Competition problem — Geometry — Maths Olympiad Prep

Library / /765 of 860

Geometry Difficulty 5.5 AIME, harder Find the answer

Let $A B C D$ be a cyclic quadrilateral, and let segments $A C$ and $B D$ intersect at $E$ . Let $W$ and $Y$ be the feet of the altitudes from $E$ to sides $D A$ and $B C$ , respectively, and let $X$ and $Z$ be the midpoints of sides $A B$ and $C D$ , respectively. Given that the area of $A E D$ is 9, the area of $B E C$ is 25, and $\angle E B C-\angle E C B=30^{\circ}$ , then compute the area of $W X Y Z$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Reflect $E$ across $D A$ to $E_{W}$ , and across $B C$ to $E_{Y}$ . As $A B C D$ is cyclic, $\triangle A E D$ and $\triangle B E C$ are similar. Thus $E_{W} A E D$ and $E B E_{Y} C$ are similar too. Now since $W$ is the midpoint of $E_{W} E, X$ is the midpoint of $A B, Y$ is the midpoint of $E E_{Y}$ , and $Z$ is the midpoint of $D C$ , we have that $W X Y Z$ is similar to $E_{W} A E D$ and $E B E_{Y} C$ . From the given conditions, we have $E W: E Y=3: 5$ and $\angle W E Y=150^{\circ}$ . Suppose $E W=3 x$ and $E Y=5 x$ . Then by the law of cosines, we have $W Y=\sqrt{34+15 \sqrt{3}} x$ . Thus, $E_{W} E: W Y=6: \sqrt{34+15 \sqrt{3}}$ . So by the similarity ratio, $[W X Y Z]=\left[E_{W} A E D\right]\left(\frac{\sqrt{34+15 \sqrt{3}}}{6}\right)^{2}=2 \cdot 9 \cdot\left(\frac{34+15 \sqrt{3}}{36}\right)=17+\frac{15}{2} \sqrt{3}$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.