Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Philippines

Problem:

Let $ABC$ be an equilateral triangle. Let $\overrightarrow{AB}$ be extended to a point $D$ such that $B$ is the midpoint of $\overline{AD}$ . A variable point $E$ is taken on the same plane such that $DE = AB$ . If the distance between $C$ and $E$ is as large as possible, what is $\angle BED$ ?

Solution

Solution:

$15^\circ$

To make $C$ and $E$ as far as possible, $C$ , $D$ , $E$ must be collinear in that order.

With $\angle ABC = 60^\circ$ , we have $\angle CBD = 120^\circ$ . Since $BC = BD$ , we then have $\angle CDB = \frac{1}{2}\left(180^\circ - 120^\circ\right) = 30^\circ$ . Finally, since $BD = DE$ , we have $\angle BED = \frac{1}{2} \cdot 30^\circ = 15^\circ$ .

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