1969 Canadian MO Problems/Problem 9

Revision as of 13:10, 28 July 2006 by 4everwise (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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 $\displaystyle \sqrt{2}$.


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.

Invalid username
Login to AoPS