# Difference between revisions of "2011 AIME II Problems/Problem 10"

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− | Problem: | + | == Problem 10 == |

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

+ | |||

+ | == Solution == | ||

+ | Let <math>E</math> and <math>F</math> be the midpoints of <math>\overline{AB}</math> and <math>\overline{CD}</math>, respectively, such that <math>\overline{BE}</math> intersects <math>\overline{CF}</math>. | ||

+ | |||

+ | Since <math>E</math> and <math>F</math> are midpoints, <math>BE = 15</math> and <math>CF = 7</math>. | ||

+ | |||

+ | <math>B</math> and <math>C</math> are located on the circumference of the circle, so <math>OB = OC = 25</math>. | ||

+ | |||

+ | The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so <math>\triangle OEB</math> and <math>\triangle OFC</math> are right triangles (with <math>\angle OEB</math> and <math>\angle OFC</math> being the right angles). By the Pythagorean Theorem, <math>OE = \sqrt{25^2 - 15^2} = 20</math>, and <math>OF = \sqrt{25^2 - 7^2} = 24</math>. | ||

+ | |||

+ | Let <math>x</math>, <math>a</math>, and <math>b</math> be lengths <math>OP</math>, <math>EP</math>, and <math>FP</math>, respectively. OEP and OFP are also right triangles, so <math>x^2 = a^2 + 20^2 \to a^2 = x^2 - 400</math>, and <math>x^2 = b^2 + 24^2 \to b^2 = x^2 - 576</math> | ||

+ | |||

+ | We are given that <math>EF</math> has length 12, so, using the Law of Cosines with <math>\triangle EPF</math>: | ||

+ | |||

+ | <math>12^2 = a^2 + b^2 - 2ab \cos (\angle EPF) = a^2 + b^2 - 2ab \cos (\angle EPO + \angle FPO)</math> | ||

+ | |||

+ | Substituting for <math>a</math> and <math>b</math>, and applying the Cosine of Sum formula: | ||

+ | |||

+ | <math>144 = (x^2 - 400) + (x^2 - 576) + 2 \sqrt{x^2 - 400} \sqrt{x^2 - 576} \left( \cos \angle EPO \cos \angle FPO - \sin \angle EPO \sin \angle FPO \right)</math> | ||

+ | |||

+ | <math>\angle EPO</math> and <math>\angle FPO</math> are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines: | ||

+ | |||

+ | <math>144 = 2x^2 - 976 + 2 \sqrt{(x^2 - 400)(x^2 - 576)} \left(\frac{\sqrt{x^2 - 400}}{x} \frac{\sqrt{x^2 - 576}}{x} - \frac{20}{x} \frac{24}{x} \right)</math> | ||

+ | |||

+ | Combine terms and multiply both sides by <math>x^2</math>: <math>144 x^2 = 2 x^4 - 976 x^2 - 2 (x^2 - 400) (x^2 - 576) + 960 \sqrt{(x^2 - 400)(x^2 - 576)</math> | ||

+ | |||

+ | Combine terms again, and divide both sides by 64: <math>13 x^2 = 7200 + 15 \sqrt{x^4 - 976 x^2 + 230400}</math> | ||

+ | |||

+ | Square both sides: <math>169 x^4 - 187000 x^2 + 51,840,000 = 225 x^4 - 219600 x^2 + 51840000</math> | ||

− | + | This reduces to <math>x^2 = \frac{4050}{7} = (OP)^2</math>; (4050 + 7) divided by 1000 has remainder <math>\fbox{057}</math>. |

## Revision as of 10:10, 7 April 2011

## Problem 10

A circle with center O has radius 25. Chord of length 30 and chord of length 14 intersect at point P. The distance between the midpoints of the two chords is 12. The quantity can be expressed as , where m and n are relatively prime positive integers. Find the remainder when m + n is divided by 1000.

## Solution

Let and be the midpoints of and , respectively, such that intersects .

Since and are midpoints, and .

and are located on the circumference of the circle, so .

The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so and are right triangles (with and being the right angles). By the Pythagorean Theorem, , and .

Let , , and be lengths , , and , respectively. OEP and OFP are also right triangles, so , and

We are given that has length 12, so, using the Law of Cosines with :

Substituting for and , and applying the Cosine of Sum formula:

and are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines:

Combine terms and multiply both sides by : $144 x^2 = 2 x^4 - 976 x^2 - 2 (x^2 - 400) (x^2 - 576) + 960 \sqrt{(x^2 - 400)(x^2 - 576)$ (Error compiling LaTeX. ! Missing } inserted.)

Combine terms again, and divide both sides by 64:

Square both sides:

This reduces to ; (4050 + 7) divided by 1000 has remainder .