ID stringlengths 8 10 | Year int64 1.98k 2.02k | Problem Number int64 1 15 | Question stringlengths 37 2.66k | Answer int64 0 997 | Part stringclasses 2 values | Solution listlengths 1 25 |
|---|---|---|---|---|---|---|
2020-I-1 | 2,020 | 1 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 547 | I | [
"[asy] size(10cm); pair A, B, C, D, F; A = (0, tan(3 * pi / 7)); B = (1, 0); C = (-1, 0); F = rotate(90/7, A) * (A - (0, 2)); D = rotate(900/7, F) * A; draw(A -- B -- C -- cycle); draw(F -- D); draw(D -- B); label(\"$A$\", A, N); label(\"$B$\", B, E); label(\"$C$\", C, W); label(\"$D$\", D, W); label(\"$E$\", F, E)... |
2020-I-2 | 2,020 | 2 | There is a unique positive real number $x$ such that the three numbers $\log_8(2x),\log_4x,$ and $\log_2x,$ in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 17 | I | [
"Since these form a geometric series, $\\frac{\\log_2{x}}{\\log_4{x}}$ is the common ratio. Rewriting this, we get $\\frac{\\log_x{4}}{\\log_x{2}} = \\log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\\frac{\\log_4{x}}{\\log_8{2x}} = 2 \\implies \\log_4{x} = 2\\log_8{2} + 2\\log_8{x} \\i... |
2020-I-3 | 2,020 | 3 | A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten. | 621 | I | [
"From the given information, $121a+11b+c=512+64b+8c+a \\implies 120a=512+53b+7c$. Since $a$, $b$, and $c$ have to be positive, $a \\geq 5$. Since we need to minimize the value of $n$, we want to minimize $a$, so we have $a = 5$. Then we know $88=53b+7c$, and we can see the only solution is $b=1$, $c=5$. Finally, $5... |
2020-I-4 | 2,020 | 4 | Let $S$ be the set of positive integers $N$ with the property that the last four digits of $N$ are $2020,$ and when the last four digits are removed, the result is a divisor of $N.$ For example, $42{,}020$ is in $S$ because $4$ is a divisor of $42{,}020.$ Find the sum of all the digits of all the numbers in $S.$ For example, the number $42{,}020$ contributes $4+2+0+2+0=8$ to this total. | 93 | I | [
"We note that any number in $S$ can be expressed as $a(10,000) + 2,020$ for some integer $a$. The problem requires that $a$ divides this number, and since we know $a$ divides $a(10,000)$, we need that $a$ divides 2020. Each number contributes the sum of the digits of $a$, as well as $2 + 0 + 2 +0 = 4$. Since $2020$... |
2020-I-5 | 2,020 | 5 | Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order. | 52 | I | [
"Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two. If we choose any of the numbers $1$ through $6$, there are five other spots to put them, so we get $6 \\cdot 5 = 30$. However, we ov... |
2020-I-6 | 2,020 | 6 | A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7$ . Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 173 | I | [
"[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P ... |
2020-I-7 | 2,020 | 7 | A club consisting of $11$ men and $12$ women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as $1$ member or as many as $23$ members. Let $N$ be the number of such committees that can be formed. Find the sum of the prime numbers that divide $N.$ | 81 | I | [
"Let $k$ be the number of women selected. Then, the number of men not selected is $11-(k-1)=12-k$. Note that the sum of the number of women selected and the number of men not selected is constant at $12$. Each combination of women selected and men not selected corresponds to a committee selection. Since choosing 12... |
2020-I-8 | 2,020 | 8 | A bug walks all day and sleeps all night. On the first day, it starts at point $O,$ faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the point $P.$ Then $OP^2=\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 103 | I | [
"[asy] size(8cm); pair O, A, B, C, D, F, G, H, I, P, X; O = (0, 0); A = (5, 0); X = (8, 0); P = (5, 5 / sqrt(3)); B = rotate(-120, A) * ((O + A) / 2); C = rotate(-120, B) * ((A + B) / 2); D = rotate(-120, C) * ((B + C) / 2); F = rotate(-120, D) * ((C + D) / 2); G = rotate(-120, F) * ((D + F) / 2); H = rotate(-120, ... |
2020-I-9 | 2,020 | 9 | Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ | 77 | I | [
"[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0... |
2020-I-10 | 2,020 | 10 | Let $m$ and $n$ be positive integers satisfying the conditions $\quad\bullet\ \gcd(m+n,210)=1,$ $\quad\bullet\ m^m$ is a multiple of $n^n,$ and $\quad\bullet\ m$ is not a multiple of $n.$ Find the least possible value of $m+n.$ | 407 | I | [
"Taking inspiration from $4^4 \\mid 10^{10}$ we are inspired to take $n$ to be $p^2$, the lowest prime not dividing $210$, or $11 \\implies n = 121$. Now, there are $242$ factors of $11$, so $11^{242} \\mid m^m$, and then $m = 11k$ for $k \\geq 22$. Now, $\\gcd(m+n, 210) = \\gcd(11+k,210) = 1$. Noting $k = 26$ is t... |
2020-I-11 | 2,020 | 11 | For integers $a,b,c$ and $d,$ let $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d.$ Find the number of ordered triples $(a,b,c)$ of integers with absolute values not exceeding $10$ for which there is an integer $d$ such that $g(f(2))=g(f(4))=0.$ | 510 | I | [
"There can be two different cases for this problem, either $f(2)=f(4)$ or not. If it is, note that by Vieta's formulas $a = -6$. Then, $b$ can be anything. However, $c$ can also be anything, as we can set the root of $g$ (not equal to $f(2) = f(4)$) to any integer, producing a possible integer value of $d$. Therefo... |
