So I was looking at the putnam problem of the day and have a question on the rules of the scenario presented:
Prove that the points of an isosceles triangle of side length 1 cannot be colored in four colors such that no two points at distance at least $2 - \sqrt 2$ from each other receive the same color.
After some initial tinkering it seems like it can occur which makes me think I'm not understanding the rules of the game. See figure 1.
Here is a link to an image of my set up.
My set up: Segments $AB$ and $BC$ have length 1, and $B$ is a right angle, which makes $AC = \sqrt 2$. $[A, D)$ is color 1, $[D, B)$ is color 2, $[B, C)$ is color 3, and $[C, A)$ is color 4.
But here it is easy find two points on $AC$ that are at least $2 – \sqrt 2$ apart. This tells me that I'm not understanding the rules of the game.
I don't want the answer, I just want to know what I'm missing.
A counterexample would need to be a coloring such that no two points at distance at least $2-\sqrt{2}$ from each other have the same color.
Your example doesn't qualify since, with your chosen coloring, all points on $[C,A)$ have the same color, and there are lots of pairs of points on $[C,A)$ which have distance at least $2-\sqrt{2}$ from each other.
Perhaps you were missing the word "cannot"?
The problem asks you to prove that for all colorings, something must be true.
You constructed a coloring where the something is true.
Thus, your coloring is not a counterexample, since the something is true, not false, and it's also not a proof, since one coloring where something is true doesn't prove that it's true for all colorings.