Difference between revisions of "2011 AIME II Problems"

Latest revision as of 07:43, 11 October 2020

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

Problem 1

Gary purchased a large beverage, but only drank $m/n$ of it, where $m$ and $n$ are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only $2/9$ as much beverage. 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 $ABC$ , $AB=20$ and $AC=11$ . 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 expressed in the form $\frac{m}{n}$ , 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 of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there?

Solution

Problem 7

Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let $m$ be the maximum number of red marbles for which such an arrangement is possible, and let $N$ be the number of ways he can arrange the $m+5$ marbles to satisfy the requirement. Find the remainder when $N$ is divided by $1000$ .

Solution

Problem 8

Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ .

Solution

Problem 9

Let $x_1$ , $x_2$ , $\dots$ , $x_6$ be nonnegative real numbers such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 1$ , and $x_1x_3x_5 + x_2x_4x_6 \ge {\frac{1}{540}}$ . Let $p$ and $q$ be relatively prime positive integers such that $\frac{p}{q}$ is the maximum possible value of $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_6 + x_5x_6x_1 + x_6x_1x_2$ . Find $p + q$ .

Solution

Problem 10

A circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000.

Solution

Problem 11

Let $M_n$ be the $n \times n$ matrix with entries as follows: for $1 \le i \le n$ , $m_{i,i} = 10$ ; for $1 \le i \le n - 1$ , $m_{i+1,i} = m_{i,i+1} = 3$ ; all other entries in $M_n$ are zero. Let $D_n$ be the determinant of matrix $M_n$ . Then $\sum_{n=1}^{\infty} \frac{1}{8D_n+1}$ can be represented as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ .

Note: The determinant of the $1 \times 1$ matrix $[a]$ is $a$ , and the determinant of the $2 \times 2$ matrix $\left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] = ad - bc$ ; for $n \ge 2$ , the determinant of an $n \times n$ matrix with first row or first column $a_1$ $a_2$ $a_3$ $\dots$ $a_n$ is equal to $a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n$ , where $C_i$ is the determinant of the $(n - 1) \times (n - 1)$ matrix formed by eliminating the row and column containing $a_i$ .

Solution

Problem 12

Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

Problem 13

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ , respectively. Given that $AB = 12$ and $\angle O_1PO_2 = 120 ^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ .

Solution

Problem 14

There are $N$ permutations $(a_1, a_2, \dots, a_{30})$ of $1, 2, \dots, 30$ such that for $m \in \{2,3,5\}$ , $m$ divides $a_{n+m} - a_n$ for all integers $n$ with $1 \le n < n+m \le 30$ . Find the remainder when $N$ is divided by 1000.

Solution

Problem 15

Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\left\lfloor\sqrt{P(x)}\right\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ , $b$ , $c$ , $d$ , and $e$ are positive integers. Find $a + b + c + d + e$ .

