Maths Olympiad Prep

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Combinatorics Difficulty 5.1 AIME, harder Prove it Soviet Union

Problem:

An equilateral triangle of side nn is divided into n2n^2 equilateral triangles of side 11. A path is drawn along the sides of the triangles which passes through each vertex just once. Prove that the path makes an acute angle at at least nn vertices.

Solution

Solution:

Figure 1

The diagram has 1+2+…+n=n(n+1)/21 + 2 + \ldots + n = n(n + 1) / 2 upright triangles and 1+2+…+n−1=n(n−1)/21 + 2 + \ldots + n - 1 = n(n - 1) / 2 upside down triangles. It has 1+2+…+n+1=(n+1)(n+2)/21 + 2 + \ldots + n + 1 = (n + 1)(n + 2) / 2 vertices. So the path must be (n+1)(n+2)/2−1=(n2+3n)/2(n + 1)(n + 2) / 2 - 1 = (n^2 + 3n) / 2 units long. Each unit length of the path is in just one upright triangle. The path cannot contain all three sides of a small triangle, or it would pass through a vertex more than once. So it must contain two sides of (n2+3n)/2−n(n+1)/2=n(n^2 + 3n) / 2 - n(n + 1) / 2 = n triangles. But if it contains two sides of a triangle, then it must make an acute angle at the vertex where they meet.

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