Difference between revisions of "2015 AIME II Problems"

Latest revision as of 19:15, 13 January 2024

2015 AIME II (Answer Key) Printable version \| AoPS Contest Collections • PDF
Instructions This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$ , inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 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 $1000$ .

Solution

Problem 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 $m+n$ .

Solution

Problem 3

Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ .

Solution

Problem 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 $p + q$ .

Solution

Problem 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 $\frac{1}{2015}$ .

Solution

Problem 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 $c$ .

Solution

Problem 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 $m+n$ .

Solution

Problem 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 $p+q$ .

Solution

Problem 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 $v^2$ .

$[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]$

Solution

Problem 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 $1, 2, \ldots, 7$ .

Solution

Problem 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 $m+n$ .

Solution

Problem 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.

Solution

Problem 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 $a_n < 0$ .

Solution

Problem 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 $2x^3+(xy)^3+2y^3$ .

Solution

Problem 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 $m+n$ .

$[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]$

Solution

2015 AIME II (Problems • Answer Key • Resources)
Preceded by 2015 AIME I Problems	Followed by 2016 AIME I Problems
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 ==Problem 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 <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+In a new school, <math>40</math> percent of the students are freshmen, <math>30</math> percent are sophomores, <math>20</math> percent are juniors, and <math>10</math> percent are seniors. All freshmen are required to take Latin, and <math>80</math> percent of sophomores, <math>50</math> percent of the juniors, and <math>20</math> percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 [[2015 AIME II Problems/Problem 2 | Solution]]
 ==Problem 3==
+Let <math>m</math> be the least positive integer divisible by <math>17</math> whose digits sum to <math>17</math>. Find <math>m</math>.
+[[2015 AIME II Problems/Problem 3 | Solution]]
 ==Problem 4==
+In an isosceles trapezoid, the parallel bases have lengths <math>\log 3</math> and <math>\log 192</math>, and the altitude to these bases has length <math>\log 16</math>. The perimeter of the trapezoid can be written in the form <math>\log 2^p 3^q</math>, where <math>p</math> and <math>q</math> are positive integers. Find <math>p + q</math>.
+[[2015 AIME II Problems/Problem 4 | Solution]]
 ==Problem 5==
+Two unit squares are selected at random without replacement from an <math>n \times n</math> grid of unit squares. Find the least positive integer <math>n</math> such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than <math>\frac{1}{2015}</math>.
+[[2015 AIME II Problems/Problem 5 | Solution]]
 ==Problem 6==
+Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form <math>P(x) = 2x^3-2ax^2+(a^2-81)x-c</math> for some positive integers <math>a</math> and <math>c</math>. Can you tell me the values of <math>a</math> and <math>c</math>?"
+After some calculations, Jon says, "There is more than one such polynomial."
+Steve says, "You're right.  Here is the value of <math>a</math>." He writes down a positive integer and asks, "Can you tell me the value of <math>c</math>?"
+Jon says, "There are still two possible values of <math>c</math>."
+Find the sum of the two possible values of <math>c</math>.
+[[2015 AIME II Problems/Problem 6 | Solution]]
 ==Problem 7==
+Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = w</math>, the area of <math>PQRS</math> can be expressed as the quadratic polynomial
+<cmath>\text{Area}(PQRS) = \alpha w - \beta \cdot w^2.</cmath>
+Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+[[2015 AIME II Problems/Problem 7 | Solution]]
 ==Problem 8==
+Let <math>a</math> and <math>b</math> be positive integers satisfying <math>\frac{ab+1}{a+b} < \frac{3}{2}</math>. The maximum possible value of <math>\frac{a^3b^3+1}{a^3+b^3}</math> is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
+[[2015 AIME II Problems/Problem 8 | Solution]]
 ==Problem 9==
+A cylindrical barrel with radius <math>4</math> feet and height <math>10</math> feet is full of water. A solid cube with side length <math>8</math> feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is <math>v</math> cubic feet. Find <math>v^2</math>.
+<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>
+[[2015 AIME II Problems/Problem 9 | Solution]]
 ==Problem 10==
+Call a permutation <math>a_1, a_2, \ldots, a_n</math> of the integers <math>1, 2, \ldots, n</math> ''quasi-increasing'' if <math>a_k \leq a_{k+1} + 2</math> for each <math>1 \leq k \leq n-1</math>. For example, <math>53421</math> and <math>14253</math> are quasi-increasing permutations of the integers <math>1, 2, 3, 4, 5</math>, but <math>45123</math> is not. Find the number of quasi-increasing permutations of the integers <math>1, 2, \ldots, 7</math>.
+[[2015 AIME II Problems/Problem 10 | Solution]]
 ==Problem 11==
+The circumcircle of acute <math>\triangle ABC</math> has center <math>O</math>. The line passing through point <math>O</math> perpendicular to <math>\overline{OB}</math> intersects lines <math>AB</math> and <math>BC</math> at <math>P</math> and <math>Q</math>, respectively. Also <math>AB=5</math>, <math>BC=4</math>, <math>BQ=4.5</math>, and <math>BP=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+[[2015 AIME II Problems/Problem 11 | Solution]]
 ==Problem 12==
+There are <math>2^{10} = 1024</math> possible <math>10</math>-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 <math>3</math> adjacent letters that are identical.
+[[2015 AIME II Problems/Problem 12 | Solution]]
 ==Problem 13==
+Define the sequence <math>a_1, a_2, a_3, \ldots</math> by <math>a_n = \sum\limits_{k=1}^n \sin{k}</math>, where <math>k</math> represents radian measure. Find the index of the 100th term for which <math>a_n < 0</math>.
+[[2015 AIME II Problems/Problem 13 | Solution]]
 ==Problem 14==
+Let <math>x</math> and <math>y</math> be real numbers satisfying <math>x^4y^5+y^4x^5=810</math> and <math>x^3y^6+y^3x^6=945</math>. Evaluate <math>2x^3+(xy)^3+2y^3</math>.
+[[2015 AIME II Problems/Problem 14 | Solution]]
 ==Problem 15==
+Circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> have radii <math>1</math> and <math>4</math>, respectively, and are externally tangent at point <math>A</math>. Point <math>B</math> is on <math>\mathcal{P}</math> and point <math>C</math> is on <math>\mathcal{Q}</math> such that <math>BC</math> is a common external tangent of the two circles. A line <math>\ell</math> through <math>A</math> intersects <math>\mathcal{P}</math> again at <math>D</math> and intersects <math>\mathcal{Q}</math> again at <math>E</math>. Points <math>B</math> and <math>C</math> lie on the same side of <math>\ell</math>, and the areas of <math>\triangle DBA</math> and <math>\triangle ACE</math> are equal. This common area is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+<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>
+[[2015 AIME II Problems/Problem 15 | Solution]]
-{{AIME box|year=2015|n=II|before=[[2014 AIME I]], [[2014 AIME II]]|after=[[2016 AIME]]}}
+{{AIME box|year=2015|n=II|before=[[2015 AIME I Problems]]|after=[[2016 AIME I Problems]]}}
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Difference between revisions of "2015 AIME II Problems"

Latest revision as of 19:15, 13 January 2024

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15