An equilateral triangle $ABC$ with side $n$ a positive integer is divided into triangular lattice consisting of unit triangles.

Each point on the lattice is colored red, white, or blue such that:

1. No point on $AB$ is colored red

2. No point on $BC$ is colored blue

3. No point on $AC$ is colored white

Prove that there exists a unit triangle whose vertices are colored with 3 different colors.

Challenge problem: Same problem but for tetrahedron and four colors.

Solution

We first state the one-dimensional version of this problem:

Given a sequence of points on a line colored red or blue, and the first point must be colored red, and the last point must be colored blue, then we have an ODD number of segments that has two differently-colored end points.

This one-dimensional version is trivial to prove.

Now we turn to the 2-dimensional version as stated in the problem.

For each side that has both red and blue endpoints, we "mark" that side.

Then for each unit triangle, we give it a score based on how many marked sides it has. Now, if we consider the "outside" area as another "surrogate unit triangle", then we can also give it a score based on how many marked sides are facing the outside.

First, note that the total score of all areas must be even, since any marked side contribute one score to two areas.

Now, applying the 1-dimensional version of the problem, we know that side AC has an odd number of marked sides. Side AB and BC does not have any marked sides. So the outside area has an odd score. Thus, there is an odd number of unit triangles that has an odd score. In order for a unit triangle to have an odd score, it must have three differently-colored vertices.

## Thursday, November 18, 2010

### Triangular (and tetrahedral) lattice

Labels:
coloring,
Combinatorics,
graph theory,
lattice,
parity,
tetrahedron,
triangle

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