Question 62
Consider a four-point configuration in the plane, every three points of which can be covered by a strip of unit width. Prove that:
a) the four points can be covered by a strip of width at most $\sqrt{2}$; and
b) if no strip of width less than $\sqrt{2}$ covers all four points, then they are the vertices of a square of side length $\sqrt{2}$.
a) the four points can be covered by a strip of width at most $\sqrt{2}$; and
b) if no strip of width less than $\sqrt{2}$ covers all four points, then they are the vertices of a square of side length $\sqrt{2}$.