# 2020 CIME I Problems/Problem 5

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem 5

Let $ABCD$ be a rectangle with sides $AB>BC$ and let $E$ be the reflection of $A$ over $\overline{BD}$. If $EC=AD$ and the area of $ECBD$ is $144$, find the area of $ABCD$.

## Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Let $O$ be the center of rectangle $ABCD$. Because $E$ is the reflection of $A$ over $\overline{BD}$ and $\angle BAD = 90$ degrees, we have $\angle BED = 90$ degrees. This means $E$ lies on the circle with diameter $\overline{BD}$, or the circumcircle of rectangle $ABCD$. We are given $EC=BC$, so by symmetry, $EC=ED$. Since the three lengths are equal and $\overarc{BD}=180$ degrees, we must have $\overarc{BC}=\overarc{CE}=\overarc{ED}=60$ degrees, so $\triangle OBC$, $\triangle OCE$, $\triangle OED$ are all equilateral. Given that the area of cyclic quadrilateral $ECBD$ is $144$, the area of $\triangle OBC$ is $\frac{1}{3} \cdot 144 = 48$. This is $\frac{1}{4}$ of the area of rectangle $ABCD$, so the answer is $48 \cdot 4 = \boxed{192}$.

The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 