Five points are placed anywhere inside (or on the boundary of) a square of side 1. Show that two of them are at distance at most \frac{\sqrt{2}}{2} \approx 0.7071 from each other, and show that this bound cannot be improved.

Five Points in a Unit Square

Issue 3 Core Published 12 Aug 2026 By The Chalkboard Desk

The problem

Five points are placed anywhere inside (or on the boundary of) a square of side . Show that two of them are at distance at most from each other, and show that this bound cannot be improved.

Points may lie on the boundary of the square, and points may coincide.
You must both prove the bound and demonstrate that it is the best possible.

Take a minute

Reveal hint 1

Five points and one number that is not five — cut the square into the right number of equal pieces first.

Reveal hint 2

The longest distance between two points of a square of side is its diagonal .

Solution

Answer, attained by putting one point at each corner and the fifth at the centre.

Derivation

Cut the square into four.
Apply the pigeonhole principle.
Bound the distance inside one subsquare.
Prove the bound is sharp.
Check the corner-to-corner distances.
Notice the partition is forced.