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Problem 1177

AIME late

Geometry Difficulty 5.1 Find the answer HMMT February · United States · 2020

Let $ABCD$ be a cyclic quadrilateral, and let segments $AC$ and $BD$ intersect at $E$ . Let $W$ and $Y$ be the feet of the altitudes from $E$ to sides $DA$ and $BC$ , respectively, and let $X$ and $Z$ be the midpoints of sides $AB$ and $CD$ , respectively. Given that the area of $AED$ is $9$ , the area of $BEC$ is $25$ , and $\angle EBC - \angle ECB = 30^{\circ}$ , then compute the area of $WXYZ$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Solution:
Reflect $E$ across $DA$ to $E_{W}$ , and across $BC$ to $E_{Y}$ . As $ABCD$ is cyclic, $\triangle AED$ and $\triangle BEC$ are similar. Thus $E_{W}AED$ and $EBE_{Y}C$ are similar too.
Now since $W$ is the midpoint of $E_{W}E$ , $X$ is the midpoint of $AB$ , $Y$ is the midpoint of $EE_{Y}$ , and $Z$ is the midpoint of $DC$ , we have that $WXYZ$ is similar to $E_{W}AED$ and $EBE_{Y}C$ .

From the given conditions, we have $EW : EY = 3 : 5$ and $\angle WEY = 150^{\circ}$ . Suppose $EW = 3x$ and $EY = 5x$ . Then by the law of cosines, we have
$WY = \sqrt{34 + 15\sqrt{3}}\, x.$
Thus, $E_{W}E : WY = 6 : \sqrt{34 + 15\sqrt{3}}$ . So by the similarity ratio,
$[WXYZ] = [E_{W}AED] \left(\frac{\sqrt{34 + 15\sqrt{3}}}{6}\right)^2 = 2 \cdot 9 \cdot \left(\frac{34 + 15\sqrt{3}}{36}\right) = 17 + \frac{15}{2}\sqrt{3}.$

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