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 |
|---|---|---|---|---|---|---|
2016-II-6 | 2,016 | 6 | For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 275 | II | [
"Note that all the coefficients of odd-powered terms is an odd number of odd degree terms multiplied together, and all coefficients of even-powered terms have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equa... |
2016-II-7 | 2,016 | 7 | Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$ . The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$ . Find the differe... | 840 | II | [
"Letting $AI=a$ and $IB=b$, we have \\[IJ^{2}=a^{2}+b^{2} \\geq 1008\\] by AM-GM inequality. Also, since $EFGH||ABCD$, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas fo... |
2016-II-8 | 2,016 | 8 | Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,$ and $61$ . | 728 | II | [
"Note that the prime factorization of the product is $3^{2}\\cdot 7 \\cdot 11 \\cdot 17 \\cdot 31 \\cdot 41 \\cdot 61$. Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$, and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \\cdot 6$ ... |
2016-II-9 | 2,016 | 9 | The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$ . | 262 | II | [
"Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$. When we get to $b_2=9$ and $a_2=91$, we have $a_4=271$ and $b_4=729$, which works, therefore, the answer is $b_3+a_3=81+181=262.",
"We have $a_k=r^{k-1}$ and $b_k=(k-1)d$. First, $b_{k-1... |
2016-II-10 | 2,016 | 10 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. F... | 43 | II | [
"[asy] import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(8cm); pair A = origin, B = (13,0), P = (4,0), Q = (7,0), T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)), S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10)); Drawing(A--B--C--cycle); D(circumcircle(A,B,C),rgb(0,0.6,1)); Draw... |
2016-II-11 | 2,016 | 11 | For positive integers $N$ and $k$ , define $N$ to be $k$ -nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $1000$ that are neither $7$ -nice nor $8$ -nice. | 749 | II | [
"We claim that an integer $N$ is only $k$-nice if and only if $N \\equiv 1 \\pmod k$. By the number of divisors formula, the number of divisors of $\\prod_{i=1}^n p_i^{a_i}$ is $\\prod_{i=1}^n (a_i+1)$. Since all the $a_i$'s are divisible by $k$ in a perfect $k$ power, the only if part of the claim follows. To show... |
2016-II-12 | 2,016 | 12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color. [asy] dr... | 732 | II | [
"Choose a section to start coloring. Assume, WLOG, that this section is color $1$. We proceed coloring clockwise around the ring. Let $f(n,C)$ be the number of ways to color the first $n$ sections (proceeding clockwise) such that the last section has color $C$. In general (except for when we complete the coloring),... |
2016-II-13 | 2,016 | 13 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The value of a square is the sum of its row number and column number. The score of an arrangement of rooks is the le... | 371 | II | [
"We casework to find the number of ways to get each possible score. Note that the lowest possible score is $2$ and the highest possible score is $7$. Let the bijective function $f(x)=\\{1,2,3,4,5,6\\} \\to \\{1,2,3,4,5,6\\}$ denote the row number of the rook for the corresponding column number. For a score of $2$, ... |
2016-II-14 | 2,016 | 14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | 450 | II | [
"The inradius of $\\triangle ABC$ is $100\\sqrt 3$ and the circumradius is $200 \\sqrt 3$. Now, consider the line perpendicular to plane $ABC$ through the circumcenter of $\\triangle ABC$. Note that $P,Q,O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, O$... |
2016-II-15 | 2,016 | 15 | For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$ . Let $x_1, x_2, ..., x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i=1$ and $\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}$ . The maximum possible value of... | 863 | II | [
"Note that \\begin{align*}\\sum_{1 \\leq i < j \\leq 216} x_ix_j &= \\frac12\\left(\\left(\\sum_{i=1}^{216} x_i\\right)^2-\\sum_{i=1}^{216} x_i^2\\right)\\\\&=\\frac12\\left(1-\\sum x_i^2\\right).\\end{align*} Substituting this into the second equation and collecting $x_i^2$ terms, we find \\[\\sum_{i=1}^{216}\\fra... |
2017-I-1 | 2,017 | 1 | Fifteen distinct points are designated on $\triangle ABC$ : the 3 vertices $A$ , $B$ , and $C$ ; $3$ other points on side $\overline{AB}$ ; $4$ other points on side $\overline{BC}$ ; and $5$ other points on side $\overline{CA}$ . Find the number of triangles with positive area whose vertices are among these $15$ points... | 390 | I | [
"Every triangle is uniquely determined by 3 points. There are $\\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\\binom{5}{3}$ invalid cases on segment $AB$, $\\binom{6}{3}$ invalid cases on segment $BC$, and $\\binom{7}{3}$ ... |
2017-I-2 | 2,017 | 2 | When each of $702$ , $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ , $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$ . | 62 | I | [
"Let's work on both parts of the problem separately. First, \\[855 \\equiv 787 \\equiv 702 \\equiv r \\pmod{m}.\\] We take the difference of $855$ and $787$, and also of $787$ and $702$. We find that they are $85$ and $68$, respectively. Since the greatest common divisor of the two differences is $17$ (and the only... |
2017-I-3 | 2,017 | 3 | For a positive integer $n$ , let $d_n$ be the units digit of $1 + 2 + \dots + n$ . Find the remainder when \[\sum_{n=1}^{2017} d_n\] is divided by $1000$ . | 69 | I | [
