Problem

Source: Romanian TST 2018 Problem 3 Day 2

Tags: combinatorial geometry, combinatorics



Consider a 4-point configuration in the plane such that every 3 points can be covered by a strip of a unit width. Prove that: 1) the four points can be covered by a strip of length at most $\sqrt2$ and 2)if no strip of length less that $\sqrt2$ covers all the four points, then the points are vertices of a square of length $\sqrt2$