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Difference between revisions of "2013 AIME I Problems"
(→‎Problem 11)
 
(11 intermediate revisions by 7 users not shown)
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== Problem 4 ==
 
== Problem 4 ==
−
In the array of 13 squares shown below, 8 squares are colored red, and the remaining 5 squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated 90 degrees around the central square is <math>\frac{1}{n}</math> , where ''n'' is a positive integer. Find ''n''.
+
In the array of <math>13</math> squares shown below, <math>8</math> squares are colored red, and the remaining <math>5</math> squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated <math>90^{\circ}</math> around the central square is <math>\frac{1}{n}</math> , where <math>n</math> is a positive integer. Find <math>n</math>.
 +
 
 +
<asy>
 +
draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0));
 +
draw((2,0)--(2,2)--(3,2)--(3,0)--(3,1)--(2,1)--(4,1)--(4,0)--(2,0));
 +
draw((1,2)--(1,4)--(0,4)--(0,2)--(0,3)--(1,3)--(-1,3)--(-1,2)--(1,2));
 +
draw((-1,1)--(-3,1)--(-3,0)--(-1,0)--(-2,0)--(-2,1)--(-2,-1)--(-1,-1)--(-1,1));
 +
draw((0,-1)--(0,-3)--(1,-3)--(1,-1)--(1,-2)--(0,-2)--(2,-2)--(2,-1)--(0,-1));
 +
size(100);</asy>
  
 
[[2013 AIME I Problems/Problem 4|Solution]]
 
[[2013 AIME I Problems/Problem 4|Solution]]
−
 
  
 
== Problem 5 ==
 
== Problem 5 ==
Line 54: Line 61:
  
 
== Problem 8 ==
 
== Problem 8 ==
−
The domain of the function <math>f(x) = \arcsin(\log_{m}(nx))</math> is a closed interval of length <math>\frac{1}{2013}</math> , where <math>m</math> and <math>n</math> are positive integers and <math>m>1</math>. Find the remainder when the smallest possible sum <math>m+n</math> is divided by 1000.
+
The domain of the function <math>f(x) = \arcsin(\log_{m}(nx))</math> is a closed interval of length <math>\frac{1}{2013}</math> , where <math>m</math> and <math>n</math> are positive integers and <math>m>1</math>. Find the remainder when the smallest possible sum <math>m+n</math> is divided by <math>1000</math>.
  
 
[[2013 AIME I Problems/Problem 8|Solution]]
 
[[2013 AIME I Problems/Problem 8|Solution]]
  
 +
==Problem 9==
 +
A paper equilateral triangle <math>ABC</math> has side length <math>12</math>. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance <math>9</math> from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.
  
−
==Problem 9==
+
<asy>
−
A paper equilateral triangle <math>ABC</math> has side length 12. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance 9 from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.
+
import cse5;
 +
size(12cm);
 +
pen tpen = defaultpen + 1.337;
 +
real a = 39/5.0;
 +
real b = 39/7.0;
 +
pair B = MP("B", (0,0), dir(200));
 +
pair A = MP("A", (9,0), dir(-80));
 +
pair C = MP("C", (12,0), dir(-20));
 +
pair K = (6,10.392);
 +
pair M = (a*B+(12-a)*K) / 12;
 +
pair N = (b*C+(12-b)*K) / 12;
 +
draw(B--M--N--C--cycle, tpen);
 +
draw(M--A--N--cycle);
 +
fill(M--A--N--cycle, mediumgrey);
 +
pair shift = (-20.13, 0);
 +
pair B1 = MP("B", B+shift, dir(200));
 +
pair A1 = MP("A", K+shift, dir(90));
 +
pair C1 = MP("C", C+shift, dir(-20));
 +
draw(A1--B1--C1--cycle, tpen);</asy>
  
 
[[2013 AIME I Problems/Problem 9|Solution]]
 
[[2013 AIME I Problems/Problem 9|Solution]]
−
 
  
 
==Problem 10==
 
==Problem 10==
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== Problem 11 ==
 
== Problem 11 ==
−
Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, ''N'', of play blocks which satisfies the conditions:
+
Ms. Math's kindergarten class has <math>16</math> registered students. The classroom has a very large number, <math>N</math>, of play blocks which satisfies the conditions:
  
−
(a) If 16, 15, or 14 students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
+
(a) If <math>16</math>, <math>15</math>, or <math>14</math> students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
  
 
(b) There are three integers <math>0 < x < y < z < 14</math> such that when <math>x</math>, <math>y</math>, or <math>z</math> students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
 
(b) There are three integers <math>0 < x < y < z < 14</math> such that when <math>x</math>, <math>y</math>, or <math>z</math> students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
  
−
Find the sum of the distinct prime divisors of the least possible value of ''N'' satisfying the above conditions.
+
Find the sum of the distinct prime divisors of the least possible value of <math>N</math> satisfying the above conditions.
  
