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, 2024

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

In triangle $ABC$ , points $M$ and $N$ are the midpoints of $AB$ and $AC$ , respectively, and points $P$ and $Q$ trisect $BC$ . Given that $A$ , $M$ , $N$ , $P$ , and $Q$ lie on a circle and $BC = 1$ , compute the area of triangle $ABC$ .

Solution

Solution:

Note that $MP \parallel AQ$ , so $AMPQ$ is an isosceles trapezoid. In particular, we have $AM = MB = BP = PQ = \frac{1}{3}$ , so $AB = \frac{2}{3}$ . Thus $ABC$ is isosceles with base $1$ and legs $\frac{2}{3}$ , and the height from $A$ to $BC$ is $\frac{\sqrt{7}}{6}$ , so the area is $\frac{\sqrt{7}}{12}$ .

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