HMMT-Feb 2013 geo, Problem 2

Let $A B C D$ be an isosceles trapezoid such that $A D=B C, A B=3$, and $C D=8$. Let $E$ be a point in the plane such that $B C=E C$ and $A E \perp E C$. Compute $A E$.

Solution: Let $r=B C=E C=A D . \triangle A C E$ has right angle at $E$, so by the Pythagorean Theorem, \[ A E^{2}=A C^{2}-C E^{2}=A C^{2}-r^{2} \] Let the height of $\triangle A C D$ at $A$ intersect $D C$ at $F$. Once again, by the Pythagorean Theorem, \[ A C^{2}=F C^{2}+A F^{2}=\left(\frac{8-3}{2}+3\right)^{2}+A D^{2}-D F^{2}=\left(\frac{11}{2}\right)^{2}+r^{2}-\left(\frac{5}{2}\right)^{2} \] Plugging into the first equation, \[ A E^{2}=\left(\frac{11}{2}\right)^{2}+r^{2}-\left(\frac{5}{2}\right)^{2}-r^{2} \] so $A E=2 \sqrt{6}$. $\fbox{\sqrt{2 \sqrt{6}}}$.

Answer: \sqrt{2 \sqrt{6}}