# 2006 iTest Problems/Problem 24

## Problem

Points $D$ and $E$ are chosen on side $BC$ of triangle $ABC$ such that $E$ is between $B$ and $D$ and $BE=1$, $ED=DC=3$. If $\angle BAD=\angle EAC=90^\circ$, the area of $ABC$ can be expressed as $\tfrac{p\sqrt q}r$, where $p$ and $r$ are relatively prime positive integers and $q$ is a positive integer not divisible by the square of any prime. Compute $p+q+r$. $[asy] import olympiad; size(200); defaultpen(linewidth(0.7)+fontsize(11pt)); pair D = origin, E = (3,0), C = (-3,0), B = (4,0); path circ1 = arc(D,3,0,180), circ2 = arc(B/2,2,0,180); pair A = intersectionpoint(circ1, circ2); draw(E--A--C--B--A--D); label("A",A,N); label("B",B,SE); label("C",C,SW); label("D",D,S); label("E",E,S); [/asy]$

## Solution

Note that if a circle has center $D$ and radius $CD$, then $CE$ is the circle's diameter because $CD = DE$. Because $\angle CAE$ is a right angle, $A$ is also on the circle with center $D$, so $DA = 3$. By the Pythagorean Theorem, $AB = \sqrt{16 - 9} = \sqrt{7}$.

The area of $ABD$ is $\tfrac12 \cdot 3 \cdot \sqrt{7} = \tfrac{3\sqrt{7}}{2}$. Let $x$ be the length of the altitude from $A$ to $BD$, so $\tfrac12 \cdot 4x = \tfrac{3\sqrt{7}}{2}$. Thus, $x = \tfrac{3\sqrt{7}}{4}$.

Since $BD$ is on line $BC$, the area of $ABC$ is equal to $\tfrac12 \cdot 7 \cdot \tfrac{3\sqrt{7}}{4} = \tfrac{21\sqrt{7}}{8}$, so $p+q+r = \boxed{36}$.