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[ "Let $\\log _{T} 8=x$. Then $T^{x}=8$. Thus the given expression equals $2^{x}-\\left(T^{x}\\right)^{\\log _{T} 2}=2^{x}-T^{x \\log _{T} 2}=$ $2^{x}-T^{\\log _{T} 2^{x}}=2^{x}-2^{x}=\\mathbf{0}$ (independent of $T$ )." ]

[ "0" ]

[ "The problem requests the value of $k$ such that $20+k+T+5=20(T+2)$, thus $k=19 T+15$. With $T=0$, it follows that $k=\\mathbf{1 5}$." ]

[ "15" ]

[ "Let $h$ be the distance between $\\overline{A R}$ and $\\overline{M L}$, and for simplicity, let $A R=M L=15 n$. Then $[A R M L]=15 n h$, and $[P Q W Z]=(1 / 2)(P Q+W Z) h$. Note that $P Q=15 n / 3=5 n$ and $W Z=15 n-3 n-3 n=9 n$. Thus $[P Q W Z]=7 n h=(7 / 15) \\cdot[A R M L]=7 T / 15$. With $T=15$, the answer is 7 ." ]

[ "7" ]

[ "Let $N=T+10$. In order for $k^{3}(k \\in \\mathbb{N})$ to be a divisor of $N$ !, the largest odd prime factor of $k$ (call it $p$ ) must be less than or equal to $N / 3$ so that there are at least three multiples of $p$ among the product of the first $N$ positive integers. If $p=3$, then the smallest possible value of $N$ is 9 , and the largest perfect cube factor of 9 ! is $2^{6} \\cdot 3^{3}$. Similarly, if $p=5$, then the smallest possible value of $N$ is 15 , and the largest perfect cube factor of 15 ! is $2^{9} \\cdot 3^{6} \\cdot 5^{3}$. With $T=7, N=17$, and the largest perfect cube factor of 17 ! is $2^{15} \\cdot 3^{6} \\cdot 5^{3}$. Thus $k^{3} \\mid 17$ ! if and only if $k \\mid 2^{5} \\cdot 3^{2} \\cdot 5^{1}$. Therefore $k=2^{x} 3^{y} 5^{z}$, where $x, y, z$ are nonnegative integers with $x \\leq 5, y \\leq 2, z \\leq 1$, yielding $6 \\cdot 3 \\cdot 2=\\mathbf{3 6}$ possible values of $k$." ]

[ "36" ]

[ "Note that the $x$-coordinates of $A$ and $M$ correspond to the two roots $r_{1}, r_{2}$ of $x^{2}+2 x-T$. If $s$ is the side length of square $A R M L$, then $A M=s \\sqrt{2}=\\left|r_{1}-r_{2}\\right|=\\sqrt{\\left(r_{1}-r_{2}\\right)^{2}}=$ $\\sqrt{\\left(r_{1}+r_{2}\\right)^{2}-4 r_{1} r_{2}}=\\sqrt{(-2)^{2}-4(-T)}=2 \\sqrt{1+T}$. Thus $[A R M L]=s^{2}=2(1+T)$. With $T=36,[A R M L]=\\mathbf{7 4}$." ]

[ "74" ]

[ "The given information implies that triangles $H E X, X A G$, and $G O H$ are congruent, hence triangle $H X G$ is equilateral. If $H X=s$, then the radius of the circle circumscribing $\\triangle H X G$ is $s / \\sqrt{3}$ so that the circle's area is $\\pi s^{2} / 3$. It remains to compute $s$. With $\\mathrm{m} \\angle H E X=120^{\\circ}$, use the Law of Cosines to find\n\n$$\n\\begin{aligned}\nH X^{2} & =H E^{2}+E X^{2}-2 H E \\cdot E X \\cdot \\cos 120^{\\circ} \\\\\n& =p^{2}+q^{2}-2 p q(-1 / 2) \\\\\n& =p^{2}+q^{2}+p q .\n\\end{aligned}\n$$\n\nUsing the answers 74 and 7 from positions 7 and 9 , respectively, conclude that $S=\\{2,7,37\\}$ and that $(p, q)=(2,7)$. Hence the foregoing yields $H X^{2}=4+49+14=67$. Thus the area of circle $\\omega$ is $\\frac{\\mathbf{6 7 \\pi}}{\\mathbf{3}}$." ]

[ "$\\frac{67 \\pi}{3}$" ]

[ "There are $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)$ ways of choosing the two people to set up and $\\left(\\begin{array}{c}n-2 \\\\ 2\\end{array}\\right)$ ways of choosing the two people to take down the campsite, so there are $\\frac{n(n-1)}{2} \\cdot \\frac{(n-2)(n-3)}{2}$ ways of choosing the four people, or $\\frac{n(n-1)(n-2)(n-3)}{4}$ ways total; call this function $C(n)$. For the least $n$ such that $\\frac{n(n-1)(n-2)(n-3)}{4} \\geq T$, as a rough approximation, note that $n-3<\\sqrt[4]{4 T}<n$. With $T=184$, the approximation becomes $n-3<\\sqrt[4]{736}<n$. Now $5^{4}=625$ while $6^{4}=1296$, so $5<n<9$. Try values starting from $n=6$ :\n\n$$\n\\begin{aligned}\n& C(6)=\\frac{6 \\cdot 5 \\cdot 4 \\cdot 3}{4}=90 \\\\\n& C(7)=\\frac{7 \\cdot 6 \\cdot 5 \\cdot 4}{4}=210 .\n\\end{aligned}\n$$\n\nThus $n=7$." ]

[ "7" ]

[ "In this case, the two parabolas are tangent exactly when the system of equations has a unique solution. (Query: Is this the case for every pair of equations representing parabolas?) So set the right sides equal to each other: $x^{2}+T x=-(x-2 T)^{2}+b$. Then $x^{2}+T x=$ $-x^{2}+4 T x-4 T^{2}+b$, or equivalently, $2 x^{2}-3 T x+4 T^{2}-b=0$. The equation has a double root when the discriminant is 0 , so set $(-3 T)^{2}-4\\left(4 T^{2}-b\\right)(2)=0$ and solve: $9 T^{2}-32 T^{2}+8 b=0$ implies $-23 T^{2}+8 b=0$, or $b=23 T^{2} / 8$. Using $T=8$ yields $b=\\mathbf{1 8 4}$." ]

[ "184" ]

[ "Using the identity $\\left(x^{2}-y^{2}\\right)^{2}+(2 x y)^{2}=\\left(x^{2}+y^{2}\\right)^{2}$, notice that $a_{2 n+1}^{2}+a_{2 n+2}^{2}=\\left(a_{2 n}^{2}-a_{2 n-1}^{2}\\right)^{2}+$ $\\left(2 a_{2 n} a_{2 n-1}\\right)^{2}=\\left(a_{2 n}^{2}+a_{2 n-1}^{2}\\right)^{2}$. So surprisingly, for all $n \\in \\mathbb{N}, a_{2 n+1}^{2}+a_{2 n+2}^{2}=1$. Thus if $n$ is even, the sum of the squares of the first $n$ terms is $n / 2$. With $T=19, T-3=16$, and the sum is 8 ." ]

[ "8" ]

[ "Using the formula $D(n)=\\frac{n(n-3)}{2}$ twice yields $D(n)-D(n-1)=\\frac{n^{2}-3 n}{2}-\\frac{n^{2}-5 n+4}{2}=\\frac{2 n-4}{2}=n-2$. So $T=n-2$, thus $n=T+2$, and with $T=17, n=19$." ]

[ "19" ]

[ "If $d$ is the common difference of the sequence, then the $n^{\\text {th }}$ term of the sequence is $a_{n}=$ $a_{16}+d(n-16)$. The values $a_{16}=13$ and $a_{30}=20$ yield $d=(20-13) /(30-16)=1 / 2$, hence $a_{n}=13+(1 / 2)(n-16)$. If $a_{n}=T$, then $n=2(T-13)+16=2 T-10$. With $T=27 / 2$, it follows that $n=\\mathbf{1 7}$." ]

[ "17" ]

[ "The surface area is given by the expression $2 \\cdot 1 \\cdot 3+2 \\cdot 1 \\cdot h+2 \\cdot 3 \\cdot h=6+8 h$. Because $6+8 h=T, h=\\frac{T-6}{8}$. With $T=114, h=108 / 8=\\mathbf{2 7} / \\mathbf{2}$." ]

[ "$\\frac{27}{2}$" ]

[ "Use sums and products of roots formulas: the desired quantity $c=(r+1)(s+1)=r s+r+s+1$. From the first equation, $r s=93$, while from the second equation, $(r+1)+(s+1)=r+s+2=$ 22. So $r s+r+s+1=93+22-1=\\mathbf{1 1 4}$." ]

[ "114" ]

[ "Write $N$ as\n\n$$\n\\begin{aligned}\n& (10,000,000-1) \\cdot 888,888 \\\\\n= & 8,888,880,000,000-888,888 \\\\\n= & 8,888,879,111,112 .\n\\end{aligned}\n$$\n\nThe sum of the digits of $N$ is 63 ." ]

[ "63" ]

[ "Any two of the triangles intersect in at most six points, because each side of one triangle can intersect the other triangle in at most two points. To count the total number of intersections among the five triangles, note that there are $\\left(\\begin{array}{l}5 \\\\ 2\\end{array}\\right)=10$ ways to select a pair of triangles, and each pair may result in 6 intersections. Thus $10 \\times 6=60$ is an upper bound.\n\nThis can be achieved, for example, by taking six equilateral triangles of equal size, centered at a single point, and rotating them different amounts so that no three sides intersect at a single point. Thus the answer is 60." ]

[ "60" ]

[ "In order for the sums of the squares of four digits to be 17 , the digits must be either $0,2,2$, and 3 , or $0,0,1$, and 4 , in some order. If the leading digit is 2 , there are $3 !=6$ possible four-digit numbers. If the leading digit is 1,3 , or 4 , there are $\\frac{3 !}{2 !}=3$ possible four-digit numbers. In total, there are $6+3 \\cdot 3=15$ four-digit integers in $S$, and the median will be the eighth least. The least eight integers in $S$, from least to greatest, are: 1004, 1040, 1400, 2023, 2032, 2203, 2230, 2302. Thus the median of $S$ is 2302." ]

[ "2302" ]

[ "Let $C L=x$. Because the quadrilaterals $E U C L$ and $L I D E$ are congruent, $\\overline{E L}$ is a diameter of the circle in which the hexagon is inscribed, so $E L=10$. Furthermore, because $\\overline{E L}$ is a diameter of the circle, it follows that the inscribed $\\angle E U L$ is a right angle, hence $U L=8$.\n\n\n\n<img_3485>\n\nUsing Ptolemy's Theorem for cyclic quadrilaterals and the fact that $\\triangle E C L$ is also a right triangle,\n\n$$\n\\begin{aligned}\n& U C \\cdot E L+E U \\cdot C L=E C \\cdot U L \\\\\n\\Longrightarrow & 6(10+x)=8 \\sqrt{100-x^{2}} \\\\\n\\Longrightarrow & 36(10+x)^{2}=64(10+x)(10-x) \\\\\n\\Longrightarrow & 6 \\sqrt{10+x}=8 \\sqrt{10-x} \\\\\n\\Longrightarrow & 36(10+x)=64(10-x) \\\\\n\\Longrightarrow & 360+36 x=640-64 x \\\\\n\\Longrightarrow & 100 x=280 \\\\\n\\Longrightarrow & x=\\frac{\\mathbf{1 4}}{\\mathbf{5}} .\n\\end{aligned}\n$$" ]

[ "$\\frac{14}{5}$" ]

[ "Any 15-letter palindrome is determined completely by its first 8 letters, because the last 7 letters must be the first 7 in reverse. Such a palindrome contains the string $A R M L$ if and only if its first 8 letters contain either $A R M L$ or $L M R A$. (The string $A R M L$ cannot cross the middle of the palindrome, because the 7th and 9th letters must be the same.) It therefore suffices to count the number of 8-letter strings consiting of letters in the ARMLLexicon that contain either ARML or LMRA.\n\nThere are 5 possible positions for $A R M L$, and likewise with $L M R A$. For each choice of position, there are four remaining letters, which can be any letter in the ARMLLexicon (here, $W, X, Y$, and $Z$ are used to denote arbitrary letters that need not be distinct). This leads to the following table:\n\n\n\n| Word | Num. Possibilities |\n| :---: | :---: |\n| ARMLWXYZ | $10^{4}$ |\n| WARMLXYZ | $10^{4}$ |\n| WXARMLYZ | $10^{4}$ |\n| WXYARMLZ | $10^{4}$ |\n| WXYZARML | $10^{4}$ |\n| LMRAWXYZ | $10^{4}$ |\n| WLMRAXYZ | $10^{4}$ |\n| WXLMRAYZ | $10^{4}$ |\n| WXYLMRAZ | $10^{4}$ |\n| WXYZLMRA | $10^{4}$ |\n\nThis gives $10 \\cdot 10^{4}$ possible words, but each word with two of ARML or LMRA (e.g., ARMLARML or $A A R M L M R A$ ) is counted twice. There are four words with two of $A R M L$ or $L M R A$ that use all 8 letters, and four possible types of words that use 7 of the 8 positions and leave one \"free space\". This leads to the following table:\n\n| Word | Num. Possibilities |\n| :---: | :---: |\n| ARMLARML | 1 |\n| LMRALMRA | 1 |\n| ARMLLMRA | 1 |\n| LMRAARML | 1 |\n| ARMLMRAW | 10 |\n| LMRARMLW | 10 |\n| WARMLMRA | 10 |\n| WLMRARML | 10 |\n\nThus the total number of desired words is $10 \\cdot 10^{4}-4 \\cdot 10-4 \\cdot 1=\\mathbf{9 9 9 5 6}$." ]

[ "99956" ]

[ "First, note that\n\n$$\n\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\leq(\\log x)^{2} \\Longrightarrow \\frac{3+\\left\\lfloor(\\log x)^{2}\\right\\rfloor}{4} \\leq \\frac{3+(\\log x)^{2}}{4}\n$$\n\nTherefore\n\n$$\n\\log x \\leq \\frac{(\\log x)^{2}+3}{4} \\Longrightarrow 0 \\leq(\\log x)^{2}-4 \\log x+3=(\\log x-1)(\\log x-3)\n$$\n\nThis implies either $\\log x \\leq 1$ or $\\log x \\geq 3$, so $0 \\leq(\\log x)^{2} \\leq 1$ or $(\\log x)^{2} \\geq 9$.\n\nIn the first case, $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=0$ or $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=1$, so $\\log x=\\frac{3}{4}$ or $\\log x=1$, hence $x=10^{3 / 4}$ or $x=10$.\n\nTo solve the second case, note that $\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\geq(\\log x)^{2}-1$, so $0 \\geq(\\log x)^{2}-4 \\log x+2$. The solutions to $t^{2}-4 t+2=0$ are $t=\\frac{4 \\pm \\sqrt{16-8}}{2}=2 \\pm \\sqrt{2}$ by the Quadratic Formula, so $2-\\sqrt{2} \\leq \\log x \\leq 2+\\sqrt{2}$. This implies that $6-4 \\sqrt{2} \\leq(\\log x)^{2} \\leq 6+4 \\sqrt{2}$, so $0 \\leq\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\leq 11$. However, this case is for $(\\log x)^{2} \\geq 9$, so the only possibilities that need to be considered are $9 \\leq\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\leq 11$.\n\n- If $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=9$, then $\\log x=3$, so $x=10^{3}$.\n- If $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=10$, then $\\log x=\\frac{13}{4}$, so $x=10^{13 / 4}$.\n- Finally, if $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=11$, then $\\log x=\\frac{7}{2}$, which yields $(\\log x)^{2}=\\frac{49}{4}>12$, so there are no solutions.\n\nThus the product of all possible values of $x$ is $y=10^{3 / 4} \\cdot 10 \\cdot 10^{13 / 4} \\cdot 10^{3}=10^{8}$, so $y=\\mathbf{8}$." ]

[ "8" ]

[ "First note that the solutions of the given equation are real because the equation's discriminant is positive. By Vieta's Formulas, $r_{1}+r_{2}=180(*)$ and $r_{1} r_{2}=8(* *)$. The expression to be computed can be written with a common denominator as\n\n$$\n\\frac{\\sqrt[3]{r_{1}^{4}}+\\sqrt[3]{r_{2}^{4}}}{\\sqrt[3]{r_{1} r_{2}}}\n$$\n\nBy $(* *)$, the denominator is equal to $\\sqrt[3]{8}=2$. To compute the numerator, first let $S_{k}=\\sqrt[3]{r_{1}^{k}}+\\sqrt[3]{r_{2}^{k}}$, so that the numerator is $S_{4}$. Then note that\n\n$$\n\\begin{aligned}\n\\left(S_{1}\\right)^{3} & =r_{1}+3 \\sqrt[3]{r_{1}^{2} r_{2}}+3 \\sqrt[3]{r_{2}^{2} r_{1}}+r_{2} \\\\\n& =\\left(r_{1}+r_{2}\\right)+3 \\sqrt[3]{r_{1} r_{2}}\\left(\\sqrt[3]{r_{1}}+\\sqrt[3]{r_{2}}\\right) \\\\\n& =180+3 \\cdot 2 \\cdot S_{1}\n\\end{aligned}\n$$\n\nwhere $(*)$ and $(* *)$ are used to substitute values into the second equality. Next note that $S_{1}^{3}-6 S_{1}-180$ can be factored as $\\left(S_{1}-6\\right)\\left(S_{1}^{2}+6 S_{1}+30\\right)$. Because the polynomial $t^{2}+6 t+30$ has no real roots, the unique real solution to $(\\dagger)$ is $S_{1}=6$, so $\\sqrt[3]{r_{1}}+\\sqrt[3]{r_{2}}=6$. Square each side of the previous equation to obtain $S_{2}+2 \\sqrt[3]{r_{1} r_{2}}=36$, hence $S_{2}=36-2 \\cdot 2$; that is, $\\sqrt[3]{r_{1}^{2}}+\\sqrt[3]{r_{2}^{2}}=32$. Again, square both sides of this equation to obtain $\\sqrt[3]{r_{1}^{4}}+2 \\sqrt[3]{r_{1}^{2} r_{2}^{2}}+\\sqrt[3]{r_{2}^{4}}=1024$, so $S_{4}+2 \\sqrt[3]{r_{1}^{2} r_{2}^{2}}=1024$, from which $S_{4}=1024-2 \\cdot 4=1016$. Thus the desired expression equals $\\frac{S_{4}}{2}=\\frac{1016}{2}=\\mathbf{5 0 8}$." ]

