Consider a square with sides of length 1. Can we find 9 points in it such that the distance between every pair of points is strictly greater than $\frac12$?
We allow the points to be on the boundary of the square. The question arose in a conversation, and we are pretty confident that there is no such arrangement, and moreover the only configuration with pairwise distances at least half is the one where we take the vertices, the midpoints and the center.
However, proving this seems nontrivial. An equivalent formulation is to look at the $\frac14$ blowup of the square and ask whether we can fit 9 circles with radius $\frac14$ in it or not.
Moreover, the pigeonhole principle seems intuitively to be too weak for proving this, because if the shapes will be disjoint then each will have area around $\frac18$ and for, say, squares diameter half enforces area exactly $\frac18$, but we can't fit 8 such squares without gaps.