$.\n\n<img_3939>" ]

[ "$\\frac{45}{2}$" ]

[ "Note that for $n>0, \\log _{4} n=\\log _{64} n^{3}$ and $\\log _{8} n=\\log _{64} n^{2}$. Adding together the three given equations and using both the preceding facts and properties of logarithms yields\n\n$$\n\\begin{aligned}\n& \\log _{4}(x y z)+\\log _{8}\\left(x^{2} y^{2} z^{2}\\right)=11 \\\\\n\\Longrightarrow & \\log _{64}(x y z)^{3}+\\log _{64}(x y z)^{4}=11 \\\\\n\\Longrightarrow & \\log _{64}(x y z)^{7}=11 \\\\\n\\Longrightarrow & 7 \\log _{64}(x y z)=11 .\n\\end{aligned}\n$$\n\nThe last equation is equivalent to $x y z=64^{11 / 7}=2^{66 / 7}$, hence the desired value of $k$ is $\\frac{\\mathbf{6 6}}{\\mathbf{7}}$." ]

[ "$\\frac{66}{7}$" ]

[ "For convenience, let $\\alpha=\\pi / 4038$. Denote by\n\n$$\n0 \\leq \\theta<2 \\pi=8076 \\alpha\n$$\n\nthe complex argument of $z$, selected uniformly at random from the interval $[0,2 \\pi)$. Then $z$ itself lies in Quadrant II if and only if\n\n$$\n2019 \\alpha=\\frac{\\pi}{2}<\\theta<\\pi=4038 \\alpha\n$$\n\nOn the other hand, $z^{2019}$ has argument 2019日, and hence it lies in Quadrant II if and only if there is some integer $k$ with\n\n$$\n\\begin{gathered}\n\\frac{\\pi}{2}+2 k \\pi<2019 \\theta<\\pi+2 k \\pi \\\\\n\\Longleftrightarrow(4 k+1) \\cdot \\frac{\\pi}{2}<2019 \\theta<(4 k+2) \\cdot \\frac{\\pi}{2} \\\\\n\\Longleftrightarrow(4 k+1) \\alpha<\\theta<(4 k+2) \\alpha .\n\\end{gathered}\n$$\n\nBecause it is also true that $2019 \\alpha<\\theta<4038 \\alpha$, the set of $\\theta$ that satisfies the conditions of the problem is the union of intervals:\n\n$$\n(2021 \\alpha, 2022 \\alpha) \\cup(2025 \\alpha, 2026 \\alpha) \\cup \\cdots \\cup(4037 \\alpha, 4038 \\alpha)\n$$\n\nThere are 505 such intervals, the $j^{\\text {th }}$ interval consisting of $(4 j+2017) \\alpha<\\theta<(4 j+2018) \\alpha$. Each interval has length $\\alpha$, so the sum of the intervals has length $505 \\alpha$. Thus the final answer is\n\n$$\n\\frac{505 \\alpha}{2 \\pi}=\\frac{505}{2 \\cdot 4038}=\\frac{\\mathbf{5 0 5}}{\\mathbf{8 0 7 6}} .\n$$" ]

[ "$\\frac{505}{8076}$" ]

[ "Let $S(n)$ denote the sum of the digits of $n$, so that solving the problem is equivalent to solving $S(n)=5 S(n+2019)$. Using the fact that $S(n) \\equiv n(\\bmod 9)$ for all $n$, it follows that\n\n$$\n\\begin{aligned}\nn & \\equiv 5(n+2019) \\equiv 5(n+3)(\\bmod 9) \\\\\n4 n & \\equiv-15(\\bmod 9) \\\\\nn & \\equiv 3(\\bmod 9)\n\\end{aligned}\n$$\n\nThen $S(n+2019) \\equiv 6(\\bmod 9)$. In particular, $S(n+2019) \\geq 6$ and $S(n) \\geq 5 \\cdot 6=30$. The latter inequality implies $n \\geq 3999$, which then gives $n+2019 \\geq 6018$. Thus if $n+2019$ were a four-digit number, then $S(n+2019) \\geq 7$. Moreover, $S(n+2019)$ can only be 7, because otherwise, $S(n)=5 S(n+2019) \\geq 40$, which is impossible (if $n$ has four digits, then $S(n)$ can be no greater than 36). So if $n+2019$ were a four-digit number, then $S(n+2019)=7$ and $S(n)=35$. But this would imply that the digits of $n$ are $8,9,9,9$ in some order, contradicting the assumption that $n+2019$ is a four-digit number. On the other hand, if $n+2019$ were a five-digit number such that $S(n+2019) \\geq 6$, then the least such value of $n+2019$ is 10005 , and indeed, this works because it corresponds to $n=\\mathbf{7 9 8 6}$, the least possible value of $n$." ]

[ "7986" ]

[ "The graph of the second equation is simply the circle of radius $\\sqrt{37}$ centered at $(1,-1)$. The first graph is more interesting, and its behavior depends on $K$.\n\n- For small values of $K$, the first equation determines a set of four circles of radius $\\sqrt{K}$ with centers at $(5,5),(5,-5),(-5,5)$, and $(-5,-5)$. Shown below are versions with $K=1, K=4$, and $K=16$.\n<img_3421>\n- However, when $K>25$, the graph no longer consists of four circles! As an example, for $K=36$, the value $x=5$ gives $(|y|-5)^{2}=36$; hence $|y|=-1$ or $|y|=6$. The first option is impossible; the graph ends up \"losing\" the portions of the upper-right circle that would cross the $x$ - or $y$-axes compared to the graph for $(x-5)^{2}+(y-5)^{2}=36$. The graph for $K=36$ is shown below.\n\n<img_3535>\n\n- As $K$ continues to increase, the \"interior\" part of the curve continues to shrink, until at $K=50$, it simply comprises the origin, and for $K>50$, it does not exist. As examples, the graphs with $K=50$ and $K=64$ are shown below.\n<img_3577>\n\n\n\nOverlay the graph of the circle of radius $\\sqrt{37}$ centered at $(1,-1)$ with the given graphs. When $K=25$, this looks like the following graph.\n\n<img_3905>\n\nNote that the two graphs intersect at $(0,5)$ and $(-5,0)$, as well as four more points (two points near the positive $x$-axis and two points near the negative $y$-axis). When $K$ is slightly greater than 25 , this drops to four intersection points. The graph for $K=27$ is shown below.\n\n<img_3751>\n\nThus for the greatest $K$ for which there are exactly two intersection points, those two intersection points should be along the positive $x$ - and negative $y$-axes. If the intersection point on the positive $x$-axis is at $(h, 0)$, then $(h-1)^{2}+(0+1)^{2}=37$ and $(h-5)^{2}+(0-5)^{2}=K$. Thus $h=7$ and $K=\\mathbf{2 9}$" ]

[ "29" ]

[ "Call the original sequence of ten squares $T=\\left(1^{2}, 2^{2}, \\ldots, 10^{2}\\right)$. A morphed sequence is one that can be obtained by morphing $T$ a finite number of times.\n\nThis solution is divided into three steps. In the first step, a characterization of the possible final morphed sequences is given. In the second step, a lower bound on the number of steps is given, and in the third step, it is shown that this bound can be achieved.\n\nStep 1. Note the following.\n\n- The sum of the elements of $T$ is $1^{2}+2^{2}+\\cdots+10^{2}=385$, and morphs are sum-preserving. So any morphed sequence has sum 385 and a mean of 38.5.\n- The sequence $T$ has positive integer terms, and morphs preserve this property. Thus any morphed sequence has positive integer terms.\n- The sequence $T$ is strictly increasing, and morphs preserve this property. Thus any morphed sequence is strictly increasing.\n\n\n\nNow if the morphed sequence is an arithmetic progression, it follows from the above three observations that it must have the form\n\n$$\n(38.5-4.5 d, 38.5-3.5 d, \\ldots, 38.5+4.5 d)\n$$\n\nwhere $d$ is an odd positive integer satisfying $38.5-4.5 d>0$. Therefore the only possible values of $d$ are $7,5,3,1$; thus there are at most four possibilities for the morphed sequence, shown in the table below. Denote these four sequences by $A, B, C, D$.\n\n| | $T$ | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $d=7:$ | $A$ | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 |\n| $d=5:$ | $B$ | 16 | 21 | 26 | 31 | 36 | 41 | 46 | 51 | 56 | 61 |\n| $d=3:$ | $C$ | 25 | 28 | 31 | 34 | 37 | 40 | 43 | 46 | 49 | 52 |\n| $d=1:$ | $D$ | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 |\n\nStep 2. Given any two sequences $X=\\left(x_{1}, \\ldots, x_{10}\\right)$ and $Y=\\left(y_{1}, \\ldots, y_{10}\\right)$ with $\\sum_{i=1}^{10} x_{i}=\\sum_{i=1}^{10} y_{i}=385$, define the taxicab distance\n\n$$\n\\rho(X, Y)=\\sum_{i=1}^{10}\\left|x_{i}-y_{i}\\right|\n$$\n\nObserve that if $X^{\\prime}$ is a morph of $X$, then $\\rho\\left(X^{\\prime}, Y\\right) \\geq \\rho(X, Y)-2$. Therefore the number of morphs required to transform $T$ into some sequence $Z$ is at least $\\frac{1}{2} \\rho(T, Z)$. Now\n\n$$\n\\frac{1}{2} \\rho(T, A)=\\frac{1}{2} \\sum_{i=1}^{10}\\left|i^{2}-7 i\\right|=56\n$$\n\nand also $\\rho(T, A)<\\min (\\rho(T, B), \\rho(T, C), \\rho(T, D))$. Thus at least 56 morphs are needed to obtain sequence $A$ (and more morphs would be required to obtain any of sequences $B, C$, or $D$ ).\n\nStep 3. To conclude, it remains to verify that one can make 56 morphs and arrive from $T$ to $A$. One of many possible constructions is given below.\n\n| $T$ | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |\n| ---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 6 morphs | 1 | 4 | 9 | 16 | 25 | 42 | 49 | 58 | 81 | 100 |\n| 2 morphs | 1 | 4 | 9 | 16 | 27 | 42 | 49 | 56 | 81 | 100 |\n| 8 morphs | 1 | 4 | 9 | 16 | 35 | 42 | 49 | 56 | 73 | 100 |\n| 10 morphs | 1 | 4 | 9 | 26 | 35 | 42 | 49 | 56 | 63 | 100 |\n| 2 morphs | 1 | 4 | 9 | 28 | 35 | 42 | 49 | 56 | 63 | 98 |\n| 12 morphs | 1 | 4 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 86 |\n| 10 morphs | 1 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 76 |\n| 6 morphs | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 |\n\nTherefore the least number of morphs needed to transform $T$ into an arithmetic progression is $\\mathbf{5 6}$." ]

