Difference between revisions of "2011 AIME II Problems"
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== Problem 1 == | == Problem 1 == | ||
− | Gary purchased a large beverage, but only drank | + | 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]] | [[2011 AIME II Problems/Problem 1|Solution]] | ||
== Problem 2 == | == Problem 2 == | ||
− | On square ABCD, point E lies on side AD and point F lies on side BC, so that <math>BE=EF=FD=30</math>. 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]] | [[2011 AIME II Problems/Problem 2|Solution]] | ||
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== Problem 4 == | == Problem 4 == | ||
− | In triangle ABC, AB= | + | 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]] | [[2011 AIME II Problems/Problem 4|Solution]] | ||
== Problem 5 == | == 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]] | [[2011 AIME II Problems/Problem 5|Solution]] | ||
== Problem 6 == | == 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 interesting 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]] | [[2011 AIME II Problems/Problem 6|Solution]] | ||
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== Problem 8 == | == Problem 8 == | ||
− | Let <math>z_1 | + | 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>. |
[[2011 AIME II Problems/Problem 8|Solution]] | [[2011 AIME II Problems/Problem 8|Solution]] | ||
== Problem 9 == | == Problem 9 == | ||
− | 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 { | + | 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]] | [[2011 AIME II Problems/Problem 9|Solution]] | ||
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== Problem 11 == | == Problem 11 == | ||
− | 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>. | + | 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} | 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 \\ | a & b \\ | ||
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== Problem 13 == | == Problem 13 == | ||
− | 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\ | + | 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]] | [[2011 AIME II Problems/Problem 13|Solution]] | ||
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== Problem 15 == | == Problem 15 == | ||
− | 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>\lfloor\sqrt{P(x)}\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 | + | 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]] | [[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}} |
Latest revision as of 07:43, 11 October 2020
2011 AIME II (Answer Key) | AoPS Contest Collections • PDF | ||
Instructions
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1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |
Contents
Problem 1
Gary purchased a large beverage, but only drank of it, where and are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only as much beverage. Find .
Problem 2
On square , point lies on side and point lies on side , so that . Find the area of the square .
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.
Problem 4
In triangle , and . The angle bisector of angle intersects at point , and point is the midpoint of . Let be the point of intersection of and the line . The ratio of to can be expressed in the form , where and are relatively prime positive integers. Find .
Problem 5
The sum of the first terms of a geometric sequence is . The sum of the first terms is . Find the sum of the first terms.
Problem 6
Define an ordered quadruple of integers as interesting if , and . How many interesting ordered quadruples are there?
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 be the maximum number of red marbles for which such an arrangement is possible, and let be the number of ways he can arrange the marbles to satisfy the requirement. Find the remainder when is divided by .
Problem 8
Let be the 12 zeroes of the polynomial . For each , let be one of or . Then the maximum possible value of the real part of can be written as where and are positive integers. Find .
Problem 9
Let , , , be nonnegative real numbers such that , and . Let and be relatively prime positive integers such that is the maximum possible value of . Find .
Problem 10
A circle with center has radius 25. Chord of length 30 and chord of length 14 intersect at point . The distance between the midpoints of the two chords is 12. The quantity can be represented as , where and are relatively prime positive integers. Find the remainder when is divided by 1000.
Problem 11
Let be the matrix with entries as follows: for , ; for , ; all other entries in are zero. Let be the determinant of matrix . Then can be represented as , where and are relatively prime positive integers. Find .
Note: The determinant of the matrix is , and the determinant of the matrix ; for , the determinant of an matrix with first row or first column is equal to , where is the determinant of the matrix formed by eliminating the row and column containing .
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 , where and are relatively prime positive integers. Find .
Problem 13
Point lies on the diagonal of square with . Let and be the circumcenters of triangles and , respectively. Given that and , then , where and are positive integers. Find .
Problem 14
There are permutations of such that for , divides for all integers with . Find the remainder when is divided by 1000.
Problem 15
Let . A real number is chosen at random from the interval . The probability that is equal to , where , , , , and are positive integers. Find .
See also
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 |
- American Invitational Mathematics Examination
- AIME Problems and Solutions
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.