[ "508" ]

[ "Let $O, O_{1}$ and $O_{2}$ be the centers, and let $r, r_{1}$ and $r_{2}$ be the radii of the circles $\\omega, \\omega_{1}$, and $\\omega_{2}$, respectively. Let $R$ be the point of tangency between $\\omega_{1}$ and $\\omega_{2}$.\n\nLet $H_{1}$ and $H_{2}$ be the projections of $O_{1}$ and $O_{2}$ onto $\\overline{A B}$. Also, let $H$ be the projection of $O_{1}$ onto $\\overline{O_{2} H_{2}}$. Note that $O H_{1}=r-r_{1}, O H_{2}=r-r_{2}, O O_{1}=r+r_{1}, O O_{2}=r+r_{2}$, and $O_{1} O_{2}=r_{1}+r_{2}$. From the Pythagorean Theorem, it follows that $O_{1} H_{1}=2 \\sqrt{r r_{1}}$ and $O_{2} H_{2}=2 \\sqrt{r r_{2}}$. Similarly, applying the Pythagorean Theorem to triangle $O_{1} H O_{2}$ yields $\\left(O_{1} H\\right)^{2}+\\left(O_{2} H\\right)^{2}=\\left(O_{1} O_{2}\\right)^{2}$, which is equivalent to\n\n$$\n\\left(2 \\sqrt{r r_{2}}-2 \\sqrt{r r_{1}}\\right)^{2}+\\left(2 r-r_{1}-r_{2}\\right)^{2}=\\left(r_{1}+r_{2}\\right)^{2}\n$$\n\nwhich yields $r^{2}=4 r_{1} r_{2}$ after simplifying.\n<img_4036>\n\n\n\nNote that $\\overline{A O} \\| \\overline{O_{2} D}$, hence $\\angle A O Q \\cong \\angle D O_{2} Q$, which implies that isosceles triangles $A O Q$ and $D O_{2} Q$ are similar. Thus $\\angle A Q O \\cong \\angle D Q O_{2}$ and therefore points $A, Q$, and $D$ are collinear. Analogously, it follows that the points $B, P$, and $C$ are collinear, as are the points $C, R$, and $D$.\n\nIn right triangle $A B D, \\overline{B Q}$ is the altitude to $\\overline{A D}$. By similarity of triangles, it follows that $D Q \\cdot D A=B D^{2}$ and $A Q \\cdot A D=A B^{2}$. Hence $B D=4 \\sqrt{10}, A B=4 \\sqrt{15}$, and $r=2 \\sqrt{15}$. Because $\\frac{D O_{2}}{A O}=\\frac{D Q}{A Q}=\\frac{2}{3}$, it follows that $r_{2}=\\frac{4}{3} \\sqrt{15}$ and $r_{1}=\\frac{3}{4} \\sqrt{15}$.\n\nNote that $A C=2 \\sqrt{r r_{1}}=3 \\sqrt{10}, B D=2 \\sqrt{r r_{2}}=4 \\sqrt{10}$, and\n\n$$\nC D^{2}=A B^{2}+(B D-A C)^{2}=(4 \\sqrt{15})^{2}+(4 \\sqrt{10}-3 \\sqrt{10})^{2}=240+10=250\n$$\n\nwhich implies that $C D=\\mathbf{5} \\sqrt{\\mathbf{1 0}}$.\n\nAlternate Solution: Conclude that $r^{2}=4 r_{1} r_{2}$, as explained above. Note that $\\angle C A Q \\cong \\angle Q D B \\cong \\angle Q R D$, using the fact that the two given lines are parallel and $\\omega_{2}$ is tangent one of them at $D$. Quadrilateral $C A Q R$ is cyclic, so apply Power of a Point to obtain $D Q \\cdot D A=D R \\cdot D C$. Because $\\frac{r_{2}}{r}=\\frac{Q D}{Q A}=\\frac{2}{3}$, conclude that $r_{2}=2 x, r=3 x$, and hence $r_{1}=\\frac{9}{8} x$. It follows that $\\frac{D R}{C R}=\\frac{r_{2}}{r_{1}}=\\frac{16}{9}$ and $D R=\\frac{16}{25} \\cdot C D$. Thus\n\n$$\nD R \\cdot D C=\\frac{16}{25} \\cdot C D^{2}=D Q \\cdot D A=8 \\cdot 20\n$$\n\nhence $C D=5 \\sqrt{10}$." ]

[ "$5 \\sqrt{10}$" ]

[ "Notice that $\\triangle A R M$ is fixed, so the number of integers that could be the perimeter of $A R M L$ is the same as the number of integers that could be the length $A L$ in $\\triangle A L M$. By the Triangle Inequality, $32-25<A L<32+25$, so $A L$ is at least 8 and no greater than 56 . The number of possible integer values for $A L$ is $56-8+1=49$." ]

[ "49" ]

[ "For brevity, $P$ will be used to represent the polynomial $P(x)$, and let $\\operatorname{deg}(P)$ represent the degree of $P$. Rewrite the given condition as follows:\n\n$$\n\\begin{aligned}\n\\frac{1}{A(x)}+\\frac{1}{B(x)}+\\frac{1}{x+10}=\\frac{1}{x} & \\Longrightarrow \\frac{A+B}{A B}=\\frac{10}{x(x+10)} \\\\\n& \\Longrightarrow A B-\\frac{x(x+10)}{10} A-\\frac{x(x+10)}{10} B=0 \\\\\n& \\Longrightarrow\\left(A-\\frac{x(x+10)}{10}\\right)\\left(B-\\frac{x(x+10)}{10}\\right)=\\frac{x^{2}(x+10)^{2}}{100} .\n\\end{aligned}\n$$\n\nBecause $A$ and $B$ are both polynomials, $A-\\frac{x(x+10)}{10}$ must be some factor $F$ of $\\frac{x^{2}(x+10)^{2}}{100}$. Furthermore, if $\\operatorname{deg}(F) \\leq 1$, then $A$ has leading coefficient $\\frac{1}{10}$, which violates the condition that $A$ has leading coefficient 1 . So\n\n\n\n$\\operatorname{deg}(F) \\geq 2$. Thus $F$ must be a nonzero constant times one of\n\n$$\n\\left\\{x^{2}, x(x+10),(x+10)^{2}, x^{2}(x+10), x(x+10)^{2}, x^{2}(x+10)^{2}\\right\\} .\n$$\n\nThe degree of $A$ determines what this constant must be.\n\n- If $\\operatorname{deg}(A) \\geq 3$, then $\\operatorname{deg}(F)=\\operatorname{deg}(A) \\geq 3$ and $F$ has leading coefficient 1 . Any such $F$ is valid.\n- If $\\operatorname{deg}(A)=2$, then $\\operatorname{deg}(F)=2$ and $F$ has leading coefficient $\\frac{9}{10}$. Again, any such $F$ is valid.\n- If $\\operatorname{deg}(A) \\leq 1$, then $\\operatorname{deg}(F)=2$ and $F$ has leading coefficient $-\\frac{1}{10}$. But not all $F$ are valid, because this does not guarantee that the leading coefficient of $A$ is 1 . Among $-\\frac{1}{10} x^{2},-\\frac{1}{10} x(x+10)$, and $-\\frac{1}{10}(x+10)^{2}$ as possible values of $F$, only $-\\frac{1}{10} x^{2}$ gives a valid $A$ with leading coefficient 1 .\n\nThus $F$ is one of\n\n$$\n\\left\\{-\\frac{1}{10} x^{2}, \\frac{9}{10} x^{2}, \\frac{9}{10} x(x+10), \\frac{9}{10}(x+10)^{2}, x^{2}(x+10), x(x+10)^{2}, x^{2}(x+10)^{2}\\right\\} .\n$$\n\nThen\n\n$$\n\\sum\\left(A(10)-\\frac{10 \\cdot 20}{10}\\right)=-\\frac{1}{10} \\cdot 10^{2}+\\frac{9}{10} \\cdot\\left(10^{2}+10 \\cdot 20+20^{2}\\right)+\\left(10^{2} \\cdot 20+10 \\cdot 20^{2}+10^{2} \\cdot 20^{2}\\right)=46620\n$$\n\nso $\\sum A(10)=7 \\cdot \\frac{10 \\cdot 20}{10}+46620=\\mathbf{4 6 7 6 0}$, as desired." ]

[ "46760" ]

[ "Let $r$ and $Q$ denote the respective radius and center of the circle whose radius is concerned. Let this circle be tangent to arc $\\widehat{P_{1} P_{2}}$ at point $P$, and let it be tangent to sides $\\overline{M N}$ and $\\overline{M R}$ at points $T_{1}$ and $T_{2}$, respectively.\n\n<img_3571>\n\nNote that $Q$ lies on diagonal $\\overline{M O}$ because it is equidistant to $\\overline{M N}$ and $\\overline{M R}$. Points $Q, P$, and $O$ must be collinear because the circles centered at $Q$ and $O$ are mutually tangent at point $P$. It therefore follows that $P$ also lies on diagonal $\\overline{M O}$. Because triangles $Q T_{1} M$ and $Q T_{2} M$ are isosceles right triangles, it follows that $M Q=r \\sqrt{2}$. Thus\n\n$$\nb \\sqrt{2}=M O=M Q+Q P+P O=r \\sqrt{2}+r+a\n$$\n\nSolving this equation yields $r=a+2 b-(a+b) \\sqrt{2}$. With $T=688, a=6$ and $b=8$, so $r=22-14 \\sqrt{2}$, hence $x+y=22+14=\\mathbf{3 6}$." ]

[ "36" ]

[ "Let $A M=a$ and $A P=b$, and let $s=\\sqrt{T}$ be the side length of square $A B C D$. Then $M B=s-a$ and $D P=s-b$. Using the right angles of $M N O P$ and complementary acute angles in triangles $A M P, B N M$, $C O N$, and $D P O$, note that\n\n$$\n\\angle A M P \\cong \\angle B N M \\cong \\angle C O N \\cong D P O\n$$\n\nAlso note that $\\mathrm{m} \\angle B M N=180^{\\circ}-\\left(90^{\\circ}+\\mathrm{m} \\angle A M P\\right)$, so it also follows that\n\n$$\n\\angle B M N \\cong \\angle C N O \\cong \\angle D O P \\cong A P M\n$$\n\n<img_3526>\n\nThus, by side-angle-side congruence, it follows that $\\triangle A M P \\cong \\triangle C O N$ and $\\triangle B N M \\cong \\triangle D P O$. Moreover, by side-angle-side similarity, it follows that $\\triangle A M P \\sim \\triangle B N M \\sim \\triangle C O N \\sim \\triangle D P O$. Thus $B N=s-b, N C=b$, $C O=a$, and $O D=s-a$. The similarity relation implies $\\frac{A M}{B N}=\\frac{A P}{B M}$, so $\\frac{a}{s-b}=\\frac{b}{s-a}$. Cross-multiplying, rearranging, and simplifying yields $s(a-b)=(a+b)(a-b)$. Thus either $a=b$ or $s=a+b$. In the case where $a=b, A M=A P=\\frac{2}{\\sqrt{2}}=\\sqrt{2}$, so $M N=(s-\\sqrt{2}) \\sqrt{2}=s \\sqrt{2}-2$. With $T=36, s=6$, and the answer is thus $6 \\sqrt{\\mathbf{2}}-\\mathbf{2}$. For completeness, it remains to verify that for this particular value of $s$, the case where $s=a+b$ is impossible. Applying the Pythagorean Theorem in $\\triangle M A P$ yields $a^{2}+b^{2}=4$. Now if $s=6=a+b$, then by squaring, it would follow that $a^{2}+b^{2}+2 a b=36 \\Longrightarrow 4+2 a b=36 \\Longrightarrow a b=16$. But the equation $a+b=a+\\frac{16}{a}=6$ has no real solutions, thus $a+b \\neq 6$. (Alternatively, note that by the Arithmetic Mean-Geometric Mean Inequality, $a+\\frac{16}{a} \\geq 2 \\sqrt{a \\cdot \\frac{16}{a}}=8>6$.)" ]

[ "$6 \\sqrt{2}-2$" ]

[ "The number of ways to choose 2 distinct letters out of 13 is $\\frac{13 \\cdot 12}{2}=78$. The probability of matching on both halves is therefore $\\frac{1}{78^{2}}=\\frac{1}{6084}$." ]

[ "$\\frac{1}{6084}$" ]

[ "The problem is equivalent to finding the least integer $n$ such that $\\frac{1}{2^{n}}<T$, or $2^{n}>\\frac{1}{T}=6084$. Because $2^{12}=4096$ and $2^{13}=8192$, the answer is $\\mathbf{1 3}$." ]

[ "13" ]

[ "The discriminant of the quadratic, $T^{2}+4 n$, must be a perfect square. Because $T$ and the discriminant have the same parity, and the leading coefficient of the quadratic is 1 , by the quadratic formula, the discriminant being a perfect square is sufficient to guarantee integer solutions. Before knowing $T$, note that $\\sqrt{4 \\cdot 2024}=$ $\\sqrt{8096}$ is slightly less than 90 because $90^{2}=8100$, and the square root must have the same parity as $T$. Because\n\n\n\n$T=13$, the square root must be greater than $\\sqrt{13^{2}+4 \\cdot 2023}=\\sqrt{8261}$, which is between 90 and 91 , so the desired square root is 91 . Hence $13^{2}+4 n=91^{2}$, so $n=\\mathbf{2 0 2 8}$." ]

[ "2028" ]

[ "First we prove that every sequence of five consecutive positive integers contains a cromulent element.\n\nProof: Consider a sequence of five consecutive integers. Exactly one number in such a sequence will be a multiple of 5 , but that number could also be a multiple of 2 and hence share a common factor with at least one other number in the sequence. There are several cases to consider, namely whether the sequence starts with an even number or an odd number.\n\nIf the sequence starts with an even number, then the second and fourth numbers are both odd, and at least one of them is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence because it is neither a multiple of 2 nor 3 and hence is at least 5 away from the nearest integer with a common factor. Thus the sequence contains a cromulent element.\n\nIf the sequence starts with an odd number, then again, it contains an odd number that is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence, thus the sequence contains a cromulent element. In fact, it contains two such numbers if the first or last number is a multiple of 3 , and if the middle number is a multiple of 3 , then all three odd elements are cromulent.\n\n\nThe minimum number is 1 and the maximum number is 2 . One example of a sequence of length 6 with one cromulent element is $5,6,7,8,9$, 10, where 7 is the cromulent element. To show that it is not possible for\n\n\na sequence of six consecutive elements to have zero cromulent elements, consider two cases. If the sequence begins with an even number, that number is not cromulent, and one of the other five elements must be cromulent by the argument in the proof above. A similar argument establishes that one element must be cromulent if the sequence of length 6 begins with an odd number (and thus ends in an even number).\n\nOne example of a sequence of length 6 with two cromulent elements is $1,2,3,4,5,6$, where 1 and 5 are both cromulent.\n\nTo prove that a sequence of length 6 cannot have three cromulent elements, consider that the cromulent elements would all have to be odd, and one of those three would be a multiple of 3 . Because one of the even elements must also be a multiple of 3 , it is not possible for all three odd elements to be cromulent." ]

[ "1,2" ]

[ "The minimum number is 1 and the maximum number is 3 . One example of a sequence of length 7 with one cromulent element is $4,5,6,7,8,9,10$, where 7 is the cromulent element. To show that it is not possible for such a sequence to have zero cromulent elements, consider two cases. If the sequence begins with an even number, then it contains three odd numbers. At most one of these is divisible by 3 , and at most one is divisible by 5 , so one of the odd numbers must be divisible by neither 3 nor 5 . This odd number differs by at most 6 from each other element of the sequence, so the only prime factors it can share with another element of the sequence are 2, 3, and 5 . Because it is divisible by none of these primes, it follows that the odd number in question is cromulent. Similarly, if the sequence begins with an odd number, then it contains four odd numbers; at most two of these are divisible by 3 , and at most one is divisible by 5 , so again, one odd number in the sequence must be divisible by neither 3 nor 5 . By the same argument, this element is cromulent.\n\nOne example of a sequence of length 7 with three cromulent elements is $1,2,3,4,5,6$, 7 , where 1,5 , and 7 are all cromulent.\n\nTo prove that a sequence of length 7 cannot have four cromulent elements, consider that the cromulent elements would all have to be odd. At least one of these four odd elements must be a multiple of 3 . Because one of the even elements must also be a multiple of 3 , it is thus not possible for all four odd elements to be cromulent." ]

[ "1,3" ]