"We see that $d_n$ appears in cycles of $20$ and the cycles are \\[1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\\] adding a total of $70$ each cycle. Since $\\left\\lfloor\\frac{2017}{20}\\right\\rfloor=100$, we know that by $2017$, there have been $100$ cycles and $7000$ has been added. This can be discarded as we're ... |
2017-I-4 | 2,017 | 4 | A pyramid has a triangular base with side lengths $20$ , $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the squa... | 803 | I | [
"Let the triangular base be $\\triangle ABC$, with $\\overline {AB} = 24$. We find that the altitude to side $\\overline {AB}$ is $16$, so the area of $\\triangle ABC$ is $(24*16)/2 = 192$. Let the fourth vertex of the tetrahedron be $P$, and let the midpoint of $\\overline {AB}$ be $M$. Since $P$ is equidistant fr... |
2017-I-5 | 2,017 | 5 | A rational number written in base eight is $\underline{a} \underline{b} . \underline{c} \underline{d}$ , where all digits are nonzero. The same number in base twelve is $\underline{b} \underline{b} . \underline{b} \underline{a}$ . Find the base-ten number $\underline{a} \underline{b} \underline{c}$ . | 321 | I | [
"First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12. $11_{12}=15_8$ $22_{12}=32_8$ $33_{12}=47_8$ $44_{12}=64_8$ $55_{12}=101_8$ We stop because ... |
2017-I-6 | 2,017 | 6 | A circle circumscribes an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the larg... | 48 | I | [
"The probability that the chord doesn't intersect the triangle is $\\frac{11}{25}$. The only way this can happen is if the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite one of the base angles is $\\frac{2x}{360}=\\frac{... |
2017-I-7 | 2,017 | 7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ . | 564 | I | [
"Let $c=6-(a+b)$, and note that $\\binom{6}{a + b}=\\binom{6}{c}$. The problem thus asks for the sum $\\binom{6}{a} \\binom{6}{b} \\binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$. Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots ... |
2017-I-8 | 2,017 | 8 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\ang... | 41 | I | [
"Noting that $\\angle OQP$ and $\\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$, if $\\overline{QR} \\leq 100$, then $\\overarc{QR}$ must be less than or equal to $60^{\\circ}... |
2017-I-9 | 2,017 | 9 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ . | 45 | I | [
"Writing out the recursive statement for $a_n, a_{n-1}, \\dots, a_{10}$ and summing them gives \\[a_n+\\dots+a_{10}=(a_n+\\cdots+a_{11})+a_{10}=100(a_{n-1}+\\dots+a_{10})+n+\\dots+10\\] Which simplifies to \\[a_n=99(a_{n-1}+\\dots+a_{10})+\\frac{1}{2}(n+10)(n-9)\\] Therefore, $a_n$ is divisible by 99 if and only if... |
2017-I-10 | 2,017 | 10 | Let $z_1 = 18 + 83i$ , $z_2 = 18 + 39i,$ and $z_3 = 78 + 99i,$ where $i = \sqrt{-1}$ . Let $z$ be the unique complex number with the properties that $\frac{z_3 - z_1}{z_2 - z_1} \cdot \frac{z - z_2}{z - z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$ . | 56 | I | [
"This problem is pretty obvious how to bash, and indeed many of the solutions below explain how to do that. But there’s no fun in that, and let’s see if we can come up with a slicker solution that will be more enjoyable. Instead of thinking of complex numbers as purely a real plus a constant times $i$, let’s graph ... |
2017-I-11 | 2,017 | 11 | Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ , $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ , $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5... | 360 | I | [
"Assume that $5 \\in \\{a_1, a_2, a_3\\}$, $m \\neq 5$, and WLOG, $\\max{(a_1, a_2, a_3)} = 5$. Then we know that the other two medians in $\\{a_1, a_2, a_3\\}$ and the smallest number of rows 1, 2, and 3 are all less than 5. But there are only 4 numbers less than 5 in $1, 2, 3, \\dots, 9$, a Contradiction. Thus, i... |
2017-I-12 | 2,017 | 12 | Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,... | 252 | I | [
"We shall solve this problem by doing casework on the lowest element of the subset. Note that the number $1$ cannot be in the subset because $1*1=1$. Let $S$ be a product-free set. If the lowest element of $S$ is $2$, we consider the set $\\{3, 6, 9\\}$. We see that 5 of these subsets can be a subset of $S$ ($\\{3\... |
2017-I-13 | 2,017 | 13 | For every $m \geq 2$ , let $Q(m)$ be the least positive integer with the following property: For every $n \geq Q(m)$ , there is always a perfect cube $k^3$ in the range $n < k^3 \leq mn$ . Find the remainder when \[\sum_{m = 2}^{2017} Q(m)\] is divided by $1000$ . | 59 | I | [
"Lemma 1: The ratio between $k^3$ and $(k+1)^3$ decreases as $k$ increases. Lemma 2: If the range $(n,mn]$ includes $y$ cubes, $(p,mp]$ will always contain at least $y-1$ cubes for all $p$ in $[n,+\\infty)$. If $m=14$, the range $(1,14]$ includes one cube. The range $(2,28]$ includes 2 cubes, which fulfills the Lem... |
2017-I-14 | 2,017 | 14 | Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$ . Find the remainder when $x$ is divided by $1000$ . | 896 | I | [
"The first condition implies \\[a^{128} = \\log_a\\log_a 2 + \\log_a 24 - 128\\] \\[128+a^{128} = \\log_a\\log_a 2^{24}\\] \\[a^{a^{128}a^{a^{128}}} = 2^{24}\\] \\[\\left(a^{a^{128}}\\right)^{\\left(a^{a^{128}}\\right)} = 2^{24} = 8^8\\] So $a^{a^{128}} = 8$. Putting each side to the power of $128$: \\[\\left(a^{12... |
2017-I-15 | 2,017 | 15 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt3$ , $5$ , and $\sqrt{37}$ , as shown, is $\tfrac{m\sqrt{p}}{n}$ , where $m$ , $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square o... | 145 | I | [
"Lemma: If $x,y$ satisfy $px+qy=1$, then the minimal value of $\\sqrt{x^2+y^2}$ is $\\frac{1}{\\sqrt{p^2+q^2}}$. Proof: Recall that the distance between the point $(x_0,y_0)$ and the line $px+qy+r = 0$ is given by $\\frac{|px_0+qy_0+r|}{\\sqrt{p^2+q^2}}$. In particular, the distance between the origin and any point... |
