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Tenth Philippine Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Philippines

Problem:

Let ABCABC be an equilateral triangle. Let AB→\overrightarrow{AB} be extended to a point DD such that BB is the midpoint of AD‾\overline{AD}. A variable point EE is taken on the same plane such that DE=ABDE = AB. If the distance between CC and EE is as large as possible, what is ∠BED\angle BED?

Solution

Solution:

15∘15^\circ

To make CC and EE as far as possible, CC, DD, EE must be collinear in that order.

With ∠ABC=60∘\angle ABC = 60^\circ, we have ∠CBD=120∘\angle CBD = 120^\circ. Since BC=BDBC = BD, we then have ∠CDB=12(180∘−120∘)=30∘\angle CDB = \frac{1}{2}\left(180^\circ - 120^\circ\right) = 30^\circ. Finally, since BD=DEBD = DE, we have ∠BED=12⋅30∘=15∘\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.