[ "For an integer $n \\geq 4$, let $S_{n}$ denote the set of real numbers $x$ that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. (Note that $S_{n}$ is nonempty, as the polynomial $x^{2}+(n-2) x+1$ has a discriminant of $(n-2)^{2}-4$, which is nonnegative for $n \\geq 4$.) Then $a_{n}=\\prod_{x \\in S_{n}} x$.\n\nSuppose that $a, b$, and $c$ are positive integers and $x$ is a real solution to $a x^{2}+b x+c=0$. Then $x$ must be nonzero. (In fact, $x$ must be negative.) Dividing the above equation by $x^{2}$ yields $a+\\frac{b}{x}+\\frac{c}{x^{2}}=0$, thus $r=\\frac{1}{x}$ is a solution to the quadratic equation $c r^{2}+b r+a=0$. This shows that $x \\in S_{n}$ if and only if $\\frac{1}{x} \\in S_{n}$.\n\nOne might then think that $a_{n}$ must equal 1, because one can presumably pair up all elements in a given $S_{n}$ into $\\left\\{x, \\frac{1}{x}\\right\\}$ pairs. But there is a (negative) value of $x$ for which $x=\\frac{1}{x}$, namely $x=-1$. Therefore the value of $a_{n}$ depends only on whether $-1 \\in S_{n}$. It is readily seen via a parity argument that $-1 \\in S_{n}$ if and only if $n$ is even. If $n=2 k$, then the polynomial $x^{2}+k x+(k-1)$ has -1 as a root. (In fact, any quadratic polynomial whose middle coefficient is $k$ and whose coefficients sum to $2 k$ will work.) But if $n=2 k+1$, then $a(-1)^{2}+b(-1)+c=a-b+c=(a+b+c)-2 b=(2 k+1)-2 b$ will be odd, and so $-1 \\notin S_{n}$.\n\nThus $a_{n}=-1$ when $n$ is even, $a_{n}=1$ when $n$ is odd, and finally,\n\n$$\n\\frac{a_{4}}{a_{5}}+\\frac{a_{5}}{a_{6}}+\\frac{a_{6}}{a_{7}}+\\cdots+\\frac{a_{2022}}{a_{2023}}=\\underbrace{(-1)+(-1)+(-1)+\\cdots+(-1)}_{2019(-1) \\mathrm{s}}=-\\mathbf{2 0 1 9} .\n$$" ]

[ "-2019" ]

[ "Assume without loss of generality that $u>v$. The condition that $(x+u)(x+v)+4$ has integer roots is equivalent to the discriminant $(u+v)^{2}-4(u v+4)=(u-v)^{2}-16$ being a perfect square. This is possible if and only if $u-v=4$ or $u-v=5$. There are $(30-4)+(30-5)=26+25=51$ such ordered pairs $(u, v)$, so the answer is\n\n$$\n\\frac{51}{\\left(\\begin{array}{c}\n30 \\\\\n2\n\\end{array}\\right)}=\\frac{\\mathbf{1 7}}{\\mathbf{1 4 5}}\n$$" ]

[ "$\\frac{17}{145}$" ]

[ "The sum of the measures of the interior angles of a convex hexagon is $(6-2)\\left(180^{\\circ}\\right)=720^{\\circ}$. Let the measures of the angles be $a, a+d, \\ldots, a+5 d$. This implies that $6 a+15 d=720 \\rightarrow 2 a+5 d=240 \\rightarrow 5 d=240-2 a$. Note that $a+5 d<180 \\rightarrow 240-a<180 \\rightarrow a>60$. By inspection, note that the least $a$ greater than 60 that produces an integer $d$ is $a=65 \\rightarrow d=22$. Thus the least possible degree-measure of the smallest angle is $65^{\\circ}$, and the hexagon has angles with degree-measures $65^{\\circ}, 87^{\\circ}, 109^{\\circ}, 131^{\\circ}, 153^{\\circ}$, and $175^{\\circ}$." ]

[ "65" ]

[ "If three distinct digits are chosen from the set of digits $\\{0,1,2, \\ldots, 9\\}$, then there is exactly one way to arrange them in decreasing order. There are $\\left(\\begin{array}{c}10 \\\\ 3\\end{array}\\right)=120$ ways to choose the first three digits and 120 ways to choose the last three digits. Thus the answer is $120 \\cdot 120=\\mathbf{1 4 4 0 0}$." ]

[ "14400" ]

[ "One can verify that no single-digit positive integer $n$ satisfies the conditions of the problem.\n\nIf $n$ has two digits, then $n+23$ cannot be a three-digit number; this can be verified by checking the numbers $n \\geq 88$, because if $n<88$, then one of the digits of $n+23$ is 0 . Therefore both $n$ and $n+23$ must be two-digit numbers, so the only possible carry for $n+23$ will occur in the tens place. If there is a carry for $n+23$, then $n=\\underline{a} \\underline{8}$ or $n=\\underline{a} \\underline{9}$, while $n+23=(a+3) 1$ or $n+23=(a+3) 2$, respectively (the case $n=\\underline{a} \\underline{7}$ is omitted because then $P(n+23)=0)$. In either case, $P(n+23)<P(n)$ because $a \\geq 1$. Otherwise, assume $n=\\underline{a} \\underline{b}$ and $n+23=(a+2)(b+3)$ is a solution to the given equation, which implies\n\n$$\n23=P(n+23)-P(n)=(a+2)(b+3)-a b=3 a+2 b+6 \\text {. }\n$$\n\nThis means $3 a+2 b=17$, which has solutions $(a, b)=(5,1),(3,4)$ as $a, b$ are digits and $b<7$. The two-digit solutions are $n=34$ or $n=51$; thus the least $n$ such that $P(n+23)=P(n)+23$ is $n=34$." ]

[ "34" ]

[ "$\\quad$ Use polynomial long division to rewrite $f(x)$ as\n\n$$\nf(x)=x^{2}-6 x+1+\\frac{1}{x^{2}+1}\n$$\n\nThe quadratic function $x^{2}-6 x+1=(x-3)^{2}-8$ has a minimum of -8 , achieved at $x=3$. The \"remainder term\" $\\frac{1}{x^{2}+1}$ is always positive. Thus $f(x)>-8$ for all $x$, so any integer value of $f(x)$ must be at least -7 .\n\nWhen $x=3$, the remainder term is less than 1 , so $f(3)$ is less than -7 . But $f(4)=-\\frac{34}{5}>-7$, so there must be some value of $x$ between 3 and 4 for which $f(x)=-7$, so the least integer value of $f(x)$ is $\\mathbf{- 7}$. The reader may note that $f(x)=-7$ when $x \\approx 2.097$ and $x \\approx 3.970$." ]

[ "-7" ]

[ "Because triangles $A B C$ and $X Y Z$ are noncongruent yet have two adjacent sides and an angle in common, the two triangles are the two possibilities in the ambiguous case of the Law of Sines. Without loss of generality, let triangle $A B C$ have obtuse angle $C$ and triangle $X Y Z$ have acute angle $Z$ so that $\\mathrm{m} \\angle C+\\mathrm{m} \\angle Z=$ $180^{\\circ}$. Place triangle $A B C$ so that $B$ and $Y$ coincide, and $C$ and $Z$ coincide. Because $\\mathrm{m} \\angle C$ and $\\mathrm{m} \\angle Z$ add up to $180^{\\circ}$, it follows that points $X, Z$, and $A$ all lie on the same line. The two triangles together then form $\\triangle A B X$, where $\\mathrm{m} \\angle B A X=\\mathrm{m} \\angle B X A=30^{\\circ}$ and $B X=A B=10$. Therefore the sum of the areas of the two triangles is equal to the area of triangle $A B X$, which is $\\frac{1}{2} \\cdot 10 \\cdot 10 \\cdot \\sin \\left(120^{\\circ}\\right)=\\frac{5 \\cdot 10 \\cdot \\sqrt{3}}{2}=\\mathbf{2 5} \\sqrt{\\mathbf{3}}$.\n\n<img_3887>\n\nFigure not drawn to scale.", "As explained above, let $\\triangle A B C$ have obtuse angle $C$ and $\\triangle X Y Z$ have acute angle $Z$. By the Law of Sines, $\\sin (\\angle C)=\\sin (\\angle Z)=\\frac{5}{9}$. This implies $\\mathrm{m} \\angle X Y Z=\\frac{5 \\pi}{6}-\\arcsin \\left(\\frac{5}{9}\\right)$ and $\\mathrm{m} \\angle A B C=$ $\\arcsin \\left(\\frac{5}{9}\\right)-\\frac{\\pi}{6}$. The areas of the triangles are $[X Y Z]=\\frac{1}{2} \\cdot 10 \\cdot 9 \\cdot \\sin \\left(\\frac{5 \\pi}{6}-\\arcsin \\left(\\frac{5}{9}\\right)\\right)$ and $[A B C]=\\frac{1}{2} \\cdot 10 \\cdot 9$. $\\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)-\\frac{\\pi}{6}\\right)$. By the angle subtraction rule, it follows that\n\n$$\n\\begin{aligned}\n\\sin \\left(\\frac{5 \\pi}{6}-\\arcsin \\left(\\frac{5}{9}\\right)\\right) & =\\sin \\left(\\frac{5 \\pi}{6}\\right) \\cos \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right)-\\cos \\left(\\frac{5 \\pi}{6}\\right) \\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right) \\text { and } \\\\\n\\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)-\\frac{\\pi}{6}\\right) & =\\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right) \\cos \\left(\\frac{\\pi}{6}\\right)-\\cos \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right) \\sin \\left(\\frac{\\pi}{6}\\right) .\n\\end{aligned}\n$$\n\nThe sum of the two sines is $\\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right)\\left(\\cos \\left(\\frac{\\pi}{6}\\right)-\\cos \\left(\\frac{5 \\pi}{6}\\right)\\right)=\\frac{5}{9} \\cdot \\sqrt{3}$ because $\\sin \\left(\\frac{\\pi}{6}\\right)=\\sin \\left(\\frac{5 \\pi}{6}\\right)$. Finally, the sum of the areas of the two triangles is $\\frac{1}{2} \\cdot 10 \\cdot 9 \\cdot \\frac{5}{9} \\cdot \\sqrt{3}=25 \\sqrt{3}$." ]

[ "$25 \\sqrt{3}$" ]

[ "One possible list is $1,1,3,7$, which has mode 1 , median 2 , and mean 3 . The sum is $1+1+3+7=12$. A list with fewer than four numbers cannot produce a median and unique mode that are distinct from each other. To see this, first note that a list with one number has the same median and mode. In a list with two numbers, the mode is not unique if the numbers are different, and if the numbers are the same, the median and mode are equal. In a list of three numbers with a unique mode, the mode must occur twice. Hence the\n\n\n\nmode is equal to the middle number of the three, which is the median. Thus a list with a median and unique mode that are different from each other must contain at least four numbers.\n\nNow suppose that a list satisfying the given conditions sums to less than 12 . The mean must be greater than 1, and because the list contains at least four numbers, the mean must be exactly 2 . The median must also be greater than 1 , and if the mode is 4 , then the sum must be greater than 12 . Thus it remains to determine if a mean of 2 with mode 1 and median 3 can be achieved with a list of four or five positive integers. However, having two 1s in the list and a median of 3 forces the remaining numbers in each case to have a sum too large for a mean of 2 . The least possible sum is therefore $\\mathbf{1 2}$." ]

[ "12" ]

[ "The problem calls for the number of ordered partitions of 17 , where two partitions are considered the same if they are cyclic permutations of each other. Because 17 is prime, each ordered partition of 17 into $n$ parts will be a cyclic permutation of exactly $n$ such partitions (including itself), unless $n=17$. (If $n=17$, then all the numbers are 1s, and there is exactly one table David can make.) By the sticks and stones method, the number of ordered partitions of 17 into $n$ nonzero parts is $\\left(\\begin{array}{c}16 \\\\ n-1\\end{array}\\right)$, and this overcounts the number of tables by a factor of $n$, except when $n=17$. Thus the number of possible tables is\n\n$$\n1+\\sum_{n=1}^{16}\\left(\\begin{array}{c}\n16 \\\\\nn-1\n\\end{array}\\right) \\cdot \\frac{1}{n}=1+\\sum_{n=1}^{16}\\left(\\begin{array}{c}\n17 \\\\\nn\n\\end{array}\\right) \\cdot \\frac{1}{17}=1+\\frac{2^{17}-2}{17}=\\mathbf{7 7 1 1}\n$$" ]

[ "7711" ]

[ "Note that $\\mathrm{m} \\angle A+\\mathrm{m} \\angle C=90^{\\circ}$ in quadrilateral $A B C D$. Because quadrilateral $A B E D$ is cyclic, it follows that $\\mathrm{m} \\angle A D E+\\mathrm{m} \\angle A B E=180^{\\circ}$. Moreover, because $\\mathrm{m} \\angle A B E+\\mathrm{m} \\angle E B C+\\mathrm{m} \\angle A D E=270^{\\circ}$, it follows that $\\angle E B C$ is a right angle. Thus $B E=\\sqrt{C E^{2}-B C^{2}}=\\sqrt{5^{2}-4^{2}}=3$. Let $\\mathrm{m} \\angle B E C=\\theta$; then $\\cos \\theta=\\frac{3}{5}$ and $\\sin \\theta=\\frac{4}{5}$.\n\n<img_3458>\n\nApplying the Law of Cosines to $\\triangle B E D$ yields\n\n$$\nB D^{2}=3^{2}+7^{2}-2 \\cdot 3 \\cdot 7 \\cos \\left(180^{\\circ}-\\theta\\right)=3^{2}+7^{2}+2 \\cdot 3 \\cdot 7 \\cos \\theta=\\frac{416}{5}\n$$\n\nThus $B D=\\frac{4 \\sqrt{26}}{\\sqrt{5}}$. Let $R$ be the circumradius of $\\triangle A B D$ and $\\triangle B E D$. Then the requested diameter is $2 R$, and\n\n\n\napplying the Law of Sines to $\\triangle B E D$ yields\n\n$$\n2 R=\\frac{B D}{\\sin \\left(180^{\\circ}-\\theta\\right)}=\\frac{B D}{\\sin \\theta}=\\frac{4 \\sqrt{26}}{\\sqrt{5}} \\cdot \\frac{5}{4}=\\sqrt{\\mathbf{1 3 0}}\n$$" ]

[ "$\\sqrt{130}$" ]

[ "Multiply each of the given parenthesized expressions by its complex conjugate to obtain\n\n$$\n\\begin{aligned}\n142^{2}+5 \\cdot 333^{2} & =\\left(57^{2}+5 \\cdot 8^{2}\\right)\\left(6^{2}+5 \\cdot 5^{2}\\right) \\\\\n& =\\left(24^{2}+5 \\cdot 1^{2}\\right)\\left(3^{2}+5 \\cdot 14^{2}\\right) \\\\\n& =\\left(a^{2}+5 b^{2}\\right)\\left(c^{2}+5 d^{2}\\right) .\n\\end{aligned}\n$$\n\nThe expression on the second line is equal to $581 \\cdot 989=7 \\cdot 83 \\cdot 23 \\cdot 43$ (one can perhaps factor 989 a little faster by noting that 23 divides $6^{2}+5 \\cdot 5^{2}=7 \\cdot 23$ but not 581 , so it must divide 989 ). Thus $a^{2}+5 b^{2}$ and $c^{2}+5 d^{2}$ must be a factor pair of this number. It is not possible to express $1,7,23,43$, or 83 in the form $x^{2}+5 y^{2}$ for integers $x, y$.\n\nLet $N=a^{2}+5 b^{2}$, and without loss of generality, assume that 7 divides $N$. From the above analysis, $N$ must be $7 \\cdot 23,7 \\cdot 43$, or $7 \\cdot 83$. By direct computation of checking all positive integers $b$ less than $\\sqrt{\\frac{N}{5}}$, the only possibilities for $(a, b)$ are:\n\n- when $N=7 \\cdot 23$, either $(9,4)$ or $(6,5)$;\n- when $N=7 \\cdot 43$, either $(16,3)$ or $(11,6)$; and\n- when $N=7 \\cdot 83$, either $(24,1)$ or $(9,10)$.\n\nNext, observe that\n\n$$\n\\frac{-142+333 \\sqrt{5} i}{a+b \\sqrt{5} i}=\\frac{(-142 a+1665 b)+(333 a+142 b) \\sqrt{5} i}{N}\n$$\n\nmust equal $c+d \\sqrt{5} i$, so $N$ must divide $-142 a+1665 b$ and $333 a+142 b$. But\n\n- 7 does not divide $333 \\cdot 9+142 \\cdot 4$ or $333 \\cdot 6+142 \\cdot 5$;\n- 43 does not divide $333 \\cdot 16+142 \\cdot 3$; and\n- 83 does not divide $333 \\cdot 9+142 \\cdot 10$.\n\nThus the only candidates are $(a, b)=(11,6)$ and $(a, b)=(24,1)$. Note that $(24,1)$ yields the second factorization given in the problem statement, which has a negative real part in one of its factors. Thus the only remaining candidate for $(a, b)$ is $(11,6)$, which yields $(c, d)=(28,15)$, thus the answer is $11+6=\\mathbf{1 7}$." ]

[ "17" ]