2017-II-1 | 2,017 | 1 | Find the number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ . | 196 | II | [
"The number of subsets of a set with $n$ elements is $2^n$. The total number of subsets of $\\{1, 2, 3, 4, 5, 6, 7, 8\\}$ is equal to $2^8$. The number of sets that are subsets of at least one of $\\{1, 2, 3, 4, 5\\}$ or $\\{4, 5, 6, 7, 8\\}$ can be found using complementary counting. There are $2^5$ subsets of $\\... |
2017-II-2 | 2,017 | 2 | Teams $T_1$ , $T_2$ , $T_3$ , and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$ , and $T_2$ plays $T_3$ . The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$ , the probability that $T_i$ wins is $\frac{i}{i+j}$ , and the out... | 781 | II | [
"There are two scenarios in which $T_4$ wins. The first scenario is where $T_4$ beats $T_1$, $T_3$ beats $T_2$, and $T_4$ beats $T_3$, and the second scenario is where $T_4$ beats $T_1$, $T_2$ beats $T_3$, and $T_4$ beats $T_2$. Consider the first scenario. The probability $T_4$ beats $T_1$ is $\\frac{4}{4+1}$, the... |
2017-II-3 | 2,017 | 3 | A triangle has vertices $A(0,0)$ , $B(12,0)$ , and $C(8,10)$ . The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 409 | II | [
"[asy] pair A,B,C,D,X,Z,P; A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4); fill(B--X--P--Z--cycle,lightgray); draw(A--B--C--cycle); dot(A); label(\"$A$\",A,SW); dot(B); label(\"$B$\",B,SE); dot(C); label(\"$C$\",C,N); draw(X--P,dashed); draw(Z--P,dashed); dot(X); label(\"$X$\",X,NE); dot(Z); label(\"$Z$\... |
2017-II-4 | 2,017 | 4 | Find the number of positive integers less than or equal to $2017$ whose base-three representation contains no digit equal to $0$ . | 222 | II | [
"Solution 1 The base-$3$ representation of $2017_{10}$ is $2202201_3$. Because any $7$-digit base-$3$ number that starts with $22$ and has no digit equal to $0$ must be greater than $2017_{10}$, all $7$-digit numbers that have no digit equal to $0$ must start with $21$ or $1$ in base $3$. Of the base-$3$ numbers th... |
2017-II-5 | 2,017 | 5 | A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189$ , $320$ , $287$ , $234$ , $x$ , and $y$ . Find the greatest possible value of $x+y$ . | 791 | II | [
"Let these four numbers be $a$, $b$, $c$, and $d$, where $a>b>c>d$. $x+y$ needs to be maximized, so let $x=a+b$ and $y=a+c$ because these are the two largest pairwise sums. Now $x+y=2a+b+c$ needs to be maximized. Notice that $2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))$. No matter how th... |
2017-II-6 | 2,017 | 6 | Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer. | 195 | II | [
"Manipulating the given expression, $\\sqrt{n^2+85n+2017}=\\frac{1}{2}\\sqrt{4n^2+340n+8068}=\\frac{1}{2}\\sqrt{(2n+85)^2+843}$. The expression under the radical must be an square number for the entire expression to be an integer, so $(2n+85)^2+843=s^2$. Rearranging, $s^2-(2n+85)^2=843$. By difference of squares, $... |
2017-II-7 | 2,017 | 7 | Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution. | 501 | II | [
"[asy] Label f; f.p=fontsize(5); xaxis(-3,3,Ticks(f,1.0)); yaxis(-3,26,Ticks(f,1.0)); real f(real x){return (x+2)^2;} real g(real x){return x*-1;} real h(real x){return x*-2;} real i(real x){return x*-3;} real j(real x){return x*8;} draw(graph(f,-2,3),green); draw(graph(g,-2,2),red); draw(graph(h,-2,1),red); draw(g... |
2017-II-8 | 2,017 | 8 | Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer. | 134 | II | [
"We start with the last two terms of the polynomial $1+n+\\frac{n^2}{2!}+\\frac{n^3}{3!}+\\frac{n^4}{4!}+\\frac{n^5}{5!}+\\frac{n^6}{6!}$, which are $\\frac{n^5}{5!}+\\frac{n^6}{6!}$. This can simplify to $\\frac{6n^5+n^6}{720}$, which can further simplify to $\\frac{n^5(6+n)}{720}$. Notice that the prime factoriza... |
2017-II-9 | 2,017 | 9 | A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one c... | 13 | II | [
"Without loss of generality, assume that the $8$ numbers on Sharon's cards are $1$, $1$, $2$, $3$, $4$, $5$, $6$, and $7$, in that order, and assume the $8$ colors are red, red, and six different arbitrary colors. There are ${8\\choose2}-1$ ways of assigning the two red cards to the $8$ numbers; we subtract $1$ bec... |
2017-II-10 | 2,017 | 10 | Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$ . Point $M$ is the midpoint of $\overline{AD}$ , point $N$ is the trisection point of $\overline{AB}$ closer to $A$ , and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$ . Point $P$ lies on the quadrilateral $BCON$ , and $\overline{BP}$ bisects ... | 546 | II | [
"[asy] pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,blue); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); label(\"$M$\",M,W); label(\... |
2017-II-11 | 2,017 | 11 | Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way). | 544 | II | [
"It is obvious that any configuration of one-way roads which contains a town whose roads all lead into it or lead out of it cannot satisfy the given. We claim that any configuration which does not have a town whose roads all lead into it or lead out of it does satisfy the given conditions. Now we show that a loop o... |
2017-II-12 | 2,017 | 12 | Circle $C_0$ has radius $1$ , and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$ . Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$ . Circle $C_2$ has radius $r^2$ and is internally tang... | 110 | II | [
"Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the ... |
2017-II-13 | 2,017 | 13 | For each integer $n\geq3$ , let $f(n)$ be the number of $3$ -element subsets of the vertices of a regular $n$ -gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$ . | 245 | II | [