2020-I-12 | 2,020 | 12 | Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$ | 270 | I | [
"As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$. Lifting the Exponent shows that \\[3=v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\\] so thus, $3^2$ divides $n$. It also shows that \\[7=v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\\] so thus, $7^5$ divides $n$. Now, setting $n = 4c$ (ne... |
2020-I-13 | 2,020 | 13 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p$ . | 36 | I | [
"Points are defined as shown. It is pretty easy to show that $\\triangle AFE \\sim \\triangle AGH$ by spiral similarity at $A$ by some short angle chasing. Now, note that $AD$ is the altitude of $\\triangle AFE$, as the altitude of $AGH$. We need to compare these altitudes in order to compare their areas. Note that... |
2020-I-14 | 2,020 | 14 | Let $P(x)$ be a quadratic polynomial with complex coefficients whose $x^2$ coefficient is $1.$ Suppose the equation $P(P(x))=0$ has four distinct solutions, $x=3,4,a,b.$ Find the sum of all possible values of $(a+b)^2.$ | 85 | I | [
"Either $P(3) = P(4)$ or not. We first see that if $P(3) = P(4)$ it's easy to obtain by Vieta's that $(a+b)^2 = 49$. Now, take $P(3) \\neq P(4)$ and WLOG $P(3) = P(a), P(4) = P(b)$. Now, consider the parabola formed by the graph of $P$. It has vertex $\\frac{3+a}{2}$. Now, say that $P(x) = x^2 - (3+a)x + c$. We not... |
2020-I-15 | 2,020 | 15 | Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ | 58 | I | [
"The following is a power of a point solution to this menace of a problem: [asy] defaultpen(fontsize(12)+0.6); size(250); pen p=fontsize(10)+gray+0.4; var phi=75.5, theta=130, r=4.8; pair A=r*dir(270-phi-57), C=r*dir(270+phi-57), B=r*dir(theta-57), H=extension(A,foot(A,B,C),B,foot(B,C,A)); path omega=circumcircle(A... |
2020-II-1 | 2,020 | 1 | Find the number of ordered pairs of positive integers $(m,n)$ such that ${m^2n = 20 ^{20}}$ . | 231 | II | [
"In this problem, we want to find the number of ordered pairs $(m, n)$ such that $m^2n = 20^{20}$. Let $x = m^2$. Therefore, we want two numbers, $x$ and $n$, such that their product is $20^{20}$ and $x$ is a perfect square. Note that there is exactly one valid $n$ for a unique $x$, which is $\\tfrac{20^{20}}{x}$. ... |
2020-II-2 | 2,020 | 2 | Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$ , and $(0,1)$ . The probability that the slope of the line determined by $P$ and the point $\left(\frac58, \frac38 \right)$ is greater than or equal to $\frac12$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 171 | II | [
"The areas bounded by the unit square and alternately bounded by the lines through $\\left(\\frac{5}{8},\\frac{3}{8}\\right)$ that are vertical or have a slope of $1/2$ show where $P$ can be placed to satisfy the condition. One of the areas is a trapezoid with bases $1/16$ and $3/8$ and height $5/8$. The other area... |
2020-II-3 | 2,020 | 3 | The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 103 | II | [
"Let $\\log _{2^x}3^{20}=\\log _{2^{x+3}}3^{2020}=n$. Based on the equation, we get $(2^x)^n=3^{20}$ and $(2^{x+3})^n=3^{2020}$. Expanding the second equation, we get $8^n\\cdot2^{xn}=3^{2020}$. Substituting the first equation in, we get $8^n\\cdot3^{20}=3^{2020}$, so $8^n=3^{2000}$. Taking the 100th root, we get $... |
2020-II-4 | 2,020 | 4 | Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$ , $B(0,12)$ , $C(16,0)$ , $A'(24,18)$ , $B'(36,18)$ , $C'(24,2)$ . A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$ . | 108 | II | [
"After sketching, it is clear a $90^{\\circ}$ rotation is done about $(x,y)$. Looking between $A$ and $A'$, $x+y=18$. Thus $90+18=108. ~mn28407",
"Because the rotation sends the vertical segment $\\overline{AB}$ to the horizontal segment $\\overline{A'B'}$, the angle of rotation is $90^\\circ$ degrees clockwise. ... |
2020-II-5 | 2,020 | 5 | For each positive integer $n$ , let $f(n)$ be the sum of the digits in the base-four representation of $n$ and let $g(n)$ be the sum of the digits in the base-eight representation of $f(n)$ . For example, $f(2020) = f(133210_{\text{4}}) = 10 = 12_{\text{8}}$ , and $g(2020) = \text{the digit sum of }12_{\text{8}} = 3$ . Let $N$ be the least value of $n$ such that the base-sixteen representation of $g(n)$ cannot be expressed using only the digits $0$ through $9$ . Find the remainder when $N$ is divided by $1000$ . | 151 | II | [
"Let's work backwards. The minimum base-sixteen representation of $g(n)$ that cannot be expressed using only the digits $0$ through $9$ is $A_{16}$, which is equal to $10$ in base 10. Thus, the sum of the digits of the base-eight representation of the sum of the digits of $f(n)$ is $10$. The minimum value for which... |
2020-II-6 | 2,020 | 6 | Define a sequence recursively by $t_1 = 20$ , $t_2 = 21$ , and \[t_n = \frac{5t_{n-1}+1}{25t_{n-2}}\] for all $n \ge 3$ . Then $t_{2020}$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 626 | II | [
"Let $t_n=\\frac{s_n}{5}$. Then, we have $s_n=\\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$. By substitution, we find $s_3 = \\frac{53}{50}$, $s_4=\\frac{103}{105\\cdot50}$, $s_5=\\frac{101}{105}$, $s_6=100$, and $s_7=105$. So $s_n$ has a period of $5$. Thus $s_{2020}=s_5=\\frac{101}{105}$. So, $\\f... |