[ "56" ]

[ "To begin, denote by $R$ the radius of $\\omega$. The semiperimeter of triangle $A B C$ is 42 , and then applying Heron's formula yields\n\n$$\n[A B C]=\\frac{14 \\cdot 30 \\cdot 40}{4 R}=\\sqrt{42 \\cdot 28 \\cdot 12 \\cdot 2}=168\n$$\n\nfrom which it follows that $R=\\frac{14 \\cdot 30 \\cdot 40}{4 \\cdot 168}=25$.\n\nNow consider the point circle with radius zero centered at $T$ in tandem with the circle $\\omega$. Because $P A=P T$, it follows that $P$ lies on the radical axis of these circles. Moreover, the midpoints of $\\overline{T B}$ and $\\overline{T C}$ lie on this radical axis as well. Thus $P$ lies on the midline of $\\triangle T B C$ that is parallel to $\\overline{B C}$.\n\n<img_3211>\n\nTo finish, let $O$ denote the center of $\\omega$ and $M$ the midpoint of $\\overline{B C}$. By considering right triangle $T B O$ with altitude $\\overline{B M}$, it follows that $M T \\cdot M O=M B^{2}$, but also $M O=\\sqrt{O B^{2}-M B^{2}}=\\sqrt{25^{2}-20^{2}}=15$, so\n\n$$\nM T=\\frac{M B^{2}}{M O}=\\frac{400}{15}=\\frac{80}{3}\n$$\n\nThus the distance from $P$ to $\\overline{B C}$ is $\\frac{1}{2} M T=\\frac{40}{3}$. Finally,\n\n$$\n[P B C]=\\frac{1}{2} \\cdot \\frac{40}{3} \\cdot B C=\\frac{\\mathbf{8 0 0}}{\\mathbf{3}}\n$$" ]

[ "$\\frac{800}{3}$" ]

[ "Adding the two given equations yields $b c+c d=c(b+d)=39$. The greatest possible value of $c$ therefore occurs when $c=\\mathbf{3 9}$ and $b+d=1$." ]

[ "39" ]

[ "Knowing that 13 of the cards are not in Emile's set, there are $\\left(\\begin{array}{c}39 \\\\ T\\end{array}\\right)$ ways for him to have chosen a set of $T$ cards. Given that Emile's set contains no clubs, the suits of the three aces are fixed (i.e., diamonds, hearts, and spades). The number of possible sets of cards in which these three aces appear is therefore $\\left(\\begin{array}{c}36 \\\\ T-3\\end{array}\\right)$. The desired probability is therefore $\\left(\\begin{array}{c}36 \\\\ T-3\\end{array}\\right) /\\left(\\begin{array}{c}39 \\\\ T\\end{array}\\right)$. With $T=39$, this probability is $1 / 1=\\mathbf{1}$, which is consistent with the fact that Emile's set contains all cards in the deck that are not clubs, hence he is guaranteed to have all three of the remaining aces." ]

[ "1" ]

[ "Let $C D=3 x$ and let $h$ be the length of the altitude between bases $\\overline{A B}$ and $\\overline{C D}$. Then $[A B C D]=3 x h$ and $[M P Q]=\\frac{1}{2} x h$. Hence $\\frac{[A B C D]}{[M P Q]}=\\mathbf{6}$. Both the position of $M$ and the ratio $\\frac{A B}{B C}=T$ are irrelevant." ]

[ "6" ]

[ "It can readily be shown that $\\log _{a} b=\\log _{a^{2}} b^{2}$. Thus it follows that $\\log _{T} \\sqrt{x-7}=\\log _{T^{2}}(x-7)$. Hence the left-hand side of the given equation is $\\log _{T^{2}}(x-7)(x-2)$ and the equation is equivalent to $(x-7)(x-2)=T^{2}$, which is equivalent to $x^{2}-9 x+14-T^{2}=0$. With $T=6$, this equation is $x^{2}-9 x-22=0 \\Longrightarrow(x-11)(x+2)=0$. Plugging $x=-2$ into the given equation leads to the first term of the left-hand side having a negative radicand and the second term having an argument of 0 . However, one can easily check that $x=\\mathbf{1 1}$ indeed satisfies the given equation." ]

[ "11" ]

[ "A general term in the expansion of $(p x+y+z)^{T-1}$ has the form $K(p x)^{a} y^{b} z^{c}$, where $a, b$, and $c$ are nonnegative integers such that $a+b+c=T-1$. Using the \"stars and bars\" approach, the number of nonnegative integral solutions to $a+b+c=T-1$ is the number of arrangements of $T-1$ stars and 2 bars in a row (the bars act has separators and the \" 2 \" arises because it is one less than the number of variables in the equation). Thus there are $\\left(\\begin{array}{c}T+1 \\\\ 2\\end{array}\\right)$ solutions. Each term will be a multiple of $p$ unless $a=0$. In this case, the number of terms that are not multiples of $p$ is the number of nonnegative integral solutions to the equation $b+c=T-1$, which is $T$ ( $b$ can range from 0 to $T-1$ inclusive, and then $c$ is fixed). Hence $M=\\left(\\begin{array}{c}T+1 \\\\ 2\\end{array}\\right)-T=\\frac{T^{2}-T}{2}$. With $T=11$, the answer is $\\mathbf{5 5}$" ]

[ "55" ]

[ "The condition that $20, T-5, K$ is an increasing geometric sequence implies that $\\frac{T-5}{20}=\\frac{K}{T-5}$, hence $K=\\frac{(T-5)^{2}}{20}$. The condition that $19, K, 4 T+11$ is an increasing arithmetic sequence implies that $K-19=4 T+11-K$, hence $K=2 T+15$. With $T=55$, each of these equations implies that $K=\\mathbf{1 2 5}$. Note that the two equations can be combined and solved without being passed a value of $T$. A quadratic equation results, and its roots are $T=55$ or $T=-5$. However, with $T=-5$, neither of the given sequences is increasing." ]

[ "125" ]

[ "In general, let cube $\\mathcal{C}_{n}$ have edge length $x$. Then the diameter of sphere $\\mathcal{S}_{n}$ is the space diagonal of $\\mathcal{C}_{n}$, which has length $x \\sqrt{3}$. This in turn is the edge length of cube $\\mathcal{C}_{n+1}$. Hence the edge lengths of $\\mathcal{C}_{1}, \\mathcal{C}_{2}, \\ldots$ form an increasing geometric sequence with common ratio $\\sqrt{3}$ and volumes of $\\mathcal{C}_{1}, \\mathcal{C}_{2}, \\ldots$ form an increasing geometric sequence with common ratio $3 \\sqrt{3}$. With $T=125$, the edge length of $\\mathcal{C}_{1}$ is 5 , so the sequence of edge lengths of the cubes is $5,5 \\sqrt{3}, 15, \\ldots$, and the respective sequence of the volumes of the cubes is $125,375 \\sqrt{3}, 3375, \\ldots$. Hence $k=3$, and the edge length of $\\mathcal{C}_{3}$ is $\\mathbf{1 5}$." ]

[ "15" ]

[ "Let $s$ be the side length of square $K E N T$; then $M E=s$. Let $J$ be the foot of the altitude from $M$ to $\\overline{K E}$. Then $\\mathrm{m} \\angle J E M=30^{\\circ}$ and $\\mathrm{m} \\angle E M J=60^{\\circ}$. Hence $M J=\\frac{s}{2}, J E=\\frac{s \\sqrt{3}}{2}$, and $K J=K E-J E=s-\\frac{s \\sqrt{3}}{2}$. Applying the Pythagorean Theorem to $\\triangle K J M$ implies that $K M^{2}=\\left(s-\\frac{s \\sqrt{3}}{2}\\right)^{2}+\\left(\\frac{s}{2}\\right)^{2}=2 s^{2}-s^{2} \\sqrt{3}$. With $s=20$, the value of $b$ is therefore $s^{2}=\\mathbf{4 0 0}$." ]

[ "400" ]

[ "According to Vieta's formulas, $a+b+c=-(-20)=20$ and $a b+b c+c a=19$. Noting that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)$, it follows that $a^{2}+b^{2}+c^{2}=20^{2}-2 \\cdot 19=\\mathbf{3 6 2}$. The value of $T$ is irrelevant." ]

