HMMT February

Geometry Difficulty 5.2 AIME, harder Find the answer

Let $AD, BE$ , and $CF$ be segments sharing a common midpoint, with $AB < AE$ and $BC < BF$ . Suppose that each pair of segments forms a $60^{\circ}$ angle, and that $AD=7, BE=10$ , and $CF=18$ . Let $K$ denote the sum of the areas of the six triangles $\triangle ABC, \triangle BCD, \triangle CDE, \triangle DEF, \triangle EFA$ , and $\triangle FAB$ . Compute $K \sqrt{3}$ .

Your answer

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

Solution

Let $M$ be the common midpoint, and let $x=7, y=10, z=18$ . One can verify that hexagon $ABCDEF$ is convex. We have $[ABC]=[ABM]+[BCM]-[ACM]=\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{x}{2} \cdot \frac{y}{2}+\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{y}{2} \cdot \frac{z}{2}-\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{x}{2} \cdot \frac{z}{2}=\frac{\sqrt{3}(xy+yz-zx)}{16}$ . Summing similar expressions for all 6 triangles, we have $K=\frac{\sqrt{3}(2xy+2yz+2zx)}{16}$ Substituting $x, y, z$ gives $K=47 \sqrt{3}$ , for an answer of 141.

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