2020-II-7 | 2,020 | 7 | Two congruent right circular cones each with base radius $3$ and height $8$ have axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 298 | II | [
"Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the ... |
2020-II-8 | 2,020 | 8 | Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$ . | 101 | II | [
"First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \\dots, {a + n(n-1)}$, where $a = n - \\frac{n(n-1)}2.$ This is certainly true for $n=1$. Suppose that it is true for $n = m-1 \\ge 1$, and note that the zeros of $f_m$ are the solutions of $|x - m| = k$, where $k$ is a ... |
2020-II-9 | 2,020 | 9 | While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break. | 90 | II | [
"There are $2^{5}-1$ intersections that we must consider if we are to perform a PIE bash on this problem. Since we don't really want to think that hard, and bashing does not take that long for this problem, we can write down half of all permutations that satisfy the conditions presented in the problem in \"lexicogr... |
2020-II-10 | 2,020 | 10 | Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$ . | 239 | II | [
"The formula for the sum of cubes, also known as Nicomachus's Theorem, is as follows: \\[1^3+2^3+3^3+\\dots+k^3=(1+2+3+\\dots+k)^2=\\left(\\frac{k(k+1)}{2}\\right)^2\\] for any positive integer $k$. So let's apply this to this problem. Let $m=n+5$. Then we have \\begin{align*} 1^3+2^3+3^3+\\dots+(m-5)^3&\\equiv 17 ... |
2020-II-11 | 2,020 | 11 | Let $P(x) = x^2 - 3x - 7$ , and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$ . David computes each of the three sums $P + Q$ , $P + R$ , and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2$ , then $R(0) = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 71 | II | [
"Let $Q(x) = x^2 + ax + 2$ and $R(x) = x^2 + bx + c$. We can write the following: \\[P + Q = 2x^2 + (a - 3)x - 5\\] \\[P + R = 2x^2 + (b - 3)x + (c - 7)\\] \\[Q + R = 2x^2 + (a + b)x + (c + 2)\\] Let the common root of $P+Q,P+R$ be $r$; $P+R,Q+R$ be $s$; and $P+Q,Q+R$ be $t$. We then have that the roots of $P+Q$ ar... |
2020-II-12 | 2,020 | 12 | Let $m$ and $n$ be odd integers greater than $1.$ An $m\times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n$ , those in the second row are numbered left to right with the integers $n + 1$ through $2n$ , and so on. Square $200$ is in the top row, and square $2000$ is in the bottom row. Find the number of ordered pairs $(m,n)$ of odd integers greater than $1$ with the property that, in the $m\times n$ rectangle, the line through the centers of squares $200$ and $2000$ intersects the interior of square $1099$ . | 248 | II | [
"Let us take some cases. Since $m$ and $n$ are odds, and $200$ is in the top row and $2000$ in the bottom, $m$ has to be $3$, $5$, $7$, or $9$. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of $< 1$. Therefore, $m < 1800 \\mod n < 1800-m$. If $m=3$,... |
2020-II-13 | 2,020 | 13 | Convex pentagon $ABCDE$ has side lengths $AB=5$ , $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ . | 60 | II | [
"Assume the incircle touches $AB$, $BC$, $CD$, $DE$, $EA$ at $P,Q,R,S,T$ respectively. Then let $PB=x=BQ=RD=SD$, $ET=y=ES=CR=CQ$, $AP=AT=z$. So we have $x+y=6$, $x+z=5$ and $y+z$=7, solve it we have $x=2$, $z=3$, $y=4$. Let the center of the incircle be $I$, by SAS we can proof triangle $BIQ$ is congruent to triang... |
2020-II-14 | 2,020 | 14 | For real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$ , and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$ . For example, $\{3\} = 0$ and $\{4.56\} = 0.56$ . Define $f(x)=x\{x\}$ , and let $N$ be the number of real-valued solutions to the equation $f(f(f(x)))=17$ for $0\leq x\leq 2020$ . Find the remainder when $N$ is divided by $1000$ . | 10 | II | [
"Note that the upper bound for our sum is $2019,$ and not $2020,$ because if it were $2020$ then the function composition cannot equal to $17.$ From there, it's not too hard to see that, by observing the function composition from right to left, $N$ is (note that the summation starts from the right to the left): \\[... |
2020-II-15 | 2,020 | 15 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ , $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$ . | 717 | II | [
"Let $O$ be the circumcenter of $\\triangle ABC$; say $OT$ intersects $BC$ at $M$; draw segments $XM$, and $YM$. We have $MT=3\\sqrt{15}$. Since $\\angle A=\\angle CBT=\\angle BCT$, we have $\\cos A=\\tfrac{11}{16}$. Notice that $AXTY$ is cyclic, so $\\angle XTY=180^{\\circ}-A$, so $\\cos XTY=-\\cos A$, and the cos... |
2021-I-1 | 2,021 | 1 | Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 97 | I | [
"For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou: $LWWWW$ $WLWWW$ $WWLWW$ $WWWLW$ $WWWWL$ We proceed with casework: Case (1): Sequences #1-4, in which Zou does not lose the last race. The probability that ... |
2021-I-2 | 2,021 | 2 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . [asy] pair A, B, C, D, E, F; A = (0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy] | 109 | I | [
"Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\\angle FGA= \\angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\\angle AFG= \\angle CDG=90 ^{\\circ}$. Therefore, by AA similarity, we know that $\\triangle AFG\\sim\\triangle CDG$. Let $AG... |
2021-I-3 | 2,021 | 3 | Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ | 50 | I | [
