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 |
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2013-I-11 | 2,013 | 11 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions: (a) If $16$ , $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and (b) There are three integers $0 < x < y < z < 14$ such that when $x$ , $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over. Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions. | 148 | I | [
"$N$ must be some multiple of $\\text{lcm}(14, 15, 16)= 2^{4}\\cdot 3\\cdot 5\\cdot 7$ ; this $lcm$ is hereby denoted $k$ and $N = qk$. $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $10$, and $12$ all divide $k$, so $x, y, z = 9, 11, 13$ We have the following three modulo equations: $nk\\equiv 3\\pmod{9}$ $nk\\equiv 3\\p... |
2013-I-12 | 2,013 | 12 | Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^o$ and $\angle Q = 60^o$ . A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$ , side $\overline{CD}$ lies on $\overline{QR}$ , and one of the remaining vertices lies on $\overline{RP}$ . There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$ , where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$ . | 21 | I | [
"First, find that $\\angle R = 45^\\circ$. Draw $ABCDEF$. Now draw $\\bigtriangleup PQR$ around $ABCDEF$ such that $Q$ is adjacent to $C$ and $D$. The height of $ABCDEF$ is $\\sqrt{3}$, so the length of base $QR$ is $2+\\sqrt{3}$. Let the equation of $\\overline{RP}$ be $y = x$. Then, the equation of $\\overline{PQ... |
2013-I-13 | 2,013 | 13 | Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ , $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$ . | 961 | I | [
"Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$. Also, the area we seek is simply the ratio $k=\\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}$, because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90. For ease, all ratios I will use to... |
2013-I-14 | 2,013 | 14 | For $\pi \le \theta < 2\pi$ , let \[P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots\] and \[Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta +\ldots\] so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 36 | I | [
"Noticing the $\\sin$ and $\\cos$ in both $P$ and $Q,$ we think of the angle addition identities: \\[\\sin(a + b) = \\sin a \\cos b + \\cos a \\sin b, \\cos(a + b) = \\cos a \\cos b - \\sin a \\sin b\\] With this in mind, we multiply $P$ by $\\sin \\theta$ and $Q$ by $\\cos \\theta$ to try and use some angle additi... |
2013-I-15 | 2,013 | 15 | Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions: (a) $0\le A<B<C\le99$ , (b) there exist integers $a$ , $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ , (c) $p$ divides $A-a$ , $B-b$ , and $C-c$ , and (d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$ . | 272 | I | [
"From condition (d), we have $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$. Condition $\\text{(c)}$ states that $p\\mid B-D-a$, $p | B-a+d$, and $p\\mid B+D-a-d$. We subtract the first two to get $p\\mid-d-D$, and we do the same for the last two to get $p\\mid 2d-D$. We subtract these two to get $p\\mid 3d$. So $... |
2013-II-1 | 2,013 | 1 | Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ , $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$ . | 275 | II | [
"There are $24 \\cdot 60=1440$ normal minutes in a day , and $10 \\cdot 100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\\frac{1440}{1000}$, which simplifies to $\\frac{36}{25}$. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\\text... |
2013-II-2 | 2,013 | 2 | Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$ . | 881 | II | [
"To simplify, we write this logarithmic expression as an exponential one. Just looking at the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means $\\log_{2^a}(\\log_{2^b}(2^{1000}))=1$ (because $2^... |
2013-II-3 | 2,013 | 3 | A large candle is $119$ centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom. Specifically, the candle takes $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn down the $k$ -th centimeter. Suppose it takes $T$ seconds for the candle to burn down completely. Then $\tfrac{T}{2}$ seconds after it is lit, the candle's height in centimeters will be $h$ . Find $10h$ . | 350 | II | [
"We find that $T=10(1+2+\\cdots +119)$. From Gauss's formula, we find that the value of $T$ is $10(7140)=71400$. The value of $\\frac{T}{2}$ is therefore $35700$. We find that $35700$ is $10(3570)=10\\cdot \\frac{k(k+1)}{2}$, so $3570=\\frac{k(k+1)}{2}$. As a result, $7140=k(k+1)$, which leads to $0=k^2+k-7140$. We... |
2013-II-4 | 2,013 | 4 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$ . | 40 | II | [
"The distance from point $A$ to point $B$ is $\\sqrt{13}$. The vector that starts at point A and ends at point B is given by $B - A = (1, 2\\sqrt{3})$. Since the center of an equilateral triangle, $P$, is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equat... |
2013-II-5 | 2,013 | 5 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$ . | 20 | II | [
"[asy] pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$D$\", D, S); label(\"$M$\", M, S); label(\"$E$\", E, S); draw(A--D); draw(A--M); draw(A--E); [/asy] Without loss o... |
2013-II-6 | 2,013 | 6 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | 282 | II | [
"Solution 1 The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\\geq1000$, so $x\\geq\\frac{999}{2}\\implies x\\geq500$. $x=500$ does not work, so $x>500$. Let $n=x-500$. By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ wi... |
2013-II-7 | 2,013 | 7 | A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting. | 945 | II | [
