Round 3

Geometry Difficulty 4.5 AIME Prove it Mongolia

Let $ABC$ be a triangle with a point $D$ on $AB$ and a point $E$ on $BC$ such that $AD = CE$ and $2DE = AC$ . Show that the circumradius of $BDE$ is equal to half the circumradius of $ABC$ .
(Proposed by Khulan Tumenbayar)

Solution

Let us choose a point $S$ on the circumcircle of $BDE$ such that $SD = SE$ . Since $\angle BES = \angle BDS$ , we see that $\angle SDA = \angle SEC$ . Because $AD = EC$ and $SD = SE$ , we have $\triangle SDA = \triangle SEC$ . It follows that $\angle ASC = \angle DSE = \angle DBE = \angle ABC$ . Hence $S$ must lie on the circumcircle of $ABC$ . Thus $\triangle DSE \sim \triangle ASC$ . From the congruency, we have $\frac{DE}{AC} = \frac{1}{2}$ . Hence the ratio of the circumradii of $BDE$ and $ABC$ is $\frac{1}{2}$ .

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