2015 AIME II Problems/Problem 7

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 $m + n = 36 + 125 = \boxed{161}$ .

Solution #2

Similar triangles can also solve the problem. First, solve for the area of the triangle. $[ABC] = 90$ . This can be done by Heron's Formula or placing an 8-15-17 right triangle on $BC$ and solving. (The 8 side would be collinear with line $AB$ )

2015 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2015 AIME II Problems/Problem 7

Contents

Problem

Solution

Solution #2

See also