"There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\\cdot\\frac{10}{60} \\cdot (x-3t)=1775$, and simplify that we get $19x-21t=355$. Now the problem is to find a reasonable integer solution. Now we know $x= \\frac{355+21t}{19}$,... |
2013-II-8 | 2,013 | 8 | A hexagon that is inscribed in a circle has side lengths $22$ , $22$ , $20$ , $22$ , $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ . | 272 | II | [
"[asy] import olympiad; import math; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label(\"$A$\",A,W);label(\"$B$\",B,NW);label(\"$C$\",C,... |
2013-II-9 | 2,013 | 9 | A $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$ . | 106 | II | [
"Firstly, we consider how many different ways possible to divide the $7\\times 1$ board. We ignore the cases of 1 or 2 pieces since we need at least one tile of each color. Three pieces: $5+1+1$, $4+2+1$, $4+1+2$, etc, $\\dbinom{6}{2}=15$ ways in total (just apply stars and bars here) Four pieces: $\\dbinom{6}{3}=2... |
2013-II-10 | 2,013 | 10 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ , $b$ , $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ . | 146 | II | [
"[asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label(\"$O$\", (0,0), S);label(\"$B$\", B, SW);label(\"$A$\", A, S); dot((0,0)); [/asy] Now we put the figure in the Cartesian plane, let the center of the circle ... |
2013-II-11 | 2,013 | 11 | Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ , and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$ . | 399 | II | [
"Any such function can be constructed by distributing the elements of $A$ on three tiers. The bottom tier contains the constant value, $c=f(f(x))$ for any $x$. (Obviously $f(c)=c$.) The middle tier contains $k$ elements $x\\ne c$ such that $f(x)=c$, where $1\\le k\\le 6$. The top tier contains $6-k$ elements such t... |
2013-II-12 | 2,013 | 12 | Let $S$ be the set of all polynomials of the form $z^3 + az^2 + bz + c$ , where $a$ , $b$ , and $c$ are integers. Find the number of polynomials in $S$ such that each of its roots $z$ satisfies either $|z| = 20$ or $|z| = 13$ . | 540 | II | [
"Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from Vieta's formulas. Case 1: $f(z)=(z-r)(z-\\omega)(z-\\omega^*)$, where $r\\in \\mathbb{R}$, $\\omega$ is nonreal, and $\\omega^*$ is the complex conjugate of omega (note t... |
2013-II-13 | 2,013 | 13 | In $\triangle ABC$ , $AC = BC$ , and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that $CE = \sqrt{7}$ and $BE = 3$ , the area of $\triangle ABC$ 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 $m+n$ . | 10 | II | [
"We can set $AE=ED=m$. Set $BD=k$, therefore $CD=3k, AC=4k$. Thereafter, by Stewart's Theorem on $\\triangle ACD$ and cevian $CE$, we get $2m^2+14=25k^2$. Also apply Stewart's Theorem on $\\triangle CEB$ with cevian $DE$. After simplification, $2m^2=17-6k^2$. Therefore, $k=1, m=\\frac{\\sqrt{22}}{2}$. Finally, note... |
2013-II-14 | 2,013 | 14 | For positive integers $n$ and $k$ , let $f(n, k)$ be the remainder when $n$ is divided by $k$ , and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$ . Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$ . | 512 | II | [
"The Pattern We can find that $20\\equiv 6 \\pmod{7}$ $21\\equiv 5 \\pmod{8}$ $22\\equiv 6 \\pmod{8}$ $23\\equiv 7 \\pmod{8}$ $24\\equiv 6 \\pmod{9}$ $25\\equiv 7 \\pmod{9}$ $26\\equiv 8 \\pmod{9}$ Observing these and we can find that the reminders are in groups of three continuous integers, considering this is tru... |
2013-II-15 | 2,013 | 15 | Let $A,B,C$ be angles of an acute triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ , $q$ , $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$ . | 222 | II | [
"Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \\sin{A}$. By the Law of Sines, we must have $CA = \\sin{B}$ and $AB = \\sin{C}$. Now let us analyze the given: \\begin{align*} \\cos^2A + \\cos^2B + 2\\sin A\\sin B\\cos C &... |
2014-I-1 | 2,014 | 1 | The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters. [asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy] | 790 | I | [
"The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of $\\dfrac{80}{3}$mm. According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is $\\sqrt{50^2+\\left(\\frac{80}{3}\\right)^2}=\\frac{170}{3}$mm. Since that on the l... |
2014-I-2 | 2,014 | 2 | An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$ . Find $N$ . | 144 | I | [
"First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$. The probability both are green is $\\frac{4}{10}\\cdot\\frac{16}{16+N}$, and the probability both are blue is $\\frac{6}{10}\\cdot\\frac{N}{16+N}$, so \\[\\frac{4}{10... |
2014-I-3 | 2,014 | 3 | Find the number of rational numbers $r$ , $0<r<1,$ such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of $1000$ . | 200 | I | [
"Let the numerator and denominator $x,y$ with $\\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\\gcd(x,y) = 1$ then $\\gcd(x,y) =\\gcd(x,1000-x)= \\gcd(x,1000-x-(-1)x)=\\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\\phi(1000) = 400$ numbers relatively p... |
2014-I-4 | 2,014 | 4 | Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 49 | I | [
"For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of $\\dfrac{1}{3}$ mi/min. Let $d$ be the length of the trains, $r_1$ be the speed of train 1 (the faster train), and $r_2$ be the speed of train 2. Consider the problem from the bikers' moving frame of ... |
2014-I-5 | 2,014 | 5 | Let the set $S = \{P_1, P_2, \dots, P_{12}\}$ consist of the twelve vertices of a regular $12$ -gon. A subset $Q$ of $S$ is called communal if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.) | 134 | I | [
"By looking at the problem and drawing a few pictures, it quickly becomes obvious that one cannot draw a circle that covers $2$ disjoint areas of the $12$-gon without including all the vertices in between those areas. In other words, in order for a subset to be communal, all the vertices in the subset must be adjac... |