"We want to find the number of positive integers $n<1000$ which can be written in the form $n = 2^a - 2^b$ for some non-negative integers $a > b \\ge 0$ (note that if $a=b$, then $2^a-2^b = 0$). We first observe $a$ must be at most 10; if $a \\ge 11$, then $2^a - 2^b \\ge 2^{10} > 1000$. As $2^{10} = 1024 \\approx ... |
2021-I-4 | 2,021 | 4 | Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | 331 | I | [
"Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \\ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \\ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first p... |
2021-I-5 | 2,021 | 5 | Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. | 31 | I | [
"Let the terms be $a-b$, $a$, and $a+b$. Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$, or $3a^2+2b^2=ab^2$. Rearranging, we get $b^2=\\frac{3a^2}{a-2}$. Simplifying further, $b^2=3a+6+\\frac{12}{a-2}$. Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\\pm1, 2, 3, 4, 6, 12$. Loo... |
2021-I-6 | 2,021 | 6 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ , $CP=60\sqrt{5}$ , $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | 192 | I | [
"First scale down the whole cube by $12$. Let point $P$ have coordinates $(x, y, z)$, point $A$ have coordinates $(0, 0, 0)$, and $s$ be the side length. Then we have the equations \\begin{align*} (s-x)^2+y^2+z^2&=\\left(5\\sqrt{10}\\right)^2, \\\\ x^2+(s-y)^2+z^2&=\\left(5\\sqrt{5}\\right)^2, \\\\ x^2+y^2+(s-z)^2&... |
2021-I-7 | 2,021 | 7 | Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\] | 63 | I | [
"It is trivial that the maximum value of $\\sin \\theta$ is $1$, is achieved at $\\theta = \\frac{\\pi}{2}+2k\\pi$ for some integer $k$. This implies that $\\sin(mx) = \\sin(nx) = 1$, and that $mx = \\frac{\\pi}{2}+2a\\pi$ and $nx = \\frac{\\pi}{2}+2b\\pi$, for integers $a, b$. Taking their ratio, we have \\[\\frac... |
2021-I-8 | 2,021 | 8 | Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions. | 57 | I | [
"We take cases for the outermost absolute value, then rearrange: \\[\\left|20|x|-x^2\\right|=c\\pm21.\\] Let $f(x)=\\left|20|x|-x^2\\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values: \\[f(x) = \\begin{cases} \\left|-20x-x^2\\right| & \\mathrm{if} \\ x \\le 0 \\begin{cases} 20x+x^2 & ... |
2021-I-9 | 2,021 | 9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | 567 | I | [
"Let $\\overline{AE}, \\overline{AF},$ and $\\overline{AG}$ be the perpendiculars from $A$ to $\\overleftrightarrow{BC}, \\overleftrightarrow{CD},$ and $\\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\\overline{AF}$ and $\\overline{BD}.$ We set $AB=x$ and $AH=y,$ as shown below. [asy... |
2021-I-10 | 2,021 | 10 | Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$ , if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$ , then \[a_{k+1} = \frac{m + 18}{n+19}.\] Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$ . | 59 | I | [
"We know that $a_{1}=\\tfrac{t}{t+1}$ when $t=2020$ so $1$ is a possible value of $j$. Note also that $a_{2}=\\tfrac{2038}{2040}=\\tfrac{1019}{1020}=\\tfrac{t}{t+1}$ for $t=1019$. Then $a_{2+q}=\\tfrac{1019+18q}{1020+19q}$ unless $1019+18q$ and $1020+19q$ are not relatively prime which happens when $q+1$ divides $1... |
2021-I-11 | 2,021 | 11 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 301 | I | [
"This solution refers to the Diagram section. By the Converse of the Inscribed Angle Theorem, if distinct points $X$ and $Y$ lie on the same side of $\\overline{PQ}$ (but not on $\\overline{PQ}$ itself) for which $\\angle PXQ=\\angle PYQ,$ then $P,Q,X,$ and $Y$ are cyclic. From the Converse of the Inscribed Angle T... |
2021-I-12 | 2,021 | 12 | Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ( $12$ -gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$ . At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 19 | I | [
"Define the distance between two frogs as the number of sides between them that do not contain the third frog. Let $E(a,b,c)$ denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are $a,b,$ and $c$ (in either clockwise or counterclockwise order). Without the ... |
2021-I-13 | 2,021 | 13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ . | 672 | I | [
"Let $O_i$ and $r_i$ be the center and radius of $\\omega_i$, and let $O$ and $r$ be the center and radius of $\\omega$. Since $\\overline{AB}$ extends to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. ... |
2021-I-14 | 2,021 | 14 | For any positive integer $a,$ $\sigma(a)$ denotes the sum of the positive integer divisors of $a$ . Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ . Find the sum of the prime factors in the prime factorization of $n$ . | 125 | I | [
"We first claim that $n$ must be divisible by $42$. Since $\\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$, we can first consider the special case where $a$ is prime and $a \\neq 0,1 \\pmod{43}$. By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists. Then $\\sigma(a^n)-1... |
2021-I-15 | 2,021 | 15 | Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ . | 285 | I | [
"Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately: The two parabolas intersect at four distinct points. By a quick sketch, we have two subconditions: The point $(-k,2... |
