2020 CIME I Problems/Problem 5

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

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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 $48 \cdot 4 = \boxed{192}$ .

2020 CIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2020 CIME I Problems/Problem 5

Problem 5

Solution