2014-I-6 | 2,014 | 6 | The graphs $y = 3(x-h)^2 + j$ and $y = 2(x-h)^2 + k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ . | 36 | I | [
"Begin by setting $x$ to 0, then set both equations to $h^2=\\frac{2013-j}{3}$ and $h^2=\\frac{2014-k}{2}$, respectively. Notice that because the two parabolas have to have positive x-intercepts, $h\\ge32$. We see that $h^2=\\frac{2014-k}{2}$, so we now need to find a positive integer $h$ which has positive integer... |
2014-I-7 | 2,014 | 7 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane.) | 100 | I | [
"Let $w = \\operatorname{cis}{(\\alpha)}$ and $z = 10\\operatorname{cis}{(\\beta)}$. Then, $\\dfrac{w - z}{z} = \\dfrac{\\operatorname{cis}{(\\alpha)} - 10\\operatorname{cis}{(\\beta)}}{10\\operatorname{cis}{\\beta}}$. Multiplying both the numerator and denominator of this fraction by $\\operatorname{cis}{(-\\beta)... |
2014-I-8 | 2,014 | 8 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base 10, where digit $a$ is not zero. Find the three-digit number $abc$ . | 937 | I | [
"We have that $N^2 - N = N(N - 1)\\equiv 0\\mod{10000}$ Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$. Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$, which is impossible for a number that is divisible by either $2$ or $5$... |
2014-I-9 | 2,014 | 9 | Let $x_1< x_2 < x_3$ be the three real roots of the equation $\sqrt{2014} x^3 - 4029x^2 + 2 = 0$ . Find $x_2(x_1+x_3)$ . | 2 | I | [
"Substituting $n$ for $2014$, we get \\[\\sqrt{n}x^3 - (1+2n)x^2 + 2 = \\sqrt{n}x^3 - x^2 - 2nx^2 + 2\\] \\[= x^2(\\sqrt{n}x - 1) - 2(nx^2 - 1) = 0\\] Noting that $nx^2 - 1$ factors as a difference of squares to \\[(\\sqrt{n}x - 1)(\\sqrt{n}x+1)\\] we can factor the left side as \\[(\\sqrt{n}x - 1)(x^2 - 2(\\sqrt{n... |
2014-I-10 | 2,014 | 10 | A disk with radius $1$ is externally tangent to a disk with radius $5$ . Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$ . That is, if the center of the smaller disk has moved to the point $D$ , and the point on the smaller disk that began at $A$ has now moved to point $B$ , then $\overline{AC}$ is parallel to $\overline{BD}$ . Then $\sin^2(\angle BEA)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 58 | I | [
"[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"$A$\",a,W,fontsize(9)); label(\"$B$\",b,NW,fontsize(9)); label(\"$C$\",c,E,fontsize(9)); label(\"$D$\",d,E,fontsize(9)); label(\"$E$... |
2014-I-11 | 2,014 | 11 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 391 | I | [
"Perform the coordinate transformation $(x, y)\\rightarrow (x+y, x-y)$. Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\\langle 1, -1 \\rangle$, $\\langle 1, 1 \\rangle$, $\\langle -1, -1 \\rangle$, $\\langle -1, 1 \\rangle$ respectively. Moreover, the transformati... |
2014-I-12 | 2,014 | 12 | Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ . | 453 | I | [
"The natural way to go is casework. And the natural process is to sort $f$ and $g$ based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: $f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2$. Note that the $1, 2$ and $1, 3$ cases are symmetrical and we need just a $*2$. Note also that the total... |
2014-I-13 | 2,014 | 13 | On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$ . [asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy] | 850 | I | [
"Let $s$ be the side length of $ABCD$, let $Q$, and $R$ be the midpoints of $\\overline{EG}$ and $\\overline{FH}$, respectively, let $S$ be the foot of the perpendicular from $Q$ to $\\overline{CD}$, let $T$ be the foot of the perpendicular from $R$ to $\\overline{AD}$. [asy] size(150); defaultpen(fontsize(10pt)); ... |
2014-I-14 | 2,014 | 14 | Let $m$ be the largest real solution to the equation \[\dfrac{3}{x-3} + \dfrac{5}{x-5} + \dfrac{17}{x-17} + \dfrac{19}{x-19} = x^2 - 11x - 4\] There are positive integers $a, b,$ and $c$ such that $m = a + \sqrt{b + \sqrt{c}}$ . Find $a+b+c$ . | 263 | I | [
"The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\\frac{3}{x-3}$, then the fraction becomes of the form $\\frac{x}{x - 3}$. A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (w... |
2014-I-15 | 2,014 | 15 | In $\triangle ABC, AB = 3, BC = 4,$ and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B, \overline{BC}$ at $B$ and $D,$ and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4},$ length $DE=\frac{a\sqrt{b}}{c},$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ . | 41 | I | [
"Since $\\angle DBE = 90^\\circ$, $DE$ is the diameter of $\\omega$. Then $\\angle DFE=\\angle DGE=90^\\circ$. But $DF=FE$, so $\\triangle DEF$ is a 45-45-90 triangle. Letting $DG=3x$, we have that $EG=4x$, $DE=5x$, and $DF=EF=\\frac{5x}{\\sqrt{2}}$. Note that $\\triangle DGE \\sim \\triangle ABC$ by SAS similarity... |
2014-II-1 | 2,014 | 1 | Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room. | 334 | II | [
"From the given information, we can see that Abe can paint $\\frac{1}{15}$ of the room in an hour, Bea can paint $\\frac{1}{15}\\times\\frac{3}{2} = \\frac{1}{10}$ of the room in an hour, and Coe can paint the room in $\\frac{1}{15}\\times 2 = \\frac{2}{15}$ of the room in an hour. After $90$ minutes, Abe has paint... |
2014-II-2 | 2,014 | 2 | Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$ . The probability that a man has none of the three risk factors given that he does not have risk factor A is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 76 | II | [
"We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram. [asy] pair A,B,C,D,E,F,G; A=(0,55); B=(60,55); C=(60,0); D=(0,0); draw(A--B--C--D--A); E=(30,35); F=(20,20); G=(40,20); draw(circle(E,15)); draw(circle(F,15)); draw(circle(G,15)); draw(\"$... |