[ "Editor's Note: It was pointed out that the conditions of the problem determine two possible values of $\\tan \\angle A B D$ : one based on $\\mathrm{m} \\angle A<90^{\\circ}$, and the other based on $\\mathrm{m} \\angle A>90^{\\circ}$. A complete solution is provided below. We thank Matthew Babbitt and Silas Johnson for their contributions to this solution.\n\n\n\nLet $A B=x, B C=y$, and $\\mathrm{m} \\angle A=\\alpha$.\n\n<img_3267>\n\nIt then follows that\n\n<img_3933>\n\n$$\n\\left[A B D^{\\prime}\\right]=\\left\\{\\begin{array}{ll}\n\\frac{x y \\sin 2 \\alpha}{2} & \\text { if } \\alpha<90^{\\circ} \\\\\n\\frac{-x y \\sin 2 \\alpha}{2} & \\text { if } \\alpha>90^{\\circ}\n\\end{array} \\quad \\text { and } \\quad\\left[B^{\\prime} C D\\right]=\\frac{x(x-y) \\sin \\alpha}{2}\\right.\n$$\n\nBecause $\\overline{B C}, \\overline{A B^{\\prime}}$, and $\\overline{D^{\\prime} C^{\\prime}}$ are all parallel, it follows that $\\triangle B C C^{\\prime}$ and $\\triangle B C D^{\\prime}$ have the same height with respect to base $\\overline{B C}$, and thus $\\left[B C C^{\\prime}\\right]=\\left[B C D^{\\prime}\\right]$. Therefore $\\left[B C D^{\\prime}\\right]=\\left[A B D^{\\prime}\\right]$, and it follows that triangles $B C D^{\\prime}$ and $A B D^{\\prime}$ have the same height with respect to base $\\overline{B D^{\\prime}}$. Thus $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$. Let $M$ be the midpoint of $\\overline{A C}$. Consider the following two cases.\n\nCase 1: Suppose that $\\alpha<90^{\\circ}$. Because $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$, it follows that $M$ lies on $\\overleftrightarrow{B D^{\\prime}}$. But $\\overleftrightarrow{B D}$ also passes through the midpoint of $\\overline{A C}$ by parallelogram properties, so it follows that $D$ must lie on $\\overline{B D^{\\prime}}$. This implies that $\\left[A B D^{\\prime}\\right]$ must also equal $\\frac{y^{2} \\sin \\alpha}{2}+\\frac{x y \\sin \\alpha}{2}=\\frac{\\left(x y+y^{2}\\right) \\sin \\alpha}{2}$.\n\nThus $x(x-y) \\sin \\alpha=x y \\sin 2 \\alpha=\\left(x y+y^{2}\\right) \\sin \\alpha$, which implies $x: y=\\sqrt{2}+1$ and $\\sin \\alpha=\\cos \\alpha=\\frac{\\sqrt{2}}{2}$. Finally, from right triangle $D^{\\prime} A B$ with legs in the ratio $1: \\sqrt{2}+1$, it follows that $\\tan (\\angle A B D)=\\tan \\left(\\angle A B D^{\\prime}\\right)=$ $\\sqrt{2}-1$.\n\nCase 2: Suppose that $\\alpha>90^{\\circ}$. The points $D$ and $D^{\\prime}$ lie on opposite sides of $\\overleftrightarrow{A B}$. Because $B C=A D^{\\prime}$ and points $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$, it follows that $A C B D^{\\prime}$ is either a parallelogram or an isosceles trapezoid. It cannot be the former because that would imply that $\\overleftrightarrow{D^{\\prime} A}\\|\\overleftrightarrow{B C}\\| \\overleftrightarrow{A D}$. Thus $A C B D^{\\prime}$ is an isosceles trapezoid. Then $\\left[B A D^{\\prime}\\right]=\\left[B M D^{\\prime}\\right]$. Because $B, M$, and $D$ are collinear and $B D: B M=2$, it follows that $\\left[B D D^{\\prime}\\right]=2 \\cdot\\left[B M D^{\\prime}\\right]$. Moreover, $\\left[B D D^{\\prime}\\right]=\\left[B A D^{\\prime}\\right]+[B A D]+\\left[D A D^{\\prime}\\right]$, so $\\left[B A D^{\\prime}\\right]=[B A D]+\\left[D A D^{\\prime}\\right]$. Thus $\\left[B A D^{\\prime}\\right]=\\frac{x y \\sin \\alpha}{2}+\\frac{y^{2} \\sin \\alpha}{2}=\\frac{\\left(x y+y^{2}\\right) \\sin \\alpha}{2}$.\n\nThus $x(x-y) \\sin \\alpha=-x y \\sin 2 \\alpha=\\left(x y+y^{2}\\right) \\sin \\alpha$, which implies $x: y=\\sqrt{2}+1, \\sin \\alpha=\\frac{\\sqrt{2}}{2}$, and $\\cos \\alpha=-\\frac{\\sqrt{2}}{2}$, so $\\alpha=135^{\\circ}$. Let $H$ be the foot of the perpendicular from $D$ to $\\overleftrightarrow{A B}$. Then $A D H$ is a $45^{\\circ}-45^{\\circ}-90^{\\circ}$ triangle with $H A=H D=\\frac{y}{\\sqrt{2}}$. Thus\n\n$$\n\\begin{aligned}\n\\tan \\angle A B D & =\\frac{D H}{B H}=\\frac{D H}{B A+A H} \\\\\n& =\\frac{y / \\sqrt{2}}{x+y / \\sqrt{2}}=\\frac{y}{x \\sqrt{2}+y} \\\\\n& =\\frac{y}{y(\\sqrt{2}+1)(\\sqrt{2})+y} \\\\\n& =\\frac{1}{(\\sqrt{2}+1)(\\sqrt{2})+1} \\\\\n& =\\frac{\\mathbf{3}-\\sqrt{\\mathbf{2}}}{\\mathbf{7}}\n\\end{aligned}\n$$", "Let $x, y$, and $\\alpha$ be as defined in the first solution. Then $C D=x$ because $A B C D$ is a parallelogram. Also note that $A B^{\\prime}=x, B^{\\prime} C^{\\prime}=y$, and $A D^{\\prime}=y$ because $A B C D$ and $A B^{\\prime} C^{\\prime} D^{\\prime}$ are congruent. Thus $D B^{\\prime}=A B^{\\prime}-A D=x-y$. Let $E$ be the intersection of $\\overleftrightarrow{A B}$ and $\\overleftrightarrow{C^{\\prime} D^{\\prime}}$, as shown in both configurations below.\n<img_3701>\n\nBecause $E$ lies on $\\overleftrightarrow{A B}$, it follows that $\\angle B^{\\prime} A E=180^{\\circ}-\\angle B A D=180^{\\circ}-\\alpha$. Thus, in quadrilateral $A B^{\\prime} C^{\\prime} E$, $\\overline{A B^{\\prime}} \\| \\overline{C^{\\prime} E}$ and $\\angle A B^{\\prime} C^{\\prime}=\\angle B^{\\prime} A E=180^{\\circ}-\\alpha$. Therefore quadrilateral $A B^{\\prime} C^{\\prime} E$ is an isosceles trapezoid. Hence $A E=B^{\\prime} C^{\\prime}=y$. It follows that $B E=B A+A E=x+y$. Therefore, from the sine area formula with respect to $\\angle C B E$,\n\n$$\n[B C E]=\\frac{1}{2} x(x+y) \\sin \\left(180^{\\circ}-\\alpha\\right)=\\frac{1}{2} x(x+y) \\sin \\alpha .\n$$\n\nBecause $\\overline{E C^{\\prime}} \\| \\overline{B C}$, it follows that $\\left[B C C^{\\prime}\\right]=[B C E]=\\frac{1}{2} x(x+y) \\sin \\alpha$. From the sine area formula with respect to $\\angle B A D^{\\prime}$ and $\\angle B^{\\prime} D C$, respectively,\n\n$$\n\\left[B A D^{\\prime}\\right]=\\frac{1}{2} x y|\\sin (2 \\alpha)|, \\quad\\left[B^{\\prime} D C\\right]=\\frac{1}{2} x(x-y) \\sin \\alpha\n$$\n\nThus\n\n$$\n\\frac{1}{2} x(x+y) \\sin \\alpha=\\frac{1}{2} x y|\\sin (2 \\alpha)|=\\frac{1}{2} x(x-y) \\sin \\alpha .\n$$\n\n\nBecause $\\overline{B C}, \\overline{A B^{\\prime}}$, and $\\overline{D^{\\prime} C^{\\prime}}$ are all parallel, it follows that $\\triangle B C C^{\\prime}$ and $\\triangle B C D^{\\prime}$ have the same height with respect to base $\\overline{B C}$, and thus $\\left[B C C^{\\prime}\\right]=\\left[B C D^{\\prime}\\right]$. Therefore $\\left[B C D^{\\prime}\\right]=\\left[A B D^{\\prime}\\right]$, and it follows that triangles $B C D^{\\prime}$ and $A B D^{\\prime}$ have the same height with respect to base $\\overline{B D^{\\prime}}$. Thus $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$. Let $M$ be the midpoint of $\\overline{A C}$. Consider the following two cases.\n\nCase 1: Suppose that $\\alpha<90^{\\circ}$. Because $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$, it follows that $M$ lies on $\\overleftrightarrow{B D^{\\prime}}$. But $\\overleftrightarrow{B D}$ also passes through the midpoint of $\\overline{A C}$ by parallelogram properties, so it follows that $D$ must lie on $\\overline{B D^{\\prime}}$. This implies that $\\left[A B D^{\\prime}\\right]$ must also equal $\\frac{y^{2} \\sin \\alpha}{2}+\\frac{x y \\sin \\alpha}{2}=\\frac{\\left(x y+y^{2}\\right) \\sin \\alpha}{2}$.\n\nThus $x(x-y) \\sin \\alpha=x y \\sin 2 \\alpha=\\left(x y+y^{2}\\right) \\sin \\alpha$, which implies $x: y=\\sqrt{2}+1$ and $\\sin \\alpha=\\cos \\alpha=\\frac{\\sqrt{2}}{2}$. Finally, from right triangle $D^{\\prime} A B$ with legs in the ratio $1: \\sqrt{2}+1$, it follows that $\\tan (\\angle A B D)=\\tan \\left(\\angle A B D^{\\prime}\\right)=$ $\\sqrt{2}-1$.\n\nCase 2: Suppose that $\\alpha>90^{\\circ}$. The points $D$ and $D^{\\prime}$ lie on opposite sides of $\\overleftrightarrow{A B}$. Because $B C=A D^{\\prime}$ and points $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$, it follows that $A C B D^{\\prime}$ is either a parallelogram or an isosceles trapezoid. It cannot be the former because that would imply that $\\overleftrightarrow{D^{\\prime} A}\\|\\overleftrightarrow{B C}\\| \\overleftrightarrow{A D}$. Thus $A C B D^{\\prime}$ is an isosceles trapezoid. Then $\\left[B A D^{\\prime}\\right]=\\left[B M D^{\\prime}\\right]$. Because $B, M$, and $D$ are collinear and $B D: B M=2$, it follows that $\\left[B D D^{\\prime}\\right]=2 \\cdot\\left[B M D^{\\prime}\\right]$. Moreover, $\\left[B D D^{\\prime}\\right]=\\left[B A D^{\\prime}\\right]+[B A D]+\\left[D A D^{\\prime}\\right]$, so $\\left[B A D^{\\prime}\\right]=[B A D]+\\left[D A D^{\\prime}\\right]$. Thus $\\left[B A D^{\\prime}\\right]=\\frac{x y \\sin \\alpha}{2}+\\frac{y^{2} \\sin \\alpha}{2}=\\frac{\\left(x y+y^{2}\\right) \\sin \\alpha}{2}$.\n\nThus $x(x-y) \\sin \\alpha=-x y \\sin 2 \\alpha=\\left(x y+y^{2}\\right) \\sin \\alpha$, which implies $x: y=\\sqrt{2}+1, \\sin \\alpha=\\frac{\\sqrt{2}}{2}$, and $\\cos \\alpha=-\\frac{\\sqrt{2}}{2}$, so $\\alpha=135^{\\circ}$. Let $H$ be the foot of the perpendicular from $D$ to $\\overleftrightarrow{A B}$. Then $A D H$ is a $45^{\\circ}-45^{\\circ}-90^{\\circ}$ triangle with $H A=H D=\\frac{y}{\\sqrt{2}}$. Thus\n\n$$\n\\begin{aligned}\n\\tan \\angle A B D & =\\frac{D H}{B H}=\\frac{D H}{B A+A H} \\\\\n& =\\frac{y / \\sqrt{2}}{x+y / \\sqrt{2}}=\\frac{y}{x \\sqrt{2}+y} \\\\\n& =\\frac{y}{y(\\sqrt{2}+1)(\\sqrt{2})+y} \\\\\n& =\\frac{1}{(\\sqrt{2}+1)(\\sqrt{2})+1} \\\\\n& =\\frac{\\mathbf{3}-\\sqrt{\\mathbf{2}}}{\\mathbf{7}}\n\\end{aligned}\n$$" ]

[ "$\\sqrt{2}-1$,$\\frac{3-\\sqrt{2}}{7}$" ]

[ "A candidate for desired number is $\\underline{2} \\underline{0} \\underline{X} \\underline{Y}$, where $X$ and $Y$ are digits and $X+Y=15$. To minimize this number, take $Y=9$. Then $X=6$, and the desired number is 2069 ." ]

[ "2069" ]

[ "Note that $|r-s|=\\sqrt{r^{2}-2 r s+s^{2}}=\\sqrt{(r+s)^{2}-4 r s}$. By Vieta's Formulas, $r+s=-(-18)$ and $r s=K$, so $|r-s|=\\sqrt{18^{2}-4 K}$. With $T=2069, K=17$, and the answer is $\\sqrt{324-68}=\\sqrt{256}=16$." ]

[ "16" ]

[ "Let $\\lfloor\\sqrt{n}\\rfloor=x$. Then $n$ can be written as $x^{2}+y$, where $y$ is an integer such that $0 \\leq y<2 x+1$. Let $m$ be the greatest perfect square less than or equal to $9 T$. Then the definition of the sequence and the previous observation imply that $A_{K}=A_{9 T}=\\sqrt{m}+(9 T-m)=\\lfloor\\sqrt{9 T}\\rfloor+\\left(9 T-\\lfloor\\sqrt{9 T}\\rfloor^{2}\\right)$. With $T=7, K=9 T=63$, $\\lfloor\\sqrt{9 T}\\rfloor=7$, and the answer is therefore $7+\\left(63-7^{2}\\right)=\\mathbf{2 1}$." ]

[ "21" ]

[ "Write $20^{T} \\cdot 23^{T}$ as $2^{2 T} \\cdot 5^{T} \\cdot 23^{T}$. This number has $K=(2 T+1)(T+1)^{2}$ positive divisors. With $T=21, K=43 \\cdot 22^{2}$. The greatest prime factor of $K$ is $\\mathbf{4 3}$." ]

[ "43" ]

[ "Using the symmetry property of binomial coefficients, the desired value of $n$ is $T-3-17=T-20$. With $T=43$, the answer is $\\mathbf{2 3}$." ]

[ "23" ]

[ "Assuming that $T$ is a positive integer, because units digits of powers of $T$ cycle in groups of at most 4, the numbers $T^{2023}$ and $T^{23}$ have the same units digit, hence the number $T^{2023}-T^{23}$ has a units digit of 0 , and the answer is thus the units digit of $T^{20}$. With $T=23$, the units digit of $23^{20}$ is the same as the units digit of $3^{20}$, which is the same as the units digit of $3^{4}=81$, so the answer is $\\mathbf{1}$." ]

[ "1" ]

[ "The probability of flipping all heads is $\\left(\\frac{1}{2}\\right)^{T}$, so the probability of flipping at least one tails is $1-\\frac{1}{2^{T}}$. With $T=3$, the desired probability is $1-\\frac{1}{8}=\\frac{7}{8}$." ]

[ "$\\frac{7}{8}$" ]

[ "The left-hand side of the given equation can be factored as $(x+m)(x+n)$. The two solutions are therefore $-m$ and $-n$, so the answer is $\\min \\{-m,-n\\}$. With $T=\\frac{7}{8}, m=7, n=8$, and $\\min \\{-7,-8\\}$ is $\\mathbf{- 8}$." ]

[ "-8" ]

[ "Note that $(-1+i)^{2}=1+2 i-1=2 i$. Thus $(-1+i)^{4}=(2 i)^{2}=-4$, and $(-1+i)^{8}=(-4)^{2}=16$. The expression $\\frac{1}{2^{T}}$ is a power of 16 if $T$ is a negative multiple of 4 . With $T=-8, \\frac{1}{2^{-8}}=2^{8}=16^{2}=\\left((-1+i)^{8}\\right)^{2}=$ $(-1+i)^{16}$, so the desired value of $k$ is $\\mathbf{1 6}$." ]

[ "16" ]

[ "By the change of base rule and a property of $\\operatorname{logs}, \\log _{4} T=\\frac{\\log _{2} T}{\\log _{2} 4}=\\frac{\\log _{2} T}{2}=\\log _{2} \\sqrt{T}$. Thus $x=\\sqrt{T}$, and with $T=16, x=4$." ]

[ "4" ]

[ "Let the side length of square base $E O J S$ be $2 x$, and let $M$ be the midpoint of $\\overline{E O}$. Then $\\overline{L M} \\perp \\overline{E O}$, and $L M=\\sqrt{(5 \\sqrt{2})^{2}-x^{2}}$ by the Pythagorean Theorem. Thus $[L E O]=\\frac{1}{2} \\cdot 2 x \\sqrt{(5 \\sqrt{2})^{2}-x^{2}}=$\n\n\n\n$x \\sqrt{(5 \\sqrt{2})^{2}-x^{2}}$. With $T=4, x=1$, and the answer is $1 \\cdot \\sqrt{50-1}=\\mathbf{7}$." ]

[ "7" ]

[ "Note that $T$ and $T+10$ have the same units digit. Because units digits of powers of $T$ cycle in groups of at most 4 , the numbers $T^{2023}$ and $(T+10)^{23}$ have the same units digit, hence the number $T^{2023}-(T+10)^{23}$ has a units digit of 0 , and the answer is thus the units digit of $(T-2)^{20}$. With $T=7$, the units digit of $5^{20}$ is 5 ." ]

[ "5" ]

[ "Draw radius $A P$ and note that $A P B$ is a right triangle with $\\mathrm{m} \\angle A P B=90^{\\circ}$. Note that $A B=R-r$ and $A P=r$, so by the Pythagorean Theorem, $B P=\\sqrt{(R-r)^{2}-r^{2}}=\\sqrt{R^{2}-2 R r}$. With $r=1$ and $R=5$, it follows that $B P=\\sqrt{\\mathbf{1 5}}$." ]

[ "$\\sqrt{15}$" ]

[ "Factor 15 ! -13 ! to obtain $13 !(15 \\cdot 14-1)=13$ ! $\\cdot 209$. The largest prime divisor of 13 ! is 13 , so continue by factoring $209=11 \\cdot 19$. Thus the largest prime divisor of 15 ! - 13 ! is 19 ." ]

[ "19" ]

