Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Find the answer

Distinct points $A, B, C, D$ are given such that triangles $A B C$ and $A B D$ are equilateral and both are of side length 10 . Point $E$ lies inside triangle $A B C$ such that $E A=8$ and $E B=3$ , and point $F$ lies inside triangle $A B D$ such that $F D=8$ and $F B=3$ . What is the area of quadrilateral $A E F D$ ?

Your answer

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

Solution

$\angle F B D+\angle A B F=\angle A B D=60^{\circ}$ . Since $E B=B F=3$ , this means that $E B F$ is an equilateral triangle of side length 3. Now we have $[A E F D]=[A E B D]-[E B F]-[F B D]=[A E B]+[A B D]-[E B F]-$ $[F B D]=[A B D]-[E B F]=\frac{\sqrt{3}}{4}\left(10^{2}-3^{2}\right)=\frac{91 \sqrt{3}}{4}$ .

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