2014-II-3 | 2,014 | 3 | A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$ . [asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy] | 720 | II | [
"When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below. [asy] pair R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.4... |
2014-II-4 | 2,014 | 4 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy $0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},$ where $a$ , $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ . | 447 | II | [
"Notice repeating decimals can be written as the following: $0.\\overline{ab}=\\frac{10a+b}{99}$ $0.\\overline{abc}=\\frac{100a+10b+c}{999}$ where a,b,c are the digits. Now we plug this back into the original fraction: $\\frac{10a+b}{99}+\\frac{100a+10b+c}{999}=\\frac{33}{37}$ Multiply both sides by $999*99.$ This ... |
2014-II-5 | 2,014 | 5 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ . | 420 | II | [
"Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$, and the remaining root of $q(x)$ is $-(r+s+1)$. The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$, and equating the two coefficients gives \\[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\\]from which $s... |
2014-II-6 | 2,014 | 6 | Charles has two six-sided dice. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 167 | II | [
"The probability that he rolls a six twice when using the fair die is $\\frac{1}{6}\\times \\frac{1}{6}=\\frac{1}{36}$. The probability that he rolls a six twice using the biased die is $\\frac{2}{3}\\times \\frac{2}{3}=\\frac{4}{9}=\\frac{16}{36}$. Given that Charles rolled two sixes, we can see that it is $16$ ti... |
2014-II-7 | 2,014 | 7 | Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$ . Find the sum of all positive integers $n$ for which $\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.$ | 21 | II | [
"First, let's split it into two cases to get rid of the absolute value sign $\\left |\\sum_{k=1}^n\\log_{10}f(k)\\right|=1 \\iff \\sum_{k=1}^n\\log_{10}f(k)=1,-1$ Now we simplify using product-sum logarithmic identites: $\\log_{10}{f(1)}+\\log_{10}{f(2)}+...+\\log_{10}{f(n)}=\\log_{10}{f(1)\\cdot f(2) \\cdot ... \\... |
2014-II-8 | 2,014 | 8 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle $D$ is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$ . | 254 | II | [
"[asy] import graph; size(7.99cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079; draw(circle((7.780000000000009,-1.320000000000002), 2.00000000... |
2014-II-9 | 2,014 | 9 | Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. | 581 | II | [
"We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \\cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $... |
2014-II-10 | 2,014 | 10 | Let $z$ be a complex number with $|z|=2014$ . Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}$ . Then the area enclosed by $P$ can be written in the form $n\sqrt{3}$ , where $n$ is an integer. Find the remainder when $n$ is divided by $1000$ . | 147 | II | [
"Note that the given equality reduces to \\[\\frac{1}{w+z} = \\frac{w+z}{wz}\\] \\[wz = {(w+z)}^2\\] \\[w^2 + wz + z^2 = 0\\] \\[\\frac{w^3 - z^3}{w-z} = 0\\] \\[w^3 = z^3, w \\neq z\\] Now, let $w = r_w e^{i \\theta_w}$ and likewise for $z$. Consider circle $O$ with the origin as the center and radius 2014 on the ... |
2014-II-11 | 2,014 | 11 | In $\triangle RED$ , $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$ . $RD=1$ . Let $M$ be the midpoint of segment $\overline{RD}$ . Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$ . Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$ . Then $AE=\frac{a-\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$ . | 56 | II | [
"Let $P$ be the foot of the perpendicular from $A$ to $\\overline{CR}$, so $\\overline{AP}\\parallel\\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\\overline{CR}$, and $\\overline{PM}\\parallel\\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \\frac{CD}{2}$. We can then us... |
2014-II-12 | 2,014 | 12 | Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1$ . Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}$ . Find $m$ . | 399 | II | [
"Note that $\\cos{3C}=-\\cos{(3A+3B)}$. Thus, our expression is of the form $\\cos{3A}+\\cos{3B}-\\cos{(3A+3B)}=1$. Let $\\cos{3A}=x$ and $\\cos{3B}=y$. Using the fact that $\\cos(3A+3B)=\\cos 3A\\cos 3B-\\sin 3A\\sin 3B=xy-\\sqrt{1-x^2}\\sqrt{1-y^2}$, we get $x+y-xy+\\sqrt{1-x^2}\\sqrt{1-y^2}=1$, or $\\sqrt{1-x^2}... |
2014-II-13 | 2,014 | 13 | Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$ , no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 28 | II | [
"Label the left shoes be $L_1,\\dots, L_{10}$ and the right shoes $R_1,\\dots, R_{10}$. Notice that there are $10!$ possible pairings. Let a pairing be \"bad\" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad. Note that, in order to have a bad pairing, the... |
2014-II-14 | 2,014 | 14 | In $\triangle ABC$ , $AB=10$ , $\measuredangle A=30^{\circ}$ , and $\measuredangle C=45^{\circ}$ . Let $H$ , $D$ , and $M$ be points on line $\overline{BC}$ such that $AH\perp BC$ , $\measuredangle BAD=\measuredangle CAD$ , and $BM=CM$ . Point $N$ is the midpoint of segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp BC$ . Then $AP^2=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 77 | II | [
"Let us just drop the perpendicular from $B$ to $AC$ and label the point of intersection $O$. We will use this point later in the problem. As we can see, $M$ is the midpoint of $BC$ and $N$ is the midpoint of $HM$ $AHC$ is a $45-45-90$ triangle, so $\\angle{HAB}=15^\\circ$. $AHD$ is $30-60-90$ triangle. $AH$ and $P... |