[ "362" ]

[ "The expression inside the absolute value bars simplifies to $K^{2}-19 K-20+19 K-K^{2}=-20$. Hence the answer is $\\mathbf{2 0}$ and the value of $K(=\\sqrt{361}=19)$ is not needed." ]

[ "20" ]

[ "Note that $[L E O]=\\frac{1}{2}(\\sin \\angle L E O) \\cdot L E \\cdot E O=\\frac{1}{2} \\cdot \\frac{1}{T} \\cdot \\frac{1}{n} \\cdot\\left(n^{3}-4 n^{2}+5 n\\right)=\\frac{n^{2}-4 n+5}{2 T}$. Because $T$ is a constant, the least possible value of $[L E O]$ is achieved when the function $f(n)=n^{2}-4 n+5$ is minimized.\n\n\n\nThis occurs when $n=-(-4) /(2 \\cdot 1)=2$, and the minimum value is $f(2)=1$. Hence the desired least possible value of $[L E O]$ is $\\frac{1}{2 T}$, and with $T=20$, this is $\\frac{1}{40}$." ]

[ "$\\frac{1}{40}$" ]

[ "Note that $x^{2}-y^{2}=(x+y)(x-y)=5(x-y)$, hence $x-y=\\frac{1}{5 T}$. Then $x+z=(x+y)+(x-y)+(z-x)=$ $5+\\frac{1}{5 T}+7=12+\\frac{1}{5 T}$. With $T=\\frac{1}{40}$, the answer is thus $12+8=\\mathbf{2 0}$." ]

[ "20" ]

[ "When $n$ is a nonnegative integer, the product of the positive divisors of $2^{n}$ is $2^{0} \\cdot 2^{1} \\cdot \\ldots \\cdot 2^{n-1} \\cdot 2^{n}=$ $2^{0+1+\\cdots+(n-1)+n}=2^{n(n+1) / 2}$. Because $T=20$ is an integer, it follows that $K=\\frac{T(T+1)}{2}=\\mathbf{2 1 0}$." ]

[ "210" ]

[ "Because dessert is optional, there are effectively $2+1=3$ dessert choices. Hence, by the Multiplication Principle, it follows that $T=7 \\cdot K \\cdot 3$, thus $K=\\frac{T}{21}$. With $T=210$, the answer is 10 ." ]

[ "10" ]

[ "In km/hr, the combined speed of Sam and Marty is $S+M$. Thus one can determine the total time they traveled and use this to determine the number of kilometers that Sam traveled. However, this is not needed, and there is a simpler approach. Suppose that Marty traveled a distance of $d$. Then because Sam's speed is $\\frac{S}{M}$ of Marty's speed, Sam will have traveled a distance of $\\frac{S}{M} \\cdot d$. Thus, together, they traveled $d+\\frac{S}{M} \\cdot d$. Setting this equal to 100 and solving yields $d=\\frac{100 M}{M+S}$. Thus Sam traveled $\\frac{S}{M} \\cdot d=\\frac{100 S}{M+S}$. With $S=15$ and $M=10$, this is equal to $60 \\mathrm{~km}$." ]

[ "60" ]

[ "Let $S$ be the set of numbers that gain an extra digit when doubled. First notice that the numbers in $S$ are precisely those whose first digit is at least 5 . Thus there are five one-digit numbers in $S, 50$ two-digit numbers in $S$, and 500 three-digit numbers in $S$. Therefore 5000 is the $556^{\\text {th }}$ smallest number in $S$, and because all four-digit numbers greater than 5000 are in $S$, the $2011^{\\text {th }}$ smallest number in $S$ is $5000+(2011-556)=\\mathbf{6 4 5 5}$." ]

[ "6455" ]

[ "Let $N$ be the point of tangency of the circle with $\\overline{A B}$ and draw $\\overline{M B}$, as shown below.\n\n<img_3520>\n\nBecause $\\triangle B M C$ and $\\triangle B M N$ are right triangles sharing a hypotenuse, and $\\overline{M N}$ and $\\overline{M C}$ are radii, $\\triangle B M C \\cong \\triangle B M N$. Thus $B N=12$ and $A N=1$. Also $\\triangle A N M \\sim \\triangle A C B$ because the right triangles share $\\angle A$, so $\\frac{N M}{A N}=\\frac{C B}{A C}$. Therefore $\\frac{r}{1}=\\frac{12}{5}$, so $r=\\frac{\\mathbf{1 2}}{\\mathbf{5}}$.", "Let $r$ denote the radius of the circle, and let $D$ be the foot of the perpendicular from $O$ to $\\overline{A B}$. Note that $\\triangle A B C \\sim \\triangle A O D$. Thus $\\frac{A B}{A O}=\\frac{B C}{D O} \\Longrightarrow \\frac{13}{5-r}=\\frac{12}{r}$, and $r=\\frac{\\mathbf{1 2}}{\\mathbf{5}}$." ]

[ "$\\frac{12}{5}$" ]

[ "Let $a$ be the third term of the geometric progression, and let $r$ be the common ratio. Then the product of the first five terms is\n\n$$\n\\left(a r^{-2}\\right)\\left(a r^{-1}\\right)(a)(a r)\\left(a r^{2}\\right)=a^{5}=32\n$$\n\nso $a=2$. Because the fourth term is $17, r=\\frac{17}{a}=\\frac{17}{2}$. The second term is $a r^{-1}=\\frac{2}{17 / 2}=\\frac{4}{17}$." ]

[ "$\\frac{4}{17}$" ]

[ "Because $A_{1} A_{2} A_{k} A_{k+1}$ is a rectangle, $n$ must be even, and moreover, $k=\\frac{n}{2}$. Also, the rectangle's diagonals meet at the center $O$ of the circumscribing circle. $O$ is also the center of the $n$-gon. The diagram below shows the case $n=16$.\n\n\n\n<img_3867>\n\nThen $\\left[A_{1} A_{2} O\\right]=\\frac{1}{4}\\left[A_{1} A_{2} A_{k} A_{k+1}\\right]=\\frac{1}{n}\\left[A_{1} A_{2} \\ldots A_{n}\\right]=60$. So $\\frac{1}{4}(6)=\\frac{1}{n}(60)$, and $n=40$." ]

[ "40" ]

[ "Let $x$ be the number of orange marbles. Then the probability of drawing an orange marble is $\\frac{x}{x+20+12}=\\frac{x}{x+32}$. If this probability equals $\\frac{1}{y}$, then $y=\\frac{x+32}{x}=1+\\frac{32}{x}$. This expression represents an integer only when $x$ is a factor of 32 , thus $x \\in\\{1,2,4,8,16,32\\}$. The corresponding $y$-values are $33,17,9,5,3$, and 2 , and their sum is $\\mathbf{6 9}$." ]

[ "69" ]

[ "From the first two equations, conclude that $d=2 a$. From the last two, $3 b=2 a$. Thus all solutions to the system will be of the form $(3 K, 2 K, c, 6 K)$ for some integer $K$. Substituting these expressions into the system, each equation now becomes $c K^{2}=2000=2^{4} \\cdot 5^{3}$. So $K^{2}$ is of the form $2^{2 m} 5^{2 n}$. There are 3 choices for $m$ and 2 for $n$, so there are 6 values for $K^{2}$, which means there are 12 solutions overall, including negative values for $K$.\n\nAlthough the problem does not require finding them, the twelve values of $K$ are $\\pm 1, \\pm 2, \\pm 4$, $\\pm 5, \\pm 10, \\pm 20$. These values yield the following quadruples $(a, b, c, d)$ :\n\n$$\n\\begin{aligned}\n& (3,2,2000,6),(-3,-2,2000,-6), \\\\\n& (6,4,500,12),(-6,-4,500,-12), \\\\\n& (12,8,125,24),(-12,-8,125,-24), \\\\\n& (15,10,80,30),(-15,-10,80,-30), \\\\\n& (30,20,20,60),(-30,-20,20,-60), \\\\\n& (60,40,5,120),(-60,-40,5,-120) .\n\\end{aligned}\n$$" ]

[ "12" ]

[ "In simplifying the numerator and denominator of the left side of the equation, notice that\n\n$$\n\\begin{aligned}\nk+(k+1)+\\cdots+k n & =\\frac{1}{2}(k n(k n+1)-k(k-1)) \\\\\n& =\\frac{1}{2}(k(n+1)(k n-k+1))\n\\end{aligned}\n$$\n\nThis identity allows the given equation to be transformed:\n\n$$\n\\begin{aligned}\n\\frac{3(n+1)(3 n-3+1)}{5(n+1)(5 n-5+1)} & =\\frac{4}{11} \\\\\n\\frac{3(n+1)(3 n-2)}{5(n+1)(5 n-4)} & =\\frac{4}{11} \\\\\n\\frac{3 n-2}{5 n-4} & =\\frac{20}{33}\n\\end{aligned}\n$$\n\nSolving this last equation yields $n=14$. Using the same identity twice more, for $n=14$ and $k=2$ and $k=4$, the desired quantity is $\\frac{2(2 n-1)}{4(4 n-3)}=\\frac{\\mathbf{2 7}}{\\mathbf{1 0 6}}$." ]

[ "$\\frac{27}{106}$" ]

