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Difference between revisions of "2014 AIME I Problems"

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==Problem 1==
 
==Problem 1==
−
The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectrangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.
+
The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.  
  
 +
<asy>
 +
size(200);
 +
defaultpen(linewidth(0.7));
 +
path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin;
 +
path laceR=reflect((75,0),(75,-240))*laceL;
 +
draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray);
 +
for(int i=0;i<=3;i=i+1)
 +
{
 +
path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5);
 +
unfill(circ1); draw(circ1);
 +
unfill(circ2); draw(circ2);
 +
}
 +
draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));</asy>
 
[[2014 AIME I Problems/Problem 1|Solution]]
 
[[2014 AIME I Problems/Problem 1|Solution]]
  
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==Problem 3==
 
==Problem 3==
−
Find the number of rational numbers <math>r,</math> <math>0<r<1,</math> such that when <math>r</math> is written as a fraction in lowest terms, the numerator and the denominator have a sum of <math>1000</math>.
+
Find the number of rational numbers <math>r</math>, <math>0<r<1,</math> such that when <math>r</math> is written as a fraction in lowest terms, the numerator and the denominator have a sum of <math>1000</math>.
  
 
[[2014 AIME I Problems/Problem 3|Solution]]
 
[[2014 AIME I Problems/Problem 3|Solution]]
  
 
==Problem 4==
 
==Problem 4==
−
Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east as <math>20</math> miles per hour, and Steve rides west at <math>20</math> miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly <math>1</math> minute to go past Jon. The westbound train takes <math>10</math> times as long as the eastbound train to go past Steve. The length of each train is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+
Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at <math>20</math> miles per hour, and Steve rides west at <math>20</math> miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly <math>1</math> minute to go past Jon. The westbound train takes <math>10</math> times as long as the eastbound train to go past Steve. The length of each train is <math>\tfrac{m}{n}</math> miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
[[2014 AIME I Problems/Problem 4|Solution]]
 
[[2014 AIME I Problems/Problem 4|Solution]]
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==Problem 7==
 
==Problem 7==
−
Let <math>w</math> and <math>z</math> be complex numbers such that <math>|w| = 1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. (Note that <math>\arg(w)</math>, for w <math>\neq 0</math>, denotes the measure of the angle that the ray from <math>0</math> to <math>w</math> makes with the positive real axis in the complex plane.
+
Let <math>w</math> and <math>z</math> be complex numbers such that <math>|w| = 1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. (Note that <math>\arg(w)</math>, for <math>w \neq 0</math>, denotes the measure of the angle that the ray from <math>0</math> to <math>w</math> makes with the positive real axis in the complex plane.)
  
 
[[2014 AIME I Problems/Problem 7|Solution]]
 
[[2014 AIME I Problems/Problem 7|Solution]]
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==Problem 11==
 
==Problem 11==
−
A token starts at the point <math>(0,0)</math> of an <math>xy</math>-coordinate grid and them makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of <math>|y|=|x|</math> is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+
A token starts at the point <math>(0,0)</math> of an <math>xy</math>-coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of <math>|y|=|x|</math> is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
[[2014 AIME I Problems/Problem 11|Solution]]
 
[[2014 AIME I Problems/Problem 11|Solution]]
Line 105: Line 118:
  
 
==Problem 15==
 
==Problem 15==
−
In <math>\triangle ABC, AB = 3, BC = 4,</math> and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B, \overline{BC}</math> at <math>B</math>and <math>D,</math> and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4},</math> length <math>DE=\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 not divisible by the square of any prime. Find <math>a+b+c</math>.
+
In <math>\triangle ABC, AB = 3, BC = 4,</math> and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B, \overline{BC}</math> at <math>B</math> and <math>D,</math> and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4},</math> length <math>DE=\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 not divisible by the square of any prime. Find <math>a+b+c</math>.
  
 
[[2014 AIME I Problems/Problem 15|Solution]]
 
[[2014 AIME I Problems/Problem 15|Solution]]
 +
 +
{{AIME box|year=2014|n=I|before=[[2013 AIME II Problems]]|after=[[2014 AIME II Problems]]}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:59, 30 November 2021

2014 AIME I (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

The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.

[asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy] Solution

Problem 2

An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$. Find $N$.

Solution

Problem 3

Find the number of rational numbers $r$, $0<r<1,$ such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of $1000$.

Solution

Problem 4

Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 5

Let the set $S = \{P_1, P_2, \dots, P_{12}\}$ consist of the twelve vertices of a regular $12$-gon. A subset $Q$ of $S$ is called communal if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)

Solution

Problem 6

The graphs $y = 3(x-h)^2 + j$ and $y = 2(x-h)^2 + k$ have y-intercepts of $2013$ and $2014$, respectively, and each graph has two positive integer x-intercepts. Find $h$.

Solution

Problem 7

Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$. Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$. The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. (Note that $\arg(w)$, for $w \neq 0$, denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane.)

Solution

Problem 8

The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base 10, where digit $a$ is not zero. Find the three-digit number $abc$.

Solution

Problem 9

Let $x_1< x_2 < x_3$ be the three real roots of the equation $\sqrt{2014} x^3 - 4029x^2 + 2 = 0$. Find $x_2(x_1+x_3)$.

Solution

Problem 10

A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\overline{AC}$ is parallel to $\overline{BD}$. Then $\sin^2(\angle BEA)=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 11

A token starts at the point $(0,0)$ of an $xy$-coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 12

Let $A=\{1,2,3,4\}$, and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$. The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m$.

Solution

Problem 13

On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$.

[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]

Solution

Problem 14

Let $m$ be the largest real solution to the equation

\[\dfrac{3}{x-3} + \dfrac{5}{x-5} + \dfrac{17}{x-17} + \dfrac{19}{x-19} = x^2 - 11x - 4\]

There are positive integers $a, b,$ and $c$ such that $m = a + \sqrt{b + \sqrt{c}}$. Find $a+b+c$.

Solution

Problem 15

In $\triangle ABC, AB = 3, BC = 4,$ and $CA = 5$. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B, \overline{BC}$ at $B$ and $D,$ and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4},$ length $DE=\frac{a\sqrt{b}}{c},$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$.

Solution

2014 AIME I (Problems • Answer Key • Resources)
Preceded by
2013 AIME II Problems 		Followed by
2014 AIME II 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. AMC logo.png

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