"Considering $n \\pmod{6}$, we have the following formulas: $n\\equiv 0$: $\\frac{n(n-4)}{2} + \\frac{n}{3}$ $n\\equiv 2, 4$: $\\frac{n(n-2)}{2}$ $n\\equiv 3$: $\\frac{n(n-3)}{2} + \\frac{n}{3}$ $n\\equiv 1, 5$: $\\frac{n(n-1)}{2}$ To derive these formulas, we note the following: Any isosceles triangle formed by th... |
2017-II-14 | 2,017 | 14 | A $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ , $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points. | 168 | II | [
"$Case \\textrm{ } 1:$ The lines are not parallel to the faces A line through the point $(a,b,c)$ must contain $(a \\pm 1, b \\pm 1, c \\pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube. We look at the one from $(1,1,1)$ to $(10,... |
2017-II-15 | 2,017 | 15 | Tetrahedron $ABCD$ has $AD=BC=28$ , $AC=BD=44$ , and $AB=CD=52$ . For any point $X$ in space, define $f(X)=AX+BX+CX+DX$ . The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ . | 682 | II | [
"Set $a=BC=28$, $b=CA=44$, $c=AB=52$. Let $O$ be the point which minimizes $f(X)$. $\\textrm{Claim 1: } O \\textrm{ is the gravity center } \\ \\tfrac {1}{4}(\\vec A + \\vec B + \\vec C + \\vec D).",
"Isosceles tetrahedron $ABCD$ or Disphenoid can be inscribed in a parallelepiped $AB'CD'C'DA'B,$ whose facial diag... |
2018-I-1 | 2,018 | 1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ . | 600 | I | [
"Let the linear factors be $(x+c)(x+d)$. Then, $a=c+d$ and $b=cd$. We know that $1\\le a\\le 100$ and $b\\ge 0$, so $c$ and $d$ both have to be non-negative However, $a$ cannot be $0$, so at least one of $c$ and $d$ must be greater than $0$, ie positive. Also, $a$ cannot be greater than $100$, so $c+d$ must be less... |
2018-I-2 | 2,018 | 2 | The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , ... | 925 | I | [
"We have these equations: $196a+14b+c=225a+15c+b=222a+37c$. Taking the last two we get $3a+b=22c$. Because $c \\neq 0$ otherwise $a \\ngtr 0$, and $a \\leq 5$, $c=1$. Then we know $3a+b=22$. Taking the first two equations we see that $29a+14c=13b$. Combining the two gives $a=4, b=10, c=1$. Then we see that $222 \\t... |
2018-I-3 | 2,018 | 3 | Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy... | 157 | I | [
"We have $2+4\\cdot 2$ cases total. The two are all red and all green. Then, you have 4 of one, 1 of other. 3 of one, 2 of other. 2 of one, 3 of other. 1 of one, 4 of other. Then flip the order, so times two. Obviously the denominator is $10\\cdot 9\\cdot 8\\cdot 7\\cdot 6$, since we are choosing a card without rep... |
2018-I-4 | 2,018 | 4 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$ . Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$ . Then $AD$ can be expressed in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Fi... | 289 | I | [
"We apply Law of Cosines on $\\angle A$ twice (one from $\\triangle ADE$ and one from $\\triangle ABC$), \\begin{align*} 12^2 &= 10^2 + 10^2 - 2(10)(10) \\cdot \\cos{A} \\\\[5pt] x^2&=x^2+(10-x)^2-2(x)(10-x)\\cdot \\cos{A} \\end{align*} Solving for $\\cos{A}$ in both equations, we get \\begin{align*} \\cos{A} &= \\... |
2018-I-5 | 2,018 | 5 | For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\] there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\] Find the product of all possible values of $K$ . | 189 | I | [
"Using the logarithmic property $\\log_{a^n}b^n = \\log_{a}b$, we note that \\[(2x+y)^2 = x^2+xy+7y^2\\] That gives \\[x^2+xy-2y^2=0\\] upon simplification and division by $3$. Factoring $x^2+xy-2y^2=0$ gives \\[(x+2y)(x-y)=0\\] Then, \\[x=y \\text{ or }x=-2y\\] From the second equation, \\[9x^2+6xy+y^2=3x^2+4xy+Ky... |
2018-I-6 | 2,018 | 6 | Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ . | 440 | I | [
"Let $a=z^{120}$. This simplifies the problem constraint to $a^6-a \\in \\mathbb{R}$. This is true if $\\text{Im}(a^6)=\\text{Im}(a)$. Let $\\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\\le\\theta<2\\pi$. We are given $\\sin\\theta = \\sin{6\\theta}$.... |
2018-I-7 | 2,018 | 7 | A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles). | 52 | I | [
"We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases. Case 1: vertices are on one base. Then we can call one of the vertices $A$ for distinction. Either the triangle can have sides $1, 1, \\sqrt{3}$ with 6 cases or $\\sqrt{3}, \\sqrt{3}, \\sqrt{3}$ with 2 cases... |
2018-I-8 | 2,018 | 8 | Let $ABCDEF$ be an equiangular hexagon such that $AB=6, BC=8, CD=10$ , and $DE=12$ . Denote $d$ the diameter of the largest circle that fits inside the hexagon. Find $d^2$ . | 147 | I | [
"First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that $EF=2, FA=16$. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length $6+8+10=24$. Then, if you drew it to scale, notice that the \"widest... |
2018-I-9 | 2,018 | 9 | Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$ , and two distinct elements of a subset have a sum of $24$ . For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets. | 210 | I | [
"This problem is tricky because it is the capital of a few \"bashy\" calculations. Nevertheless, the process is straightforward. Call the set $\\{a, b, c, d\\}$. Note that there are only two cases: 1 where $a + b = 16$ and $c + d = 24$ or 2 where $a + b = 16$ and $a + c = 24$. Also note that there is no overlap bet... |
2018-I-10 | 2,018 | 10 | The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a cou... | 4 | I | [
"We divide this up into casework. The \"directions\" the bug can go are $\\text{Clockwise}$, $\\text{Counter-Clockwise}$, and $\\text{Switching}$. Let an $I$ signal going clockwise (because it has to be in the inner circle), an $O$ signal going counter-clockwise, and an $S$ switching between inner and outer circles... |
