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Difference between revisions of "2011 AIME I Problems"
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== Problem 1 ==
 
== Problem 1 ==
−
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is <math>k\%</math> acid. From jar C, <math>\frac{m}{n}</math> liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>.
+
Jar <math>A</math> contains four liters of a solution that is <math>45\%</math> acid. Jar <math>B</math> contains five liters of a solution that is <math>48\%</math> acid. Jar <math>C</math> contains one liter of a solution that is <math>k\%</math> acid. From jar <math>C</math>, <math>\frac{m}{n}</math> liters of the solution is added to jar <math>A</math>, and the remainder of the solution in jar <math>C</math> is added to jar B. At the end both jar <math>A</math> and jar <math>B</math> contain solutions that are <math>50\%</math> acid. Given that <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>.
  
 
[[2011 AIME I Problems/Problem 1|Solution]]
 
[[2011 AIME I Problems/Problem 1|Solution]]
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== Problem 7 ==
 
== Problem 7 ==
−
Find the number of positive integers <math>m</math> for which there exist nonnegative integers <math>x_0</math>, <math>x_1</math>, \dots, <math>x_{2011}</math> such that
+
Find the number of positive integers <math>m</math> for which there exist nonnegative integers <math>x_0</math>, <math>x_1</math> , <math>\dots</math> , <math>x_{2011}</math> such that
 
<cmath>m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.</cmath>
 
<cmath>m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.</cmath>
  
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<center><asy>
 
<center><asy>
 
unitsize(1 cm);
 
unitsize(1 cm);
−
 
 
pair translate;
 
pair translate;
 
pair[] A, B, C, U, V, W, X, Y, Z;
 
pair[] A, B, C, U, V, W, X, Y, Z;
−
 
 
A[0] = (1.5,2.8);
 
A[0] = (1.5,2.8);
 
B[0] = (3.2,0);
 
B[0] = (3.2,0);
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Y[0] = (0.69*B[0] + 0.31*C[0]);
 
Y[0] = (0.69*B[0] + 0.31*C[0]);
 
Z[0] = (0.69*B[0] + 0.31*A[0]);
 
Z[0] = (0.69*B[0] + 0.31*A[0]);
−
 
 
translate = (7,0);
 
translate = (7,0);
 
A[1] = (1.3,1.1) + translate;
 
A[1] = (1.3,1.1) + translate;
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Y[1] = Y[0] + translate;
 
Y[1] = Y[0] + translate;
 
Z[1] = Z[0] + translate;
 
Z[1] = Z[0] + translate;
−
 
 
draw (A[0]--B[0]--C[0]--cycle);
 
draw (A[0]--B[0]--C[0]--cycle);
 
draw (U[0]--V[0],dashed);
 
draw (U[0]--V[0],dashed);
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draw (W[1]--C[1]--X[1]);
 
draw (W[1]--C[1]--X[1]);
 
draw (Y[1]--B[1]--Z[1]);
 
draw (Y[1]--B[1]--Z[1]);
−
 
 
dot("$A$",A[0],N);
 
dot("$A$",A[0],N);
 
dot("$B$",B[0],SE);
 
dot("$B$",B[0],SE);
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dot("$X$",X[1],dir(-70));
 
dot("$X$",X[1],dir(-70));
 
dot("$Y$",Y[1],dir(250));
 
dot("$Y$",Y[1],dir(250));
−
dot("$Z$",Z[1],NE);
+
dot("$Z$",Z[1],NE);</asy></center>
−
</asy></center>
 
  
 
[[2011 AIME I Problems/Problem 8|Solution]]
 
[[2011 AIME I Problems/Problem 8|Solution]]
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== Problem 11 ==
 
== Problem 11 ==
−
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by 1000. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by 1000.
+
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by 1000. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by 1000.
  
 
[[2011 AIME I Problems/Problem 11|Solution]]
 
[[2011 AIME I Problems/Problem 11|Solution]]
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== Problem 13 ==
 
== Problem 13 ==
−
A cube with side length 10 is suspended above a plane.  The vertex closest to the plane is labeled <math>A</math>. The three vertices adjacent to vertex <math>A</math> are at heights 10, 11, and 12 above the plane.  The distance from vertex <math>A</math> to the plane can be expressed as <math>\frac{r - \sqrt{s}}{t}</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers. Find <math>r + s + t</math>.
+
A cube with side length 10 is suspended above a plane.  The vertex closest to the plane is labeled <math>A</math>. The three vertices adjacent to vertex <math>A</math> are at heights 10, 11, and 12 above the plane.  The distance from vertex <math>A</math> to the plane can be expressed as <math>\frac{r - \sqrt{s}}{t}</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers. Find <math>r + s + t</math>.
  
