HMMT February

Geometry Difficulty 4.6 AIME Find the answer

Equilateral triangles $A B F$ and $B C G$ are constructed outside regular pentagon $A B C D E$ . Compute $\angle F E G$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We have $\angle F E G=\angle A E G-\angle A E F$ . Since $E G$ bisects $\angle A E D$ , we get $\angle A E G=54^{\circ}$ . Now, $\angle E A F=108^{\circ}+60^{\circ}=168^{\circ}$ . Since triangle $E A F$ is isosceles, this means $\angle A E F=6^{\circ}$ , so the answer is $54^{\circ}-6^{\circ}=48^{\circ}$ .

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