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, 2021

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a trapezoid with $AB \parallel CD$ , $AB=5$ , $BC=9$ , $CD=10$ , and $DA=7$ . Lines $BC$ and $DA$ intersect at point $E$ . Let $M$ be the midpoint of $CD$ , and let $N$ be the intersection of the circumcircles of $\triangle BMC$ and $\triangle DMA$ (other than $M$ ). If $EN^{2}=\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ , compute $100a+b$ .

Proposed by: Milan Haiman

Solution

Solution:

From $\triangle EAB \sim \triangle EDC$ with length ratio $1:2$ , we have $EA=7$ and $EB=9$ . This means that $A$ , $B$ , $M$ are the midpoints of the sides of $\triangle ECD$ . Let $N'$ be the circumcenter of $\triangle ECD$ . Since $N'$ is on the perpendicular bisectors of $EC$ and $CD$ , we have $\angle N'MC = \angle N'BC = 90^{\circ}$ . Thus $N'$ is on the circumcircle of $\triangle BMC$ . Similarly, $N'$ is on the circumcircle of $\triangle DMA$ . As $N' \neq M$ , we must have $N' = N$ . So it suffices to compute $R^{2}$ , where $R$ is the circumradius of $\triangle ECD$ .

We can compute $K = [\triangle ECD]$ to be $21\sqrt{11}$ from Heron's formula, giving
$R = \frac{10 \cdot 14 \cdot 18}{4K} = \frac{30}{\sqrt{11}}$
So $R^{2} = \frac{900}{11}$ , and the final answer is $90011$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.