[ "Let $f(x)=a(x-b)^{2}+c$. The graph of $f$ is symmetric about $x=b$, so the graph of $y=f(f(x))$ is also symmetric about $x=b$. If $b \\neq 5$, then $2 b-5$, the reflection of 5 across $b$, must be a zero of $f(f(x))$. Because $f(f(x))$ has exactly one zero, $b=5$.\n\nBecause $f(2)=0$ and $f$ is symmetric about $x=5$, the other zero of $f$ is $x=8$. Because the zeros of $f$ are at 2 and 8 and $f(5)$ is a zero of $f$, either $f(5)=2$ or $f(5)=8$. The following argument shows that $f(5)=8$ is impossible. Because $f$ is continuous, if $f(5)=8$, then $f\\left(x_{0}\\right)=2$ for some $x_{0}$ in the interval $2<x_{0}<5$. In that case, $f\\left(f\\left(x_{0}\\right)\\right)=0$, so 5 would not be a unique zero of $f(f(x))$. Therefore $f(5)=2$ and $c=2$. Setting $f(2)=0$ yields the equation $a(2-5)^{2}+2=0$, so $a=-\\frac{2}{9}$, and $f(0)=-\\frac{\\mathbf{3 2}}{\\mathbf{9}}$." ]

[ "$-\\frac{32}{9}$" ]

[ "Let $a$ represent the middle $\\left(8^{\\text {th }}\\right)$ term of the sequence, and let $d$ be the difference. Then the terms of the sequence are $a-7 d, a-6 d, \\ldots, a+6 d, a+7 d$, their sum is $15 a$, and the difference between the largest and the smallest terms is $14 d$. The largest $d$ such that $14 d \\leq 999$ is $d=71$. Thus the largest possible value for $a$ is $999-7 \\cdot 71=502$. The maximal sum of the sequence is therefore $15 a=\\mathbf{7 5 3 0}$." ]

[ "7530" ]

[ "The points $O, A, B, C$ all lie on $\\omega_{2}$ in some order. There are two possible cases to consider: either $B$ is outside circle $\\omega_{1}$, or it is inside the circle, as shown below.\n\n\n\n<img_3962>\n\nThe following argument shows that the first case is impossible. By the Triangle Inequality on $\\triangle A B O$, the radius $r_{1}$ of circle $\\omega_{1}$ must be at least 20 . But because $B$ is outside $\\omega_{1}, B O>r_{1}$, which is impossible, because $B O=17$. So $B$ must be inside the circle.\n\nConstruct point $D$ on minor arc $A O$ of circle $\\omega_{2}$, so that $A D=O B$ (and therefore $\\left.D O=B C\\right)$.\n\n<img_3873>\n\nBecause $A, D, O, B$ all lie on $\\omega_{2}$, Ptolemy's Theorem applies to quadrilateral $A D O B$.\n\n\n\n<img_3934>\n\nTherefore $A D \\cdot O B+O D \\cdot A B=A O \\cdot D B=r_{1}^{2}$. Substituting $A D=O B=17, D O=B C=7$, and $A B=37$ yields $r_{1}^{2}=37 \\cdot 7+17^{2}=548$. Thus the area of $\\omega_{1}$ is $\\mathbf{5 4 8 \\pi}$." ]

[ "$548 \\pi$" ]

[ "$8^{n}=2^{3 n}$ and $16^{32}=2^{128}$. Therefore $4<3 n<128$, and $2 \\leq n \\leq 42$. Thus there are 41 such integers $n$." ]

[ "41" ]

[ "If $T$ has more than one digit when written in base $b$, then $b \\leq T$. If $T$ has fewer than three digits when written in base $b$, then $b^{2}>T$, or $b>\\sqrt{T}$. So the desired set of bases $b$ is $\\{b \\mid \\sqrt{T}<b \\leq T\\}$. When $T=41,\\lfloor\\sqrt{T}\\rfloor=6$ and so $6<b \\leq 41$. There are $41-6=\\mathbf{3 5}$ such integers." ]

[ "35" ]

[ "Let $A C=b$ and $B C=a$. Then $a^{2}+b^{2}=1600$ and $|a-b|=T-1$. Squaring the second equation yields $a^{2}+b^{2}-2 a b=(T-1)^{2}$, so $1600-2 a b=(T-1)^{2}$. Hence the area of the triangle is $\\frac{1}{2} a b=\\frac{1600-(T-1)^{2}}{4}=400-\\frac{(T-1)^{2}}{4}$ or $400-\\left(\\frac{T-1}{2}\\right)^{2}$, which for $T=35$ yields $400-289=\\mathbf{1 1 1}$." ]

[ "111" ]

[ "The identity $\\log _{b^{n}} x=\\frac{1}{n} \\log _{b} x$ yields $\\log _{2} x=10$. Then $\\log _{2} \\sqrt{x}=\\log _{2} x^{1 / 2}=\\frac{1}{2} \\log _{2} x=5$.", "Use the definition of $\\log$ to obtain $x=(\\sqrt{2})^{20}=\\left(2^{1 / 2}\\right)^{20}=2^{10}$. Thus $\\log _{2} \\sqrt{x}=\\log _{2} 2^{5}=\\mathbf{5}$.", "Use the change of base formula to obtain $\\frac{\\log x}{\\log \\sqrt{2}}=20$, so $\\log x=$ $20 \\log \\sqrt{2}=20 \\log 2^{1 / 2}=10 \\log 2$. Thus $x=2^{10}$, and $\\log _{2} \\sqrt{x}=\\log _{2} 2^{5}=5$." ]

[ "5" ]

[ "Because Hannah has only two coins, the only ways she can get more heads than Otto are if she gets 1 (and he gets 0 ), or she gets 2 (and he gets either 1 or 0 ).\n\nThe probability of Hannah getting exactly one head is $\\frac{1}{2}$. The probability of Otto getting no heads is $\\frac{1}{2^{T}}$. So the probability of both events occurring is $\\frac{1}{2^{T+1}}$.\n\nThe probability of Hannah getting exactly two heads is $\\frac{1}{4}$. The probability of Otto getting no heads is still $\\frac{1}{2^{T}}$, but the probability of getting exactly one head is $\\frac{T}{2^{T}}$, because there are $T$ possibilities for which coin is heads. So the probability of Otto getting either 0 heads or 1 head is $\\frac{1+T}{2^{T}}$, and combining that with Hannah's result yields an overall probability of $\\frac{1+T}{2^{T+2}}$.\n\nThus the probability that Hannah flips more heads than Otto is $\\frac{1}{2^{T+1}}+\\frac{1+T}{2^{T+2}}=\\frac{3+T}{2^{T+2}}$. For $T=5$, the value is $\\frac{8}{128}=\\frac{1}{16}$, giving an answer of $1+16=\\mathbf{1 7}$." ]

[ "17" ]

[ "Let $x$ be the number of twenty-edwah bills and $y$ be the number of eighty-edwah bills. Then\n\n$$\n\\begin{aligned}\n\\frac{20 x+80 y}{x+y} & =2 T \\\\\n20 x+80 y & =2 T x+2 T y \\\\\n(80-2 T) y & =(2 T-20) x\n\\end{aligned}\n$$\n\n\n\nIn the case where $T=17$ (and hence $2 T=34$ ), this equation reduces to $46 y=14 x$, or $23 y=7 x$. Because 23 and 7 are relatively prime, $23 \\mid x$ and $7 \\mid y$. Therefore the pair that yields the smallest possible value is $(x, y)=(23,7)$. Then there are $23+7=30$ bills worth a total of $23 \\cdot 20+7 \\cdot 80=460+560=1020$ edwahs, and $1020 / 30=34$, as required. The answer is $\\mathbf{1 0 2 0}$.", "Consider the equation $\\frac{20 x+80 y}{x+y}=2 T$ derived in the first solution. The identity $\\frac{20 x+80 y}{x+y}=20+\\frac{60 y}{x+y}$ yields the following:\n\n$$\n\\begin{aligned}\n\\frac{60 y}{x+y} & =2 T-20 \\\\\n\\frac{30 y}{x+y} & =T-10 \\\\\nx+y & =\\frac{30 y}{T-10} .\n\\end{aligned}\n$$\n\nFor the smallest value of $x+y$, both $x+y$ and $y$ will be relatively prime. Thus the smallest value of $x+y$ is $\\frac{30}{\\operatorname{gcd}(T-10,30)}$, which occurs when $y=\\frac{T-10}{\\operatorname{gcd}(T-10,30)}$. Substituting $T=17$, the numbers $T-10=7$ and 30 are relatively prime, so $y=7$ and $x=23$, for a total of $\\mathbf{1 0 2 0}$ edwahs." ]

[ "1020" ]

[ "Let the radii be $p, q, r$ respectively. Looking at a cross-section of the spheres through $\\overline{P Q}$ perpendicular to the plane, the points $P^{\\prime}, P, Q, Q^{\\prime}$ form a right trapezoid with $\\overline{P^{\\prime} P} \\perp \\overline{P^{\\prime} Q^{\\prime}}$ and $\\overline{Q^{\\prime} Q} \\perp \\overline{P^{\\prime} Q^{\\prime}}$. Draw $\\overline{P M}$ perpendicular to $\\overline{Q Q^{\\prime}}$ as shown.\n\n<img_3907>\n\nThen $P P^{\\prime}=M Q^{\\prime}=p$ and $Q M=q-p$, while $P Q=p+q$ and $P M=P^{\\prime} Q^{\\prime}$. By the Pythagorean Theorem, $(q-p)^{2}+P^{\\prime} Q^{\\prime 2}=(p+q)^{2}$, so $q=\\frac{\\left(P^{\\prime} Q^{\\prime}\\right)^{2}}{4 p}$. Thus $4 p q=P^{\\prime} Q^{\\prime 2}=12^{2}$. Similarly, $4 p r=P^{\\prime} R^{\\prime 2}=6^{2}$ and $4 q r=Q^{\\prime} R^{\\prime 2}=12^{2}$. Dividing the first equation by the third shows that $p=r$ (which can also be inferred from the symmetry of $\\triangle P^{\\prime} Q^{\\prime} R^{\\prime}$ ) and the equation $p r=9$ yields 3 as their common value; substitute in either of the other two equations to obtain $q=12$. Therefore the sides of $\\triangle P Q R$ are $P Q=Q R=12+3=15$ and $P R=6$. The altitude to $\\overline{P R}$ has length $\\sqrt{15^{2}-3^{2}}=6 \\sqrt{6}$, so the triangle's area is $\\frac{1}{2}(6)(6 \\sqrt{6})=\\mathbf{1 8} \\sqrt{\\mathbf{6}}$." ]