2018-I-11 | 2,018 | 11 | Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$ . | 195 | I | [
"Note that the given condition is equivalent to $3^n \\equiv 1 \\pmod{143^2}$ and $143=11\\cdot 13$. Because $\\gcd(11^2, 13^2) = 1$, the desired condition is equivalent to $3^n \\equiv 1 \\pmod{121}$ and $3^n \\equiv 1 \\pmod{169}$. If $3^n \\equiv 1 \\pmod{121}$, one can see the sequence $1, 3, 9, 27, 81, 1, 3, 9... |
2018-I-12 | 2,018 | 12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. ... | 683 | I | [
"The question asks us for the probability that a randomly chosen subset of the set of the first 18 positive integers has the property that the sum of its elements is divisible by 3. Note that the total number of subsets is $2^{18}$ because each element can either be in or not in the subset. To find the probability,... |
2018-I-13 | 2,018 | 13 | Let $\triangle ABC$ have side lengths $AB=30$ , $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ . | 126 | I | [
"First note that \\[\\angle I_1AI_2 = \\angle I_1AX + \\angle XAI_2 = \\frac{\\angle BAX}2 + \\frac{\\angle CAX}2 = \\frac{\\angle A}2\\] is a constant not depending on $X$, so by $[AI_1I_2] = \\tfrac12(AI_1)(AI_2)\\sin\\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$. Let $a = BC$, $b = AC$, $c = AB$, and $\... |
2018-I-14 | 2,018 | 14 | Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$ . From any vertex of the heptagon except $E$ , the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$ , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that ... | 351 | I | [
"This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line $E \\leftrightarrow P_4 \\leftrightarrow P_5 \\leftrightarrow S \\leftrightarrow P_1 \\leftrightarrow P_2 \\leftrightarrow P_3 \\leftrightarrow E$. We can count the number of left/r... |
2018-I-15 | 2,018 | 15 | David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varp... | 59 | I | [
"Suppose our four sides lengths cut out arc lengths of $2a$, $2b$, $2c$, and $2d$, where $a+b+c+d=180^\\circ$. Then, we only have to consider which arc is opposite $2a$. These are our three cases, so \\[\\varphi_A=a+c\\] \\[\\varphi_B=a+b\\] \\[\\varphi_C=a+d\\] Our first case involves quadrilateral $ABCD$ with $\\... |
2018-II-1 | 2,018 | 1 | Points $A$ , $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running t... | 800 | II | [
"We know that in the same amount of time given, Ina will run twice the distance of Eve, and Paul would run quadruple the distance of Eve. Let's consider the time it takes for Paul to meet Eve: Paul would've run 4 times the distance of Eve, which we can denote as $d$. Thus, the distance between $B$ and $C$ is $4d+d=... |
2018-II-2 | 2,018 | 2 | Let $a_{0} = 2$ , $a_{1} = 5$ , and $a_{2} = 8$ , and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4(a_{n-1} + a_{n-2} + a_{n-3})$ is divided by $11$ . Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$ . | 112 | II | [
"When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern. After computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at $a_{0}$. $a_{0} = 2$, $a_{1} = 5$, $a_{2} = 8$, $a_{3} = 5$, $a_{4} = ... |
2018-II-3 | 2,018 | 3 | Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube. | 371 | II | [
"The first step is to convert $36_{b}$ and $27_{b}$ into base-10 numbers. Then, we can write \\[36_{b} = 3b + 6\\] and \\[27_{b} = 2b + 7\\]. It should also be noted that $8 \\leq b < 1000$. Because there are less perfect cubes than perfect squares for the restriction we are given on $b$, it is best to list out all... |
2018-II-4 | 2,018 | 4 | In equiangular octagon $CAROLINE$ , $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$ . The self-intersecting octagon $CORNELIA$ encloses six non-overlapping triangular regions. Let $K$ be the area enclosed by $CORNELIA$ , that is, the total area of the six triangular regions. Then $K = \frac{a}{b}$ , where ... | 23 | II | [
"We can draw $CORNELIA$ and introduce some points. The diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1. In order to find the area of $CORNELIA$, we need to find 4 times the area of $\\bigtriangleup$$ACY$ and 2 times the area of $\\bigtriangleup$$YZW$. Using s... |
2018-II-5 | 2,018 | 5 | Suppose that $x$ , $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ , $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ . | 74 | II | [
"The First (pun intended) thing to notice is that $xy$ and $zx$ have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is $e^{\\frac{3\\pi i}{2}}$. Anyway this results in getting that $\\left(\\... |
2018-II-6 | 2,018 | 6 | A real number $a$ is chosen randomly and uniformly from the interval $[-20, 18]$ . The probability that the roots of the polynomial \[x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2\] are all real can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 37 | II | [
"The polynomial we are given is rather complicated, so we could use Rational Root Theorem to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, $x = 1, -1, 2, -2$ are all possible rational roots. Upon plugging these roots into the polynomial, $x = -2$ and $x = 1$ make the polynomial e... |
2018-II-7 | 2,018 | 7 | Triangle $ABC$ has side lengths $AB = 9$ , $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{... | 20 | II | [
"For each $k$ between $2$ and $2450$, the area of the trapezoid with $\\overline{P_kQ_k}$ as its bottom base is the difference between the areas of two triangles, both similar to $\\triangle{ABC}$. Let $d_k$ be the length of segment $\\overline{P_kQ_k}$. The area of the trapezoid with bases $\\overline{P_{k-1}Q_{k-... |
2018-II-8 | 2,018 | 8 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ , $(x + 2, y)$ , $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ . | 556 | II | [