2014-II-15 | 2,014 | 15 | For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k$ . Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2$ . Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0$ . Find the smallest positive integer $t$ such that $x_t=2090$ . | 149 | II | [
"Note that for any $x_i$, for any prime $p$, $p^2 \\nmid x_i$. This provides motivation to translate $x_i$ into a binary sequence $y_i$. Let the prime factorization of $x_i$ be written as $p_{a_1} \\cdot p_{a_2} \\cdot p_{a_3} \\cdots$, where $p_i$ is the $i$th prime number. Then, for every $p_{a_k}$ in the prime f... |
2015-I-1 | 2,015 | 1 | The expressions $A$ = $1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39$ and $B$ = $1 + 2 \times 3 + 4 \times 5 + \cdots + 36 \times 37 + 38 \times 39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers $A$ and $B$ . | 722 | I | [
"We have \\[|A-B|=|1+3(4-2)+5(6-4)+ \\cdots + 37(38-36)-39(1-38)|\\]\\[\\implies |2(1+3+5+7+ \\cdots +37)-1-39(37)|\\]\\[\\implies |361(2)-1-39(37)|=|722-1-1443|=|-722|\\implies 722\\]",
"We see that $A=(1\\times 2)+(3\\times 4)+(5\\times 6)+\\cdots +(35\\times 36)+(37\\times 38)+39$ and $B=1+(2\\times 3)+(4\\tim... |
2015-I-2 | 2,015 | 2 | The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 139 | I | [
"One of the best ways to solve this problem is to use PIE, or the Principle of Inclusion and Exclusion. To start off, we know that the denominator, or the total ways to pick $3$ officials, is $\\binom{9}{3} = 84$. Now, we find the number of ways that at least $2$ officials are from the same country and then subtrac... |
2015-I-3 | 2,015 | 3 | There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$ . | 307 | I | [
"Let the positive integer mentioned be $a$, so that $a^3 = 16p+1$. Note that $a$ must be odd, because $16p+1$ is odd. Rearrange this expression and factor the left side (this factoring can be done using $(a^3-b^3) = (a-b)(a^2+a b+b^2)$ or synthetic divison once it is realized that $a = 1$ is a root): \\begin{align*... |
2015-I-4 | 2,015 | 4 | Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$ . | 507 | I | [
"Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\\sqrt3)$ and $E=(18,2\\sqrt3)$ as well. Therefore, $M=(9,\\sqrt3)$ and $N=(14,4\\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives \\[x=\\dfrac 1 2 |16\\sqrt3+36\\sqrt3+0-(0+14\\sqrt3+64\\s... |
2015-I-5 | 2,015 | 5 | In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday, Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random, and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 341 | I | [
"The order of the days doesn't matter, therefore we can solve backwards. Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is $\\dfrac{1}{9}$. Assuming this, then let the first sock be arbitrary; the probability that the second sock does not match is $\\dfrac{6}{7}.$ The only \"h... |
2015-I-6 | 2,015 | 6 | Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$ . Find the degree measure of $\angle BAG$ . [asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("$A$",A,dir(0)); label("$B$",B,dir(75)); label("$C$",C,dir(90)); label("$D$",D,dir(105)); label("$E$",E,dir(180)); label("$F$",F,dir(225)); label("$G$",G,dir(260)); label("$H$",H,dir(280)); label("$I$",I,dir(315));[/asy] | 58 | I | [
"Let $O$ be the center of the circle with $ABCDE$ on it. Let $x$ be the degree measurement of $\\overarc{ED}=\\overarc{DC}=\\overarc{CB}=\\overarc{BA}$ in circle $O$ and $y$ be the degree measurement of $\\overarc{EF}=\\overarc{FG}=\\overarc{GH}=\\overarc{HI}=\\overarc{IA}$ in circle $C$. $\\angle ECA$ is, therefor... |
2015-I-7 | 2,015 | 7 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$ . [asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy] | 539 | I | [
"Let us find the proportion of the side length of $KLMN$ and $FJGH$. Let the side length of $KLMN=y$ and the side length of $FJGH=x$. [asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F... |
2015-I-8 | 2,015 | 8 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ . | 695 | I | [
"You know whatever $n$ is, it has to have 3 digits, because if it had only two, the maximum of $s(n)$ is 18 and because in AIME all answers are up to three digits. Now let $n=100a_2+10a_1+a_0$ So first we know, $a_2+a_1+a_0=20$. Okay now we have to split into cases based on which digit gets carried. This meaning, w... |
2015-I-9 | 2,015 | 9 | Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$ . Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$ . Find the number of such sequences for which $a_n=0$ for some $n$ . | 494 | I | [
"Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the cri... |
2015-I-10 | 2,015 | 10 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ . | 72 | I | [
"Let $f(x)$ = $ax^3+bx^2+cx+d$. Since $f(x)$ is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing $12$ and $-12$, it is easy to see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$; otherwise more ben... |
2015-I-11 | 2,015 | 11 | Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$ . | 108 | I | [
"Let $D$ be the midpoint of $\\overline{BC}$. Then by SAS Congruence, $\\triangle ABD \\cong \\triangle ACD$, so $\\angle ADB = \\angle ADC = 90^o$. Now let $BD=y$, $AB=x$, and $\\angle IBD = \\dfrac{\\angle ABD}{2} = \\theta$. Then $\\mathrm{cos}{(\\theta)} = \\dfrac{y}{8}$ and $\\mathrm{cos}{(2\\theta)} = \\dfrac... |
2015-I-12 | 2,015 | 12 | Consider all 1000-element subsets of the set $\{ 1, 2, 3, ... , 2015 \}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ . | 431 | I | [
"Let $M$ be the desired mean. Then because $\\dbinom{2015}{1000}$ subsets have 1000 elements and $\\dbinom{2015 - i}{999}$ have $i$ as their least element, \\begin{align*} \\binom{2015}{1000} M &= 1 \\cdot \\binom{2014}{999} + 2 \\cdot \\binom{2013}{999} + \\dots + 1016 \\cdot \\binom{999}{999} \\\\ &= \\binom{2014... |
