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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Let $ABCDEF$ be a regular hexagon of area $1$ . Let $M$ be the midpoint of $DE$ . Let $X$ be the intersection of $AC$ and $BM$ , let $Y$ be the intersection of $BF$ and $AM$ , and let $Z$ be the intersection of $AC$ and $BF$ . If $[P]$ denotes the area of polygon $P$ for any polygon $P$ in the plane, evaluate $[BXC] + [AYF] + [ABZ] - [MXZY]$ .

Solution

Solution:

Let $O$ be the center of the hexagon. The desired area is $[ABCDEF] - [ACDM] - [BFEM]$ . Note that $[ADM] = [ADE]/2 = [ODE] = [ABC]$ , where the last equation holds because $\sin 60^{\circ} = \sin 120^{\circ}$ . Thus, $[ACDM] = [ACD] + [ADM] = [ACD] + [ABC] = [ABCD]$ , but the area of $ABCD$ is half the area of the hexagon. Similarly, the area of $[BFEM]$ is half the area of the hexagon, so the answer is zero.

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