 
[[2013 AIME I Problems/Problem 11|Solution]]
 
[[2013 AIME I Problems/Problem 11|Solution]]
Line 95: Line 121:
 
For <math>\pi \le \theta < 2\pi</math>, let
 
For <math>\pi \le \theta < 2\pi</math>, let
  
−
<math>\begin{align*}</math>
+
<cmath> P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots
−
<math>P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots</math>
+
</cmath>
−
<math>\end{align*}</math>
 
  
 
and
 
and
  
−
<math>\begin{align*}</math>
+
<cmath> Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta
−
<math>Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots</math>
+
+\ldots </cmath>
−
<math>\end{align*}</math>
 
  
 
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
Line 110: Line 134:
  
 
==Problem 15==
 
==Problem 15==
−
Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions (a) <math>0\le A<B<C\le99</math>, (b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>, (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and (d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</math>.
+
Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions:
 +
 
 +
(a) <math>0\le A<B<C\le99</math>,  
 +
 
 +
(b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>,  
 +
 
 +
(c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and  
 +
 
 +
(d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</math>.
  
 
[[2013 AIME I Problems/Problem 15|Solution]]
 
[[2013 AIME I Problems/Problem 15|Solution]]
 +
 +
{{AIME box|year=2013|n=I|before=[[2012 AIME II Problems]]|after=[[2013 AIME II Problems]]}}
 +
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:07, 24 February 2024

2013 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 AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend bicycling?

Solution

Problem 2

Find the number of five-digit positive integers, $n$, that satisfy the following conditions:

    (a) the number $n$ is divisible by $5,$
    (b) the first and last digits of $n$ are equal, and
    (c) the sum of the digits of $n$ is divisible by $5.$

Solution

Problem 3

Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$

Solution


Problem 4

In the array of $13$ squares shown below, $8$ squares are colored red, and the remaining $5$ squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated $90^{\circ}$ around the central square is $\frac{1}{n}$ , where $n$ is a positive integer. Find $n$.

[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)); draw((2,0)--(2,2)--(3,2)--(3,0)--(3,1)--(2,1)--(4,1)--(4,0)--(2,0)); draw((1,2)--(1,4)--(0,4)--(0,2)--(0,3)--(1,3)--(-1,3)--(-1,2)--(1,2)); draw((-1,1)--(-3,1)--(-3,0)--(-1,0)--(-2,0)--(-2,1)--(-2,-1)--(-1,-1)--(-1,1)); draw((0,-1)--(0,-3)--(1,-3)--(1,-1)--(1,-2)--(0,-2)--(2,-2)--(2,-1)--(0,-1)); size(100);[/asy]

Solution

Problem 5

The real root of the equation $8x^3-3x^2-3x-1=0$ can be written in the form $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.

Solution


Problem 6

Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution


Problem 7

A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$.

Solution


Problem 8

The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$. Find the remainder when the smallest possible sum $m+n$ is divided by $1000$.

Solution

Problem 9

A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.

[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]

Solution

Problem 10

There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$. For each possible combination of $a$ and $b$, let ${p}_{a,b}$ be the sum of the zeros of $P(x)$. Find the sum of the ${p}_{a,b}$'s for all possible combinations of $a$ and $b$.

Solution


Problem 11

Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$, of play blocks which satisfies the conditions:

(a) If $16$, $15$, or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and

(b) There are three integers $0 < x < y < z < 14$ such that when $x$, $y$, or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.

Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.

Solution

Problem 12

Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^o$ and $\angle Q = 60^o$. A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$, side $\overline{CD}$ lies on $\overline{QR}$, and one of the remaining vertices lies on $\overline{RP}$. There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$, where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$.

Solution

Problem 13

Triangle $AB_0C_0$ has side lengths $AB_0 = 12$, $B_0C_0 = 17$, and $C_0A = 25$. For each positive integer $n$, points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$, respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$. The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$, where $p$ and $q$ are relatively prime positive integers. Find $q$.

Solution

Problem 14

For $\pi \le \theta < 2\pi$, let

\[P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots\]

and

\[Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta +\ldots\]

so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$. Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 15

Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions:

(a) $0\le A<B<C\le99$,

(b) there exist integers $a$, $b$, and $c$, and prime $p$ where $0\le b<a<c<p$,

(c) $p$ divides $A-a$, $B-b$, and $C-c$, and

(d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$.

Solution

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
2012 AIME II Problems 		Followed by
2013 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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