Difference between revisions of "2011 AIME II Problems"
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{{AIME Problems|year=2011|n=II}} | {{AIME Problems|year=2011|n=II}} | ||
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== Problem 1 == | == Problem 1 == | ||
− | Gary purchased a large | + | 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 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]] | [[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]] | ||
== Problem 7 == | == 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 | + | 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]] | [[2011 AIME II Problems/Problem 7|Solution]] | ||
== 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 \frac{1}{540}</math>. Let <math>p</math> and <math>q</math> be | + | 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>. |
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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 | + | 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]] | ||
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+ | == 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 08:43, 11 October 2020
2011 AIME II (Answer Key) | AoPS Contest Collections • PDF | ||
Instructions
| ||
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.