1969 Canadian MO Problems/Problem 9

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Problem

Show that for any quadrilateral inscribed in a circle of radius $\displaystyle 1,$ the length of the shortest side is less than or equal to $\displaystyle \sqrt{2}$.

Solution

Let $\displaystyle a,b,c,d$ be the sides and $\displaystyle e,f$ be the diagonals. By Ptolemy's theorem, $\displaystyle ab+cd = ef$. However, the diameter is the longest possible diagonal, so $\displaystyle e,f \le 2$ and $\displaystyle ab+cd \le 4$.

If $\displaystyle a,b,c,d > \sqrt{2}$, then $\displaystyle ab+cd > 4,$ which is impossible. Proof by contradiction.


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