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Selection tests for the Balkan Mathematical Olympiad, 2013

Geometry Difficulty 7.3 National Olympiad, round 2 Prove it Saudi Arabia

$A B C D E F$ is an equiangular hexagon of perimeter $21$ . Given that $A B = 3$ , $C D = 4$ , and $E F = 5$ , compute the area of hexagon $A B C D E F$ .

Solution

We extend sides $F A$ and $B C$ to intersect at $A'$ , and sides $B C$ and $D E$ to intersect at $B'$ , and sides $D E$ and $F A$ to intersect at $C'$ .

Triangles $A' B' C'$ , $A' B A$ , $B' D C$ , and $C' F E$ are equilateral with side lengths $11$ , $3$ , $4$ , and $5$ respectively. Therefore, the area of hexagon $A B C D E F$ is

$\frac{\sqrt{3}}{4} \cdot 11^{2} - \frac{\sqrt{3}}{4} \cdot 3^{2} - \frac{\sqrt{3}}{4} \cdot 4^{2} - \frac{\sqrt{3}}{4} \cdot 5^{2} = \frac{71 \sqrt{3}}{4}.$

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