2011 AIME I Problems/Problem 2

Problem

In rectangle $ABCD$ , $AB=12$ and $BC=10$ . Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE=9$ , $DF=8$ , $\overline{BE}||\overline{DF}$ , $\overline{EF}||\overline{AB}$ , and line $BE$ intersects segment $\overline{AD}$ . The length $EF$ can be expressed in the form $m\sqrt{n}-p$ , where $m$ , $n$ , and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n+p$ .

Solution

Let us call the point where $\overline{EF}$ intersects $\overline{AD}$ point $G$ , and the point where $\overline{EF}$ intersects $\overline{BC}$ point $H$ . Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are similar. This implies that $\frac{BH}{GD} = \frac{9}{8}$ . Since $BC=10$ , $BH+GD=BH+HC=BC=10$ . ( $HC$ is the same as $GD$ because they are opposite sides of a rectangle.) Now, we have a system: