Problem source

Two points $P$ and $Q$ lie within, or on the boundary of, a square of side 1cm, one corner of which is the point $O$. Show that the length of at least one of the lines $OP,PQ$ and $QO$ must be less than or equal to $(\sqrt{6}-\sqrt{2})$ cm.

Solution source

First note that $(\sqrt{6} - \sqrt{2})^2 = 8 - 2\sqrt{12} = 8 - 4\sqrt{3}$ and since $49 > 16 \times 3$ $\sqrt{6}-\sqrt{2} > 1$.

Therefore we can assume without loss of generality that $P$ and $Q$ both do not lie on the same side as each other, a side containing $O$, otherwise one of those lengths would be $1 \text{ cm} < (\sqrt{6}-\sqrt{2}) \text{ cm} $.

Let $O = (0,0)$, $P = (1,x)$, $Q = (y,1)$, then our lengths squared are:

$1 + x^2, 1 + y^2, (1-x)^2+(1-y)^2$. 

To maximise the length of the smallest side, each side should be equal in length (otherwise we could increase the length of the smallest side by moving the point between the shortest side and the longest side (without affecting the other side).

Therefore $x = y$ and $1+x^2 = 2(1-x)^2 \Rightarrow x^2-4x+1 = 0 \Rightarrow x = 2 - \sqrt{3} $. Therefore the distances are all $\sqrt{1+7-4\sqrt{3}} = \sqrt{8-4\sqrt{3}} = (\sqrt{6}-\sqrt{2}) \text{ cm}$