 
[[2011 AIME I Problems/Problem 13|Solution]]
 
[[2011 AIME I Problems/Problem 13|Solution]]
  
 
== Problem 14 ==
 
== Problem 14 ==
−
Let <math>A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8</math> be a regular octagon.  Let <math>M_1</math>, <math>M_3</math>, <math>M_5</math>, and <math>M_7</math> be the midpoints of sides <math>\overline{A_1 A_2}</math>, <math>\overline{A_3 A_4}</math>, <math>\overline{A_5 A_6}</math>, and <math>\overline{A_7 A_8}</math>, respectively.  For <math>i = 1</math>, 3, 5, 7, ray <math>R_i</math> is constructed from <math>M_i</math> towards the interior of the octagon such that <math>R_1 \perp R_3</math>, <math>R_3 \perp R_5</math>, <math>R_5 \perp R_7</math>, and <math>R_7 \perp R_1</math>.  Pairs of rays <math>R_1</math> and <math>R_3</math>, <math>R_3</math> and <math>R_5</math>, <math>R_5</math> and <math>R_7</math>, and <math>R_7</math> and <math>R_1</math> meet at <math>B_1</math>, <math>B_3</math>, <math>B_5</math>, <math>B_7</math> respectively.  If <math>B_1 B_3 = A_1 A_2</math>, then <math>\cos 2 \angle A_3 M_3 B_1</math> can be written in the form <math>m - \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers.  Find <math>m + n</math>.
+
Let <math>A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8</math> be a regular octagon.  Let <math>M_1</math>, <math>M_3</math>, <math>M_5</math>, and <math>M_7</math> be the midpoints of sides <math>\overline{A_1 A_2}</math>, <math>\overline{A_3 A_4}</math>, <math>\overline{A_5 A_6}</math>, and <math>\overline{A_7 A_8}</math>, respectively.  For <math>i = 1, 3, 5, 7</math>, ray <math>R_i</math> is constructed from <math>M_i</math> towards the interior of the octagon such that <math>R_1 \perp R_3</math>, <math>R_3 \perp R_5</math>, <math>R_5 \perp R_7</math>, and <math>R_7 \perp R_1</math>.  Pairs of rays <math>R_1</math> and <math>R_3</math>, <math>R_3</math> and <math>R_5</math>, <math>R_5</math> and <math>R_7</math>, and <math>R_7</math> and <math>R_1</math> meet at <math>B_1</math>, <math>B_3</math>, <math>B_5</math>, <math>B_7</math> respectively.  If <math>B_1 B_3 = A_1 A_2</math>, then <math>\cos 2 \angle A_3 M_3 B_1</math> can be written in the form <math>m - \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers.  Find <math>m + n</math>.
  
 
[[2011 AIME I Problems/Problem 14|Solution]]
 
[[2011 AIME I Problems/Problem 14|Solution]]
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== See also ==
 
== See also ==
 +
{{AIME box|year=2011|n=I|before=[[2010 AIME II Problems]]|after=[[2011 AIME II Problems]]}}
 
* [[American Invitational Mathematics Examination]]
 
* [[American Invitational Mathematics Examination]]
 
* [[AIME Problems and Solutions]]
 
* [[AIME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 21:21, 28 December 2023

2011 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

Jar $A$ contains four liters of a solution that is $45\%$ acid. Jar $B$ contains five liters of a solution that is $48\%$ acid. Jar $C$ contains one liter of a solution that is $k\%$ acid. From jar $C$, $\frac{m}{n}$ liters of the solution is added to jar $A$, and the remainder of the solution in jar $C$ is added to jar B. At the end both jar $A$ and jar $B$ contain solutions that are $50\%$ acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$.

Solution

Problem 2

In rectangle $ABCD$, $AB = 12$ and $BC = 10$. Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$, $DF = 8$, $\overline{BE} \parallel \overline{DF}$, $\overline{EF} \parallel \overline{AB}$, and line $BE$ intersects segment $\overline{AD}$. The length $EF$ can be expressed in the form $m \sqrt{n} - p$, where $m$, $n$, and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n + p$.

Solution

Problem 3

Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A = (24,-1)$, and let $M$ be the line perpendicular to line $L$ that contains the point $B = (5,6)$. The original coordinate axes are erased, and line $L$ is made the $x$-axis and line $M$ the $y$-axis. In the new coordinate system, point $A$ is on the positive $x$-axis, and point $B$ is on the positive $y$-axis. The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system. Find $\alpha + \beta$.

