Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Find the answer

Problem 7.8. Given an isosceles triangle $ABC (AB = BC)$ . On the ray $BA$ beyond point $A$ , point $E$ is marked, and on side $BC$ , point $D$ is marked. It is known that

$\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.$

Find the length of segment $AE$ , if $DC = 9$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Answer: 4.

Solution. Mark point $K$ on ray $B C$ such that $B E=B K$ . Then $A E=C K$ as well.

Notice that triangles $A C E$ and $C A K$ are congruent by two sides ( $A E=C K, A C$ - common side) and the angle between them ( $\angle C A E=\angle A C K$ - adjacent to the equal base angles of the isosceles triangle). Therefore, $A K=C E=13$ and $\angle A K C=\angle A E C=60^{\circ}$ .

In triangle $A D K$ , the angles at vertices $D$ and $K$ are $60^{\circ}$ , so it is equilateral, and $D K=A K=A D=13$ . Therefore, $A E=C K=D K-D C=13-9=4$ .

## 8th grade

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