[ "$18 \\sqrt{6}$" ]

[ "By the definition of $f$,\n\n$$\nf(f(x))=f(x)+(f(x))^{2}+(f(x))^{4}+(f(x))^{8}+\\cdots\n$$\n\nConsider this series term by term. The first term, $f(x)$, contains no $x^{10}$ terms, so its contribution is 0 . The second term, $(f(x))^{2}$, can produce terms of $x^{10}$ in two ways: as $x^{2} \\cdot x^{8}$ or as $x^{8} \\cdot x^{2}$. So its contribution is 2 .\n\nNow consider the third term:\n\n$$\n\\begin{aligned}\n(f(x))^{4}= & f(x) \\cdot f(x) \\cdot f(x) \\cdot f(x) \\\\\n= & \\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\\cdots\\right) \\cdot\\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\\cdots\\right) \\cdot \\\\\n& \\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\\cdots\\right) \\cdot\\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\\cdots\\right) .\n\\end{aligned}\n$$\n\nEach $x^{10}$ term in the product is the result of multiplying four terms whose exponents sum to 10 , one from each factor of $f(x)$. Thus this product contains a term of $x^{10}$ for each quadruple\n\n\n\nof nonnegative integers $(i, j, k, l)$ such that $2^{i}+2^{j}+2^{k}+2^{l}=10$; the order of the quadruple is relevant because rearrangements of the integers correspond to choosing terms from different factors. Note that none of the exponents can exceed 2 because $2^{3}+2^{0}+2^{0}+2^{0}>10$. Therefore $i, j, k, l \\leq 2$. Considering cases from largest values to smallest yields two basic cases. First, $10=4+4+1+1=2^{2}+2^{2}+2^{0}+2^{0}$, which yields $\\frac{4 !}{2 ! \\cdot 2 !}=6$ ordered quadruples. Second, $10=4+2+2+2=2^{2}+2^{1}+2^{1}+2^{1}$, which yields 4 ordered quadruples. Thus the contribution of the $(f(x))^{4}$ term is $6+4=10$.\n\nThe last term to consider is $f(x)^{8}$, because $(f(x))^{n}$ contains no terms of degree less than $n$. An analogous analysis to the case of $(f(x))^{4}$ suggests that the expansion of $(f(x))^{8}$ has an $x^{10}$ term for every ordered partition of 10 into a sum of eight powers of two. Up to order, there is only one such partition: $2^{1}+2^{1}+2^{0}+2^{0}+2^{0}+2^{0}+2^{0}+2^{0}$, which yields $\\frac{8 !}{6 ! \\cdot 2 !}=28$ ordered quadruples.\n\nTherefore the coefficient of $x^{10}$ is $2+10+28=\\mathbf{4 0}$." ]

[ "40" ]

[ "Consider the expansion of $(1.002)^{10}$ as $(1+0.002)^{10}$. Using the Binomial Theorem yields the following:\n\n$$\n(1+0.002)^{10}=1+\\left(\\begin{array}{c}\n10 \\\\\n1\n\\end{array}\\right)(0.002)+\\left(\\begin{array}{c}\n10 \\\\\n2\n\\end{array}\\right)(0.002)^{2}+\\left(\\begin{array}{c}\n10 \\\\\n3\n\\end{array}\\right)(0.002)^{3}+\\cdots+(0.002)^{10} .\n$$\n\nHowever, when $k>3$, the terms $\\left(\\begin{array}{c}10 \\\\ k\\end{array}\\right)(0.002)^{k}$ do not affect the final answer, because $0.002^{4}=$ $0.000000000016=\\frac{16}{10^{12}}$, and the maximum binomial coefficient is $\\left(\\begin{array}{c}10 \\\\ 5\\end{array}\\right)=252$, so\n\n$$\n\\left(\\begin{array}{c}\n10 \\\\\n4\n\\end{array}\\right)(0.002)^{4}+\\left(\\begin{array}{c}\n10 \\\\\n5\n\\end{array}\\right)(0.002)^{5}+\\cdots+(0.002)^{10}<\\frac{252 \\cdot 16}{10^{12}}+\\frac{252 \\cdot 16}{10^{12}}+\\cdots+\\frac{252 \\cdot 16}{10^{12}},\n$$\n\nwhere the right side of the inequality contains seven terms, giving an upper bound of $\\frac{7 \\cdot 252 \\cdot 16}{10^{12}}$. The numerator is approximately 28000 , but $\\frac{28000}{10^{12}}=2.8 \\times 10^{-8}$. So even when multiplied by $100000=10^{5}$, these terms contribute at most $3 \\times 10^{-3}$ to the value of the expression before rounding.\n\nThe result of adding the first four terms $(k=0$ through $k=3)$ and multiplying by 100,000 is given by the following sum:\n\n$$\n100000+10(200)+45(0.4)+120(0.0008)=100000+2000+18+0.096=102018.096 .\n$$\n\nThen the desired quantity is $\\lfloor 102018.096\\rfloor=\\mathbf{1 0 2 , 0 1 8}$." ]

[ "102018" ]

[ "The common ratio in the geometric sequence $1, x, y$ is $\\frac{x}{1}=x$, so $y=x^{2}$. The arithmetic sequence $x, y, 3$ has a common difference, so $y-x=3-y$. Substituting $y=x^{2}$ in the equation yields\n\n$$\n\\begin{aligned}\nx^{2}-x & =3-x^{2} \\\\\n2 x^{2}-x-3 & =0\n\\end{aligned}\n$$\n\nfrom which $x=\\frac{3}{2}$ or -1 . The respective values of $y$ are $y=x^{2}=\\frac{9}{4}$ or 1 . Thus the possible values of $x+y$ are $\\frac{15}{4}$ and 0 , so the answer is $\\frac{\\mathbf{1 5}}{\\mathbf{4}}$." ]

[ "$\\frac{15}{4}$" ]

[ "The requirement that $a_{n}$ be the smallest positive value of $n-a_{k}^{2}$ for $k<n$ is equivalent to determining the largest value of $a_{k}$ such that $a_{k}^{2}<n$. For $n=3$, use either $a_{1}=a_{2}=1$ to find $a_{3}=3-1^{2}=2$. For $n=4$, the strict inequality eliminates $a_{3}$, so $a_{4}=4-1^{2}=3$, but $a_{3}$ can be used to compute $a_{5}=5-2^{2}=1$. In fact, until $n=10$, the largest allowable prior value of $a_{k}$ is $a_{3}=2$, yielding the values $a_{6}=2, a_{7}=3, a_{8}=4, a_{9}=5$. In general, this pattern continues: from $n=m^{2}+1$ until $n=(m+1)^{2}$, the values of $a_{n}$ increase from 1 to $2 m+1$. Let $S_{m}=1+2+\\cdots+(2 m+1)$. Then the problem reduces to computing $S_{0}+S_{1}+\\cdots+S_{6}+1$, because $a_{49}=49-6^{2}$ while $a_{50}=50-7^{2}=1 . S_{m}=\\frac{(2 m+1)(2 m+2)}{2}=2 m^{2}+3 m+1$, so\n\n$$\n\\begin{aligned}\nS_{0}+S_{1}+S_{2}+S_{3}+S_{4}+S_{5}+S_{6} & =1+6+15+28+45+66+91 \\\\\n& =252\n\\end{aligned}\n$$\n\nTherefore the desired sum is $252+1=\\mathbf{2 5 3}$." ]

[ "253" ]

[ "Write $253_{b} \\cdot 341_{b}=\\left(2 b^{2}+5 b+3\\right)\\left(3 b^{2}+4 b+1\\right)=6 b^{4}+23 b^{3}+31 b^{2}+17 b+3$. Compare the coefficients in this polynomial to the digits in the numeral $\\underline{7} \\underline{4} \\underline{X} \\underline{Y} \\underline{Z}$. In the polynomial, the coefficient of $b^{4}$ is 6 , so there must be a carry from the $b^{3}$ place to get the $7 b^{4}$ in the numeral. After the carry, there should be no more than 4 left for the coefficient of $b^{3}$ as only one $b$ is carried. Therefore $23-b \\leq 4$ or $b \\geq 19$. By comparing digits, note that $Z=3$. Then\n\n$$\n\\begin{aligned}\n6 b^{4}+23 b^{3}+31 b^{2}+17 b & =\\underline{7} \\underline{4} \\underline{X} \\underline{Y} \\underline{0} \\\\\n& =7 b^{4}+4 b^{3}+X \\cdot b^{2}+Y \\cdot b\n\\end{aligned}\n$$\n\nBecause $b>0$, this equation can be simplified to\n\n$$\nb^{3}+X \\cdot b+Y=19 b^{2}+31 b+17\n$$\n\nThus $Y=17$ and $b^{2}+X=19 b+31$, from which $b(b-19)=31-X$. The expression on the left side is positive (because $b>19$ ) and the expression on the right side is at most 31 (because $X>0$ ), so the only possible solution is $b=20, X=11$. The answer is 20 ." ]

