# Difference between revisions of "User:Rowechen"

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Here's the AIME compilation I will be doing: | Here's the AIME compilation I will be doing: | ||

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− | + | == Problem 1 == | |

− | == Problem | + | Maya lists all the positive divisors of <math>2010^2</math>. She then randomly selects two distinct divisors from this list. Let <math>p</math> be the probability that exactly one of the selected divisors is a perfect square. The probability <math>p</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>. |

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− | [[ | + | [[2010 AIME I Problems/Problem 1|Solution]] |

− | ==Problem | + | ==Problem 5== |

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− | [[ | + | Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>. |

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+ | |||

+ | [[2014 AIME II Problems/Problem 5|Solution]] | ||

+ | ==Problem 6== | ||

+ | Let <math>N</math> be the number of complex numbers <math>z</math> with the properties that <math>|z|=1</math> and <math>z^{6!}-z^{5!}</math> is a real number. Find the remainder when <math>N</math> is divided by <math>1000</math>. | ||

+ | |||

+ | [[2018 AIME I Problems/Problem 6 | Solution]] | ||

== Problem 7 == | == Problem 7 == | ||

− | + | Triangle <math>ABC</math> has <math>AB=21</math>, <math>AC=22</math> and <math>BC=20</math>. Points <math>D</math> and <math>E</math> are located on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, such that <math>\overline{DE}</math> is parallel to <math>\overline{BC}</math> and contains the center of the inscribed circle of triangle <math>ABC</math>. Then <math>DE=m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | |

− | [[ | + | [[2001 AIME I Problems/Problem 7|Solution]] |

− | == Problem | + | == Problem 9 == |

− | + | Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | |

− | [[ | + | [[2001 AIME II Problems/Problem 9|Solution]] |

== Problem 9 == | == Problem 9 == | ||

− | + | Let <math> ABC </math> be a triangle with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 rectangle. A segment is drawn to divide triangle <math> ABC </math> into a triangle <math> U_1 </math> and a trapezoid <math> V_1 </math> and another segment is drawn to divide rectangle <math> DEFG </math> into a triangle <math> U_2 </math> and a trapezoid <math> V_2 </math> such that <math> U_1 </math> is similar to <math> U_2 </math> and <math> V_1 </math> is similar to <math> V_2. </math> The minimum value of the area of <math> U_1 </math> can be written in the form <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math> | |

− | [[ | + | [[2004 AIME I Problems/Problem 9|Solution]] |

− | == Problem | + | == Problem 12 == |

− | + | Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is <math>8: 7</math>. Find the minimum possible value of their common perimeter. | |

− | [[ | + | [[2010 AIME II Problems/Problem 12|Solution]] |

− | == Problem | + | == 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. | |

− | [[ | + | [[2011 AIME I Problems/Problem 12|Solution]] |

== Problem 10 == | == Problem 10 == | ||

− | + | 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 == | ||

− | + | A frog begins at <math>P_0 = (0,0)</math> and makes a sequence of jumps according to the following rule: from <math>P_n = (x_n, y_n),</math> the frog jumps to <math>P_{n+1},</math> which may be any of the points <math>(x_n + 7, y_n + 2),</math> <math>(x_n + 2, y_n + 7),</math> <math>(x_n - 5, y_n - 10),</math> or <math>(x_n - 10, y_n - 5).</math> There are <math>M</math> points <math>(x, y)</math> with <math>|x| + |y| \le 100</math> that can be reached by a sequence of such jumps. Find the remainder when <math>M</math> is divided by <math>1000.</math> | |

− | < | + | [[2012 AIME I Problems/Problem 11|Solution]] |

− | < | + | == 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>. |

− | + | [[2011 AIME I Problems/Problem 14|Solution]] | |

+ | == 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. Find <math>a + b + c + d + e</math>. | ||

+ | [[2011 AIME II Problems/Problem 15|Solution]] | ||

+ | == Problem 13 == | ||

+ | Equilateral <math>\triangle ABC</math> has side length <math>\sqrt{111}</math>. There are four distinct triangles <math>AD_1E_1</math>, <math>AD_1E_2</math>, <math>AD_2E_3</math>, and <math>AD_2E_4</math>, each congruent to <math>\triangle ABC</math>, | ||

+ | with <math>BD_1 = BD_2 = \sqrt{11}</math>. Find <math>\sum_{k=1}^4(CE_k)^2</math>. | ||

− | [[ | + | [[2012 AIME II Problems/Problem 13|Solution]] |

− | == Problem | + | == Problem 14 == |

− | + | In a group of nine people each person shakes hands with exactly two of the other people from the group. Let <math>N</math> be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when <math>N</math> is divided by <math>1000</math>. | |

− | [[ | + | [[2012 AIME II Problems/Problem 14|Solution]] |

== Problem 14 == | == Problem 14 == | ||

− | + | For <math>\pi \le \theta < 2\pi</math>, let | |

− | + | <cmath> P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots | |

− | + | </cmath> | |

− | + | ||

+ | and | ||

− | + | <cmath> Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta | |

− | + | +\ldots </cmath> | |

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− | + | so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | |

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− | [[ | + | [[2013 AIME I Problems/Problem 14|Solution]] |

## Revision as of 08:15, 29 May 2020

Hey how did you get to this page? If you aren't me then I have to say hello. If you are me then I must be pretty conceited to waste my time looking at my own page. If you aren't me, seriously, how did you get to this page? This is pretty cool. Well, nice meeting you! I'm going to stop wasting my time typing this up and do some math. Gtg. Bye.

Here's the AIME compilation I will be doing:

## Contents

## Problem 1

Maya lists all the positive divisors of . She then randomly selects two distinct divisors from this list. Let be the probability that exactly one of the selected divisors is a perfect square. The probability can be expressed in the form , where and are relatively prime positive integers. Find .

## Problem 5

Real numbers and are roots of , and and are roots of . Find the sum of all possible values of .

## Problem 6

Let be the number of complex numbers with the properties that and is a real number. Find the remainder when is divided by .

## Problem 7

Triangle has , and . Points and are located on and , respectively, such that is parallel to and contains the center of the inscribed circle of triangle . Then , where and are relatively prime positive integers. Find .

## Problem 9

Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is , where and are relatively prime positive integers. Find .

## Problem 9

Let be a triangle with sides 3, 4, and 5, and be a 6-by-7 rectangle. A segment is drawn to divide triangle into a triangle and a trapezoid and another segment is drawn to divide rectangle into a triangle and a trapezoid such that is similar to and is similar to The minimum value of the area of can be written in the form where and are relatively prime positive integers. Find

## Problem 12

Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is . Find the minimum possible value of their common perimeter.

## Problem 12

Six men and some number of women stand in a line in random order. Let 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 does not exceed 1 percent.

## 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

A frog begins at and makes a sequence of jumps according to the following rule: from the frog jumps to which may be any of the points or There are points with that can be reached by a sequence of such jumps. Find the remainder when is divided by

## Problem 14

Let be a regular octagon. Let , , , and be the midpoints of sides , , , and , respectively. For , ray is constructed from towards the interior of the octagon such that , , , and . Pairs of rays and , and , and , and and meet at , , , respectively. If , then can be written in the form , where and are positive integers. Find .

## 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 .

## Problem 13

Equilateral has side length . There are four distinct triangles , , , and , each congruent to , with . Find .

## Problem 14

In a group of nine people each person shakes hands with exactly two of the other people from the group. Let be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when is divided by .

## Problem 14

For , let

and

so that . Then where and are relatively prime positive integers. Find .