Solution

Problem 4

In triangle $ABC$, $AB = 125$, $AC = 117$, and $BC = 120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$.

Solution

Problem 5

The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements.

Solution

Problem 6

Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$, where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

Solution

Problem 7

Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$, $x_1$ , $\dots$ , $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\]

Solution

Problem 8

In triangle $ABC$, $BC = 23$, $CA = 27$, and $AB = 30$. Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$, points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$, and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$. In addition, the points are positioned so that $\overline{UV} \parallel \overline{BC}$, $\overline{WX} \parallel \overline{AB}$, and $\overline{YZ} \parallel \overline{CA}$. Right angle folds are then made along $\overline{UV}$, $\overline{WX}$, and $\overline{YZ}$. The resulting figure is placed on a level floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\frac{k \sqrt{m}}{n}$, where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k + m + n$.

[asy] unitsize(1 cm); pair translate; pair[] A, B, C, U, V, W, X, Y, Z; A[0] = (1.5,2.8); B[0] = (3.2,0); C[0] = (0,0); U[0] = (0.69*A[0] + 0.31*B[0]); V[0] = (0.69*A[0] + 0.31*C[0]); W[0] = (0.69*C[0] + 0.31*A[0]); X[0] = (0.69*C[0] + 0.31*B[0]); Y[0] = (0.69*B[0] + 0.31*C[0]); Z[0] = (0.69*B[0] + 0.31*A[0]); translate = (7,0); A[1] = (1.3,1.1) + translate; B[1] = (2.4,-0.7) + translate; C[1] = (0.6,-0.7) + translate; U[1] = U[0] + translate; V[1] = V[0] + translate; W[1] = W[0] + translate; X[1] = X[0] + translate; Y[1] = Y[0] + translate; Z[1] = Z[0] + translate; draw (A[0]--B[0]--C[0]--cycle); draw (U[0]--V[0],dashed); draw (W[0]--X[0],dashed); draw (Y[0]--Z[0],dashed); draw (U[1]--V[1]--W[1]--X[1]--Y[1]--Z[1]--cycle); draw (U[1]--A[1]--V[1],dashed); draw (W[1]--C[1]--X[1]); draw (Y[1]--B[1]--Z[1]); dot("$A$",A[0],N); dot("$B$",B[0],SE); dot("$C$",C[0],SW); dot("$U$",U[0],NE); dot("$V$",V[0],NW); dot("$W$",W[0],NW); dot("$X$",X[0],S); dot("$Y$",Y[0],S); dot("$Z$",Z[0],NE); dot(A[1]); dot(B[1]); dot(C[1]); dot("$U$",U[1],NE); dot("$V$",V[1],NW); dot("$W$",W[1],NW); dot("$X$",X[1],dir(-70)); dot("$Y$",Y[1],dir(250)); dot("$Z$",Z[1],NE);[/asy]

Solution

Problem 9

Suppose $x$ is in the interval $[0,\pi/2]$ and $\log_{24 \sin x} (24 \cos x) = \frac{3}{2}$. Find $24 \cot^2 x$.

Solution

Problem 10

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular $n$-gon determine an obtuse triangle is $\frac{93}{125}$. Find the sum of all possible values of $n$.

Solution

Problem 11

Let $R$ be the set of all possible remainders when a number of the form $2^n$, $n$ a nonnegative integer, is divided by 1000. Let $S$ be the sum of the elements in $R$. Find the remainder when $S$ is divided by 1000.

Solution

Problem 12

Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent.

Solution

Problem 13

A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r - \sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r + s + t$.

Solution

Problem 14

Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$, $M_3$, $M_5$, and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$, $\overline{A_3 A_4}$, $\overline{A_5 A_6}$, and $\overline{A_7 A_8}$, respectively. For $i = 1, 3, 5, 7$, ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$, $R_3 \perp R_5$, $R_5 \perp R_7$, and $R_7 \perp R_1$. Pairs of rays $R_1$ and $R_3$, $R_3$ and $R_5$, $R_5$ and $R_7$, and $R_7$ and $R_1$ meet at $B_1$, $B_3$, $B_5$, $B_7$ respectively. If $B_1 B_3 = A_1 A_2$, then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$, where $m$ and $n$ are positive integers. Find $m + n$.

Solution

Problem 15

For some integer $m$, the polynomial $x^3 - 2011x + m$ has the three integer roots $a$, $b$, and $c$. Find $|a| + |b| + |c|$.

Solution

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
2010 AIME II Problems 		Followed by
2011 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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