Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it Brazil

In the following diagram, ABCDE is a regular pentagon and MNP is an equilateral triangle. Find $\angle CMD$ .

Solution

First notice that $BE \parallel CD$ , so $BEM$ is an equilateral triangle. So $BE = EM$ and $\angle BEM = 60^\circ$ .

Since $ABCDE$ is a regular pentagon, $\angle BEC = 36^\circ$ , so $\angle MEC = \angle BEM + \angle BEC = 60^\circ + 36^\circ = 96^\circ$ .

We also have $CE = BE = EM$ , so triangle $CEM$ is isosceles and $\angle CME = \frac{180^\circ - \angle MEC}{2} = 42^\circ$ .

Hence $\angle BMC = 60^\circ - 42^\circ = 18^\circ$ and, by symmetry, $\angle EMD = 18^\circ$ .

Finally, $\angle CMD = 60^\circ - 2 \cdot 18^\circ = 24^\circ$ .

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