1969 Canadian MO Problems/Problem 9
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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.
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 8 | 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | Followed by Problem 10 |
