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Difference between revisions of "2015 AIME II Problems/Problem 7"

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==Solution==
 
==Solution==
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If <math>\omega = 25</math>,  the area of rectangle <math>PQRS</math> is <math>0</math>, so
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<cmath>\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0</cmath>
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and <math>\alpha = 25\beta</math>.  If <math>\omega = \frac{25}{2}</math>, we can reflect <math>APQ</math> over PQ, <math>PBS</math> over <math>PS</math>, and <math>QCR</math> over <math>QR</math> to completely cover rectangle <math>PQRS</math>, so the area of <math>PQRS</math> is half the area of the triangle.  Using Heron's formula, since <math>s = \frac{12 + 17 + 25}{2} = 27</math>,
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<cmath> [ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90</cmath>
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so
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<cmath>45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}</cmath>
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and
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<cmath>\beta = \frac{180}{625} = \frac{36}{125}</cmath>
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so the answer is <math>m + n = 36 + 125 = \boxed{161}</math>.

Revision as of 00:08, 27 March 2015

Problem

Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = w$, the area of $PQRS$ can be expressed as the quadratic polynomial

Area($PQRS$) = $\alpha w - \beta \cdot w^2$.

Then the coefficient $\beta = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

If $\omega = 25$, the area of rectangle $PQRS$ is $0$, so

\[\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0\]

and $\alpha = 25\beta$. If $\omega = \frac{25}{2}$, we can reflect $APQ$ over PQ, $PBS$ over $PS$, and $QCR$ over $QR$ to completely cover rectangle $PQRS$, so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$,

\[[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90\]

so

\[45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}\]

and

\[\beta = \frac{180}{625} = \frac{36}{125}\]

so the answer is $m + n = 36 + 125 = \boxed{161}$.

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