TITLE: Putnam 1990: Problem A-4
QUESTION [9 upvotes]: I was told this lovely puzzle recently which I thought people here might enjoy. 
Consider a paper punch that can be centered at any point
of the plane and that, when operated, removes from the
plane precisely those points whose distance from the
center is irrational. How many punches are needed to
remove every point?

REPLY [3 votes]: Three punches suffice if they are made at collinear points $A,B,C$ such that $B$ is the midpoint of $AC$ and the square of the length of $AB$ is irrational. For any point $P$, Apollonius's theorem ensures that
$$
PA^2+PC^2-2\cdot PB^2=2\cdot AB^2.
$$
As the right hand side is irrational, it follows that one of the distances $PA,PB,PC$ is irrational. Therefore the point $P$ has been removed.
Note that Apollonius's theorem applies even when the triangle $\Delta PAC$ is degenerate.