Maths Olympiad Prep

Track / Stage 6 / 365 of 400 #1365 of 1964

Problem 1365

National olympiad, first round
Geometry Difficulty 6.8 Find the answer

A polyhedron has faces that all either triangles or squares. No two square faces share an edge, and no two triangular faces share an edge. What is the ratio of the number of triangular faces to the number of square faces?

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

1. Let s s be the number of square faces and t t be the number of triangular faces.
2. According to the problem, no two square faces share an edge, and no two triangular faces share an edge. This implies that each edge of the polyhedron is shared by one triangular face and one square face.
3. Each square face has 4 edges, and each triangular face has 3 edges. Since each edge is shared by one triangular face and one square face, the total number of edges contributed by the square faces must equal the total number of edges contributed by the triangular faces.
4. Therefore, the total number of edges contributed by the square faces is 4s 4s , and the total number of edges contributed by the triangular faces is 3t 3t .
5. Since these edges are shared, we have the equation:
4s=3t 4s = 3t
6. To find the ratio of the number of triangular faces to the number of square faces, we solve for ts \frac{t}{s} :
ts=43 \frac{t}{s} = \frac{4}{3}
7. Therefore, the ratio of the number of triangular faces to the number of square faces is 4:3 4:3 .

The final answer is 4:3\boxed{4:3}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.