2015-I-13 | 2,015 | 13 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ . | 91 | I | [
"Let $x = \\cos 1^\\circ + i \\sin 1^\\circ$. Then from the identity \\[\\sin 1 = \\frac{x - \\frac{1}{x}}{2i} = \\frac{x^2 - 1}{2 i x},\\] we deduce that (taking absolute values and noticing $|x| = 1$) \\[|2\\sin 1| = |x^2 - 1|.\\] But because $\\csc$ is the reciprocal of $\\sin$ and because $\\sin z = \\sin (180^... |
2015-I-14 | 2,015 | 14 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer. | 483 | I | [
"Let $n\\ge 2$ and define $a(n) = \\left\\lfloor \\sqrt n \\right\\rfloor$. For $2\\le n \\le 1000$, we have $1\\le a(n)\\le 31$. For $a^2 \\le x < (a+1)^2$ we have $y=ax$. Thus $A(n+1)-A(n)=a(n+\\tfrac 12) = \\Delta_n$ (say), and $\\Delta_n$ is an integer if $a$ is even; otherwise $\\Delta_n$ is an integer plus $\... |
2015-I-15 | 2,015 | 15 | A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$ , and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\text{o}$ . The block is then sliced in half along the plane that passes through point $A$ , point $B$ , and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$ , where $a$ , $b$ , and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$ . [asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy] | 53 | I | [
"Label the points where the plane intersects the top face of the cylinder as $C$ and $D$, and the center of the cylinder as $O$, such that $C,O,$ and $A$ are collinear. Let $N$ be the center of the bottom face, and $M$ the midpoint of $\\overline{AB}$. Then $ON=4$, $MN=3$ (because of the 120 degree angle), and so $... |
2015-II-1 | 2,015 | 1 | Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$ . | 131 | II | [
"If $N$ is $22$ percent less than one integer $k$, then $N=\\frac{78}{100}k=\\frac{39}{50}k$. In addition, $N$ is $16$ percent greater than another integer $m$, so $N=\\frac{116}{100}m=\\frac{29}{25}m$. Therefore, $k$ is divisible by $50$ and $m$ is divisible by $25$. Setting these two equal, we have $\\frac{39}{50... |
2015-II-2 | 2,015 | 2 | In a new school, $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 25 | II | [
"We see that $40\\% \\cdot 100\\% + 30\\% \\cdot 80\\% + 20\\% \\cdot 50\\% + 10\\% \\cdot 20\\% = 76\\%$ of students are learning Latin. In addition, $30\\% \\cdot 80\\% = 24\\%$ of students are sophomores learning Latin. Thus, our desired probability is $\\dfrac{24}{76}=\\dfrac{6}{19}$ and our answer is $6+19=025... |
2015-II-3 | 2,015 | 3 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ . | 476 | II | [
"The three-digit integers divisible by $17$, and their digit sum: \\[\\begin{array}{c|c} m & s(m)\\\\ \\hline 102 & 3 \\\\ 119 & 11\\\\ 136 & 10\\\\ 153 & 9\\\\ 170 & 8\\\\ 187 & 16\\\\ 204 & 6\\\\ 221 & 5\\\\ 238 & 13\\\\ 255 & 12\\\\ 272 & 11\\\\ 289 & 19\\\\ 306 & 9\\\\ 323 & 8\\\\ 340 & 7\\\\ 357 & 15\\\\ 374 &... |
2015-II-4 | 2,015 | 4 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ . | 18 | II | [
"Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. Let the point where an altitude intersects the larger base be $E$, where $E$ is closer to $D$. Subtract the two bases and divide to find that $ED$ is $\\log 8$. The altitude can be expressed as $\\frac{4}{3} \\log 8$. Therefore, the tw... |
2015-II-5 | 2,015 | 5 | Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$ . | 90 | II | [
"Call the given grid \"Grid A\". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions $(n-1) \\times (n-1)$. There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of e... |
2015-II-6 | 2,015 | 6 | Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?" After some calculations, Jon says, "There is more than one such polynomial." Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?" Jon says, "There are still two possible values of $c$ ." Find the sum of the two possible values of $c$ . | 440 | II | [
"We call the three roots (some may be equal to one another) $x_1$, $x_2$, and $x_3$. Using Vieta's formulas, we get $x_1+x_2+x_3 = a$, $x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}$, and $x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2}$. Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \\cdot x_... |
2015-II-7 | 2,015 | 7 | Triangle $ABC$ has side lengths $AB = 12$ , $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = w$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[\text{Area}(PQRS) = \alpha w - \beta \cdot w^2.\] Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 161 | II | [
"If $\\omega = 25$, the area of rectangle $PQRS$ is $0$, so \\[\\alpha\\omega - \\beta\\omega^2 = 25\\alpha - 625\\beta = 0\\] and $\\alpha = 25\\beta$. If $\\omega = \\frac{25}{2}$, we can reflect $APQ$ over $PQ$, $PBS$ over $PS$, and $QCR$ over $QR$ to completely cover rectangle $PQRS$, so the area of $PQRS$ is h... |
2015-II-8 | 2,015 | 8 | Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$ . The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 36 | II | [
"Let us call the quantity $\\frac{a^3b^3+1}{a^3+b^3}$ as $N$ for convenience. Knowing that $a$ and $b$ are positive integers, we can legitimately rearrange the given inequality so that $a$ is by itself, which makes it easier to determine the pairs of $(a, b)$ that work. Doing so, we have \\[\\frac{ab+1}{a+b} < \\fr... |
2015-II-9 | 2,015 | 9 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$ . [asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy] | 384 | II | [
"Our aim is to find the volume of the part of the cube submerged in the cylinder. In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equil... |
2015-II-10 | 2,015 | 10 | Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, $53421$ and $14253$ are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but $45123$ is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$ . | 486 | II | [