"We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$. This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$, recording down the number of ways to get to each point recursively. $(0,0): ... |
2018-II-9 | 2,018 | 9 | Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$ , and partition the octago... | 184 | II | [
"We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\\frac{23}{2})$. Recall that the centroid is $\\frac{1}{3}$ way up each medi... |
2018-II-10 | 2,018 | 10 | Find the number of functions $f(x)$ from $\{1, 2, 3, 4, 5\}$ to $\{1, 2, 3, 4, 5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1, 2, 3, 4, 5\}$ . | 756 | II | [
"Just to visualize solution 1. If we list all possible $(x,f(x))$, from ${1,2,3,4,5}$ to ${1,2,3,4,5}$ in a specific order, we get $5 \\cdot 5 = 25$ different $(x,f(x))$ 's. Namely: \\[(1,1) (1,2) (1,3) (1,4) (1,5)\\] \\[(2,1) (2,2) (2,3) (2,4) (2,5)\\] \\[(3,1) (3,2) (3,3) (3,4) (3,5)\\] \\[(4,1) (4,2) (4,3) (4,4)... |
2018-II-11 | 2,018 | 11 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ . | 461 | II | [
"If the first number is $6$, then there are no restrictions. There are $5!$, or $120$ ways to place the other $5$ numbers. If the first number is $5$, $6$ can go in four places, and there are $4!$ ways to place the other $4$ numbers. $4 \\cdot 4! = 96$ ways. If the first number is $4$, .... 4 6 _ _ _ _ $\\implies$ ... |
2018-II-12 | 2,018 | 12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ , $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ . | 112 | II | [
"Let $AP=x$ and let $PC=\\rho x$. Let $[ABP]=\\Delta$ and let $[ADP]=\\Lambda$. We easily get $[PBC]=\\rho \\Delta$ and $[PCD]=\\rho\\Lambda$. We are given that $[ABP] +[PCD] = [PBC]+[ADP]$, which we can now write as \\[\\Delta + \\rho\\Lambda = \\rho\\Delta + \\Lambda \\qquad \\Longrightarrow \\qquad \\Delta -\\La... |
2018-II-13 | 2,018 | 13 | Misha rolls a standard, fair six-sided die until she rolls 1-2-3 in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 647 | II | [
"Let $P_n$ be the probability of getting consecutive $1,2,3$ rolls in $n$ rolls and not rolling $1,2,3$ prior to the nth roll. Let $x = P_3+P_5+...=1-(P_4+P_6+..)$. Following Solution 2, one can see that \\[P_{n+1}=P_{n}-\\frac{P_{n-2}}{6^3}\\] for all positive integers $n \\ge 5$. Summing for $n=5,7,...$ gives \\[... |
2018-II-14 | 2,018 | 14 | The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ , $PB = 4$... | 227 | II | [
"Let the sides $\\overline{AB}$ and $\\overline{AC}$ be tangent to $\\omega$ at $Z$ and $W$, respectively. Let $\\alpha = \\angle BAX$ and $\\beta = \\angle AXC$. Because $\\overline{PQ}$ and $\\overline{BC}$ are both tangent to $\\omega$ and $\\angle YXC$ and $\\angle QYX$ subtend the same arc of $\\omega$, it fol... |
2018-II-15 | 2,018 | 15 | Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$ , $f(6) = 12$ , and \[|x - y| \leq |f(x) - f(y)| \leq 3|x - y|\] for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$ . | 185 | II | [
"First, suppose $x = y + 1$. Then, the inequality becomes $1 \\leq |f(y + 1) - f(y)| \\leq 3$. In other words, the (positive) difference between consecutive function values is $1$, $2$, or $3$. Let $d_k := f(k) - f(k - 1)$. Note that $f(x) - f(y) = \\sum_{k=y+1}^{x} d_k.$ Thus, $\\sum_{k=1}^{6} d_k = f(6) - f(0) = ... |
2019-I-1 | 2,019 | 1 | Consider the integer \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\] Find the sum of the digits of $N$ . | 342 | I | [
"Let's express the number in terms of $10^n$. We can obtain $(10-1)+(10^2-1)+(10^3-1)+\\cdots+(10^{321}-1)$. By the commutative and associative property, we can group it into $(10+10^2+10^3+\\cdots+10^{321})-321$. We know the former will yield $1111....10$, so we only have to figure out what the last few digits are... |
2019-I-2 | 2,019 | 2 | Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers... | 29 | I | [
"$B-J \\ne 0$ because $B \\ne J$, so the probability that $B-J < 0$ is $\\frac{1}{2}$ by symmetry. The probability that $B-J = 1$ is $\\frac{19}{20 \\times 19} = \\frac{1}{20}$ because there are 19 pairs: $(B,J) = (2,1), \\ldots, (20,19)$. The probability that $B-J \\ge 2$ is $1-\\frac{1}{2}-\\frac{1}{20} = \\frac{... |
2019-I-3 | 2,019 | 3 | In $\triangle PQR$ , $PR=15$ , $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$ . | 120 | I | [
"We know the area of the hexagon $ABCDEF$ to be $\\triangle PQR- \\triangle PAF- \\triangle BCQ- \\triangle RED$. Since $PR^2+RQ^2=PQ^2$, we know that $\\triangle PRQ$ is a right triangle. Thus the area of $\\triangle PQR$ is $150$. Another way to compute the area is \\[\\frac12 \\cdot PQ\\cdot RQ \\sin \\angle PQR... |
2019-I-4 | 2,019 | 4 | A soccer team has $22$ available players. A fixed set of $11$ players starts the game, while the other $11$ are available as substitutes. During the game, the coach may make as many as $3$ substitutions, where any one of the $11$ players in the game is replaced by one of the substitutes. No player removed from the game... | 122 | I | [
"There are $0-3$ substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for $0$ subs is $1$, and the ways to reorganize after $n$ subs is the product of the number of new subs ($12-n$) and the players that can be ejected ($11$). ... |
2019-I-5 | 2,019 | 5 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ , $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\tfrac{1}{3}$ , independently of its previous moves. Th... | 252 | I | [
"One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \\[P(x,y) = \\frac{1}{3} P(x-1,y) + \\frac{1}{3} P(x,y-1) + \\frac{1}{3} P(x-1,y-1)\\] for $x,y \\geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero. W... |