[ "Proceed in two steps: first, determine the possible sets of side lengths for the squares; then determine which arrangement of squares produces the largest perimeter. Let the side lengths of the squares be positive integers $m \\geq n \\geq p$. Then $m^{2}+n^{2}+p^{2}=41$, so $m \\leq 6$, and because $3^{2}+3^{2}+3^{2}<41$, it follows that $m>3$. If $m=6$, then $n^{2}+p^{2}=5$, so $n=2$ and $p=1$. If $m=5$, then $n^{2}+p^{2}=16$, which has no positive integral solutions. If $m=4$, then $n^{2}+p^{2}=25$, which is possible if $n=4$ and $p=3$. So the two possible sets of values are $m=6, n=2, p=1$ or $m=4, n=4, p=3$.\n\nFirst consider $m=6, n=2, p=1$. Moving counterclockwise around the origin, one square is between the other two; by symmetry, it suffices to consider only the three possibilities for this \"middle\" square. If the middle square is the 6-square, then each of the other two squares has a side that is a subset of a side of the 6 -square. To compute the total perimeter, add the perimeters of the three squares and subtract twice the lengths of the shared segments (because they contribute 0 to the perimeter). Thus the total perimeter is $4 \\cdot 6+4 \\cdot 2+4 \\cdot 1-2 \\cdot 2-2 \\cdot 1=30$. If the middle square is the 2 -square, then one of its sides is a subset of the 6 -square's side, and one of its sides is a superset of the 1 -square's side, for a total perimeter of $4 \\cdot 6+4 \\cdot 2+4 \\cdot 1-2 \\cdot 2-2 \\cdot 1=$ 30. But if the middle square is the 1-square, then two of its sides are subsets of the other squares' sides, and the total perimeter is $4 \\cdot 6+4 \\cdot 2+4 \\cdot 1-2 \\cdot 1-2 \\cdot 1=32$.\n\nIf $m=4, n=4$, and $p=3$, similar logic to the foregoing suggests that the maximal perimeter is obtained when the smallest square is between the other two, yielding a total perimeter of $4 \\cdot 4+4 \\cdot 4+4 \\cdot 3-2 \\cdot 3-2 \\cdot 3=32$. Either of the other two arrangements yields a total perimeter of $4 \\cdot 4+4 \\cdot 4+4 \\cdot 3-2 \\cdot 3-2 \\cdot 4=30$. So the maximum perimeter is $\\mathbf{3 2}$.", "Let the side lengths be $a, b$, and $c$, and let $P$ be the perimeter. If the $a \\times a$ square is placed in between the other two (going either clockwise or counterclockwise around the origin), then\n\n$$\nP=3 b+|b-a|+2 a+|c-a|+3 c \\text {. }\n$$\n\nTo obtain a more symmetric expression, note that for any real numbers $x$ and $y$,\n\n$$\n|x-y|=\\max \\{x, y\\}-\\min \\{x, y\\}=x+y-2 \\min \\{x, y\\}\n$$\n\nUsing this identity,\n\n$$\nP=4 a+4 b+4 c-2 \\min \\{a, b\\}-2 \\min \\{a, c\\} .\n$$\n\nThus $P$ is the sum of the perimeters of the three, less twice the overlaps. To maximize $P$, choose $a$ to be the smallest of the three, which leads to $P=4 b+4 c$.\n\n\n\nAs in the first solution, the two possible sets of values are $c=6, b=2, a=1$ and $c=b=4$, $a=3$.\n\nIn the first case, the maximum length of the boundary is $P=4 \\cdot 2+4 \\cdot 6=32$, and in the second case it is $P=4 \\cdot 4+4 \\cdot 4=32$. So the maximum perimeter is $\\mathbf{3 2}$." ]

[ "32" ]

[ "Draw auxiliary segment $\\overline{O B}$, as shown in the diagram below.\n\n<img_4031>\n\nTriangle $O A B$ is equilateral, so $\\mathrm{m} \\angle O A B=60^{\\circ}$. Then $\\triangle M A P$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle with hypotenuse $A M=1 / 2$. Thus $A P=1 / 4$ and $M P=\\sqrt{3} / 4$, so\n\n$$\n\\begin{aligned}\n{[M A P] } & =\\frac{1}{2}\\left(\\frac{1}{4}\\right)\\left(\\frac{\\sqrt{3}}{4}\\right) \\\\\n& =\\frac{\\sqrt{3}}{\\mathbf{3 2}} .\n\\end{aligned}\n$$" ]

[ "$\\frac{\\sqrt{3}}{32}$" ]

[ "Subtract from both sides and regroup to obtain $p^{2}-2 p-\\left(q^{2}+6 q\\right)=8$. Completing both squares yields $(p-1)^{2}-(q+3)^{2}=0$. The left side is a difference of two squares; factor to obtain $((p-1)+(q+3))((p-1)-(q+3))=0$, whence $(p+q+2)(p-q-4)=0$. For positive primes $p$ and $q$, the first factor $p+q+2$ must also be positive. Therefore the second factor $p-q-4$ must be zero, hence $p-4=q$. Now look for primes starting with 97 and working downward. If $p=97$, then $q=93$, which is not prime; if $p=89$, then $q=85$, which is also not prime. But if $p=83$, then $q=79$, which is prime. Thus the largest possible value of $p+q$ is $83+79=\\mathbf{1 6 2}$." ]

[ "162" ]

[ "Let the four zeros be $p \\leq q \\leq r \\leq s$. The coefficient of $x^{3}$ is 0 , so $p+q+r+s=0$. The mean of four numbers in arithmetic progression is the mean of the middle two numbers, so $q=-r$. Then the common difference is $r-q=r-(-r)=2 r$, so $s=r+2 r=3 r$ and $p=q-2 r=-3 r$. Therefore the four zeros are $-3 r,-r, r, 3 r$. The product of\n\n\n\nthe zeros is $9 r^{4}$; referring to the original polynomial and using the product of roots formula gives $9 r^{4}=225$. Thus $r=\\sqrt{5}$, the zeros are $-3 \\sqrt{5},-\\sqrt{5}, \\sqrt{5}, 3 \\sqrt{5}$, and the polynomial can be factored as $(x-\\sqrt{5})(x+\\sqrt{5})(x-3 \\sqrt{5})(x+3 \\sqrt{5})$. Expanding this product yields $\\left(x^{2}-5\\right)\\left(x^{2}-45\\right)=x^{4}-50 x^{2}+225$, so $j=-50$.", "Proceed as in the original solution, finding the values $-3 \\sqrt{5},-\\sqrt{5}, \\sqrt{5}$, and $3 \\sqrt{5}$ for the zeros. By the sums and products of roots formulas, the coefficient of $x^{2}$ is the sum of all six possible products of pairs of roots:\n\n$$\n(-3 \\sqrt{5})(-\\sqrt{5})+(-3 \\sqrt{5})(\\sqrt{5})+(-3 \\sqrt{5})(3 \\sqrt{5})+(-\\sqrt{5})(\\sqrt{5})+(-\\sqrt{5})(3 \\sqrt{5})+(\\sqrt{5})(3 \\sqrt{5})\n$$\n\nObserving that some of these terms will cancel yields the simpler expression\n\n$$\n(-3 \\sqrt{5})(3 \\sqrt{5})+(-\\sqrt{5})(\\sqrt{5})=-45+-5=-50\n$$" ]

[ "-50" ]

[ "Inverting the problem, the goal is to find seven positive integers $a<b<c<d<e<f<g$ and a positive integer $n$ such that $a^{8}, b^{7}, c^{6}, \\ldots, g^{2} \\leq n$ and $n<(a+1)^{8},(b+1)^{7}, \\ldots,(g+1)^{2}$. Proceed by cases starting with small values of $a$.\n\nIf $a=1$, then because $n<(a+1)^{8}, n<256$. But because $n \\geq(a+3)^{5}, n \\geq 4^{5}=1024$. So it is impossible for $a$ to be 1 .\n\nIf $a=2$, then $a^{8}=256$ and $(a+1)^{8}=6561$, so $256 \\leq n<6561$. Then $b \\geq 3 \\Rightarrow b^{7} \\geq 2187$ and $c \\geq 4 \\Rightarrow c^{6} \\geq 4096$. So $n \\geq 4096$. Because $(3+1)^{7}=16384$ and $(4+1)^{6}=15625$, the condition $n<6561$ found previously guarantees that $\\lfloor\\sqrt[7]{n}\\rfloor=3$ and $\\lfloor\\sqrt[6]{n}\\rfloor=4$. Notice that if $4096 \\leq n<6561$, then $\\lfloor\\sqrt[5]{n}\\rfloor=5,\\lfloor\\sqrt[4]{n}\\rfloor=8$, and $\\lfloor\\sqrt[3]{n}\\rfloor \\geq 16$. In fact, $\\lfloor\\sqrt[3]{4096}\\rfloor=2^{4}=16$ and $\\lfloor\\sqrt{4096}\\rfloor=2^{6}=64$. So the desired value of $n$ is 4096 ." ]

[ "4096" ]

[ "If $n$ is even and $n \\leq 2012$, then $n$ !! $\\mid 2012$ !! trivially, while if $n>2012,2012$ !! $<n$ !!, so $n$ !! cannot divide 2012!!. Thus there are a total of 1006 even values of $n$ such that $n$ !! | 2012!!. If $n$ is odd and $n<1006$, then $n$ !! | 2012!!. To show this, rearrange the terms of 2012!! and factor:\n\n$$\n\\begin{aligned}\n2012 ! ! & =2 \\cdot 4 \\cdot 6 \\cdots 2010 \\cdot 2012 \\\\\n& =(2 \\cdot 6 \\cdot 10 \\cdots 2010)(4 \\cdot 8 \\cdot 12 \\cdots 2012) \\\\\n& =2^{503}(1 \\cdot 3 \\cdot 5 \\cdots 1005)(4 \\cdot 8 \\cdot 12 \\cdots 2012)\n\\end{aligned}\n$$\n\nHowever, the condition $n<1006$ is not necessary, only sufficient, because $n$ !! also divides 2012 if $1007 \\cdot 1009 \\cdots n \\mid(4 \\cdot 8 \\cdot 12 \\cdots 2012)$. (The factor of $2^{503}$ is irrelevant because all the factors on the left side are odd.) The expression $(4 \\cdot 8 \\cdot 12 \\cdots 2012)$ can be factored as $4^{503}(1 \\cdot 2 \\cdot 3 \\cdot \\cdots 503)=4^{503} \\cdot 503$ !. Examining the numbers $1007,1009, \\ldots$ in sequence shows that 1007 is satisfactory, because $1007=19 \\cdot 53$. On the other hand, 1009 is prime, so it cannot be a factor of $4^{503} \\cdot 503$ !. Thus the largest possible odd value of $n$ is 1007 , and there are 504 odd values of $n$ altogether. The total is $1006+504=\\mathbf{1 5 1 0}$." ]

[ "1510" ]

[ "As is usual, let $\\arg z$ refer to measure of the directed angle whose vertex is the origin, whose initial ray passes through 1 (i.e., the point $(1,0)$ ), and whose terminal ray passes through $z$. Then $\\arg 1 / z=-\\arg z$. Using the formula $a b \\sin \\gamma$ for the area of the parallelogram with sides $a$ and $b$ and included angle $\\gamma$ yields the equation\n\n$$\n\\frac{35}{37}=|z| \\cdot\\left|\\frac{1}{z}\\right| \\cdot \\sin (2 \\arg z)\n$$\n\nHowever, $|1 / z|=1 /|z|$, so the right side simplifies to $\\sin (2 \\arg z)$.\n\nTo compute the length $c$ of the diagonal from 0 to $z+1 / z$, use the Law of Cosines and the fact that consecutive angles of a parallelogram are supplementary:\n\n$$\n\\begin{aligned}\nc^{2} & =|z|^{2}+\\left|\\frac{1}{z}\\right|^{2}-2|z| \\cdot\\left|\\frac{1}{z}\\right| \\cos (\\pi-2 \\arg z) \\\\\n& =|z|^{2}+\\left|\\frac{1}{z}\\right|^{2}-2 \\cos (\\pi-2 \\arg z) \\\\\n& =|z|^{2}+\\left|\\frac{1}{z}\\right|^{2}+2 \\cos (2 \\arg z) .\n\\end{aligned}\n$$\n\nThis expression separates into two parts: the first, $|z|^{2}+|1 / z|^{2}$, is independent of the argument (angle) of $z$, while the second, $2 \\cos (2 \\arg z)$, is determined by the condition that $\\sin (2 \\arg z)=$ 35/37. The minimum value of $|z|^{2}+|1 / z|^{2}$ is 2 , as can be shown by the Arithmetic MeanGeometric Mean inequality applied to $|z|^{2}$ and $|1 / z|^{2}$ :\n\n$$\n|z|^{2}+|1 / z|^{2} \\geq 2 \\sqrt{|z|^{2} \\cdot|1 / z|^{2}}=2\n$$\n\nThe value of $\\cos (2 \\arg z)$ is given by the Pythagorean Identity:\n\n$$\n\\cos (2 \\arg z)= \\pm \\sqrt{1-\\left(\\frac{35}{37}\\right)^{2}}= \\pm \\sqrt{1-\\frac{1225}{1369}}= \\pm \\sqrt{\\frac{144}{1369}}= \\pm \\frac{12}{37}\n$$\n\nBecause the goal is to minimize the diagonal's length, choose the negative value to obtain\n\n$$\nd^{2}=2-2 \\cdot \\frac{12}{37}=\\frac{50}{37}\n$$", "Using polar coordinates, write\n\n$$\nz=r(\\cos \\theta+i \\sin \\theta)\n$$\n\nso that\n\n$$\n\\frac{1}{z}=r^{-1}(\\cos \\theta-i \\sin \\theta)\n$$\n\nWithout loss of generality, assume that $z$ is in the first quadrant, so that $\\theta>0$. Then the angle between the sides $\\overline{0 z}$ and $\\overline{0 z^{-1}}$ is $2 \\theta$, and the side lengths are $r$ and $r^{-1}$, so the area of the parallelogram is\n\n$$\n\\frac{35}{37}=r \\cdot r^{-1} \\cdot \\sin (2 \\theta)=\\sin 2 \\theta\n$$\n\nNote that $0<\\theta<\\pi / 2$, so $0<2 \\theta<\\pi$, and there are two values of $\\theta$ that satisfy this equation. Adding the expressions for $z$ and $z^{-1}$ and calculating the absolute value yields\n\n$$\n\\begin{aligned}\n\\left|z+\\frac{1}{z}\\right|^{2} & =\\left(r+r^{-1}\\right)^{2} \\cos ^{2} \\theta+\\left(r-r^{-1}\\right)^{2} \\sin ^{2} \\theta \\\\\n& =\\left(r^{2}+r^{-2}\\right)\\left(\\cos ^{2} \\theta+\\sin ^{2} \\theta\\right)+2 r \\cdot r^{-1}\\left(\\cos ^{2} \\theta-\\sin ^{2} \\theta\\right) \\\\\n& =r^{2}+r^{-2}+2 \\cos 2 \\theta .\n\\end{aligned}\n$$\n\nMinimize the terms involving $r$ using the Arithmetic-Geometric Mean inequality:\n\n$$\nr^{2}+r^{-2} \\geq 2 \\sqrt{r^{2} \\cdot r^{-2}}=2\n$$\n\nwith equality when $r^{2}=r^{-2}$, that is, when $r=1$. For the term involving $\\theta$, recall that there are two possible values:\n\n$$\n\\cos 2 \\theta= \\pm \\sqrt{1-\\sin ^{2} 2 \\theta}= \\pm \\sqrt{\\frac{37^{2}-35^{2}}{37^{2}}}= \\pm \\frac{\\sqrt{(37+35)(37-35)}}{37}= \\pm \\frac{12}{37}\n$$\n\nTo minimize this term, take the negative value, yielding\n\n$$\nd^{2}=2-2 \\cdot \\frac{12}{37}=\\frac{\\mathbf{5 0}}{\\mathbf{3 7}}\n$$", "If $z=x+y i$, then compute $1 / z$ by rationalizing the denominator:\n\n$$\n\\frac{1}{z}=\\frac{x-y i}{x^{2}+y^{2}}=\\frac{x}{x^{2}+y^{2}}+\\frac{-y}{x^{2}+y^{2}} i\n$$\n\nThe area of the parallelogram is given by the absolute value of the $2 \\times 2$ determinant\n\n$$\n\\left|\\begin{array}{cc}\nx & y \\\\\nx /\\left(x^{2}+y^{2}\\right) & -y /\\left(x^{2}+y^{2}\\right)\n\\end{array}\\right|=\\frac{1}{x^{2}+y^{2}}\\left|\\begin{array}{cc}\nx & y \\\\\nx & -y\n\\end{array}\\right|=\\frac{-2 x y}{x^{2}+y^{2}}\n$$\n\n\n\nThat is,\n\n$$\n\\frac{2 x y}{x^{2}+y^{2}}=\\frac{35}{37}\n$$\n\nCalculation shows that\n\n$$\n\\left|z+\\frac{1}{z}\\right|^{2}=\\left(x+\\frac{x}{x^{2}+y^{2}}\\right)^{2}+\\left(y-\\frac{y}{x^{2}+y^{2}}\\right)^{2}=\\left(x^{2}+y^{2}\\right)+\\frac{1}{x^{2}+y^{2}}+2\\left(\\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\\right) .\n$$\n\nAs in the previous solution, the sum of the first two terms is at least 2 , when $x^{2}+y^{2}=1$. The trick for relating the third term to the area is to express both the third term and the area in terms of the ratio\n\n$$\nt=\\frac{y}{x} .\n$$\n\nIndeed,\n\n$$\n\\frac{2 x y}{x^{2}+y^{2}}=\\frac{2 t}{1+t^{2}} \\quad \\text { and } \\quad \\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\\frac{1-t^{2}}{1+t^{2}}=\\frac{(1+t)(1-t)}{1+t^{2}}\n$$\n\nAs in the previous solution, assume without loss of generality that $z$ is in the first quadrant, so that $t>0$. As found above,\n\n$$\n\\frac{2 t}{1+t^{2}}=\\frac{35}{37}\n$$\n\nIt is not difficult to solve for $t$ using the quadratic formula, but the value of $t$ is not needed to solve the problem. Observe that\n\n$$\n\\frac{(1 \\pm t)^{2}}{1+t^{2}}=1 \\pm \\frac{2 t}{1+t^{2}}=1 \\pm \\frac{35}{37},\n$$\n\nso that\n\n$$\n\\left(\\frac{1-t^{2}}{1+t^{2}}\\right)^{2}=\\frac{(1+t)^{2}}{1+t^{2}} \\cdot \\frac{(1-t)^{2}}{1+t^{2}}=\\frac{72}{37} \\cdot \\frac{2}{37}=\\left(\\frac{12}{37}\\right)^{2}\n$$\n\nIn order to minimize $d$, take the negative square root, leading to\n\n$$\nd^{2}=2+2 \\cdot \\frac{1-t^{2}}{1+t^{2}}=2-\\frac{24}{37}=\\frac{\\mathbf{5 0}}{\\mathbf{3 7}}\n$$" ]

