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Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Consider an equilateral triangular grid $G$ with 20 points on a side, where each row consists of points spaced 1 unit apart. More specifically, there is a single point in the first row, two points in the second row, ..., and 20 points in the last row, for a total of 210 points. Let $S$ be a closed non-selfintersecting polygon which has 210 vertices, using each point in $G$ exactly once. Find the sum of all possible values of the area of $S$ .

Solution

Solution:

Imagine deforming the triangle lattice such that now it looks like a lattice of 45-45-90 right triangles with legs of length 1. Note that by doing this, the area has multiplied by $\frac{2}{\sqrt{3}}$ , so we need to readjust our answer on the isosceles triangle lattice by a factor of $\frac{\sqrt{3}}{2}$ at the end. By Pick's Theorem, the area in the new lattice is given by $I + \frac{P}{2} - 1 = 0 + 105 - 1 = 104$ . Therefore, the answer is $104 \cdot \frac{\sqrt{3}}{2} = 52 \sqrt{3}$ .

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