2019-I-6 | 2,019 | 6 | In convex quadrilateral $KLMN$ , side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ , $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find ... | 90 | I | [
"Let $\\angle MKN=\\alpha$ and $\\angle LNK=\\beta$. Let $P$ be the project of $L$ onto line $NK$. Note $\\angle KLP=\\beta$. Then, $KP=28\\sin\\beta=8\\cos\\alpha$. Furthermore, $KN=\\frac{65}{\\sin\\alpha}=\\frac{28}{\\sin\\beta} \\Rightarrow 65\\sin\\beta=28\\sin\\alpha$. Dividing the equations gives \\[\\frac{6... |
2019-I-7 | 2,019 | 7 | There are positive integers $x$ and $y$ that satisfy the system of equations \[\log_{10} x + 2 \log_{10} (\gcd(x,y)) = 60\] \[\log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) = 570.\] Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not neces... | 880 | I | [
"Add the two equations to get that $\\log x+\\log y+2(\\log(\\gcd(x,y))+2(\\log(\\text{lcm}(x,y)))=630$. Then, we use the theorem $\\log a+\\log b=\\log ab$ to get the equation, $\\log (xy)+2(\\log(\\gcd(x,y))+\\log(\\text{lcm}(x,y)))=630$. Using the theorem that $\\gcd(x,y) \\cdot \\text{lcm}(x,y)=x\\cdot y$, alon... |
2019-I-8 | 2,019 | 8 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 67 | I | [
"We can substitute $y = \\sin^2{x}$. Since we know that $\\cos^2{x}=1-\\sin^2{x}$, we can do some simplification. This yields $y^5+(1-y)^5=\\frac{11}{36}$. From this, we can substitute again to get some cancellation through binomials. If we let $z=\\frac{1}{2}-y$, we can simplify the equation to: \\[\\left(\\frac{1... |
2019-I-9 | 2,019 | 9 | Let $\tau(n)$ denote the number of positive integer divisors of $n$ . Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$ . | 540 | I | [
"In order to obtain a sum of $7$, we must have: either a number with $5$ divisors (a fourth power of a prime) and a number with $2$ divisors (a prime), or a number with $4$ divisors (a semiprime or a cube of a prime) and a number with $3$ divisors (a square of a prime). (No integer greater than $1$ can have fewer t... |
2019-I-10 | 2,019 | 10 | For distinct complex numbers $z_1,z_2,\dots,z_{673}$ , the polynomial \[(x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3\] can be expressed as $x^{2019} + 20x^{2018} + 19x^{2017}+g(x)$ , where $g(x)$ is a polynomial with complex coefficients and with degree at most $2016$ . The value of \[\left| \sum_{1 \le j <k \le 673} z_jz_k... | 352 | I | [
"In order to begin this problem, we must first understand what it is asking for. The notation \\[\\left| \\sum_{1 \\le j <k \\le 673} z_jz_k \\right|\\] simply asks for the absolute value of the sum of the product of the distinct unique roots of the polynomial taken two at a time or \\[(z_1z_2+z_1z_3+ \\dots + z_1z... |
2019-I-11 | 2,019 | 11 | In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose th... | 20 | I | [
"Let the tangent circle be $\\omega$. Some notation first: let $BC=a$, $AB=b$, $s$ be the semiperimeter, $\\theta=\\angle ABC$, and $r$ be the inradius. Intuition tells us that the radius of $\\omega$ is $r+\\frac{2rs}{s-a}$ (using the exradius formula). However, the sum of the radius of $\\omega$ and $\\frac{rs}{s... |
2019-I-12 | 2,019 | 12 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ , $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ . | 230 | I | [
"Notice that we must have \\[\\frac{f(f(z))-f(z)}{f(z)-z}=-\\frac{f(f(z))-f(z)}{z-f(z)}\\in i\\mathbb R .\\]However, $f(t)-t=t(t-20)$, so \\begin{align*} \\frac{f(f(z))-f(z)}{f(z)-z}&=\\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\\\ &=\\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\\\ &=(z-19)(z+1)\\\\ &=(z-9)^2-100. \\end{align*} ... |
2019-I-13 | 2,019 | 13 | Triangle $ABC$ has side lengths $AB=4$ , $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a... | 32 | I | [
"[asy] unitsize(20); pair A, B, C, D, E, F, X, O1, O2; A = (0, 0); B = (4, 0); C = intersectionpoints(circle(A, 6), circle(B, 5))[0]; D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0); F = intersectionpoints(circle(D, 2), circle(E, 7))[1]; X = extension(A, E, C, F); O1 = circumcenter(C, A, D); O2 = circumc... |
2019-I-14 | 2,019 | 14 | Find the least odd prime factor of $2019^8 + 1$ . | 97 | I | [
"We know that $2019^8 \\equiv -1 \\pmod{p}$ for some prime $p$. We want to find the smallest odd possible value of $p$. By squaring both sides of the congruence, we find $2019^{16} \\equiv 1 \\pmod{p}$. Since $2019^{16} \\equiv 1 \\pmod{p}$, the order of $2019$ modulo $p$ is a positive divisor of $16$. However, if ... |
2019-I-15 | 2,019 | 15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersec... | 65 | I | [
"[asy] size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1... |
2019-II-1 | 2,019 | 1 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB=9,BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 59 | II | [
"[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair E = (-6,0); draw(A--B--C--cycle); draw(B--D--A); label(\"$A$\",A,dir(-120)); label(\"$B$\",B,dir(-60)); label(\"$C$\",C,dir(60)); label(\"$D$\",D,dir(120)); label(\"$E$\",E,dir(-135)); label(\"$9$\",(A+B)/2,dir(-90)); label(... |
2019-II-2 | 2,019 | 2 | Lily pads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , wher... | 107 | II | [
"Let $P_n$ be the probability the frog visits pad $7$ starting from pad $n$. Then $P_7 = 1$, $P_6 = \\frac12$, and $P_n = \\frac12(P_{n + 1} + P_{n + 2})$ for all integers $1 \\leq n \\leq 5$. Working our way down, we find \\[P_5 = \\frac{3}{4}\\] \\[P_4 = \\frac{5}{8}\\] \\[P_3 = \\frac{11}{16}\\] \\[P_2 = \\frac{... |
2019-II-3 | 2,019 | 3 | Find the number of $7$ -tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following system of equations: \[abc=70\] \[cde=71\] \[efg=72.\] | 96 | II | [