[ "$\\frac{50}{37}$" ]

[ "Call each $1 \\times 1 \\times 1$ cube a cubelet. Then four cubelets are each painted on one face, and the other four cubelets are completely unpainted and can be ignored. For each painted cubelet, the painted face can occur in six positions, of which three are hidden from the outside, so the probability that a particular painted cubelet has no paint showing is $3 / 6=1 / 2$. Thus the probability that all four painted cubelets have no paint showing is $(1 / 2)^{4}=\\frac{1}{\\mathbf{1 6}}$." ]

[ "$\\frac{1}{16}$" ]

[ "Let $E$ be the point where the other trisector of $\\angle B$ intersects side $\\overline{A C}$. Let $A B=B C=a$, and let $B D=B E=d$. Draw $X$ on $\\overline{B C}$ so that $B X=d$. Then $C X=7$.\n\n<img_3688>\n\nThe placement of point $X$ guarantees that $\\triangle B E X \\cong \\triangle B D E$ by Side-Angle-Side. Therefore $\\angle B X E \\cong \\angle B E X \\cong \\angle B D E$, and so $\\angle C X E \\cong \\angle A D B \\cong \\angle C E B$. By Angle-Angle, $\\triangle C E X \\sim \\triangle C B E$. Let $E X=c$ and $E C=x$. Then comparing ratios of corresponding sides yields\n\n$$\n\\frac{c}{d}=\\frac{7}{x}=\\frac{x}{d+7}\n$$\n\nUsing the right proportion, $x^{2}=7(d+7)$. Because $d$ is an integer, $x^{2}$ is an integer, so either $x$ is an integer or irrational. The following argument shows that $x$ cannot be irrational. Applying the Angle Bisector Theorem to $\\triangle B C D$ yields $D E=c=\\frac{d}{d+7} \\cdot x$. Then $A C=2 x+c=$ $x\\left(2+\\frac{d}{d+7}\\right)$. Because the expression $\\left(2+\\frac{d}{d+7}\\right)$ is rational, $A C$ will not be an integer if $x$ is irrational.\n\nHence $x$ is an integer, and because $x^{2}$ is divisible by $7, x$ must also be divisible by 7 . Let $x=7 k$ so that $d=c k$. Rewrite the original proportion using $7 k$ for $x$ and $c k$ for $d$ :\n\n$$\n\\begin{aligned}\n\\frac{c}{d} & =\\frac{x}{d+7} \\\\\n\\frac{c}{c k} & =\\frac{7 k}{c k+7} \\\\\n7 k^{2} & =c k+7 \\\\\n7 k & =c+\\frac{7}{k} .\n\\end{aligned}\n$$\n\n\n\nBecause the left side of this last equation represents an integer, $7 / k$ must be an integer, so either $k=1$ or $k=7$. The value $k=1$ gives the extraneous solution $c=0$. So $k=7$, from which $c=48$. Then $d=336$ and $A C=2 x+c=2 \\cdot 49+48=\\mathbf{1 4 6}$." ]

[ "146" ]

[ "In base 7, the value of $r$ must be $0.656565 \\ldots=0 . \\overline{65}_{7}$. Then $100_{7} \\cdot r=65 . \\overline{65}_{7}$, and $\\left(100_{7}-1\\right) r=$ $65_{7}$. In base $10,65_{7}=6 \\cdot 7+5=47_{10}$ and $100_{7}-1=7^{2}-1=48_{10}$. Thus $r=47 / 48$, and $p+q=95$." ]

[ "95" ]

[ "Because $\\triangle A B C$ is isosceles with $A B=A C, \\mathrm{~m} \\angle A B C=U^{\\circ}$ and $\\mathrm{m} \\angle B A C=(180-2 U)^{\\circ}$. Therefore $\\mathrm{m} \\angle M A C=\\left(\\frac{180-2 U}{3}\\right)^{\\circ}=\\left(60-\\frac{2}{3} U\\right)^{\\circ}$. Then $\\left(60-\\frac{2}{3} U\\right)+U+T=180$, so $\\frac{1}{3} U=$ $120-T$ and $U=3(120-T)$. Substituting $T=95$ yields $U=\\mathbf{7 5}$." ]

[ "75" ]

[ "With $n$ students, Wash Ed. can choose slide-rule oilers in $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)=\\frac{n(n-1)}{2}$ ways. With $n+2$ students, there would be $\\left(\\begin{array}{c}n+2 \\\\ 2\\end{array}\\right)=\\frac{(n+2)(n+1)}{2}$ ways of choosing the oilers. The difference is $\\frac{(n+2)(n+1)}{2}-\\frac{n(n-1)}{2}=T$. Simplifying yields $\\frac{\\left(n^{2}+3 n+2\\right)-\\left(n^{2}-n\\right)}{2}=2 n+1=T$, so $n=\\frac{T-1}{2}$. Because $T=75, n=37$." ]

[ "37" ]

[ "The first angle is $123^{\\circ}$, which is in Quadrant II, the second $\\left(246^{\\circ}\\right)$ is in Quadrant III, and the third is in Quadrant I, because $3 \\cdot 123^{\\circ}=369^{\\circ} \\equiv 9^{\\circ} \\bmod 360^{\\circ}$. The missing quadrant is IV, which is $270^{\\circ}-246^{\\circ}=24^{\\circ}$ away from the second angle in the sequence. Because $3 \\cdot 123^{\\circ} \\equiv 9^{\\circ} \\bmod 360^{\\circ}$, the terminal ray of the $(n+3)^{\\mathrm{rd}}$ angle is rotated $9^{\\circ}$ counterclockwise from the $n^{\\text {th }}$ angle. Thus three full cycles are needed to reach Quadrant IV starting from the second angle: the fifth angle is $255^{\\circ}$, the eighth angle is $264^{\\circ}$, and the eleventh angle is $273^{\\circ}$. So $n=11$." ]

[ "11" ]

[ "There are 9 valid one-digit plates. For a two-digit plate to be valid, it has to be of the form $\\underline{A} \\underline{B}$, where $B \\in\\{2, \\ldots, 9\\}$, and either $A \\in\\{2, \\ldots, 9\\}$ with $A \\neq B$ or $A=1$. So there are 8 ways to choose $B$ and $8-1+1=8$ ways to choose $A$, for a total of $8 \\cdot 8=64$ plates. In general, moving from the last digit to the first, if there are $k$ ways to choose digit $n$, then there are $k-1$ ways to choose digit $n-1$ from the same set of possibilities as digit $n$ had, plus one additional way, for a total of $k-1+1=k$ choices for digit $n-1$. So if a license plate has $d$ digits, there are $10-d$ choices for the last digit and for each digit before it, yielding $(10-d)^{d}$ possible $d$-digit plates. Using $d=T-3=8$, there are $2^{8}=\\mathbf{2 5 6}$ plates." ]

[ "256" ]

[ "The first inequality states that the point $(x, y)$ is outside the circle centered at the origin with radius $\\sqrt{T}$, while the second inequality states that $(x, y)$ is inside the tilted square centered at the origin with diagonal $2 \\sqrt{2 T}$. The area of the square is $4 \\cdot \\frac{1}{2}(\\sqrt{2 T})^{2}=4 T$, while the area of the circle is simply $\\pi T$, so the area of $\\mathcal{R}$ is $4 T-\\pi T=\\mathbf{1 0 2 4}-\\mathbf{2 5 6 \\pi}$." ]

[ "$1024-256 \\pi$" ]

[ "Let the feet of the altitudes from $A$ and $B$ be $E$ and $D$, respectively, and let $F$ and $G$ be the intersection points of the angle bisectors with $\\overline{A C}$ and $\\overline{B C}$, respectively, as shown below.\n\n<img_3386>\n\nThen $\\mathrm{m} \\angle G A E=6^{\\circ}$ and $\\mathrm{m} \\angle D B F=18^{\\circ}$. Suppose $\\mathrm{m} \\angle F B C=x^{\\circ}$ and $\\mathrm{m} \\angle C A G=y^{\\circ}$. So $\\mathrm{m} \\angle C A E=(y+6)^{\\circ}$ and $\\mathrm{m} \\angle C B D=(x+18)^{\\circ}$. Considering right triangle $B D C$, $\\mathrm{m} \\angle C=90^{\\circ}-(x+18)^{\\circ}=(72-x)^{\\circ}$, while considering right triangle $A E C, \\mathrm{~m} \\angle C=$ $90^{\\circ}-(y+6)^{\\circ}=(84-y)^{\\circ}$. Thus $84-y=72-x$ and $y-x=12$. Considering $\\triangle A B E$, $\\mathrm{m} \\angle E A B=(y-6)^{\\circ}$ and $\\mathrm{m} \\angle E B A=2 x^{\\circ}$, so $(y-6)+2 x=90$, or $2 x+y=96$. Solving the system yields $x=28, y=40$. Therefore $\\mathrm{m} \\angle A=80^{\\circ}$ and $\\mathrm{m} \\angle B=56^{\\circ}$, so $\\mathrm{m} \\angle C=44^{\\circ}$.", "From right triangle $A B E, 90^{\\circ}=\\left(\\frac{1}{2} A-6^{\\circ}\\right)+B$, and from right triangle $A B D, 90^{\\circ}=\\left(\\frac{1}{2} B-18^{\\circ}\\right)+A$. Adding the two equations gives $180^{\\circ}=\\frac{3}{2}(A+B)-24^{\\circ}$, so $A+B=\\frac{2}{3} \\cdot 204^{\\circ}=136^{\\circ}$ and $C=180^{\\circ}-(A+B)=44^{\\circ}$." ]

[ "$44^{\\circ}$" ]

[ "Let $r$ be the common root. Then $r^{2}+b r+c=r^{2}+c r+b \\Rightarrow b r-c r=b-c$. So either $b=c$ or $r=1$. In the latter case, $1+b+c=0$, so $c=-1-b$.\n\nThere are 41 ordered pairs where $b=c$. If $c=-1-b$ and $-20 \\leq b \\leq 20$, then $-21 \\leq c \\leq 19$. Therefore there are 40 ordered pairs $(b,-1-b)$ where both terms are in the required intervals. Thus there are $41+40=\\mathbf{8 1}$ solutions." ]

[ "81" ]

[ "The rolls that add up to 20 are $17+3,16+4,15+5,14+6,13+7,12+8,11+9$, and $10+10$. Accounting for order, the probability of $17+3$ is $\\frac{1}{2} \\cdot \\frac{1}{32}+\\frac{1}{32} \\cdot \\frac{1}{2}=2 \\cdot \\frac{1}{2} \\cdot \\frac{1}{32}=\\frac{32}{1024}$. The combination $10+10$ has probability $\\frac{1}{32} \\cdot \\frac{1}{32}=\\frac{1}{1024}$; the other six combinations have probability $2 \\cdot \\frac{1}{32} \\cdot \\frac{1}{32}=\\frac{2}{1024}$, for a total of $\\frac{32+1+6 \\cdot 2}{1024}=\\frac{45}{1024}$ (again, accounting for two possible orders per combination). The rolls that add up to 12 are $1+11,2+10,3+9,4+8,5+7,6+6$, all\n\n\n\nof which have probability $2 \\cdot \\frac{1}{32} \\cdot \\frac{1}{32}=\\frac{2}{1024}$ except the last, which has probability $\\left(\\frac{1}{32}\\right)^{2}$, for a total of $\\frac{11}{1024}$. Thus the probability of either sum appearing is $\\frac{45}{1024}+\\frac{11}{1024}=\\frac{56}{1024}=\\frac{\\mathbf{7}}{\\mathbf{1 2 8}}$." ]

[ "$\\frac{7}{128}$" ]

[ "Begin by partitioning $\\{2,3, \\ldots, 50\\}$ into the subsets\n\n$$\n\\begin{aligned}\nA & =\\{2,4,8,16,32\\} \\\\\nB & =\\{3,9,27\\} \\\\\nC & =\\{5,25\\} \\\\\nD & =\\{6,36\\} \\\\\nE & =\\{7,49\\} \\\\\nF & =\\text { all other integers between } 2 \\text { and } 50, \\text { inclusive. }\n\\end{aligned}\n$$\n\nIf $\\log _{b} a$ is rational, then either $a$ and $b$ are both members of one of the sets $A, B, C, D$, or $E$, or $a=b \\in F$ (see note below for proof). Then the number of possible ordered pairs is\n\n$$\n\\begin{aligned}\n|A|^{2}+|B|^{2}+|C|^{2}+|D|^{2}+|E|^{2}+|F| & =25+9+4+4+4+35 \\\\\n& =\\mathbf{8 1}\n\\end{aligned}\n$$" ]

[ "81" ]

[ "Condition on the number $n$ of A's that appear in the word; $n$ is at least two, because of the requirement that $\\mathbf{A}$ occur more often than any other letter, and $n$ is at most 4 , because of the requirement that there be at least two distinct letters. In the case $n=4$, there are 3 choices for the other letter, and 5 choices for where to place it, for a total of 15 possibilities. In the case $n=3$, there are two possibilities to consider: either a second letter occurs twice, or there are two distinct letters besides A. If a second letter occurs twice, there are 3 choices\n\n\n\nfor the other letter, and $\\frac{5 !}{3 ! \\cdot 2 !}=10$ ways to arrange the three A's and two non-A's, for their locations, for a total of 30 choices. If there are two distinct letters besides $A$, then there are $\\left(\\begin{array}{l}3 \\\\ 2\\end{array}\\right)=3$ ways to pick the two letters, and $\\frac{5 !}{3 ! \\cdot 1 ! \\cdot 1 !}=20$ ways to arrange them, for a total of 60 words. Thus there are a combined total of 90 words when $n=3$. In the case $n=2$, no other letter can occur twice, so all the letters R, M, L, must appear in the word; they can be arranged in $\\frac{5 !}{2 ! \\cdot 1 ! \\cdot 1 ! \\cdot 1 !}=60$ ways. The total number of words satisfying the conditions is therefore $15+90+60=\\mathbf{1 6 5}$." ]

[ "165" ]

[ "Let $d$ be the common difference of the sequence. Then $a_{a_{2}}=a_{1}+\\left(a_{2}-1\\right) d=100 \\Rightarrow\\left(a_{2}-1\\right) d=$ 90. But $a_{2}=a_{1}+d=10+d$, so $(9+d) d=90$. Solving the quadratic yields $d=-15$ or $d=6$, but the requirement that $a_{i}$ be positive for all $i$ rules out the negative value, so $d=6$ and $a_{n}=10+(n-1) \\cdot 6$. Thus $a_{3}=10+2(6)=22$, and $a_{a_{3}}=a_{22}=10+21(6)=136$. Finally, $a_{a_{a_{3}}}=a_{136}=10+135(6)=\\mathbf{8 2 0}$." ]

[ "820" ]

[ "First, note that both graphs are symmetric about the $y$-axis, so $C$ and $D$ must be reflections of $B$ and $A$, respectively, across the $y$-axis. Thus $x_{C}=-x_{B}$ and $y_{C}=y_{B}$, so $B C=2 x_{C}$. For $x<0$, the equations become $y=x^{2}+x-12$ and $y=-x-k$; setting the $x$-expressions equal to each other yields the equation $x^{2}+2 x+(k-12)=0$, from which $x=-1 \\pm \\sqrt{13-k}$. Therefore $x_{B}=-1+\\sqrt{13-k}$ and $B C=2-2 \\sqrt{13-k}$. (Note that the existence of two distinct negative values of $-1 \\pm \\sqrt{13-k}$ forces $12<k \\leq 13$.)\n\nThus the $x$-coordinates of the four points are\n\n$$\n\\begin{aligned}\n& x_{A}=-1-\\sqrt{13-k} \\\\\n& x_{B}=-1+\\sqrt{13-k} \\\\\n& x_{C}=1-\\sqrt{13-k} \\\\\n& x_{D}=1+\\sqrt{13-k} .\n\\end{aligned}\n$$\n\nTo compute $y_{A}$, use the second equation $y=|x|-k$ to obtain $y_{A}=1+\\sqrt{13-k}-k=$ $(1-k)+\\sqrt{13-k}$; similarly, $y_{B}=(1-k)-\\sqrt{13-k}$. Therefore\n\n$$\n\\begin{aligned}\nA B & =\\sqrt{\\left(x_{B}-x_{A}\\right)^{2}+\\left(y_{B}-y_{A}\\right)^{2}} \\\\\n& =\\sqrt{(2 \\sqrt{13-k})^{2}+(-2 \\sqrt{13-k})^{2}} \\\\\n& =2 \\sqrt{2(13-k)}\n\\end{aligned}\n$$\n\nBecause $A B=B C, 2 \\sqrt{2(13-k)}=2-2 \\sqrt{13-k}$. Let $u=\\sqrt{13-k}$; then $2 \\sqrt{2} u=2-2 u$, from which $u=\\frac{2}{2+2 \\sqrt{2}}=\\frac{1}{1+\\sqrt{2}}$, which equals $\\sqrt{2}-1$ by rationalizing the denominator. Thus\n\n$$\n13-k=(\\sqrt{2}-1)^{2}=3-2 \\sqrt{2}, \\text { so } k=\\mathbf{1 0}+\\mathbf{2} \\sqrt{\\mathbf{2}} \\text {. }\n$$\n\nBecause $10+2 \\sqrt{2} \\approx 12.8$, the value of $k$ determined algebraically satisfies the inequality $12<k \\leq 13$ observed above.", "Let $C=(a, b)$. Because $C$ and $D$ lie on a line with slope 1 , $D=(a+h, b+h)$ for some $h>0$. Because both graphs are symmetric about the $y$-axis, the other two points of intersection are $A=(-a-h, b+h)$ and $B=(-a, b)$, and $a>0$.\n\nIn terms of these coordinates, the distances are $A B=C D=\\sqrt{2} h$ and $B C=2 a$. Thus the condition $A B=B C=C D$ holds if and only if $\\sqrt{2} h=2 a$, or $h=\\sqrt{2} a$.\n\nThe foregoing uses the condition that $C$ and $D$ lie on a line of slope 1 , so now use the remaining equation and subtract:\n\n$$\n\\begin{aligned}\nb & =a^{2}-a-12 \\\\\nb+h & =(a+h)^{2}-(a+h)-12 \\\\\nh & =2 a h+h^{2}-h\n\\end{aligned}\n$$\n\nBecause the points are distinct, $h \\neq 0$. Dividing by $h$ yields $2-2 a=h=\\sqrt{2} a$. Thus $a=\\frac{2}{2+\\sqrt{2}}=2-\\sqrt{2}$.\n\nFinally, because $C$ lies on the two graphs, $b=a^{2}-a-12=-8-3 \\sqrt{2}$ and $k=a-b=$ $10+2 \\sqrt{2}$." ]

