Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Find the answer

Let $n$ be the answer to this problem. Hexagon $ABCDEF$ is inscribed in a circle of radius 90. The area of $ABCDEF$ is $8n$ , $AB=BC=DE=EF$ , and $CD=FA$ . Find the area of triangle $ABC$ .

Your answer

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

Solution

Let $O$ be the center of the circle, and let $OB$ intersect $AC$ at point $M$ ; note $OB$ is the perpendicular bisector of $AC$ . Since triangles $ABC$ and $DEF$ are congruent, $ACDF$ has area $6n$ , meaning that $AOC$ has area $3n/2$ . It follows that $\frac{BM}{OM}=\frac{2}{3}$ . Therefore $OM=54$ and $MB=36$ , so by the Pythagorean theorem, $MA=\sqrt{90^{2}-54^{2}}=72$ . Thus, $ABC$ has area $72 \cdot 36=2592$ .

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