2021-II-1 | 2,021 | 1 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | 550 | II | [
"Recall that the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\\frac{101+999}{2}=550 which is the final answer. ~ math31415926535",
"For any palindrome $\\underline{ABA},$ note ... |
2021-II-2 | 2,021 | 2 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ , $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ . [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy] | 336 | II | [
"By angle chasing, we conclude that $\\triangle AGF$ is a $30^\\circ\\text{-}30^\\circ\\text{-}120^\\circ$ triangle, and $\\triangle BED$ is a $30^\\circ\\text{-}60^\\circ\\text{-}90^\\circ$ triangle. Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\\triangle BED,$ we have $DE=\\... |
2021-II-3 | 2,021 | 3 | Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$ . | 80 | II | [
"Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3,$ so WLOG $x_3=3,$ we will multiply by $5$ afterward since any of $x_1, x_2, \\ldots, x_5$ would be $3,$ after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible ... |
2021-II-4 | 2,021 | 4 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | 330 | II | [
"By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$... |
2021-II-5 | 2,021 | 5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ . | 736 | II | [
"We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the $3$rd side is between $6$ and $14$, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than t... |
2021-II-6 | 2,021 | 6 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | 454 | II | [
"By PIE, $|A|+|B|-|A \\cap B| = |A \\cup B|$. Substituting into the equation and factoring, we get that $(|A| - |A \\cap B|)(|B| - |A \\cap B|) = 0$, so therefore $A \\subseteq B$ or $B \\subseteq A$. WLOG $A\\subseteq B$, then for each element there are $3$ possibilities, either it is in both $A$ and $B$, it is in... |
2021-II-7 | 2,021 | 7 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ . | 145 | II | [
"From the fourth equation we get $d=\\frac{30}{abc}.$ Substitute this into the third equation and you get $abc + \\frac{30(ab + bc + ca)}{abc} = abc - \\frac{120}{abc} = 14$. Hence $(abc)^2 - 14(abc)-120 = 0$. Solving, we get $abc = -6$ or $abc = 20$. From the first and second equation, we get $ab + bc + ca = ab-3c... |
2021-II-8 | 2,021 | 8 | An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly $8$ moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 49 | II | [
"For all positive integers $k,$ let $N(k,\\mathrm{BB})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the bottom face. $N(k,\\mathrm{BT})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the... |
2021-II-9 | 2,021 | 9 | Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$ . | 295 | II | [
"This solution refers to the Remarks section. By the Euclidean Algorithm, we have \\[\\gcd\\left(2^m+1,2^m-1\\right)=\\gcd\\left(2,2^m-1\\right)=1.\\] We are given that $\\gcd\\left(2^m+1,2^n-1\\right)>1.$ Multiplying both sides by $\\gcd\\left(2^m-1,2^n-1\\right)$ gives \\begin{align*} \\gcd\\left(2^m+1,2^n-1\\rig... |
2021-II-10 | 2,021 | 10 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 335 | II | [
"This solution refers to the Diagram section. As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\\mathcal{P}.$ Let $\\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$... |
2021-II-11 | 2,021 | 11 | A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no. However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$ . | 258 | II | [
"Note that $\\operatorname{lcm}(6,7)=42.$ It is clear that $42\\not\\in S$ and $84\\not\\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$ In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple,... |
2021-II-12 | 2,021 | 12 | A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 47 | II | [
"Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX =... |
2021-II-13 | 2,021 | 13 | Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$ . | 797 | II | [
"Recall that $1000$ divides this expression if $8$ and $125$ both divide it. It should be fairly obvious that $n \\geq 3$; so we may break up the initial condition into two sub-conditions. (1) $5^n \\equiv n \\pmod{8}$. Notice that the square of any odd integer is $1$ modulo $8$ (proof by plugging in $1^2,3^2,5^2,7... |
2021-II-14 | 2,021 | 14 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 592 | II | [
"In this solution, all angle measures are in degrees. Let $M$ be the midpoint of $\\overline{BC}$ so that $\\overline{OM}\\perp\\overline{BC}$ and $A,G,M$ are collinear. Let $\\angle ABC=13k,\\angle BCA=2k$ and $\\angle XOY=17k.$ Note that: Since $\\angle OGX = \\angle OAX = 90,$ quadrilateral $OGAX$ is cyclic by t... |
2021-II-15 | 2,021 | 15 | Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases} \sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$ . | 258 | II | [
"Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2<n<(k+1)^2$ for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$. Note that this formula also returns the correct value when $n=(k+1)^2$, but not when $n=k^2$... |
2022-I-1 | 2,022 | 1 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | 116 | I | [
"Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial. Note that \\begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\\\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \\end{alignat*} so the slope of $R(x)$ is $\\frac{106-108}{20-16}=-\\frac12.$ ... |
