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, 2005

Geometry Difficulty 4.9 AIME Prove it Baltic Way

Problem:

Let the medians of the triangle $A B C$ meet at $M$ . Let $D$ and $E$ be different points on the line $B C$ such that $D C = C E = A B$ , and let $P$ and $Q$ be points on the segments $B D$ and $B E$ , respectively, such that $2 B P = P D$ and $2 B Q = Q E$ . Determine $\angle P M Q$ .

Solution

Solution:

Draw the parallelogram $A B C A'$ , with $A A' \parallel B C$ . Then $M$ lies on $B A'$ , and $B M = \frac{1}{3} B A'$ . So $M$ is on the homothetic image (centre $B$ , dilation $1 / 3$ ) of the circle with centre $C$ and radius $A B$ , which meets $B C$ at $D$ and $E$ . The image meets $B C$ at $P$ and $Q$ . So $\angle P M Q = 90^{\circ}$ .

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