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Difference between revisions of "2014 AIME II Problems"

(Created page with "{{AIME Problems|year=2014|n=II}} ==Problem 1== Solution ==Problem 2== Solution ==Problem 3== [[2014...")
 
(→‎Problem 15: Geq, not greater than)
 
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==Problem 1==
 
==Problem 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.
  
 
[[2014 AIME II Problems/Problem 1|Solution]]
 
[[2014 AIME II Problems/Problem 1|Solution]]
Line 7: Line 9:
 
==Problem 2==
 
==Problem 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 <math>\frac{1}{3}</math>. The probability that a man has none of the three risk factors given that he does not have risk factor A is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
  
 
[[2014 AIME II Problems/Problem 2|Solution]]
 
[[2014 AIME II Problems/Problem 2|Solution]]
  
 
==Problem 3==
 
==Problem 3==
 +
 +
A rectangle has sides of length <math>a</math> 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 <math>a</math> 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 <math>a</math> parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find <math>a^2</math>.
 +
 +
 +
<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);
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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>
  
 
[[2014 AIME II Problems/Problem 3|Solution]]
 
[[2014 AIME II Problems/Problem 3|Solution]]
  
 
==Problem 4==
 
==Problem 4==
 +
 +
The repeating decimals <math>0.abab\overline{ab}</math> and <math>0.abcabc\overline{abc}</math> satisfy
 +
 +
<math>0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},</math>
 +
 +
where <math>a</math>, <math>b</math>, and <math>c</math> are (not necessarily distinct) digits. Find the three digit number <math>abc</math>.
 +
  
 
[[2014 AIME II Problems/Problem 4|Solution]]
 
[[2014 AIME II Problems/Problem 4|Solution]]
  
 
==Problem 5==
 
==Problem 5==
 +
 +
Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>.
 +
  
 
[[2014 AIME II Problems/Problem 5|Solution]]
 
[[2014 AIME II Problems/Problem 5|Solution]]
  
 
==Problem 6==
 
==Problem 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 <math>\frac{2}{3}</math> and each of the other five sides has probability <math>\frac{1}{15}</math>. 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 <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
 +
  
 
[[2014 AIME II Problems/Problem 6|Solution]]
 
[[2014 AIME II Problems/Problem 6|Solution]]
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==Problem 7==
 
==Problem 7==
  
 +
Let <math>f(x)=(x^2+3x+2)^{\cos(\pi x)}</math>. Find the sum of all positive integers <math>n</math> for which
 +
<math>\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.</math>
  
 
[[2014 AIME II Problems/Problem 7|Solution]]
 
[[2014 AIME II Problems/Problem 7|Solution]]
  
 
==Problem 8==
 
==Problem 8==
 +
 +
Circle <math>C</math> with radius 2 has diameter <math>\overline{AB}</math>. Circle <math>D</math> is internally tangent to circle <math>C</math> at <math>A</math>. Circle <math>E</math> is internally tangent to circle <math>C</math>, externally tangent to circle <math>D</math>, and tangent to <math>\overline{AB}</math>. The radius of circle <math>D</math> is three times the radius of circle <math>E</math>, and can be written in the form <math>\sqrt{m}-n</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m+n</math>.
  
  
Line 37: Line 95:
  
 
==Problem 9==
 
==Problem 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.
  
 
[[2014 AIME II Problems/Problem 9|Solution]]
 
[[2014 AIME II Problems/Problem 9|Solution]]
  
 
==Problem 10==
 
==Problem 10==
 +
 +
Let <math>z</math> be a complex number with <math>|z|=2014</math>. Let <math>P</math> be the polygon in the complex plane whose vertices are <math>z</math> and every <math>w</math> such that <math>\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}</math>. Then the area enclosed by <math>P</math> can be written in the form <math>n\sqrt{3}</math>, where <math>n</math> is an integer. Find the remainder when <math>n</math> is divided by <math>1000</math>.
 +
  
 
[[2014 AIME II Problems/Problem 10|Solution]]
 
[[2014 AIME II Problems/Problem 10|Solution]]
−
 
  
 
==Problem 11==
 
==Problem 11==
−
 
+
In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math> RD=1</math>. Let <math>M</math> be the midpoint of segment <math>\overline{RD}</math>. Point <math>C</math> lies on side <math>\overline{ED}</math> such that <math>\overline{RC}\perp\overline{EM}</math>. Extend segment <math>\overline{DE}</math> through <math>E</math> to point <math>A</math> such that <math>CA=AR</math>. Then <math>AE=\frac{a-\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer. Find <math>a+b+c</math>.
  
