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, 2024

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Let $\mathcal{P}$ be a regular 10-gon in the coordinate plane. Mark computes the number of distinct $x$ coordinates that vertices of $\mathcal{P}$ take. Across all possible placements of $\mathcal{P}$ in the plane, compute the sum of all possible answers Mark could get.

Solution

Solution:

10 distinct coordinates

5 distinct coordinates

6 distinct coordinates

Let $\mathcal{P}$ have vertices $P_{1} P_{2} \ldots P_{10}$ . If no two vertices have the same $x$ -coordinate, then Mark gets $10$ .

Otherwise, two vertices $P_{i}$ and $P_{j}$ have the same $x$ -coordinate. Then $P_{k}$ and $P_{i+j-k}$ also have the same $x$ -coordinate (indices taken modulo $10$ ), as $P_{i} P_{j} \parallel P_{k} P_{i+j-k}$ .

If $i+j$ is odd, the ten vertices of $\mathcal{P}$ pair up into $5$ pairs of the form $\left(P_{k}, P_{i+j-k}\right)$ , so Mark gets $5$ . If $i+j$ is even, then the vertices $P_{\frac{i+j}{2}}$ and $P_{\frac{i+j}{2}+5}$ do not pair up, and the remaining $8$ vertices form $4$ pairs, so Mark gets $6$ .

Thus, the answer is $10+5+6=21$ .

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