2022-I-2 | 2,022 | 2 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | 227 | I | [
"We are given that \\[100a + 10b + c = 81b + 9c + a,\\] which rearranges to \\[99a = 71b + 8c.\\] Taking both sides modulo $71,$ we have \\begin{align*} 28a &\\equiv 8c \\pmod{71} \\\\ 7a &\\equiv 2c \\pmod{71}. \\end{align*} The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$ Therefore, the requested thre... |
2022-I-3 | 2,022 | 3 | In isosceles trapezoid $ABCD,$ parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650,$ respectively, and $AD=BC=333.$ The angle bisectors of $\angle A$ and $\angle D$ meet at $P,$ and the angle bisectors of $\angle B$ and $\angle C$ meet at $Q.$ Find $PQ.$ | 242 | I | [
"We have the following diagram: [asy] /* Made by MRENTHUSIASM , modified by Cytronical */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorp... |
2022-I-4 | 2,022 | 4 | Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | 834 | I | [
"We rewrite $w$ and $z$ in polar form: \\begin{align*} w &= e^{i\\cdot\\frac{\\pi}{6}}, \\\\ z &= e^{i\\cdot\\frac{2\\pi}{3}}. \\end{align*} The equation $i \\cdot w^r = z^s$ becomes \\begin{align*} e^{i\\cdot\\frac{\\pi}{2}} \\cdot \\left(e^{i\\cdot\\frac{\\pi}{6}}\\right)^r &= \\left(e^{i\\cdot\\frac{2\\pi}{3}}\\... |
2022-I-5 | 2,022 | 5 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D.$ | 550 | I | [
"Define $m$ as the number of minutes they swim for. Let their meeting point be $A$. Melanie is swimming against the current, so she must aim upstream from point $A$, to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she m... |
2022-I-6 | 2,022 | 6 | Find the number of ordered pairs of integers $(a,b)$ such that the sequence \[3,4,5,a,b,30,40,50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | 228 | I | [
"Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$. Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $a$ and $b$ can never be $20$. Since $a < b$, there are ${24 - 2 \\choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions i... |
2022-I-7 | 2,022 | 7 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 289 | I | [
"To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i} \\geq \\frac{1}{7\\cdot8\\cdot9}.$ If we minimize the numerator, then $a \\cdot b \\cdot c - d \\cdot e \\cdot f = 1.$ Note that $a \\cdo... |
2022-I-8 | 2,022 | 8 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$ | 378 | I | [
"We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\\omega_A$ is the incircle of $\\triangle AB'C'$. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir... |
2022-I-9 | 2,022 | 9 | Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement \[\textbf{R B B Y G G Y R O P P O}\] is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 247 | I | [
"Consider this position chart: \\[\\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\\] Since there has to be an even number of spaces between each pair of the same color, spots $1$, $3$, $5$, $7$, $9$, and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configuration... |
2022-I-10 | 2,022 | 10 | Three spheres with radii $11,$ $13,$ and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A,$ $B,$ and $C,$ respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560.$ Find $AC^2.$ | 756 | I | [
"This solution refers to the Diagram section. We let $\\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third, as shown below: [asy] size(400)... |
2022-I-11 | 2,022 | 11 | Let $ABCD$ be a parallelogram with $\angle BAD < 90^\circ.$ A circle tangent to sides $\overline{DA},$ $\overline{AB},$ and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ,$ as shown. Suppose that $AP=3,$ $PQ=9,$ and $QC=16.$ Then the area of $ABCD$ can be expressed in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ [asy] defaultpen(linewidth(0.6)+fontsize(11)); size(8cm); pair A,B,C,D,P,Q; A=(0,0); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy] | 150 | I | [
"Let's redraw the diagram, but extend some helpful lines. [asy] size(10cm); pair A,B,C,D,EE,F,P,Q,O; A=(0,0); EE = (24,15); F = (30,0); O = (10.5,7.5); label(\"$A$\", A, SW); B=(6,15); label(\"$B$\", B, NW); C=(30,15); label(\"$C$\", C, NE); D=(24,0); label(\"$D$\", D, SE); P=(5.2,2.6); label(\"$P$\", (5.8,2.6), N)... |
2022-I-12 | 2,022 | 12 | For any finite set $X,$ let $|X|$ denote the number of elements in $X.$ Define \[S_n = \sum |A \cap B|,\] where the sum is taken over all ordered pairs $(A,B)$ such that $A$ and $B$ are subsets of $\{1,2,3,\ldots,n\}$ with $|A|=|B|.$ For example, $S_2 = 4$ because the sum is taken over the pairs of subsets \[(A,B) \in \left\{(\emptyset,\emptyset),(\{1\},\{1\}),(\{1\},\{2\}),(\{2\},\{1\}),(\{2\},\{2\}),(\{1,2\},\{1,2\})\right\},\] giving $S_2 = 0+1+0+0+1+2=4.$ Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p+q$ is divided by $1000.$ | 245 | I | [
"Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$. The problem states that for $n=2, S_n$ is $4$. Let's try it out for $n=3$. Let's perform casework on the number of elements in $A, B$. $\\textbf{Case 1:} |A| = |B| = 1$ In this case, the only possible equivalenci... |
2022-I-13 | 2,022 | 13 | Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.\overline{3636} = \frac{4}{11}$ and $0.\overline{1230} = \frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$ | 392 | I | [