"The simple recurrence can be found. When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$, before $n - 2$, and at the very end. Ex. Inserting 4 into the string 123: 4 can go before the 2 (1423), before the 3 (1243), and at the ... |
2015-II-11 | 2,015 | 11 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ , $BC=4$ , $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 23 | II | [
"Solution 1 Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ . Notice that $\\triangle{OMB} \\sim \\triangle{QOB}$ because both are right triangles, and $\\angle{OBQ} \\cong \\angle{OBM}$. By $\\frac{MB}{BO}=\\frac{BO}{BQ}$, $MB = r\\left(\\frac{r}{4.5}\\right) = \\fr... |
2015-II-12 | 2,015 | 12 | There are $2^{10} = 1024$ possible $10$ -letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than $3$ adjacent letters that are identical. | 548 | II | [
"Let $a_{n}$ be the number of ways to form $n$-letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical. Note that, at the end of each $n$-letter string, there are $3$ possibilities for the last letter chain: it must be either $1$, $2$, or $3$ letters long. Removing this last ch... |
2015-II-13 | 2,015 | 13 | Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ . | 628 | II | [
"If $n = 1$, $a_n = \\sin(1) > 0$. Then if $n$ satisfies $a_n < 0$, $n \\ge 2$, and \\[a_n = \\sum_{k=1}^n \\sin(k) = \\cfrac{1}{\\sin{1}} \\sum_{k=1}^n\\sin(1)\\sin(k) = \\cfrac{1}{2\\sin{1}} \\sum_{k=1}^n\\cos(k - 1) - \\cos(k + 1) = \\cfrac{1}{2\\sin(1)} [\\cos(0) + \\cos(1) - \\cos(n) - \\cos(n + 1)].\\] Since ... |
2015-II-14 | 2,015 | 14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ . | 89 | II | [
"The expression we want to find is $2(x^3+y^3) + x^3y^3$. Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x^3+y^3)=945$, respectively. Dividing the latter by the former equation yields $\\frac{x^2-xy+y^2}{xy} = \\frac{945}{810}$. Adding 3 to both sides and simplifying yields $\\frac{(x+y)^2}{xy} = \\f... |
2015-II-15 | 2,015 | 15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ such that $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . [asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | 129 | II | [
"Let $M$ be the intersection of $\\overline{BC}$ and the common internal tangent of $\\mathcal P$ and $\\mathcal Q.$ We claim that $M$ is the circumcenter of right $\\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\\overline{BC},$ imp... |
2016-I-1 | 2,016 | 1 | For $-1<r<1$ , let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\] Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$ . Find $S(a)+S(-a)$ . | 336 | I | [
"The sum of an infinite geometric series is $\\frac{a}{1-r}\\rightarrow \\frac{12}{1\\mp a}$. The product $S(a)S(-a)=\\frac{144}{1-a^2}=2016$. $\\frac{12}{1-a}+\\frac{12}{1+a}=\\frac{24}{1-a^2}$, so the answer is $\\frac{2016}{6}=336."
] |
2016-I-2 | 2,016 | 2 | Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 71 | I | [
"It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc. Then there are $21^2=441$ possible rolls. There are $2\\cdot(1\\cdot 6+2\\cdot 5+3\\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $\\frac{56}{441}=\\frac{8}{63}$. The answer is $8+63=071 See also 2006 AMC 1... |
2016-I-3 | 2,016 | 3 | A regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated. [asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy] | 810 | I | [
"Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to go to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place. Then, there are 2 paths... |
2016-I-4 | 2,016 | 4 | A right prism with height $h$ has bases that are regular hexagons with sides of length $12$ . A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60^\circ$ . Find $h^2$ . | 108 | I | [
"Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$, and let $D$ be the last vertex of the triangular pyramid. Then $\\angle CAB = 120^\\circ$. Let $X$ be the foot of the altitude from $A$ to $\\overline{BC}$. Then since $\\triangle ABX$ is a $30-60-90$ triangle, $AX = 6$. Since the dihedral an... |
2016-I-5 | 2,016 | 5 | Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$ . | 53 | I | [
"Let $d$ be the number of days Anh reads for. Because the difference between the number of pages and minutes Anh reads for each day stays constant and is an integer, $d$ must be a factor of the total difference, which is $374-319=55$. Also note that the number of pages Anh reads is $dn+\\frac{d(d-1)}{2}$. Similarly... |
2016-I-6 | 2,016 | 6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $\overline{AB}$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC= \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 13 | I | [
"Suppose we label the angles as shown below. [asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(... |
2016-I-7 | 2,016 | 7 | For integers $a$ and $b$ consider the complex number \[\frac{\sqrt{ab+2016}}{ab+100}-\left(\frac{\sqrt{|a+b|}}{ab+100}\right)i.\] Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number. | 103 | I | [
"We consider two cases: Case 1: $ab \\ge -2016$. In this case, if \\[0 = \\text{Im}\\left({\\frac{\\sqrt{ab+2016}}{ab+100}-\\left({\\frac{\\sqrt{|a+b|}}{ab+100}}\\right)i}\\right) = -\\frac{\\sqrt{|a+b|}}{ab+100}\\] then $ab \\ne -100$ and $|a + b| = 0 = a + b$. Thus $ab = -a^2$ so $a^2 < 2016$. Thus $a = -44,-43, ... |
2016-I-8 | 2,016 | 8 | For a permutation $p = (a_1,a_2,\ldots,a_9)$ of the digits $1,2,\ldots,9$ , let $s(p)$ denote the sum of the three $3$ -digit numbers $a_1a_2a_3$ , $a_4a_5a_6$ , and $a_7a_8a_9$ . Let $m$ be the minimum value of $s(p)$ subject to the condition that the units digit of $s(p)$ is $0$ . Let $n$ denote the number of permutations $p$ with $s(p) = m$ . Find $|m - n|$ . | 162 | I | [
