Library / Harvard-MIT November Tournament /22 of 74

Harvard-MIT November Tournament

Geometry Difficulty 4.8 AIME Prove it United States

Problem:
A polyhedron has faces that are 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?

Solution

Solution:
Let $s$ be the number of square faces and $t$ be the number of triangular faces. Every edge is adjacent to exactly one square face and one triangular face. Therefore, the number of edges is equal to $4s$ , and it is also equal to $3t$ . Thus $4s = 3t$ and $\frac{t}{s} = \frac{4}{3}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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