Solution

@@ Line 1: / Line 1: @@
 {{AIME Problems|year=2011|n=II}}
-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''.
+Gary purchased a large beverage, but only drank <math>m/n</math> of it, where <math>m</math> and <math>n</math> are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only <math>2/9</math> as much beverage. Find <math>m+n</math>.
 [[2011 AIME II Problems/Problem 1|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.
+On square <math>ABCD</math>, point <math>E</math> lies on side <math>AD</math> and point <math>F</math> lies on side <math>BC</math>, so that <math>BE=EF=FD=30</math>. Find the area of the square <math>ABCD</math>.
 [[2011 AIME II Problems/Problem 2|Solution]]
@@ Line 19: / Line 17: @@
 == 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.
+In triangle <math>ABC</math>, <math>AB=20</math> and <math>AC=11</math>. The angle bisector of angle <math>A</math> intersects <math>BC</math> at point <math>D</math>, and point <math>M</math> is the midpoint of <math>AD</math>. Let <math>P</math> be the point of intersection of <math>AC</math> and the line <math>BM</math>. The ratio of <math>CP</math> to <math>PA</math> can be expressed in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 [[2011 AIME II Problems/Problem 4|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.
+The sum of the first <math>2011</math> terms of a geometric sequence is <math>200</math>. The sum of the first <math>4022</math> terms is <math>380</math>. Find the sum of the first <math>6033</math> terms.
 [[2011 AIME II Problems/Problem 5|Solution]]
 == Problem 6 ==
-Define an ordered quadruple (a, b, c, d) as interesting if <math>1 \le a<b<c<d \le 10</math>, and a+d>b+c. How many ordered quadruples are there?
+Define an ordered quadruple of integers <math>(a, b, c, d)</math> as ''interesting'' if <math>1 \le a<b<c<d \le 10</math>, and <math> a+d>b+c </math>. How many interesting ordered quadruples are there?
 [[2011 AIME II Problems/Problem 6|Solution]]
 == Problem 7 ==
-Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal tot he number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let ''m'' be the maximum number of red marbles for which such an arrangement is possible, and let N be the number of ways he can arrange the ''m''+5 marbles to satisfy the requirement. Find the remainder when N is divided by 1000.
+Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let <math>m</math> be the maximum number of red marbles for which such an arrangement is possible, and let <math>N</math> be the number of ways he can arrange the <math>m+5</math> marbles to satisfy the requirement. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 [[2011 AIME II Problems/Problem 7|Solution]]
-== Problem 8* ==
+== Problem 8 ==
-In triangle <math>ABC</math>, <math>BC = 23</math>, <math>CA = 27</math>, and <math>AB = 30</math>.  Points <math>V</math> and <math>W</math> are on <math>\overline{AC}</math> with <math>V</math> on <math>\overline{AW}</math>, points <math>X</math> and <math>Y</math> are on <math>\overline{BC}</math> with <math>X</math> on <math>\overline{CY}</math>, and points <math>Z</math> and <math>U</math> are on <math>\overline{AB}</math> with <math>Z</math> on <math>\overline{BU}</math>.  In addition, the points are positioned so that <math>\overline{UV} \parallel \overline{BC}</math>, <math>\overline{WX} \parallel \overline{AB}</math>, and <math>\overline{YZ} \parallel \overline{CA}</math>.  Right angle folds are then made along <math>\overline{UV}</math>, <math>\overline{WX}</math>, and <math>\overline{YZ}</math>.  The resulting figure is placed on a level floor to make a table with triangular legs.  Let <math>h</math> be the maximum possible height of a table constructed from triangle <math>ABC</math> whose top is parallel to the floor.  Then <math>h</math> can be written in the form <math>\frac{k \sqrt{m}}{n}</math>, where <math>k</math> and <math>n</math> are relatively prime positive integers and <math>m</math> is a positive integer that is not divisible by the square of any prime.  Find <math>k + m + n</math>.
+Let <math>z_1,z_2,z_3,\dots,z_{12}</math> be the 12 zeroes of the polynomial <math>z^{12}-2^{36}</math>. For each <math>j</math>, let <math>w_j</math> be one of <math>z_j</math> or <math>i z_j</math>. Then the maximum possible value of the real part of <math>\sum_{j=1}^{12} w_j</math> can be written as <math>m+\sqrt{n}</math> where <math>m</math> and <math>n</math> are positive integers. Find <math>m+n</math>.
-<center><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></center>
 [[2011 AIME II Problems/Problem 8|Solution]]
-== Problem 9* ==
+== Problem 9 ==
-Suppose <math>x</math> is in the interval <math>[0,\pi/2]</math> and <math>\log_{24 \sin x} (24 \cos x) = \frac{3}{2}</math>.  Find <math>24 \cot^2 x</math>.
