2011 AIME II Problems

2011 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

Note: All questions with a star after the problem number are not yet the correct problem, as I copy/pasted the format from the 2011 AIME I page.

Problem 1

Gary purchased a large bevarage, but only drank m/n of it, where m and n are relatively prime positive integers. If he had purchased half as much and drank twice as much, he would have wasted only 2/9 as much bevarage. Find m+n.

Solution

Problem 2

On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD.

Solution

Problem 3

The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.

Solution

Problem 4

In triangle ABC, AB=(20/11)AC. The angle bisector of angle A intersects BC at point D, and point M is the midpoint of AD. Let P be the point of intersection of AC and the line BM. The ratio of CP to PA can be expresses in the form m/n, where m and n are relatively prime positive integers. Find m+n.

Solution

Problem 5

The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 terms.

Solution

Problem 6

Define an ordered quadruple (a, b, c, d) as interesting if $1 \le a<b<c<d \le 10$ , and a+d>b+c. How many ordered quadruples are there?

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 $k + m + n$ .

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);
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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 $m + n$ .