"$0.\\overline{abcd}=\\frac{abcd}{9999} = \\frac{x}{y}$, $9999=9\\times 11\\times 101$. Then we need to find the number of positive integers $x$ that (with one of more $y$ such that $y|9999$) can meet the requirement $1 \\leq {x}\\cdot\\frac{9999}{y} \\leq 9999$. Make cases by factors of $x$. (A venn diagram of cas... |
2022-I-14 | 2,022 | 14 | Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^\circ.$ Find the perimeter of $\triangle ABC.$ | 459 | I | [
"Denote $BC = a$, $CA = b$, $AB = c$. Let the splitting line of $\\triangle ABC$ through $M$ (resp. $N$) crosses $\\triangle ABC$ at another point $X$ (resp. $Y$). WLOG, we assume $c \\leq b$. $\\textbf{Case 1}$: $a \\leq c \\leq b$. We extend segment $AB$ to $D$, such that $BD = a$. We extend segment $AC$ to $E$, ... |
2022-I-15 | 2,022 | 15 | Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 33 | I | [
"First, let define a triangle with side lengths $\\sqrt{2x}$, $\\sqrt{2z}$, and $l$, with altitude from $l$'s equal to $\\sqrt{xz}$. $l = \\sqrt{2x - xz} + \\sqrt{2z - xz}$, the left side of one equation in the problem. Let $\\theta$ be angle opposite the side with length $\\sqrt{2x}$. Then the altitude has length ... |
2022-II-1 | 2,022 | 1 | Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | 154 | II | [
"Let $x$ be the number of people at the party before the bus arrives. We know that $x\\equiv 0\\pmod {12}$, as $\\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \\equiv 0 \\pmod{25}$, as $\\frac{11}{25}$ of the people at the party are adults after the bus arri... |
2022-II-2 | 2,022 | 2 | Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac23$ . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac34$ . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 125 | II | [
"Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label. This problem can be solved by using $2$ cases. $\\textbf{Case 1:}$ $C$'s opponent for the semifinal is $A$ The probability... |
2022-II-3 | 2,022 | 3 | A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 21 | II | [
"Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\\frac{6}{\\sqrt{2}} = 3\\sqrt{2}$). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$. B... |
2022-II-4 | 2,022 | 4 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 112 | II | [
"Define $a$ to be $\\log_{20x} (22x) = \\log_{2x} (202x)$, what we are looking for. Then, by the definition of the logarithm, \\[\\begin{cases} (20x)^{a} &= 22x \\\\ (2x)^{a} &= 202x. \\end{cases}\\] Dividing the first equation by the second equation gives us $10^a = \\frac{11}{101}$, so by the definition of logs, ... |
2022-II-5 | 2,022 | 5 | Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | 72 | II | [
"Let $a$, $b$, and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$. \\[a - b = p_1\\] \\[b - c = p_2\\] \\[a - c = p_3\\] $p_3 = a - c = a - b + b - c = p_1 + p_2$. Because $p_3$ is the sum of two primes, $p_1$ and $p_2$, $p_1$ or $p_2$ must be $2$. Let $p_1 = 2$, then $p_3 = p_2 + 2$... |
2022-II-6 | 2,022 | 6 | Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 841 | II | [
"To find the greatest value of $x_{76} - x_{16}$, $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $x_{76} = x_{77} = x_{78} = \\dots = x_{100} > 0$. If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \\dots = x_{1} < 0$. The other ... |
2022-II-7 | 2,022 | 7 | A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles. | 192 | II | [
"[asy] //Created by isabelchen size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/... |
2022-II-9 | 2,022 | 9 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ . [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy] | 244 | II | [
"We can use recursion to solve this problem: 1. Fix 7 points on $\\ell_A$, then put one point $B_1$ on $\\ell_B$. Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\\ell_B$. For example, $f(1) = 6$ because there are 6 regions. 2. Now, put the second p... |
2022-II-10 | 2,022 | 10 | Find the remainder when \[\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}\] is divided by $1000$ . | 4 | II | [
"We first write the expression as a summation. \\begin{align*} \\sum_{i=3}^{40} \\binom{\\binom{i}{2}}{2} & = \\sum_{i=3}^{40} \\binom{\\frac{i \\left( i - 1 \\right)}{2}}{2} \\\\ & = \\sum_{i=3}^{40} \\frac{\\frac{i \\left( i - 1 \\right)}{2} \\left( \\frac{i \\left( i - 1 \\right)}{2}- 1 \\right)}{2} \\\\ & = \\f... |
2022-II-11 | 2,022 | 11 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | 180 | II | [
"[asy] defaultpen(fontsize(12)+0.6); size(300); pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H); draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp-... |
2022-II-12 | 2,022 | 12 | Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that \[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\] Find the least possible value of $a+b.$ | 23 | II | [
"Denote $P = \\left( x , y \\right)$. Because $\\frac{x^2}{a^2}+\\frac{y^2}{a^2-16} = 1$, $P$ is on an ellipse whose center is $\\left( 0 , 0 \\right)$ and foci are $\\left( - 4 , 0 \\right)$ and $\\left( 4 , 0 \\right)$. Hence, the sum of distance from $P$ to $\\left( - 4 , 0 \\right)$ and $\\left( 4 , 0 \\right)$... |
2022-II-13 | 2,022 | 13 | There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ . | 220 | II | [