[ "20" ]

[ "Notice first that all intersections of the two graphs occur in the interval $0 \\leq x \\leq \\frac{5 \\pi}{2}$, because the maximum value of $10 \\pi \\sin ^{2} x$ is $10 \\pi$ (at odd multiples of $\\frac{\\pi}{2}$ ), and $4 x>10 \\pi$ when $x>\\frac{5 \\pi}{2}$. The graphs are shown below.\n\n<img_3576>\n\nWithin that interval, both graphs are symmetric about the point $A=\\left(\\frac{5 \\pi}{4}, 5 \\pi\\right)$. For the case of $y=10 \\pi \\sin ^{2} x$, this symmetry can be seen by using the power-reducing identity $\\sin ^{2} x=$ $\\frac{1-\\cos 2 x}{2}$. Then the equation becomes $y=5 \\pi-5 \\pi \\cos 2 x$, which has amplitude $5 \\pi$ about the line $y=5 \\pi$, and which crosses the line $y=5 \\pi$ for $x=\\frac{\\pi}{4}, \\frac{3 \\pi}{4}, \\frac{5 \\pi}{4}, \\ldots$ Label the points of intersection $A, B, C, D, E, F$, and $O$ as shown. Then $\\overline{A B} \\cong \\overline{A C}, \\overline{B D} \\cong \\overline{C E}$, and $\\overline{O D} \\cong \\overline{E F}$. Thus\n\n$$\n\\begin{aligned}\nB D+A C+E F & =O D+D B+B A \\\\\n& =O A .\n\\end{aligned}\n$$\n\nBy the Pythagorean Theorem,\n\n$$\n\\begin{aligned}\nO A & =\\sqrt{\\left(\\frac{5 \\pi}{4}\\right)^{2}+(5 \\pi)^{2}} \\\\\n& =\\frac{5 \\pi}{4} \\sqrt{1^{2}+4^{2}} \\\\\n& =\\frac{5 \\pi}{\\mathbf{4}} \\sqrt{\\mathbf{1 7}}\n\\end{aligned}\n$$" ]

[ "$\\frac{5 \\pi}{4} \\sqrt{17}$" ]

[ "Let $\\mathrm{m} \\angle B=\\alpha$. Then the sum of the measures of the angles in the hexagon is:\n\n$$\n\\begin{aligned}\n720^{\\circ} & =\\mathrm{m} \\angle A+\\mathrm{m} \\angle C+\\mathrm{m} \\angle E+\\mathrm{m} \\angle D+\\mathrm{m} \\angle B+\\mathrm{m} \\angle F \\\\\n& =10 \\alpha+5 \\alpha+5 \\alpha+2 \\alpha+\\alpha+\\alpha=24 \\alpha .\n\\end{aligned}\n$$\n\n\n\nThus $30^{\\circ}=\\alpha$ and $\\mathrm{m} \\angle A=300^{\\circ}$, so the exterior angle at $A$ has measure $60^{\\circ}=\\mathrm{m} \\angle D$. Further, because $A B=C D$ and $D E=A F$, it follows that $\\triangle C D E \\cong \\triangle B A F$. Thus\n\n$$\n[A B C D E F]=[A B C E F]+[C D E]=[A B C E F]+[A B F]=[B C E F] .\n$$\n\n<img_3798>\n\nTo compute $[B C E F]$, notice that because $\\mathrm{m} \\angle D=60^{\\circ}, \\triangle C D E$ is equilateral. In addition,\n\n$$\n\\begin{aligned}\n150^{\\circ} & =\\mathrm{m} \\angle B C D \\\\\n& =\\mathrm{m} \\angle B C E+\\mathrm{m} \\angle D C E=\\mathrm{m} \\angle B C E+60^{\\circ} .\n\\end{aligned}\n$$\n\nTherefore $\\mathrm{m} \\angle B C E=90^{\\circ}$. Similarly, because the hexagon is symmetric, $\\mathrm{m} \\angle C E F=90^{\\circ}$, so quadrilateral $B C E F$ is actually a square with side length 3 . Thus $C E=\\frac{B E}{\\sqrt{2}}=\\frac{3}{\\sqrt{2}}$, and $[A B C D E F]=[B C E F]=\\frac{9}{2}$.\n\nAlternate Solution: Calculate the angles of the hexagon as in the first solution. Then proceed as follows.\n\nFirst, $A B C D E F$ can be partitioned into four congruent triangles. Because the hexagon is equilateral and $\\mathrm{m} \\angle A B C=\\mathrm{m} \\angle A F E=30^{\\circ}$, it follows that $\\triangle A B C$ and $\\triangle A F E$ are congruent isosceles triangles whose base angles measure $75^{\\circ}$. Next, $\\mathrm{m} \\angle A B C+\\mathrm{m} \\angle B C D=30^{\\circ}+150^{\\circ}=$ $180^{\\circ}$, so $\\overline{A B} \\| \\overline{C D}$. Because these two segments are also congruent, quadrilateral $A B C D$ is a parallelogram. In particular, $\\triangle C D A \\cong \\triangle A B C$. Similarly, $\\triangle E D A \\cong \\triangle A F E$.\n\nNow let $a=A C=A E$ be the length of the base of these isosceles triangles, and let $b=A B$ be the length of the other sides (or of the equilateral hexagon). Because the four triangles are congruent, $[A B C D E F]=[A B C]+[A C D]+[A D E]+[A E F]=4[A B C]=4 \\cdot \\frac{1}{2} b^{2} \\sin 30^{\\circ}=b^{2}$. Applying the Law of Cosines to $\\triangle A B C$ gives $a^{2}=b^{2}+b^{2}-2 b^{2} \\cos 30^{\\circ}=(2-\\sqrt{3}) b^{2}$. Because $4-2 \\sqrt{3}=(\\sqrt{3}-1)^{2}$, this gives $a=\\left(\\frac{\\sqrt{3}-1}{\\sqrt{2}}\\right) b$. Using the given length $B E=3$ and applying the Law of Cosines to $\\triangle A B E$ gives\n\n$$\n\\begin{aligned}\n9 & =a^{2}+b^{2}-2 a b \\cos 135^{\\circ} \\\\\n& =a^{2}+b^{2}+\\sqrt{2} a b \\\\\n& =(2-\\sqrt{3}) b^{2}+b^{2}+(\\sqrt{3}-1) b^{2} \\\\\n& =2 b^{2} .\n\\end{aligned}\n$$\n\nThus $[A B C D E F]=b^{2}=\\frac{9}{2}$." ]

[ "$\\frac{9}{2}$" ]

[ "Let $A=(0,5)$ and $B=(12,0)$, and let $C=(1,-1)$. First compute the distance sum: $d(A, C)+d(B, C)=19$. Notice that if $P=(x, y)$ is on the segment from $(0,-1)$ to $(12,-1)$, then $d(A, P)+d(B, P)$ is constant. This is because if $0<x<12$,\n\n$$\n\\begin{aligned}\nd(A, P)+d(B, P) & =|0-x|+|5-(-1)|+|12-x|+|0-(-1)| \\\\\n& =x+6+(12-x)+1 \\\\\n& =19\n\\end{aligned}\n$$\n\nSimilarly, $d(A, P)+d(P, B)=19$ whenever $P$ is on the segment from $(0,6)$ to $(12,6)$. If $P$ is on the segment from $(13,0)$ to $(13,5)$, then $P$ 's coordinates are $(13, y)$, with $0 \\leq y \\leq 5$, and thus\n\n$$\n\\begin{aligned}\nd(A, P)+d(B, P) & =|0-13|+|5-y|+|12-13|+|0-y| \\\\\n& =13+(5-y)+1+y \\\\\n& =19\n\\end{aligned}\n$$\n\nSimilarly, $d(A, P)+d(P, B)=19$ whenever $P$ is on the segment from $(-1,0)$ to $(-1,5)$.\n\nFinally, if $P$ is on the segment from $(12,-1)$ to $(13,0)$, then $d(A, P)+d(B, P)$ is constant:\n\n$$\n\\begin{aligned}\nd(A, P)+d(B, P) & =|0-x|+|5-y|+|12-x|+|0-y| \\\\\n& =x+(5-y)+(x-12)+(-y) \\\\\n& =2 x-2 y-7\n\\end{aligned}\n$$\n\nand because the line segment has equation $x-y=13$, this expression reduces to\n\n$$\n\\begin{aligned}\nd(A, P)+d(B, P) & =2(x-y)-7 \\\\\n& =2(13)-7 \\\\\n& =19\n\\end{aligned}\n$$\n\nSimilarly, $d(A, P)+d(B, P)=19$ on the segments joining $(13,5)$ and $(12,6),(0,6)$ and $(-1,5)$, and $(-1,0)$ to $(0,-1)$. The shape of the \"ellipse\" is given below.\n\n<img_3562>\n\n\n\nThe simplest way to compute the polygon's area is to subtract the areas of the four corner triangles from that of the enclosing rectangle. The enclosing rectangle's area is $14 \\cdot 7=98$, while each triangle has area $\\frac{1}{2} \\cdot 1 \\cdot 1=\\frac{1}{2}$. Thus the area is $98-4 \\cdot \\frac{1}{2}=\\mathbf{9 6}$." ]