+Let <math>x_1</math>, <math>x_2</math>, <math>\dots</math>, <math>x_6</math> be nonnegative real numbers such that <math>x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 1</math>, and <math>x_1x_3x_5 + x_2x_4x_6 \ge {\frac{1}{540}}</math>. Let <math>p</math> and <math>q</math> be relatively prime positive integers such that <math>\frac{p}{q}</math> is the maximum possible value of <math>x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_6 + x_5x_6x_1 + x_6x_1x_2</math>. Find <math>p + q</math>.
 [[2011 AIME II Problems/Problem 9|Solution]]
-== Problem 10* ==
+== Problem 10 ==
-The probability that a set of three distinct vertices chosen at random from among the vertices of a regular <math>n</math>-gon determine an obtuse triangle is <math>\frac{93}{125}</math>.  Find the sum of all possible values of <math>n</math>.
+A circle with center <math>O</math> has radius 25. Chord <math>\overline{AB}</math> of length 30 and chord <math>\overline{CD}</math> of length 14 intersect at point <math>P</math>. The distance between the midpoints of the two chords is 12. The quantity <math>OP^2</math> can be represented as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find the remainder when <math>m + n</math> is divided by 1000.
 [[2011 AIME II Problems/Problem 10|Solution]]
-== 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>M_n</math> be the <math>n \times n</math> matrix with entries as follows: for <math>1 \le i \le n</math>, <math>m_{i,i} = 10</math>; for <math>1 \le i \le n - 1</math>, <math>m_{i+1,i} = m_{i,i+1} = 3</math>; all other entries in <math>M_n</math> are zero. Let <math>D_n</math> be the determinant of matrix <math>M_n</math>. Then <math>\sum_{n=1}^{\infty} \frac{1}{8D_n+1}</math> can be represented as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p + q</math>.
+Note: The determinant of the <math>1 \times 1</math> matrix <math>[a]</math> is <math>a</math>, and the determinant of the <math>2 \times 2</math> matrix <math>\left[ {\begin{array}{cc}
+ a & b  \\
+ c & d  \\
+  \end{array} } \right] = ad - bc</math>; for <math>n \ge 2</math>, the determinant of an <math>n \times n</math> matrix with first row or first column <math>a_1</math> <math>a_2</math> <math>a_3</math> <math>\dots</math> <math>a_n</math> is equal to <math>a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n</math>, where <math>C_i</math> is the determinant of the <math>(n - 1) \times (n - 1)</math> matrix formed by eliminating the row and column containing <math>a_i</math>.
 [[2011 AIME II Problems/Problem 11|Solution]]
-== Problem 12* ==
+== Problem 12 ==
-Six men and some number of women stand in a line in random order.  Let <math>p</math> 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 <math>p</math> does not exceed 1 percent.
+Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
 [[2011 AIME II Problems/Problem 12|Solution]]
-== 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>.
+Point <math>P</math> lies on the diagonal <math>AC</math> of square <math>ABCD</math> with <math>AP > CP</math>. Let <math>O_1</math> and <math>O_2</math> be the circumcenters of triangles <math>ABP</math> and <math>CDP</math>, respectively. Given that <math>AB = 12</math> and <math>\angle O_1PO_2 = 120 ^{\circ}</math>, then <math>AP = \sqrt{a} + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers. Find <math>a + b</math>.
 [[2011 AIME II 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, 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>.
+There are <math>N</math> permutations <math>(a_1, a_2, \dots, a_{30})</math> of <math>1, 2, \dots, 30</math> such that for <math>m \in \{2,3,5\}</math>, <math>m</math> divides <math>a_{n+m} - a_n</math> for all integers <math>n</math> with <math>1 \le n < n+m \le 30</math>. Find the remainder when <math>N</math> is divided by 1000.
 [[2011 AIME II Problems/Problem 14|Solution]]
-== Problem 15* ==
+== Problem 15 ==
-For some integer <math>m</math>, the polynomial <math>x^3 - 2011x + m</math> has the three integer roots <math>a</math>, <math>b</math>, and <math>c</math>.  Find <math>|a| + |b| + |c|</math>.
+Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\left\lfloor\sqrt{P(x)}\right\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>.
 [[2011 AIME II Problems/Problem 15|Solution]]
+== See also ==
+{{AIME box|year=2011|n=II|before=[[2011 AIME I Problems]]|after=[[2012 AIME I Problems]]}}
+* [[American Invitational Mathematics Examination]]
+* [[AIME Problems and Solutions]]
+* [[Mathematics competition resources]]
+{{MAA Notice}}

2011 AIME II (Problems • Answer Key • Resources)
Preceded by 2011 AIME I Problems	Followed by 2012 AIME I Problems
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "2011 AIME II Problems"

Latest revision as of 07:43, 11 October 2020

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also