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

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== Problem 10 ==
 
== Problem 10 ==
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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.
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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 [[midpoint]]s 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.
  
 
== Solution ==
 
== Solution ==
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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>
 
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>
  
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We are given that <math>EF</math> has length 12, so, using the Law of Cosines with <math>\triangle EPF</math>:
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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>
 
<math>12^2 = a^2 + b^2 - 2ab \cos (\angle EPF) = a^2 + b^2 - 2ab \cos (\angle EPO + \angle FPO)</math>
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Square both sides: <math>169 x^4 - 187000 x^2 + 51,840,000 = 225 x^4 - 219600 x^2 + 51840000</math>
 
Square both sides: <math>169 x^4 - 187000 x^2 + 51,840,000 = 225 x^4 - 219600 x^2 + 51840000</math>
  
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This reduces to <math>x^2 = \frac{4050}{7} = (OP)^2</math>; (4050 + 7) divided by 1000 has remainder <math>\fbox{057}</math>.
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This reduces to <math>x^2 = \frac{4050}{7} = (OP)^2</math>; (4050 + 7) divided by 1000 has remainder <math>\boxed{057}</math>.
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==See also==
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{{AIME box|year=2011|n=II|num-b=9|num-a=11}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 21:32, 22 August 2011

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

Let $E$ and $F$ be the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively, such that $\overline{BE}$ intersects $\overline{CF}$.

Since $E$ and $F$ are midpoints, $BE = 15$ and $CF = 7$.

$B$ and $C$ are located on the circumference of the circle, so $OB = OC = 25$.

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

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

We are given that $EF$ has length 12, so, using the Law of Cosines with $\triangle EPF$:

$12^2 = a^2 + b^2 - 2ab \cos (\angle EPF) = a^2 + b^2 - 2ab \cos (\angle EPO + \angle FPO)$

Substituting for $a$ and $b$, and applying the Cosine of Sum formula:

$144 = (x^2 - 400) + (x^2 - 576) + 2 \sqrt{x^2 - 400} \sqrt{x^2 - 576} \left( \cos \angle EPO \cos \angle FPO - \sin \angle EPO \sin \angle FPO \right)$

$\angle EPO$ and $\angle FPO$ are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines:

$144 = 2x^2 - 976 + 2 \sqrt{(x^2 - 400)(x^2 - 576)} \left(\frac{\sqrt{x^2 - 400}}{x} \frac{\sqrt{x^2 - 576}}{x} - \frac{20}{x} \frac{24}{x} \right)$

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

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

Square both sides: $169 x^4 - 187000 x^2 + 51,840,000 = 225 x^4 - 219600 x^2 + 51840000$

This reduces to $x^2 = \frac{4050}{7} = (OP)^2$; (4050 + 7) divided by 1000 has remainder $\boxed{057}$.

See also

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