"Because $0 < x < 1$, we have \\begin{align*} P \\left( x \\right) & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\binom{6}{a} x^{2310a} \\left( - 1 \\right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\\\ & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\su... |
2022-II-14 | 2,022 | 14 | For positive integers $a$ , $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ , $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$ . | 188 | II | [
"Notice that we must have $a = 1$; otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it can have all the values from $1$ to $c-1$ cents. Plus $\\lfloor \\frac{999}{c} \\rfloor$ ... |
2022-II-15 | 2,022 | 15 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ , $O_1O_2 = 15$ , $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy] | 140 | II | [
"First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$. Let points $A'$ and $B'$ be the reflections of $A$ and $B$, respectively, about the perpendicular bisector of $\\overline{O_1O_2}$. Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furt... |
2023-I-1 | 2,023 | 1 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 191 | I | [
"For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions. First, there are $\\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\\cdot8\\cdot6\\cdot4\\cdot2$ ways to pla... |
2023-I-2 | 2,023 | 2 | Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$ | 881 | I | [
"Denote $x = \\log_b n$. Hence, the system of equations given in the problem can be rewritten as \\begin{align*} \\sqrt{x} & = \\frac{1}{2} x , \\\\ bx & = 1 + x . \\end{align*} Solving the system gives $x = 4$ and $b = \\frac{5}{4}$. Therefore, \\[n = b^x = \\frac{625}{256}.\\] Therefore, the answer is $625 + 256 ... |
2023-I-3 | 2,023 | 3 | A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect. | 607 | I | [
"In this solution, let $\\boldsymbol{n}$-line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$-line points. There are $\\binom{40}{2}=780$ pairs of lines. Among them: The $3$-line points account for $3\\cdot\\binom32=9$ pairs of lines. The $4$-line points account for $4\\cdo... |
2023-I-4 | 2,023 | 4 | The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f,$ where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$ | 12 | I | [
"We first rewrite $13!$ as a prime factorization, which is $2^{10}\\cdot3^5\\cdot5^2\\cdot7\\cdot11\\cdot13.$ For the fraction to be a square, it needs each prime to be an even power. This means $m$ must contain $7\\cdot11\\cdot13$. Also, $m$ can contain any even power of $2$ up to $2^{10}$, any odd power of $3$ up... |
2023-I-5 | 2,023 | 5 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 106 | I | [
"Ptolemy's theorem states that for cyclic quadrilateral $WXYZ$, $WX\\cdot YZ + XY\\cdot WZ = WY\\cdot XZ$. We may assume that $P$ is between $B$ and $C$. Let $PA = a$, $PB = b$, $PC = c$, $PD = d$, and $AB = s$. We have $a^2 + c^2 = AC^2 = 2s^2$, because $AC$ is a diameter of the circle. Similarly, $b^2 + d^2 = 2s^... |
2023-I-6 | 2,023 | 6 | Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 51 | I | [
"We break the problem into stages, one for each card revealed, then further into cases based on the number of remaining unrevealed cards of each color. Since expected value is linear, the expected value of the total number of correct card color guesses across all stages is the sum of the expected values of the numb... |
2023-I-7 | 2,023 | 7 | Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$ . | 49 | I | [
"$n$ can either be $0$ or $1$ (mod $2$). Case 1: $n \\equiv 0 \\pmod{2}$ Then, $n \\equiv 2 \\pmod{4}$, which implies $n \\equiv 1 \\pmod{3}$ and $n \\equiv 4 \\pmod{6}$, and therefore $n \\equiv 3 \\pmod{5}$. Using CRT, we obtain $n \\equiv 58 \\pmod{60}$, which gives $16$ values for $n$. Case 2: $n \\equiv 1 \\pm... |
2023-I-8 | 2,023 | 8 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | 125 | I | [
"This solution refers to the Diagram section. Let $O$ be the incenter of $ABCD$ for which $\\odot O$ is tangent to $\\overline{DA},\\overline{AB},$ and $\\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\\overleftrightarrow{DA},\\o... |
2023-I-9 | 2,023 | 9 | Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$ | 738 | I | [
"Plugging $2$ and $m$ into the expression for $p(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$. Rearranging, we have \\[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So we only n... |
2023-I-10 | 2,023 | 10 | There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$ . (Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .) | 944 | I | [
"Define $\\left\\{ x \\right\\} = x - \\left\\lfloor x \\right\\rfloor$. First, we bound $U$. We establish an upper bound of $U$. We have \\begin{align*} U & \\leq \\sum_{n=1}^{2023} \\frac{n^2 - na}{5} \\\\ & = \\frac{1}{5} \\sum_{n=1}^{2023} n^2 - \\frac{a}{5} \\sum_{n=1}^{2023} n \\\\ & = \\frac{1012 \\cdot 2023... |
2023-I-11 | 2,023 | 11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | 235 | I | [
"Define $f(x)$ to be the number of subsets of $\\{1, 2, 3, 4, \\ldots x\\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair. Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditio... |
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