This week’s Express is an oldie but a goodie:
Consider the infinite points in the coordinate plane, and suppose that each point is one of two colors: red or blue. It turns out there must be at least two points of the same color that are a distance 1 apart.
Why? Draw any equilateral triangle with side length 1. All three vertices are a distance 1 from each other, and at least two of them must be the same color, whether red or blue.
Now suppose every point in the plane is one of three colors: red, green or blue. Once again, it turns out there must be at least two points of the same color that are a distance 1 apart.
How can you show this is true using just seven points in the plane?
Consider two equilateral triangles of side length 1, sharing two vertices and an edge, such that vertices on the same edge are of different colors.
WLOG they will always look like this, the points $\sqrt{3}$ units apart must be the same color.
Consider the following equilateral triangle.
$α = 2 \cdot \sin ^{-1} \left(\dfrac{\dfrac{1}{2}}{\sqrt{3}}\right) \approx 33.5573^{\circ}$
Consider the diamond flipped or rotated $33.5573^{\circ}$ about the top point. The two bottom points must be the same color and are 1 unit apart.
