Maths Olympiad Prep

Library / /523 of 1394

, 2020

Combinatorics Difficulty 5.2 AIME, harder Prove it United States

Problem:

How many ways can the vertices of a cube be colored red or blue so that the color of each vertex is the color of the majority of the three vertices adjacent to it?

Solution

Solution:

If all vertices of the cube are of the same color, then there are 2 ways. Otherwise, look at a red vertex. Since it must have at least 2 red neighbors, there is a face of the cube containing 3 red vertices. The last vertex on this face must also be red. Similarly, all vertices on the opposite face must be blue. Thus, all vertices on one face of the cube are red while the others are blue. Since a cube has 6 faces, the answer is 2+6=82+6=8.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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