1969 Canadian MO Problems/Problem 9
Problem
Show that for any quadrilateral inscribed in a circle of radius the length of the shortest side is less than or equal to
.
Solution
Let be the edge-lengths and
be the lengths of the diagonals of the quadrilateral. By Ptolemy's Theorem,
. However, each diagonal is a chord of the circle and so must be shorter than the diameter:
and thus
.
If , then
which is impossible. Thus, at least one of the sides must have length less than
, so certainly the shortest side must.