 
[[2014 AIME II Problems/Problem 11|Solution]]
 
[[2014 AIME II Problems/Problem 11|Solution]]
  
 
==Problem 12==
 
==Problem 12==
 +
Suppose that the angles of <math>\triangle ABC</math> satisfy <math>\cos(3A)+\cos(3B)+\cos(3C)=1</math>. Two sides of the triangle have lengths 10 and 13. There is a positive integer <math>m</math> so that the maximum possible length for the remaining side of <math>\triangle ABC</math> is <math>\sqrt{m}</math>. Find <math>m</math>.
  
 
[[2014 AIME II Problems/Problem 12|Solution]]
 
[[2014 AIME II Problems/Problem 12|Solution]]
  
 
==Problem 13==
 
==Problem 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 <math>k<5</math>, no collection of <math>k</math> pairs made by the child contains the shoes from exactly <math>k</math> of the adults is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 +
  
 
[[2014 AIME II Problems/Problem 13|Solution]]
 
[[2014 AIME II Problems/Problem 13|Solution]]
Line 61: Line 125:
 
==Problem 14==
 
==Problem 14==
  
 +
In <math>\triangle ABC</math>, <math>AB=10</math>, <math>\measuredangle A=30^{\circ}</math>, and <math>\measuredangle C=45^{\circ}</math>. Let <math>H</math>, <math>D</math>, and <math>M</math> be points on line <math>\overline{BC}</math> such that <math>AH\perp BC</math>, <math>\measuredangle BAD=\measuredangle CAD</math>, and <math>BM=CM</math>. Point <math>N</math> is the midpoint of segment <math>HM</math>, and point <math>P</math> is on ray <math>AD</math> such that <math>PN\perp BC</math>. Then <math>AP^2=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
[[2014 AIME II Problems/Problem 14|Solution]]
 
[[2014 AIME II Problems/Problem 14|Solution]]
Line 66: Line 131:
 
==Problem 15==
 
==Problem 15==
  
 +
For any integer <math>k\geq 1</math>, let <math>p(k)</math> be the smallest prime which does not divide <math>k</math>. Define the integer function <math>X(k)</math> to be the product of all primes less than <math>p(k)</math> if <math>p(k)>2</math>, and <math>X(k)=1</math> if <math>p(k)=2</math>. Let <math>\{x_n\}</math> be the sequence defined by <math>x_0=1</math>, and <math>x_{n+1}X(x_n)=x_np(x_n)</math> for <math>n\geq 0</math>. Find the smallest positive integer <math>t</math> such that <math>x_t=2090</math>.
  
 
[[2014 AIME II Problems/Problem 15|Solution]]
 
[[2014 AIME II Problems/Problem 15|Solution]]
 +
 +
{{AIME box|year=2014|n=II|before=[[2014 AIME I Problems]]|after=[[2015 AIME I Problems]]}}
 +
{{MAA Notice}}

Latest revision as of 23:04, 28 December 2021

2014 AIME II (Answer Key)
Printable version | AoPS Contest Collections • PDF

Instructions

  1. 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.
  2. 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

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.

Solution

Problem 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 $p+q$.

Solution

Problem 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]

Solution

Problem 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 $abc$.


Solution

Problem 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 $|b|$.


Solution

Problem 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 $p+q$.


Solution

Problem 7

Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$. Find the sum of all positive integers $n$ for which $\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.$

Solution

Problem 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 $m+n$.


Solution

Problem 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.

Solution

Problem 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 $1000$.


Solution

Problem 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 $a+b+c$.

Solution

Problem 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 $m$.

Solution

Problem 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 $m+n$.


Solution

Problem 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 $m+n$.

Solution

Problem 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 $x_t=2090$.

Solution

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
2014 AIME I Problems 		Followed by
2015 AIME I Problems
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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