Problem. If every point in the plane ($\mathbb{R}^2$) is colored either red, yellow, green, or blue, show that some two points are a distance of either $1$ or $\sqrt{3}$ apart and have the same color.
Attempt. By placing the $5$ points on the vertices of a regular pentagon with sidelength 1, we can use the pigeonhole principle to see that two of the vertices must have the same color. Thus, there must be two points of the same color which are either distance $1$ apart or $\phi=\frac{1+\sqrt{5}}{2}$ apart. This is because the length of the diagonal of a regular pentagon is the golden ratio $\phi$. But I cannot get $\sqrt{3}$ as one of the two possible distances. I tried working with a regular hexagon but could not succeed there either.
Source. I saw this on a problem set for a Putnam exam preparation (it is Problem 3 in that sheet).
Bonus question(s). What is special about $\sqrt{3}$? In other words, if $\alpha > 0$ is any other real number with $\alpha\neq 1$, can we show that there must exist two points of the same color that are either $1$ apart or $\alpha$ apart? If not, is there a characterization of such values of $\alpha$? Of course, the hypothesis in the bonus question is the same as before, namely each point of the plane is colored by one of the $4$ distinct colors.
Thank you for your help!
Let $P$ be an arbitrary regular hexagon with side length $1$. Note that there are at least two vertices of $P$ given the same colour. We shall consider some cases:
In this case, there are at least three vertices $A,B,C$ which are given the same colour. Two of these vertices are required to not be opposite to each other and hence are at a distance of $1$ or $\sqrt{3}$ from each other. Hence, we are done.
The colour of the centre of $P$ matches with that of one of the vertices and is at a distance $1$ from all vertices.
Since non-opposite vertices are at a distance of $1$ or $\sqrt{3}$ apart, suppose that opposite vertices are coloured using the same colour and the centre is coloured using the colour which is not used to colour any of the vertices (as otherwise, we are done).
By the above assumption, in any regular hexagon of side length $1$, opposite vertices are given the same colour (again, otherwise, we are done). This implies that any two points in $\mathbb{R}^2$ which are at a distance $2$ apart must be given the same colour as one can draw a hexagon with those two as opposite vertices. Hence, the circle consisting of points $(x,y)$ such that $x^2+y^2=4$ is monochromatic and one can find two points on this circle which are at a distance $1$ apart.