[ "$10+2 \\sqrt{2}$" ]

[ "There are $6 !=720$ permutations of the zeros, so the average value is the sum, $S$, divided by 720. Setting any particular zero as $A$ leaves $5 !=120$ ways to permute the other five zeros, so over the 720 permutations, each zero occupies the $A$ position 120 times. Similarly, fixing any ordered pair $(B, C)$ of zeros allows $4 !=24$ permutations of the other four zeros, and $B C=C B$ means that each value of $B C$ occurs 48 times. Finally, fixing any ordered triple $(D, E, F)$ allows $3 !=6$ permutations of the other variables, and there are $3 !=6$ equivalent arrangements within each product $D E F$, so that the product of any three zeros occurs 36 times within the sum. Let $S_{1}=A+B+C+D+E+F$ (i.e., the sum of the zeros taken singly), $S_{2}=A B+A C+\\cdots+A F+B C+\\cdots+E F$ (i.e., the sum of the zeros taken two at a time), and $S_{3}=A B C+A B D+\\cdots+D E F$ be the sum of the zeros three at a time. Then $S=120 S_{1}+48 S_{2}+36 S_{3}$. Using the sums and products of roots formulas, $S_{1}=-2 / 1=-2$, $S_{2}=3 / 1=3$, and $S_{3}=-5 / 1=-5$. Thus $S=120(-2)+48(3)+36(-5)=-276$. The average value is thus $-\\frac{276}{720}=-\\frac{\\mathbf{2 3}}{\\mathbf{6 0}}$." ]

[ "$-\\frac{23}{60}$" ]

[ "Let $\\alpha=3+\\sqrt{5}$ and $\\beta=3-\\sqrt{5}$, so that $N=\\left\\lfloor\\alpha^{34}\\right\\rfloor$, and let $M=\\alpha^{34}+\\beta^{34}$. When the binomials in $M$ are expanded, terms in which $\\sqrt{5}$ is raised to an odd power have opposite signs, and so cancel each other out. Therefore $M$ is an integer. Because $0<\\beta<1,0<\\beta^{34}<1$, and so $M-1<\\alpha^{34}<M$. Therefore $M-1=N$. Note that $\\alpha$ and $\\beta$ are the roots of $x^{2}=6 x-4$. Therefore $\\alpha^{n+2}=6 \\alpha^{n+1}-4 \\alpha^{n}$ and $\\beta^{n+2}=6 \\beta^{n+1}-4 \\beta^{n}$. Hence $\\alpha^{n+2}+\\beta^{n+2}=$ $6\\left(\\alpha^{n+1}+\\beta^{n+1}\\right)-4\\left(\\alpha^{n}+\\beta^{n}\\right)$. Thus the sequence of numbers $\\left\\{\\alpha^{n}+\\beta^{n}\\right\\}$ satisfies the recurrence relation $c_{n+2}=6 c_{n+1}-4 c_{n}$. All members of the sequence are determined by the initial values $c_{0}$ and $c_{1}$, which can be computed by substituting 0 and 1 for $n$ in the expression $\\alpha^{n}+\\beta^{n}$, yielding $c_{0}=(3+\\sqrt{5})^{0}+(3-\\sqrt{5})^{0}=2$, and $c_{1}=(3+\\sqrt{5})^{1}+(3-\\sqrt{5})^{1}=6$. Then\n\n$$\n\\begin{aligned}\n& c_{2}=(3+\\sqrt{5})^{2}+(3-\\sqrt{5})^{2}=6 c_{1}-4 c_{0}=36-8=28 \\\\\n& c_{3}=(3+\\sqrt{5})^{3}+(3-\\sqrt{5})^{3}=6 c_{2}-4 c_{1}=168-24=144\n\\end{aligned}\n$$\n\nand because the final result is only needed modulo 100, proceed using only remainders modulo 100.\n\n\n\n| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $c_{n} \\bmod 100$ | 6 | 28 | 44 | 52 | 36 | 8 | 4 | 92 | 36 | 48 | 44 | 72 | 56 | 48 | 64 | 92 | 96 |\n\n\n| $n$ | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $c_{n} \\bmod 100$ | 8 | 64 | 52 | 56 | 28 | 44 | 52 | 36 | 8 | 4 | 92 | 36 | 48 | 44 | 72 | 56 | 48 |\n\nThus $N$ leaves a remainder of $48-1=\\mathbf{4 7}$ when divided by 100 .", "As in the previous solution, let $\\alpha=3+\\sqrt{5}$ and $\\beta=3-\\sqrt{5}$, so that $N=\\alpha^{34}+\\beta^{34}-1$ as argued above.\n\nA straightforward way to compute powers of $\\alpha$ and $\\beta$ is by successive squaring. Paying attention to just the last two digits of the integer parts yields the following values:\n\n$$\n\\begin{aligned}\n\\alpha^{2} & =14+6 \\sqrt{5} \\\\\n\\alpha^{4} & =196+180+168 \\sqrt{5} \\equiv 76+68 \\sqrt{5} ; \\\\\n\\alpha^{8} & \\equiv 96+36 \\sqrt{5} \\\\\n\\alpha^{16} & \\equiv 96+12 \\sqrt{5} \\\\\n\\alpha^{32} & \\equiv 36+4 \\sqrt{5} \\\\\n\\alpha^{34}=\\alpha^{2} \\cdot \\alpha^{32} & \\equiv 24+72 \\sqrt{5} .\n\\end{aligned}\n$$\n\nSimilarly, replacing $\\sqrt{5}$ with $-\\sqrt{5}$ yields $\\beta^{34} \\equiv 24-72 \\sqrt{5}$. Thus\n\n$$\nN \\equiv(24+72 \\sqrt{5})+(24-72 \\sqrt{5})-1 \\equiv 47(\\bmod 100)\n$$", "As in the previous solutions, let $\\alpha=3+\\sqrt{5}$ and $\\beta=3-\\sqrt{5}$, so that $N=\\alpha^{34}+\\beta^{34}-1$ as argued above.\n\nNow consider the binomial expansions more carefully:\n\n$$\n\\begin{aligned}\n\\alpha^{34} & =3^{34}+\\left(\\begin{array}{c}\n34 \\\\\n1\n\\end{array}\\right) 3^{33} \\sqrt{5}+\\left(\\begin{array}{c}\n34 \\\\\n2\n\\end{array}\\right) 3^{32} \\cdot 5+\\left(\\begin{array}{c}\n34 \\\\\n3\n\\end{array}\\right) 3^{31} \\cdot 5 \\sqrt{5}+\\cdots+\\left(\\begin{array}{c}\n34 \\\\\n33\n\\end{array}\\right) 3 \\cdot 5^{16} \\sqrt{5}+5^{17} \\\\\n\\beta^{34} & =3^{34}-\\left(\\begin{array}{c}\n34 \\\\\n1\n\\end{array}\\right) 3^{33} \\sqrt{5}+\\left(\\begin{array}{c}\n34 \\\\\n2\n\\end{array}\\right) 3^{32} \\cdot 5-\\left(\\begin{array}{c}\n34 \\\\\n3\n\\end{array}\\right) 3^{31} \\cdot 5 \\sqrt{5}+\\cdots-\\left(\\begin{array}{c}\n34 \\\\\n33\n\\end{array}\\right) 3 \\cdot 5^{16} \\sqrt{5}+5^{17} \\\\\nN & =2\\left(3^{34}+\\left(\\begin{array}{c}\n34 \\\\\n2\n\\end{array}\\right) 3^{32} \\cdot 5+\\cdots+\\left(\\begin{array}{c}\n34 \\\\\n32\n\\end{array}\\right) 3^{2} \\cdot 5^{16}+5^{17}\\right)-1 .\n\\end{aligned}\n$$\n\nThe following argument shows that every term that is summarized by the ellipsis $(\\cdots)$ in the expression for $N$ is a multiple of 50 . First, each such term has the form $\\left(\\begin{array}{l}34 \\\\ 2 k\\end{array}\\right) 3^{34-2 k} 5^{k}$, where $2 \\leq k \\leq 15$.\n\nThus it is enough to show that the binomial coefficient is even. Because $\\left(\\begin{array}{l}34 \\\\ 2 k\\end{array}\\right)=\\left(\\begin{array}{c}34 \\\\ 34-2 k\\end{array}\\right)$, it is enough to check this for $2 \\leq k \\leq 8$. Keep track of powers of 2 : $\\left(\\begin{array}{c}34 \\\\ 2\\end{array}\\right)$ is an integer, so\n\n\n\n$\\left(\\begin{array}{c}34 \\\\ 4\\end{array}\\right)=\\left(\\begin{array}{c}34 \\\\ 2\\end{array}\\right) \\cdot \\frac{32 \\cdot 31}{3 \\cdot 4}$ is a multiple of $2^{3} ;\\left(\\begin{array}{c}34 \\\\ 6\\end{array}\\right)=\\left(\\begin{array}{c}34 \\\\ 4\\end{array}\\right) \\cdot \\frac{30 \\cdot 29}{5 \\cdot 6}$ is also a multiple of $2^{3} ;\\left(\\begin{array}{c}34 \\\\ 8\\end{array}\\right)=\\left(\\begin{array}{c}34 \\\\ 6\\end{array}\\right) \\cdot \\frac{28 \\cdot 27}{7 \\cdot 8}$ is a multiple of $2^{2}$; and so on.\n\nIt can also be shown that the sum of the last two terms is a multiple of 50. Again, there are plenty of factors of 5 , so it is enough to note that both terms are odd, because $\\left(\\begin{array}{l}34 \\\\ 32\\end{array}\\right)=\\frac{34 \\cdot 33}{1 \\cdot 2}=$ $17 \\cdot 33$.\n\nThanks to the initial factor of 2 in the expression for $N$ (outside the parentheses), the previous paragraphs show that $N \\equiv 2\\left(3^{34}+\\left(\\begin{array}{c}34 \\\\ 2\\end{array}\\right) 3^{32} \\cdot 5\\right)-1(\\bmod 100)$.\n\nNow consider the powers of 3 . Because $3^{4}=81$, we find that $3^{8}=80^{2}+2 \\cdot 80+1 \\equiv$ $61(\\bmod 100), 3^{12} \\equiv 41(\\bmod 100), 3^{16} \\equiv 21(\\bmod 100)$, and $3^{20} \\equiv 1(\\bmod 100)$. (Note: those familiar with Euler's generalization of Fermat's Little Theorem will recognize this as an example, because $\\phi(25)=25-5=20$.) Therefore $3^{32}=3^{20} \\cdot 3^{12} \\equiv 41(\\bmod 100)$ and $3^{34}=3^{2} \\cdot 3^{32} \\equiv 69(\\bmod 100)$.\n\nFinally, $N \\equiv 2(69+17 \\cdot 33 \\cdot 41 \\cdot 5)-1 \\equiv 2 \\cdot 69+10 \\cdot(17 \\cdot 33 \\cdot 41)-1 \\equiv 38+10-1 \\equiv \\mathbf{4 7}$ $(\\bmod 100)$." ]

[ "47" ]

[ "Focus on $\\triangle P B C$. Either $P B=P C$ or $P B=B C$ or $P C=B C$.\n\nIf $P B=P C$, then $P$ lies on the perpendicular bisector $l$ of side $\\overline{B C}$. Considering now $\\triangle P A B$, if $P A=P B$, then $P A=P C$, and $P$ must be the circumcenter of $\\triangle A B C$; call this location $P_{1}$. If $P A=A B$, then $P A=A C$, and $P, B, C$ all lie on a circle with center $A$ and radius $A B$. There are two intersection points of that circle with $l$, one on each arc with endpoints $B$ and $C$; label the one on the major arc $P_{2}$ and on the minor $\\operatorname{arc} P_{3}$. Finally, if $P B=A B$, then $P B=A C$ by the transitive property and $P C=A C$ by the perpendicular bisector theorem, so $P B A C$ is a rhombus; $P$ is the reflection of $A$ across $\\overline{B C}$. Call this point $P_{4}$.\n\nIf $P B=B C$, then $P$ must lie on the circle centered at $B$ with radius $B C$. Considering $\\triangle P A B$, if $P A=A B$, then $P$ lies on the circle centered at $A$ with radius $A B$. Now $\\odot A$ and $\\odot B$ intersect at two points, but one of them is $C$, so the other intersection must be the location of $P$, which is $P_{5}$. The condition $P B=A B$ is impossible, because it implies that $A B=B C$, which is false because in $\\triangle A B C, \\mathrm{~m} \\angle C>\\mathrm{m} \\angle A=20^{\\circ}$, so $A B>B C$. The third possibility for $\\triangle P A B$ is that $P A=P B$, implying that the perpendicular bisector of $\\overline{A B}$ intersects $\\odot B$, which only occurs if $B C / A B \\geq 1 / 2$ (although if $B C / A B=1 / 2$, the triangle is degenerate). But $B C / A B=2 \\cos 80^{\\circ}$, and the given approximation $\\cos 80^{\\circ} \\approx 0.17$ implies that $B C / A B \\approx 0.34$. Hence the perpendicular bisector of $\\overline{A B}$ does not intersect $\\odot B$. Thus the assumption $P B=B C$ yields only one additional location for $P, P_{5}$. Similarly, $P C=B C$ yields exactly one more location, $P_{6}$, for a total of $\\mathbf{6}$ points. All six points, and their associated triangles, are pictured below.\n\n\n\n<img_3810>" ]

[ "6" ]

