1969 Canadian MO Problems/Problem 9

Problem

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

Solution

Let $a,b,c,d$ be the edge-lengths and $e,f$ be the lengths of the diagonals of the quadrilateral. By Ptolemy's Theorem, $ab+cd = ef$. However, each diagonal is a chord of the circle and so must be shorter than the diameter: $e,f \le 2$ and thus $ab+cd \le 4$.

If $a,b,c,d > \sqrt{2}$, then $ab+cd > 4,$ which is impossible. Thus, at least one of the sides must have length less than $\sqrt 2$, so certainly the shortest side must.

1969 Canadian MO (Problems)
Preceded by
Problem 8
1 2 3 4 5 6 7 8 Followed by
Problem 10


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