[ "96" ]

[ "Substituting yields $\\frac{512}{f(2)}=2 f(2) \\Rightarrow(f(2))^{2}=256 \\Rightarrow f(2)=16$. Therefore $f(1)=\\frac{1}{32}$. Using the recursion, $f(3)=\\frac{1}{2}$ and $f(4)=8$. So $f(f(4))=f(8)$. Continue to apply the recursion:\n\n$$\nf(5)=4, \\quad f(6)=32, \\quad f(7)=128, \\quad f(8)=\\mathbf{4 0 9 6} .\n$$\n\nAlternate Solution: Let $g(n)=\\log _{2} f(n)$. Then $g(n)=g(n-1)+g(n-2)$, with initial conditions $g(1)=g(2)-9$ and $-g(1)=1+g(2)$. From this, $g(1)=-5$ and $g(2)=4$, and from the recursion,\n\n$$\ng(3)=-1, \\quad g(4)=3\n$$\n\nso $f(4)=2^{g(4)}=8$. Continue to apply the recursion:\n\n$$\ng(5)=2, \\quad g(6)=5, \\quad g(7)=7, \\quad g(8)=12\n$$\n\nBecause $g(f(4))=12$, it follows that $f(f(4))=2^{12}=\\mathbf{4 0 9 6}$." ]

[ "4096" ]

[ "When the coefficients of a polynomial $f$ are reversed to form a new polynomial $g$, the zeros of $g$ are the reciprocals of the zeros of $f: r$ is a zero of $f$ if and only if $r^{-1}$ is a zero of $g$. In this case, the two polynomials have the same zeros; that is, whenever $r$ is a zero of either, so must be $r^{-1}$. Furthermore, both $r$ and $r^{-1}$ must be real as well as integers, so $r= \\pm 1$. As the only zeros are \\pm 1 , and the greatest common divisor of all the coefficients is 1 , the polynomial must have leading coefficient 1 or -1 . Thus\n\n$$\n\\begin{aligned}\nf(x) & = \\pm(x \\pm 1)(x \\pm 1) \\cdots(x \\pm 1) \\\\\n& = \\pm(x+1)^{k}(x-1)^{n-k}\n\\end{aligned}\n$$\n\nIf $A_{n}$ is the number of such degree $n$ polynomials, then there are $n+1$ choices for $k, 0 \\leq k \\leq n$. Thus $A_{n}=2(n+1)$. The number of such degree $n$ polynomials for $1 \\leq n \\leq 7$ is the sum:\n\n$$\nA_{1}+A_{2}+\\ldots+A_{7}=2(2+3+\\ldots+8)=2 \\cdot 35=\\mathbf{7 0}\n$$" ]

[ "70" ]

[ "Label the sides of the polygon, in order, $s_{0}, s_{1}, \\ldots, s_{15}$. First note that two sides of the polygon intersect at a vertex if and only if the sides are adjacent. So the sides chosen must be nonconsecutive. Second, if nonparallel sides $s_{i}$ and $s_{j}$ are extended, the angle of intersection is determined by $|i-j|$, as are the lengths of the extended portions of the segments. In other words, the spacing of the extended sides completely determines the shape of the triangle. So the problem reduces to selecting appropriate spacings, that is, finding integers $a, b, c \\geq 2$ whose sum is 16 . However, diametrically opposite sides are parallel, so (for example) the sides $s_{3}$ and $s_{11}$ cannot both be used. Thus none of $a, b, c$ may equal 8 . Taking $s_{0}$ as the first side, the second side would be $s_{0+a}=s_{a}$, and the third side would be $s_{a+b}$, with $c$ sides between $s_{a+b}$ and $s_{0}$. To eliminate reflections and rotations, specify additionally that $a \\geq b \\geq c$. The allowable partitions are in the table below.\n\n| $a$ | $b$ | $c$ | triangle |\n| :---: | :---: | :---: | :---: |\n| 12 | 2 | 2 | $s_{0} s_{12} s_{14}$ |\n| 11 | 3 | 2 | $s_{0} s_{11} s_{14}$ |\n| 10 | 4 | 2 | $s_{0} s_{10} s_{14}$ |\n| 10 | 3 | 3 | $s_{0} s_{10} s_{13}$ |\n| 9 | 5 | 2 | $s_{0} s_{9} s_{14}$ |\n| 9 | 4 | 3 | $s_{0} s_{9} s_{13}$ |\n| 7 | 7 | 2 | $s_{0} s_{7} s_{14}$ |\n| 7 | 6 | 3 | $s_{0} s_{7} s_{13}$ |\n| 7 | 5 | 4 | $s_{0} s_{7} s_{12}$ |\n| 6 | 6 | 4 | $s_{0} s_{6} s_{12}$ |\n| 6 | 5 | 5 | $s_{0} s_{6} s_{11}$ |\n\nThus there are $\\mathbf{1 1}$ distinct such triangles." ]

[ "11" ]

[ "In the diagram below, $O$ is the center of both squares $A_{1} A_{2} A_{3} A_{4}$ and $B_{1} B_{2} B_{3} B_{4}$. Let $P_{1}, P_{2}, P_{3}, P_{4}$ and $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ be the intersections of the sides of the squares as shown. Let $H_{A}$ be on $\\overline{A_{3} A_{4}}$ so that $\\angle A_{3} H_{A} O$ is right. Similarly, let $H_{B}$ be on $\\overline{B_{3} B_{4}}$ such that $\\angle B_{3} H_{B} O$ is right. Then the angle by which $B_{1} B_{2} B_{3} B_{4}$ was rotated is $\\angle H_{A} O H_{B}$. Extend $\\overline{O H_{B}}$ to meet $\\overline{A_{3} A_{4}}$ at $M$.\n\n<img_3949>\n\nBoth $\\triangle H_{A} O M$ and $\\triangle H_{B} P_{3} M$ are right triangles sharing acute $\\angle M$, so $\\triangle H_{A} O M \\sim \\triangle H_{B} P_{3} M$. By an analogous argument, both triangles are similar to $\\triangle B_{3} P_{3} Q_{3}$. Thus $\\mathrm{m} \\angle Q_{3} P_{3} B_{3}=\\theta$. Now let $B_{3} P_{3}=x, B_{3} Q_{3}=y$, and $P_{3} Q_{3}=z$. By symmetry, notice that $B_{3} P_{3}=B_{2} P_{2}$ and that $P_{3} Q_{3}=P_{2} Q_{3}$. Thus\n\n$$\nx+y+z=B_{3} Q_{3}+Q_{3} P_{2}+P_{2} B_{2}=B_{2} B_{3}=3 .\n$$\n\nBy the Pythagorean Theorem, $x^{2}+y^{2}=z^{2}$. Therefore\n\n$$\n\\begin{aligned}\nx+y & =3-z \\\\\nx^{2}+y^{2}+2 x y & =9-6 z+z^{2} \\\\\n2 x y & =9-6 z .\n\\end{aligned}\n$$\n\nThe value of $x y$ can be determined from the areas of the four triangles $\\triangle B_{i} P_{i} Q_{i}$. By symmetry, these four triangles are congruent to each other. Their total area is the area not in both squares, i.e., $9-8=1$. Thus $\\frac{x y}{2}=\\frac{1}{4}$, so $2 x y=1$. Applying this result to the above equation,\n\n$$\n\\begin{aligned}\n1 & =9-6 z \\\\\nz & =\\frac{4}{3}\n\\end{aligned}\n$$\n\n\n\nThe desired quantity is $\\sin \\theta+\\cos \\theta=\\frac{x}{z}+\\frac{y}{z}$, and\n\n$$\n\\begin{aligned}\n\\frac{x}{z}+\\frac{y}{z} & =\\frac{x+y+z}{z}-\\frac{z}{z} \\\\\n& =\\frac{3}{z}-1 \\\\\n& =\\frac{\\mathbf{5}}{\\mathbf{4}}\n\\end{aligned}\n$$" ]

[ "$\\frac{5}{4}$" ]

[ "In order for the conditions of the problem to be satisfied, $M$ and $N$ must both be divisible by 3 , but not by 9 . Thus the largest possible value of $A$ is 5 , and the largest possible value of $B$ is 7 , so $A+B=\\mathbf{1 2}$." ]

[ "12" ]

[ "From the angle sum formula, $d^{\\circ}=\\frac{180^{\\circ} \\cdot(T-2)}{T}$. With $T=12, d=\\mathbf{1 5 0}$." ]

[ "150" ]

[ "$\\quad$ Distributing, $(r+1)(s+1)=r s+(r+s)+1=\\frac{F_{k+2}}{F_{k}}+\\left(-\\frac{F_{k+1}}{F_{k}}\\right)+1=\\frac{F_{k+2}-F_{k+1}}{F_{k}}+1=\\frac{F_{k}}{F_{k}}+1=\\mathbf{2}$." ]

[ "2" ]

[ "Multiplying, $(-T-i)(i-T)=-(i+T)(i-T)=-\\left(i^{2}-T^{2}\\right)=1+T^{2}$. With $T=2,1+T^{2}=\\mathbf{5}$." ]

[ "5" ]

