2017 AIME II Problems

$\textbf{Problem 1}$ 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 $\{4, 5, 6, 7, 8\}$ .

$\textbf{Problem 1 Solution}$ 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 complimentary counting. There are $2^5$ subsets of $\{1, 2, 3, 4, 5\}$ and $2^5$ subsets of $\{4, 5, 6, 7, 8\}$ . It is easy to make the mistake of assuming there are $2^5+2^5$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ , but the $2^2$ subsets of $\{4, 5\}$ are overcounted. There are $2^5+2^5-2^2$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ , so there are $2^8-(2^5+2^5-2^2)$ subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$ . $2^8-(2^5+2^5-2^2)=\boxed{196}$ .

$\textbf{Problem 2}$ Theams $T_1$ , $T_2$ , $T_3$ , 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 outcomes of all the matches are independent. The probability that $T_4$ will be the champion is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

$\textbf{Problem 2 Solution}$ 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 $T_3$ . Consider the first scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{4+1}$ , the probability $T_3$ beats $T_2$ is $\frac{3}{3+2}$ , and the probability $T_4$ beats $T_3$ is $\frac{4}{4+3}$ . Therefore the first scenario happens with probability $\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}$ . Consider the second scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{1+4}$ , the probability $T_2$ beats $T_3$ is $\frac{2}{2+3}$ , and the probability $T_4$ beats $T_2$ is $\frac{4}{4+2}$ . Therefore the second scenario happens with probability $\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}$ . By summing these two probabilities, the probability that $T_4$ wins is $\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}+\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}$ . Because this expression is equal to $\frac{256}{525}$ , the answer is $256+525=\boxed{781}$ .

$\textbf{Problem 3}$ 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 $p+q$ .

$\textbf{Problem 3 Solution}$ [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(" $Z$ ",Z,S); dot(P); label(" $P$ ",P,NW); [/asy] A diagram is above. The set of all points closer to point $B$ than to point $A$ lie to the right of the perpendicular bisector of $AB$ (line $PZ$ in the diagram), and the set of all points closer to point $B$ than to point $C$ lie below the perpendicular bisector of $BC$ (line $PX$ in the diagram). Therefore, the set of points inside the triangle that are closer to $B$ than to either vertex $A$ or vertex $C$ is bounded by quadrilateral $BXPZ$ . Because $X$ is the midpoint of $BC$ and $Z$ is the midpoint of $AB$ , $X=(10,5)$ and $Z=(6,0)$ . The coordinates of point $P$ is the solution to the system of equations defined by lines $PX$ and $PZ$ . Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope $m$ is $-\frac{1}{m}$ , the equation for line $PX$ is $y=\frac{2}{5}x+1$ and the equation for line $PZ$ is $x=6$ . The solution of this system is $P=\left(6,\frac{17}{5}\right)$ . Using the shoelace formula on quadrilateral $BXPZ$ and triangle $ABC$ , the area of quadrilateral $BXPZ$ is $\frac{109}{5}$ and the area of triangle $ABC$ is $60$ . Finally, the probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to vertex $A$ or vertex $C$ is the ratio of the area of quadrilateral $BXPZ$ to the area of $ABC$ , which is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$ . The answer is $109+300=\boxed{409}$ .

$\textbf{Problem 4}$ Find the number of positive integers less than or equal to $2017$ whose base-three representation contains no digit equal to $0$ .

$\textbf{Problem 4 Solution}$ 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 that have no digit equal to $0$ , there are $2^5$ $7$ -digit numbers that start with $21$ , $2^6$ $7$ -digit numbers that start with $1$ , $2^6$ $6$ -digit numbers, $2^5$ $5$ -digit numbers, $2^4$ $4$ -digit numbers, $2^3$ $3$ -digit numbers, $2^2$ $2$ -digit numbers, and $2^1$ $1$ -digit numbers. Summing these up, the answer is $2^5+2^6+2^6+2^5+2^4+2^3+2^2+2^1=\boxed{222}$ .

$\textbf{Problem 5}$ 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 $x+y$ .

$\textbf{Problem 5 Solution}$ 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 $2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))$ . No matter how the numbers $189$ , $320$ , $287$ , and $234$ are assigned to the values $a+d$ , $b+c$ , $b+d$ , and $c+d$ , the sum $(a+d)+(b+c)+(b+d)+(c+d)$ will always be $189+320+287+234$ . Therefore we need to maximize $3((a+c)+(b+d))-(189+320+287+234)$ . The maximum value of $(a+c)+(b+d)$ is achieved when we let $a+c$ and $b+d$ be $320$ and $287$ because these are the two largest pairwise sums besides $x$ and $y$ . Therefore, the maximum possible value of $x+y=3(320+287)-(189+320+287+234)=\boxed{791}$ .

$\textbf{Problem 6}$ Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer.

$\textbf{Problem 6 Solution}$ 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, $(s-(2n+85))(s+(2n+85))=1\times843=3\times281$ . It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, $2n+85$ is found to be $421$ and $139$ . The two values of $n$ that satisfy one of the equations are $168$ and $27$ . Summing these together, the answer is $168+27=\boxed{195}$ .

$\textbf{Problem 7}$ 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.

$\textbf{Problem 7 Solution}$

$\boxed{501}$

$\textbf{Problem 8}$ 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.

$\textbf{Problem 8 Solution}$ $\boxed{134}$

$\textbf{Problem 9}$ A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven solors. 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 cardf with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number if $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

$\textbf{Problem 9 Solution}$ $\boxed{013}$

$\textbf{Problem 10}$ 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 the area of $BCON$ . Find the area of $\triangle CDP$ .

$\textbf{Problem 10 Solution}$ $\boxed{546}$

$\textbf{Problem 11}$ Five towns are connected by a system of raods. 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).

$\textbf{Problem 11 Solution}$ $\boxed{544}$

$\textbf{Problem 12}$ 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 tangent to $C_1$ at point $A_1$ . In this way a sequence of circles $C_1,C_2,C_3,\cdots$ and a sequence of points on the circles $A_1,A_2,A_3,\cdots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$ , and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$ , as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$ , the distance from the center $C_0$ to $B$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . [asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(" $A_0$ ",(125,0),E); dot((25,100)); label(" $A_1$ ",(25,100),SE); dot((-55,20)); label(" $A_2$ ",(-55,20),E); [/asy]

$\textbf{Problem 12 Solution}$ $\boxed{110}$

$\textbf{Problem 13}$ For each integer $n\geq3$ , let $f(n)$ be the number of $3$ -element subsets of the vertices of the regular $n$ -gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$ .

$\textbf{Problem 13 Solution}$ $\boxed{245}$

$\textbf{Problem 14}$ 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.

$\textbf{Problem 14 Solution}$

$\textbf{Problem 15}$ 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 $m+n$ .

$\textbf{Problem 15 Solution}$

Page

Toolbox

Search

2017 AIME II Problems