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

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== 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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== Solution ==
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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>.
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Since <math>E</math> and <math>F</math> are midpoints, <math>BE = 15</math> and <math>CF = 7</math>.
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<math>B</math> and <math>C</math> are located on the circumference of the circle, so <math>OB = OC = 25</math>.
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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>.
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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>
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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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<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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Substituting for <math>a</math> and <math>b</math>, and applying the Cosine of Sum formula:
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<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>
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<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:
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<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>
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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>
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Combine terms again, and divide both sides by 64: <math>13 x^2 = 7200 + 15 \sqrt{x^4 - 976 x^2 + 230400}</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>
  
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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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 $\overline{AB}$ 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 expressed 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. ! Missing } inserted.)

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 $\fbox{057}$.

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