Question

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 $\text{Area}(PQRS) = \alpha w - \beta \cdot w^2.$ Then the coefficient $\beta = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

By Heron's formula with $s = 27,$ the area of $ABC$ is $\sqrt{27 \cdot 2 \cdot 10 \cdot 15} = \sqrt{8100} = 90,$ so the altitude from $A$ to $\overline{BC}$ has length $h = \frac{2 \cdot 90}{25} = \frac{36}{5}.$

Since $\overline{PQ} \parallel \overline{BC},$ triangle $APQ$ is similar to $ABC$ with ratio $\frac{w}{25},$ so the distance from $A$ down to line $PQ$ is $\frac{w}{25}h,$ and the rectangle's height is $PS = h - \frac{w}{25}h.$ The area is $w \cdot h\left(1 - \frac{w}{25}\right) = \frac{36}{5}\,w - \frac{36}{125}\,w^2.$

Thus $\beta = \frac{36}{125},$ and $m + n = 36 + 125 = 161.$

2015 AIME II Problem 7

Solution:

Problem 7 in Other Years