[ "Let $N=20^{4} \\cdot 11^{T}=2^{8} \\cdot 5^{4} \\cdot 11^{T}$. If $m \\mid N$, then $m=2^{a} \\cdot 5^{b} \\cdot 11^{c}$ where $a, b$, and $c$ are nonnegative integers such that $a \\leq 8, b \\leq 4$, and $c \\leq T$. If $m$ is a perfect cube, then $a, b$, and $c$ must be divisible by 3 . So $a=0,3$, or $6 ; b=0$ or 3 , and $c \\in\\{0,3, \\ldots, 3 \\cdot\\lfloor T / 3\\rfloor\\}$. There are a total of $3 \\cdot 2 \\cdot(\\lfloor T / 3\\rfloor+1)$ possible values of $m$. For $T=5,\\lfloor T / 3\\rfloor+1=2$, so the number of possible values of $m$ is $\\mathbf{1 2}$." ]

[ "12" ]

[ "The general sequence looks like $x, x+d,-(x+2 d), x+3 d, x+4 d,-(x+5 d), \\ldots$ The sum of the first three terms is $x-d$; the sum of the second three terms is $x+2 d$; the sum of the third three terms is $x+5 d$, and so on. Thus the sequence of sums of terms $3 k-2,3 k-1$, and $3 k$ is an arithmetic sequence. Notice that $x=d=0.5$ and so $x-d=0$. If there are $n$ triads of terms of the original sequence, then their common difference is 1.5 and their sum is $n \\cdot\\left(\\frac{0+0+(n-1) \\cdot 1.5}{2}\\right) \\cdot T=72 \\sqrt{2}$, so $K=72$, and $n=24$. Thus the desired sum is 414." ]

[ "414" ]

[ "Let $P=(1,1), Q=(17,1)$, and $R=(B, 7)$ be the vertices of the triangle, and let $X=(B, 1)$ be the foot of the perpendicular from $R$ to $\\overleftrightarrow{P Q}$. Let $M=(A-1,1)$ and let $\\ell$ be the vertical line through $M$; then the problem is to determine the fraction of the area of $\\triangle P Q R$ that lies to the right of $\\ell$.\n\nNote that $B \\geq 0$ and $A \\geq 0$ because they are digit sums of integers. Depending on their values, the line $\\ell$ might intersect any two sides of the triangle or none at all. Each case\n\n\n\nrequires a separate computation. There are two cases where the computation is trivial. First, when $\\ell$ passes to the left of or through the leftmost vertex of $\\triangle P Q R$, which occurs when $A-1 \\leq \\min (B, 1)$, the probability is 1 . Second, when $\\ell$ passes to the right of or through the rightmost vertex of $\\triangle P Q R$, which occurs when $A-1 \\geq \\max (B, 17)$, the probability is 0 . The remaining cases are as follows.\n\nCase 1: The line $\\ell$ intersects $\\overline{P Q}$ and $\\overline{P R}$ when $1 \\leq A-1 \\leq 17$ and $A-1 \\leq B$.\n\nCase 2: The line $\\ell$ intersects $\\overline{P Q}$ and $\\overline{Q R}$ when $1 \\leq A-1 \\leq 17$ and $A-1 \\geq B$.\n\nCase 3: The line $\\ell$ intersects $\\overline{P R}$ and $\\overline{Q R}$ when $17 \\leq A-1 \\leq B$.\n\nNow proceed case by case.\n\nCase 1: Let $T$ be the point of intersection of $\\ell$ and $\\overline{P R}$. Then the desired probability is $[M Q R T] /[P Q R]=1-[P M T] /[P Q R]$. Since $\\triangle P M T \\sim \\triangle P X R$ and the areas of similar triangles are proportional to the squares of corresponding sides, $[P M T] /[P X R]=(P M / P X)^{2}$. Since $\\triangle P X R$ and $\\triangle P Q R$ both have height $X R$, their areas are proportional to their bases: $[P X R] /[P Q R]=P X / P Q$. Taking the product, $[P M T] /[P Q R]=(P M / P X)^{2}(P X / P Q)=$ $\\frac{P M^{2}}{P X \\cdot P Q}=\\frac{(A-2)^{2}}{(B-1)(17-1)}$, and the final answer is\n\n$$\n\\frac{[M Q R T]}{[P Q R]}=1-\\frac{[P M T]}{[P Q R]}=1-\\frac{(A-2)^{2}}{16(B-1)}\n$$\n\nCase 2: Let $U$ be the point of intersection of $\\ell$ and $\\overline{Q R}$. A similar analysis to the one in the previous case yields\n\n$$\n\\frac{[M Q U]}{[P Q R]}=\\frac{[M Q U]}{[X Q R]} \\cdot \\frac{[X Q R]}{[P Q R]}=\\left(\\frac{M Q}{X Q}\\right)^{2} \\frac{X Q}{P Q}=\\frac{(18-A)^{2}}{16(17-B)}\n$$\n\nCase 3: Let $T$ be the point of intersection of $\\ell$ and $\\overline{P R}$ and let $U$ be the point of intersection of $\\ell$ and $\\overline{Q R}$ as in the previous cases. Let $S$ be the point on $\\overline{P R}$ such that $\\overline{Q S} \\perp \\overline{P Q}$. Then $\\triangle T U R \\sim \\triangle S Q R$, so the areas of these two triangles are proportional to the squares of the corresponding altitudes $M X$ and $Q X$. Thinking of $\\overleftrightarrow{P R}$ as the common base, $\\triangle S Q R$ and $\\triangle P Q R$ have a common altitude, so the ratio of their areas is $S R / P R$. Since $\\triangle P Q S \\sim$ $\\triangle P X R, P S / P R=P Q / P X$ and so $\\frac{S R}{P R}=1-\\frac{P S}{P R}=1-\\frac{P Q}{P X}=\\frac{Q X}{P X}$. Therefore the desired probability is\n\n$$\n\\frac{[T U R]}{[P Q R]}=\\frac{[T U R]}{[S Q R]} \\cdot \\frac{[S Q R]}{[P Q R]}=\\left(\\frac{M X}{Q X}\\right)^{2} \\frac{Q X}{P X}=\\frac{(B-A+1)^{2}}{(B-17)(B-1)}\n$$\n\nUsing the answers from positions 7 and $9, A=4+1+4=9$ and $B=2+7=9$. The first case applies, so the probability is\n\n$$\n1-\\frac{(9-2)^{2}}{16(9-1)}=1-\\frac{49}{128}=\\frac{\\mathbf{7 9}}{\\mathbf{1 2 8}}\n$$" ]

[ "$\\frac{79}{128}$" ]

[ "Note that in general, $\\log _{b} c=\\log _{b^{n}} c^{n}$. Using this identity yields $\\log _{2} x^{T}=\\log _{2^{2}}\\left(x^{T}\\right)^{2}=$ $\\log _{4} x^{2 T}$. Thus the left hand side of the given equation simplifies to $\\log _{4} x^{2 T-1}$. Express each side in base 64: $\\log _{4} x^{2 T-1}=\\log _{64} x^{6 T-3}=\\log _{64} x^{2 k}=\\log _{8} x^{k}$. Thus $k=3 T-\\frac{3}{2}$. With $T=9.5, k=\\mathbf{2 7}$." ]

[ "27" ]

[ "Let $x$ be the length of the shorter base of the trapezoid. The area of the trapezoid is $\\frac{1}{2} \\cdot 2$. $(x+x+3)=T+1$, so $x=\\frac{T}{2}-1$. Drop perpendiculars from each vertex of the shorter base to the longer base, and note that by symmetry, the feet of these perpendiculars lie $\\frac{3}{2}=1.5$ units away from their nearest vertices of the trapezoid. Hence the congruent sides have length $\\sqrt{1.5^{2}+2^{2}}=2.5$. With $T=16, x=7$, and the desired sum of the lengths is $\\mathbf{9 . 5}$." ]

[ "9.5" ]

[ "The expected number of heads for Susan is $\\frac{T}{2}$. If Leo flips his coin $N$ times, the expected number of heads for Leo is $\\frac{N}{3}$. Thus $\\frac{N}{3}>\\frac{T}{2}$, so $N>\\frac{3 T}{2}$. With $T=10$, the smallest possible value of $N$ is $\\mathbf{1 6}$." ]

[ "16" ]

[ "Working together, Dennis and Edward take $\\frac{48}{2}=24$ minutes to mow a lawn. When the three of them work together, it takes them $\\frac{24}{2}=12$ minutes to mow a lawn. Thus they can mow 5 lawns per hour. With $T=1$, they can mow $5 \\cdot 2=\\mathbf{1 0}$ lawns in 2 hours." ]

[ "10" ]

[ "Note that $\\sin \\frac{(5-T) \\pi}{2}=\\cos \\left(\\frac{\\pi}{2}-\\frac{(5-T) \\pi}{2}\\right)=\\cos \\left(\\frac{T \\pi}{2}-2 \\pi\\right)=\\cos \\frac{T \\pi}{2}$. Thus the desired quantity is $\\sin ^{2} \\frac{T \\pi}{2}+\\cos ^{2} \\frac{T \\pi}{2}=\\mathbf{1}$ (independent of $T$ )." ]

[ "1" ]

[ "Squaring each side gives $20+\\sqrt{T+x}=25$, thus $\\sqrt{T+x}=5$, and $x=25-T$. With $T=11$, $x=14$." ]

[ "14" ]

[ "Using the angle sum formula, $180^{\\circ} \\cdot(n-2)=180^{\\circ} \\cdot 3+180^{\\circ} \\cdot 6=180^{\\circ} \\cdot 9$. Thus $n-2=9$, and $n=11$." ]

[ "11" ]