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 $\overarc{AB}=180$ 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