[ "Let $f(x)=\\left\\lfloor\\frac{x}{3}\\right\\rfloor+\\lceil 3 x\\rceil$. Observe that $f(x+3)=f(x)+1+9=f(x)+10$. Let $g(x)=f(x)-\\frac{10}{3} x$. Then $g$ is periodic, because $g(x+3)=f(x)+10-\\frac{10 x}{3}-\\frac{10 \\cdot 3}{3}=g(x)$. The graph of $g$ is shown below:\n\n<img_3987>\n\nBecause $g(x)$ is the (vertical) distance between the graph of $y=f(x)$ and the line $y=\\frac{10}{3} x$, the fact that $g$ is periodic implies that $f$ always stays within some fixed distance $D$ of the line $y=\\frac{10}{3} x$. On the other hand, because $\\frac{10}{3}>\\sqrt{11}$, the graph of $y=\\frac{10}{3} x$ gets further and further away from the graph of $y=\\sqrt{11} x$ as $x$ increases. Because the graph of $y=f(x)$ remains near $y=\\frac{10}{3} x$, the graph of $y=f(x)$ drifts upward from the line $y=\\sqrt{11} x$.\n\nFor each integer $n$, define the open interval $I_{n}=\\left(\\frac{n-1}{3}, \\frac{n}{3}\\right)$. In fact, $f$ is constant on $I_{n}$, as the following argument shows. For $x \\in I_{n}, \\frac{n}{9}-\\frac{1}{9}<\\frac{x}{3}<\\frac{n}{9}$. Because $n$ is an integer, there are no integers between $\\frac{n}{9}-\\frac{1}{9}$ and $\\frac{n}{9}$, so $\\left\\lfloor\\frac{x}{3}\\right\\rfloor$ is constant; similarly, $\\lceil 3 x\\rceil$ is constant on the same intervals. Let $l_{n}$ be the value of $f$ on the interval $I_{n}$, and let $L_{n}=f\\left(\\frac{n}{3}\\right)$, the value at the right end of the interval $I_{n}$. If $n$ is not a multiple of 9 , then $l_{n}=L_{n}$, because as $x$ increases from $n-\\varepsilon$ to $n$, the floor function does not increase. This means that $f$ is actually constant on the half-closed interval $\\left(\\frac{n-1}{3}, \\frac{n}{3}\\right]$. If neither $n$ nor $n+1$ are multiples of 9 , then $l_{n+1}=l_{n}+1$. However if $n$ is a multiple of 9 , then $L_{n}=l_{n}+1$ and $l_{n+1}=L_{n}+1$. (The value of $f(x)$ increases when $x$ increases from $n-\\varepsilon$ to $n$, as well as going from $n$ to $n+\\varepsilon$.)\n\nHence on each interval of the form $(3 n-3,3 n)$, the graph of $f$ looks like 9 steps of height 1 and width $\\frac{1}{3}$, all open on the left and closed on the right except for the last step, which is open on both ends. Between the intervals $(3 n-3,3 n)$ and $(3 n, 3 n+3), f(x)$ increases by 2 , with $f(3 n)$ halfway between steps. This graph is shown below:\n\n\n\n<img_3187>\n\nOn each interval $(3 n-3,3 n)$, the average rate of change is $3<\\sqrt{11}$, so the steps move down relative $y=\\sqrt{11} x$ within each interval. At the end of each interval, the graph of $f$ rises relative to $y=\\sqrt{11} x$. Thus the last intersection point between $f(x)$ and $\\sqrt{11} x$ will be on the ninth step of one of these intervals. Suppose this intersection point lies in the interval $(3 k-3,3 k)$. The ninth step is of height $10 k-1$. Set $x=3 k-r$, where $r<\\frac{1}{3}$. Then the solution is the largest $k$ for which\n\n$$\n\\begin{aligned}\n10 k-1 & =\\sqrt{11}(3 k-r) \\quad\\left(0<r<\\frac{1}{3}\\right) \\\\\nk(10-3 \\sqrt{11}) & =1-\\sqrt{11} r<1 \\\\\nk & <\\frac{1}{10-3 \\sqrt{11}}=10+3 \\sqrt{11}<20 .\n\\end{aligned}\n$$\n\nBecause $0<19(10-3 \\sqrt{11})<1, k=19$ implies a value of $r$ between 0 and $\\frac{1}{\\sqrt{11}}$. And because $\\frac{1}{\\sqrt{11}}<\\frac{1}{3}$\n\n$$\nx=3 k-r=\\frac{10 k-1}{\\sqrt{11}}=\\frac{\\mathbf{1 8 9} \\sqrt{\\mathbf{1 1}}}{\\mathbf{1 1}}\n$$\n\nis the largest solution to $f(x)=\\sqrt{11} x$.", "Let $x$ be the largest real number for which $\\left\\lfloor\\frac{x}{3}\\right\\rfloor+\\lceil 3 x\\rceil=\\sqrt{11} x$. Because the left-hand side of this equation is an integer, it is simpler to work with $n=\\sqrt{11} x$ instead of $x$. The equation becomes\n\n$$\n\\left\\lfloor\\frac{n}{3 \\sqrt{11}}\\right\\rfloor+\\left\\lceil\\frac{3 n}{\\sqrt{11}}\\right\\rceil=n\n$$\n\n\n\nA little bit of computation shows that $\\frac{1}{3 \\sqrt{11}}+\\frac{3}{\\sqrt{11}}>1$, so the equation cannot hold for large values of $n$. To make this explicit, write\n\n$$\n\\left\\lfloor\\frac{n}{3 \\sqrt{11}}\\right\\rfloor=\\frac{n}{3 \\sqrt{11}}-r \\quad \\text { and } \\quad\\left\\lceil\\frac{3 n}{\\sqrt{11}}\\right\\rceil=\\frac{3 n}{\\sqrt{11}}+s\n$$\n\nwhere $r$ and $s$ are real numbers between 0 and 1. (If $n \\neq 0$, then $r$ and $s$ are strictly between 0 and 1.) Then\n\n$$\n\\begin{aligned}\n1>r-s & =\\left(\\frac{n}{3 \\sqrt{11}}-\\left\\lfloor\\frac{n}{3 \\sqrt{11}}\\right\\rfloor\\right)-\\left(\\left\\lceil\\frac{3 n}{\\sqrt{11}}\\right\\rceil-\\frac{3 n}{\\sqrt{11}}\\right) \\\\\n& =\\left(\\frac{n}{3 \\sqrt{11}}+\\frac{3 n}{\\sqrt{11}}\\right)-\\left(\\left\\lfloor\\frac{n}{3 \\sqrt{11}}\\right\\rfloor+\\left\\lceil\\frac{3 n}{\\sqrt{11}}\\right\\rceil\\right) \\\\\n& =n\\left(\\frac{1}{3 \\sqrt{11}}+\\frac{3}{\\sqrt{11}}-1\\right),\n\\end{aligned}\n$$\n\nso $n<1 /\\left(\\frac{1}{3 \\sqrt{11}}+\\frac{3}{\\sqrt{11}}-1\\right)=99+30 \\sqrt{11}=198.45 \\ldots$\n\nUse trial and error with $n=198,197,196, \\ldots$, to find the value of $n$ that works. Computing the first row of the following table to three decimal digits, and computing both $\\frac{1}{3 \\sqrt{11}}$ and $\\frac{3}{\\sqrt{11}}$ to the same degree of accuracy, allows one to calculate the remaining rows with acceptable round-off errors.\n\n| $n$ | $n /(3 \\sqrt{11})$ | $3 n / \\sqrt{11}$ |\n| :---: | :---: | :---: |\n| | | |\n| 198 | 19.900 | 179.098 |\n| 197 | 19.799 | 178.193 |\n| 196 | 19.699 | 177.289 |\n| 195 | 19.598 | 176.384 |\n| 194 | 19.498 | 175.480 |\n| 193 | 19.397 | 174.575 |\n| 192 | 19.297 | 173.671 |\n| 191 | 19.196 | 172.766 |\n| 190 | 19.096 | 171.861 |\n| 189 | 18.995 | 170.957 |\n\nBecause $n=189=18+171$, the final answer is $x=\\frac{\\mathbf{1 8 9} \\sqrt{\\mathbf{1 1}}}{\\mathbf{1 1}}$." ]

[ "$\\frac{189 \\sqrt{11}}{11}$" ]

[ "Note that $x$ and $z$ can each be minimized by making $y$ as large as possible, so set $y=$ $\\operatorname{lcm}(12,20)=4$. Then $x=5, z=3$, and $x+z=\\mathbf{8}$." ]

[ "8" ]

[ "The midpoint of $\\overline{A B}$ is $\\left(\\frac{T}{2}, 11\\right)$, and the slope of $\\overleftrightarrow{A B}$ is $\\frac{12}{T-2}$. Thus the perpendicular bisector of $\\overline{A B}$ has slope $\\frac{2-T}{12}$ and passes through the point $\\left(\\frac{T}{2}, 11\\right)$. Thus the equation of the perpendicular bisector of $\\overline{A B}$ is $y=\\left(\\frac{2-T}{12}\\right) x+\\left(11-\\frac{2 T-T^{2}}{24}\\right)$. Plugging $y=3$ into this equation and solving for $x$ yields $x=\\frac{96}{T-2}+\\frac{T}{2}$. With $T=8$, it follows that $x=\\frac{96}{6}+\\frac{8}{2}=16+4=\\mathbf{2 0}$." ]

[ "20" ]

[ "The sum of the digits of $N$ must be a multiple of 3 , and the alternating sum of the digits must be a multiple of 11 . Because the number of digits of $N$ is fixed, the minimum $N$ will have the alternating sum of its digits equal to 0 , and therefore the sum of the digits of $N$ will be even, so it must be 6 . Thus if $T$ is even, then $N=1 \\underbrace{0 \\ldots .02}_{T-30^{\\prime} \\mathrm{s}}$, and if $T$ is odd, then $N=1 \\underbrace{0 \\ldots 0}_{T-30^{\\prime} \\mathrm{s}} 32$. Either way, the product of the last two digits of $N$ is 6 (independent of $T$ )." ]

[ "6" ]

[ "Because $\\overline{\\bar{z}}=z$, it follows that $f_{n}(z)=z$ when $n$ is odd, and $f_{n}(z)=\\bar{z}$ when $n$ is even. Taking $z=a+b i$, where $a$ and $b$ are real, it follows that $\\sum_{k=1}^{5} k f_{k}(z)=15 a+3 b i$. Thus $a=\\frac{T}{15}, b=\\frac{T}{3}$, and $|z|=\\sqrt{a^{2}+b^{2}}=\\frac{|T| \\sqrt{26}}{15}$. With $T=15$, the answer is $\\sqrt{\\mathbf{2 6}}$." ]

[ "$\\sqrt{26}$" ]

[ "If the prime factorization of $a b$ is $p_{1}^{e_{1}} p_{2}^{e_{2}} \\ldots p_{k}^{e_{k}}$, where the $p_{i}$ 's are distinct primes and the $e_{i}$ 's are positive integers, then in order for $\\operatorname{gcd}(a, b)$ to equal 1 , each $p_{i}$ must be a divisor of exactly one of $a$ or $b$. Thus the desired number of ordered pairs is $2^{k}$ because there are 2 choices for each prime divisor (i.e., $p_{i} \\mid a$ or $p_{i} \\mid b$ ). With $T=\\sqrt{26}$, it follows that $(\\sqrt{26})^{20} \\cdot 210^{12}=\\left(2^{10} \\cdot 13^{10}\\right) \\cdot 210^{12}=2^{22} \\cdot 3^{12} \\cdot 5^{12} \\cdot 7^{12} \\cdot 13^{10}$. Thus there are five distinct prime divisors, and the answer is $2^{5}=\\mathbf{3 2}$." ]

[ "32" ]

[ "Using $\\sin ^{2} \\theta+\\cos ^{2} \\theta=1$ gives $\\cos ^{2} \\theta=\\frac{64}{T^{2}}$, so to maximize the sum, take $\\cos \\theta=\\frac{8}{|T|}$. Using the formula for the sum of an infinite geometric series gives $\\frac{8 /|T|}{1-8 /|T|}=\\frac{8}{|T|-8}$. With $T=32$, the answer is $\\frac{8}{24}=\\frac{1}{3}$." ]

[ "$\\frac{1}{3}$" ]

[ "By the Pythagorean Theorem, half the diagonal of the square is $\\sqrt{n^{2}-(n-m)^{2}}=\\sqrt{2 m n-m^{2}}$. Thus the diagonal of the square is $2 \\sqrt{2 m n-m^{2}}$, and the square's area is $4 m n-2 m^{2}$. With $T=\\frac{9}{17}, m=9, n=17$, and the answer is 450 ." ]

[ "450" ]

[ "Note that there are 9 days in July in which a person could be a Leo (July 23-31). Let the woman (born before the $d^{\\text {th }}$ day of July) be called Carol, and let the man (born after the $d^{\\text {th }}$ day of July) be called John, and consider the possible values of $d$. If $d \\leq 21$, then Carol will not be a Leo, and the probability that John is a Leo is $\\frac{9}{31-d}$. If $d=22$ or 23 , then the probability is 1 . If $d \\geq 24$, then John will be a Leo, and Carol will not be a Leo with probability $1-\\frac{d-23}{d-1}$. With $T=-14$, the first case applies, and the desired probability is $\\frac{\\mathbf{9}}{\\mathbf{1 7}}$." ]

[ "$\\frac{9}{17}$" ]

[ "Note that $4^{8 !}=2^{2 \\cdot 8 !}$, thus $\\log _{2} 4^{8 !}=2 \\cdot 8$ !. Similarly, $\\log _{4} 2^{8 !}=\\frac{8 !}{2}$. Thus $2 \\cdot 8 !+\\frac{8 !}{2}=$ $6 !\\left(2 \\cdot 7 \\cdot 8+7 \\cdot \\frac{8}{2}\\right)=6 ! \\cdot 140$. Thus $140=T x$, and with $T=-10, x=\\mathbf{- 1 4}$." ]

[ "-14" ]

[ "Divide each side of the second equation by 2 and equate coefficients to obtain $5 b-T-a=$ $\\frac{T}{2}+4 a-1$ and $T+1=-5 b$. Thus $b=\\frac{T+1}{-5}$, and plugging this value into the first equation yields $a=-\\frac{T}{2}$. With $T=20$, the answer is $\\mathbf{- 1 0}$." ]

[ "-10" ]

[ "Because $T$ workers produce 9 widgets in 1 hour, 1 worker will produce $\\frac{9}{T}$ widgets in 1 hour. Thus 1 worker will produce $\\frac{36}{T}$ widgets in 4 hours. In order to produce $\\frac{720}{T}$ widgets in 4 hours, it will require $\\frac{720 / T}{36 / T}=\\mathbf{2 0}$ workers (independent of $T$ )." ]

[ "20" ]

[ "Let $R$ be the remainder when $T$ is divided by 11 . Note that the alternating sum of the digits of the number must be divisible by 11 . This sum will be congruent $\\bmod 11$ to $B-A+A-R=$ $B-R$, thus $B=R$. Because $A$ 's value is irrelevant, to maximize $A+B$, set $A=9$ to yield $A+B=9+R$. For $T=2018, R=5$, and the answer is $9+5=\\mathbf{1 4}$." ]

[ "14" ]

[ "Note that $365=7 \\cdot 52+1$. Thus over the next few years after 2012 , the day of the week for April $1^{\\text {st }}$ will advance by one day in a non-leap year, and it will advance by two days in a leap year. Thus in six years, the day of the week will have rotated a complete cycle, and the answer is 2018 ." ]

[ "2018" ]

[ "Let $x, 2 x$, and $4 x$ be the ages of the children $p$ years ago. Then $x+2 x+4 x=p$, so $7 x=p$. Since $p$ is prime, $x=1$. Thus the sum of the children's current ages is $(1+7)+(2+7)+(4+7)=\\mathbf{2 8}$." ]

[ "28" ]

[ "Because $N<100,5 \\cdot N<500$. Since no primes end in 4, it follows that $5 \\cdot N<400$, hence $N \\leq 79$. The reverses of $5 \\cdot 79=395,4 \\cdot 79=316$, and 79 are 593,613 , and 97 , respectively. All three of these numbers are prime, thus 79 is the largest two-digit integer $N$ for which $N$, $4 \\cdot N$, and $5 \\cdot N$ are all reverse primes." ]

[ "79" ]

[ "Let $r$ and $b$ be the number of students wearing red and blue jerseys, respectively. Then either we choose two blues and one red or one blue and two reds. Thus\n\n$$\n\\begin{aligned}\n& \\left(\\begin{array}{l}\nb \\\\\n2\n\\end{array}\\right)\\left(\\begin{array}{l}\nr \\\\\n1\n\\end{array}\\right)+\\left(\\begin{array}{l}\nb \\\\\n1\n\\end{array}\\right)\\left(\\begin{array}{l}\nr \\\\\n2\n\\end{array}\\right)=25 \\\\\n\\Rightarrow & \\frac{r b(b-1)}{2}+\\frac{b r(r-1)}{2}=25 \\\\\n\\Rightarrow & r b((r-1)+(b-1))=50 \\\\\n\\Rightarrow & r b(r+b-2)=50 .\n\\end{aligned}\n$$\n\nNow because $r, b$, and $r+b-2$ are positive integer divisors of 50 , and $r, b \\geq 2$, we have only a few possibilities to check. If $r=2$, then $b^{2}=25$, so $b=5$; the case $r=5$ is symmetric. If $r=10$, then $b(b+8)=5$, which is impossible. If $r=25$, then $b(b+23)=2$, which is also impossible. So $\\{r, b\\}=\\{2,5\\}$, and $r+b=7$." ]

[ "7" ]

[ "We observe that $\\frac{1}{B E}+\\frac{1}{B N}=\\frac{B E+B N}{B E \\cdot B N}$. The product in the denominator suggests that we compare areas. Let $[B E N]$ denote the area of $\\triangle B E N$. Then $[B E N]=\\frac{1}{2} B E \\cdot B N$, but because $P R=P S=60$, we can also write $[B E N]=[B E P]+[B N P]=\\frac{1}{2} \\cdot 60 \\cdot B E+\\frac{1}{2} \\cdot 60 \\cdot B N$. Therefore $B E \\cdot B N=60(B E+B N)$, so $\\frac{1}{B E}+\\frac{1}{B N}=\\frac{B E+B N}{B E \\cdot B N}=\\frac{1}{\\mathbf{6 0}}$. Note that this value does not depend on the length of the hypotenuse $\\overline{E N}$; for a given location of point $P, \\frac{1}{B E}+\\frac{1}{B N}$ is invariant.", "Using similar triangles, we have $\\frac{E R}{P R}=\\frac{P S}{S N}=\\frac{B E}{B N}$, so $\\frac{B E-60}{60}=$ $\\frac{60}{B N-60}=\\frac{B E}{B N}$ and $B E^{2}+B N^{2}=221^{2}$. Using algebra, we find that $B E=204, B N=85$, and $\\frac{1}{204}+\\frac{1}{85}=\\frac{1}{60}$." ]

[ "$\\frac{1}{60}$" ]

[ "If $y=\\log _{a}\\left(\\log _{a} x\\right)$, then $a^{a^{y}}=x$. Let $y=\\log _{2}\\left(\\log _{2} x\\right)=\\log _{4}\\left(\\log _{4} x\\right)$. Then $2^{2^{y}}=4^{4^{y}}=$ $\\left(2^{2}\\right)^{\\left(2^{2}\\right)^{y}}=2^{2^{2 y+1}}$, so $2 y+1=y, y=-1$, and $x=\\sqrt{\\mathbf{2}}$. (This problem is based on one submitted by ARML alum James Albrecht, 1986-2007.)", "Raise 4 (or $2^{2}$ ) to the power of both sides to get $\\left(\\log _{2} x\\right)^{2}=\\log _{4} x$. By the change of base formula, $\\frac{(\\log x)^{2}}{(\\log 2)^{2}}=\\frac{\\log x}{2 \\log 2}$, so $\\log x=\\frac{\\log 2}{2}$, thus $x=2^{1 / 2}=\\sqrt{\\mathbf{2}}$.", "Let $x=4^{a}$. The equation then becomes $\\log _{2}(2 a)=\\log _{4} a$. Raising 4 to the power of each side, we get $4 a^{2}=a$. Since $a \\neq 0$, we get $4 a=1$, thus $a=\\frac{1}{4}$ and $x=\\sqrt{2}$." ]

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