"As 71 is prime, $c$, $d$, and $e$ must be 1, 1, and 71 (in some order). However, since $c$ and $e$ are divisors of 70 and 72 respectively, the only possibility is $(c,d,e) = (1,71,1)$. Now we are left with finding the number of solutions $(a,b,f,g)$ satisfying $ab = 70$ and $fg = 72$, which separates easily into t... |
2019-II-4 | 2,019 | 4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 187 | II | [
"Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes). Case 1 (easy): Four 5's are rolled. This has probability $\\frac{1}{6^4}$ of occurring. Case 2: Two 5's are rolle... |
2019-II-5 | 2,019 | 5 | Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table und... | 520 | II | [
"Suppose first that the chairs weren't numbered, and rotation doesn't matter. Then if we let the ambassador-advisor pairs act as a single person, we essentially want to pick $4$ seats out of $8$ seats, and fixing one pair gives $7 \\cdot 6 \\cdot 5$ ways. However, now we want to number the seats, in which there are... |
2019-II-6 | 2,019 | 6 | In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$ . | 216 | II | [
"Using change of base on the second equation to base b, \\[\\frac{\\log x}{\\log \\log_{b}{x}}=54\\] \\[\\log x = 54 \\cdot \\log\\log_{b}{x}\\] \\[x = (\\log_{b}{x})^{54}\\] Note by dolphin7 - you could also just rewrite the second equation in exponent form. Substituting this into the $\\sqrt x$ of the first equat... |
2019-II-7 | 2,019 | 7 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,4... | 715 | II | [
"Let the points of intersection of $\\ell_A, \\ell_B,\\ell_C$ with $\\triangle ABC$ divide the sides into consecutive segments $BD,DE,EC,CF,FG,GA,AH,HI,IB$. Furthermore, let the desired triangle be $\\triangle XYZ$, with $X$ closest to side $BC$, $Y$ closest to side $AC$, and $Z$ closest to side $AB$. Hence, the de... |
2019-II-8 | 2,019 | 8 | The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019,$ and $f\left(\tfrac{1+\sqrt3i}{2}\right)=2015+2019\sqrt3i$ . Find the remainder when $f(1)$ is divided by $1000$ . | 53 | II | [
"We have $\\frac{1+\\sqrt{3}i}{2} = \\omega$ where $\\omega = e^{\\frac{i\\pi}{3}}$ is a primitive 6th root of unity. Then we have \\begin{align*} f(\\omega) &= a\\omega^{2018} + b\\omega^{2017} + c\\omega^{2016}\\\\ &= a\\omega^2 + b\\omega + c \\end{align*} We wish to find $f(1) = a+b+c$. We first look at the rea... |
2019-II-9 | 2,019 | 9 | Call a positive integer $n$ $k$ - pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of the positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$ . | 472 | II | [
"Every 20-pretty integer can be written in form $n = 2^a 5^b k$, where $a \\ge 2$, $b \\ge 1$, $\\gcd(k,10) = 1$, and $d(n) = 20$, where $d(n)$ is the number of divisors of $n$. Thus, we have $20 = (a+1)(b+1)d(k)$, using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, ... |
2019-II-10 | 2,019 | 10 | There is a unique angle $\theta$ between $0^\circ$ and $90^\circ$ such that for nonnegative integers $n,$ the value of $\tan(2^n\theta)$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $... | 547 | II | [
"Note that if $\\tan \\theta$ is positive, then $\\theta$ is in the first or third quadrant, so $0^{\\circ} < \\theta < 90^{\\circ} \\pmod{180^{\\circ}}$. Furthermore, the only way $\\tan{\\left(2^{n}\\theta\\right)}$ can be positive for all $n$ that are multiples of $3$ is when: \\[2^0\\theta \\equiv 2^3\\theta \\... |
2019-II-11 | 2,019 | 11 | Triangle $ABC$ has side lengths $AB=7,BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac{m}{n},$ where $m$ ... | 11 | II | [
"[asy] unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label(\"$A$\",A,dir(105)); label(\"$B$\",B,dir(-135)); label(\"$C$\",C,dir(-75)); dot((2.68,2.25)); label(\"$K$\",(2.68,2... |
2019-II-12 | 2,019 | 12 | For $n\ge1$ call a finite sequence $(a_1,a_2,\ldots,a_n)$ of positive integers progressive if $a_i<a_{i+1}$ and $a_i$ divides $a_{i+1}$ for $1\le i\le n-1$ . Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360.$ | 47 | II | [
"If the first term is $x$, then dividing through by $x$, we see that we can find the number of progressive sequences whose sum is $\\frac{360}{x} - 1$, and whose first term is not 1. If $a(k)$ denotes the number of progressive sequences whose sum is $k$ and whose first term is not 1, then we can express the answer ... |
2019-II-13 | 2,019 | 13 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | 504 | II | [
"The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, \\(1\\), and assume the side length of the octagon is \\(2\\). Let \\(r\\) denote the radius of the circle, \\(O\\) be the center of the circle. Then: \\[r^2= 1^2 + \\left(\\sqrt{2}+1\\right)^2= 4... |
2019-II-14 | 2,019 | 14 | Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5,n,$ and $n+1$ cents, $91$ cents is the greatest postage that cannot be formed. | 71 | II | [
"By the Chicken McNugget theorem, the least possible value of $n$ such that $91$ cents cannot be formed satisfies $5n - (5 + n) = 91 \\implies n = 24$, so $n$ must be at least $24$. For a value of $n$ to work, we must not only be unable to form the value $91$, but we must also be able to form the values $92$ throug... |
2019-II-15 | 2,019 | 15 | In acute triangle $ABC,$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ , $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ ca... | 574 | II | [
"First we have $a\\cos A=PQ=25$, and $(a\\cos A)(c\\cos C)=(a\\cos C)(c\\cos A)=AP\\cdot PB=10(25+15)=400$ by PoP. Similarly, $(a\\cos A)(b\\cos B)=15(10+25)=525,$ and dividing these each by $a\\cos A$ gives $b\\cos B=21,c\\cos C=16$. It is known that the sides of the orthic triangle are $a\\cos A,b\\cos B,c\\cos C... |
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