"To minimize $s(p)$, the numbers $1$, $2$, and $3$ (which sum to $6$) must be in the hundreds places. For the units digit of $s(p)$ to be $0$, the numbers in the ones places must have a sum of either $10$ or $20$. However, since the tens digit contributes more to the final sum $s(p)$ than the ones digit, and we are... |
2016-I-9 | 2,016 | 9 | Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$ . This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$ . Find the maximum possible area of $AQRS$ . | 744 | I | [
"Note that if angle $BAC$ is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where $A$ is obtuse. Therefore, angle A is acute. Let angle $CAS=n$ and angle $BAQ=m$. Then, $\\overline{AS}=31\\cos(n)$ and $\\overline{AQ}=40\\cos(m)$. Then t... |
2016-I-10 | 2,016 | 10 | A strictly increasing sequence of positive integers $a_1$ , $a_2$ , $a_3$ , $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ , $a_{2k}$ , $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ , $a_{2k+1}$ , $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$ . | 504 | I | [
"We first create a similar sequence where $a_1=1$ and $a_2=2$. Continuing the sequence, \\[1, 2,4,6,9,12,16,20,25,30,36,42,49,\\cdots\\] Here we can see a pattern; every second term (starting from the first) is a square, and every second term (starting from the third) is the end of a geometric sequence. This can be... |
2016-I-11 | 2,016 | 11 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 109 | I | [
"Plug in $x=1$ to get $(1-1)P(1+1) = 0 = (1+2)P(1) \\Rightarrow P(1) = 0$. Plug in $x=0$ to get $(0-1)P(0+1) = (0+2)P(0)\\Rightarrow P(0) = -\\frac{1}{2}P(1) = 0$. Plug in $x=-1$ to get $(-1-1)P(-1+1) = (-1+2)P(-1)\\Rightarrow (-2)P(0)=P(-1)\\Rightarrow P(-1) = 0$. So $P(x) = x(x-1)(x+1)Q(x)$ for some polynomial $Q... |
2016-I-12 | 2,016 | 12 | Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes. | 132 | I | [
"$m(m-1)$ is the product of two consecutive integers, so it is always even. Thus $m(m-1)+11$ is odd and never divisible by $2$. Thus any prime $p$ that divides $m^2-m+11$ must divide $4m^2-4m+44=(2m-1)^2+43$. We see that $(2m-1)^2\\equiv -43\\pmod{p}$. We can verify that $-43$ is not a perfect square mod $p$ for ea... |
2016-I-13 | 2,016 | 13 | Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$ . A fence is located at the horizontal line $y = 0$ . On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where $y=0$ , with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where $y < 0$ . Freddy starts his search at the point $(0, 21)$ and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river. | 273 | I | [
"Clearly Freddy's $x$-coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$-coordinate. Observe that $E(24)=0$, and \\[E(y)=1+\\frac{E(y+1)+E(y-1)+2E(y)}{4}\\] for all $y$ such that $1\\le y\\le 23$. Also note that $E(0)=1+\\frac... |
2016-I-14 | 2,016 | 14 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ . | 574 | I | [
"First note that $1001 = 143 \\cdot 7$ and $429 = 143 \\cdot 3$ so every point of the form $(7k, 3k)$ is on the line. Then consider the line $l$ from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$. Translate the line $l$ so that $(7k, 3k)$ is now the origin. There is one square and one circle that intersect the line around $... |
2016-I-15 | 2,016 | 15 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ , $Y$ , $D$ are collinear, $XC = 67$ , $XY = 47$ , and $XD = 37$ . Find $AB^2$ . | 270 | I | [
"Using the radical axis theorem, the lines $\\overline{AD}, \\overline{BC}, \\overline{XY}$ are all concurrent at one point, call it $F$. Now recall by Miquel's theorem in $\\triangle FDC$ the fact that quadrilaterals $DAXY$ and $CBXY$ are cyclic implies $FAXB$ is cyclic as well. Denote $\\omega_{3}\\equiv(FAXB)$ a... |
2016-II-1 | 2,016 | 1 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially. | 108 | II | [
"Let $r$ be the common ratio, where $r>1$. We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$. We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$, so we have $3ar=432,$ or $ar=144$, which is how much Betty had. Now we have $144+\\dfrac{144}{r}+144r=444$, or $144(r+\\dfrac{1}{r})=300$,... |
2016-II-2 | 2,016 | 2 | There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a+b$ . | 107 | II | [
"Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\\dfrac{3}{5}x+\\dfrac{2}{5}2x = \\dfrac{3}{10} \\implies \\dfrac{7}{5}x=\\dfrac{3}{10}$ $\\implies x=\\dfrac{3}{14}$. Therefore, the probability that it doesn't rain on either day is $\\left(1-\\dfrac{3}{14}\\... |
2016-II-3 | 2,016 | 3 | Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4. \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$ . | 265 | II | [
"First, we get rid of logs by taking powers: $xyz-3+\\log_5 x=2^{5}=32$, $xyz-3+\\log_5 y=3^{4}=81$, and $(xyz-3+\\log_5 z)=4^{4}=256$. Adding all the equations up and using the $\\log {xy}=\\log {x}+\\log{y}$ property, we have $3xyz+\\log_5{xyz} = 378$, so we have $xyz=125$. Solving for $x,y,z$ by substituting $12... |
2016-II-4 | 2,016 | 4 | An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box. | 180 | II | [
"By counting the number of green cubes $2$ different ways, we have $12a=20b$, or $a=\\dfrac{5}{3} b$. Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \\times b \\times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \\times ... |
2016-II-5 | 2,016 | 5 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$ . | 182 | II | [
"Do note that by counting the area in 2 ways, the first altitude is $x = \\frac{ab}{c}$. By similar triangles, the common ratio is $\\rho = \\frac{a}{c}$ for each height, so by the geometric series formula, we have \\begin{align} 6p=\\frac{x}{1-\\rho} = \\frac{ab}{c-a}. \\end{align} Writing $p=a+b+c$ and clearing d... |
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