Difference between revisions of "1969 Canadian MO Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | Let <math>\displaystyle a,b,c,d</math> be the | + | Let <math>\displaystyle a,b,c,d</math> be the [[side]]s and <math>\displaystyle e,f</math> be the [[diagonal]]s of the [[quadrilateral]]. By [[Ptolemy's Theorem]], <math>\displaystyle ab+cd = ef</math>. However, the [[diameter]] is the longest possible diagonal, so <math>\displaystyle e,f \le 2</math> and <math>\displaystyle ab+cd \le 4</math>. |
| − | If <math>\displaystyle a,b,c,d > \sqrt{2}</math>, then <math>\displaystyle ab+cd > 4,</math> which is impossible. | + | If <math>\displaystyle a,b,c,d > \sqrt{2}</math>, then <math>\displaystyle ab+cd > 4,</math> which is impossible. Thus, at least one of the sides must have length less than <math>\sqrt 2</math>, so certainly the shortest side must. |
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Revision as of 12:52, 28 July 2006
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 sides and 
be the diagonals of the quadrilateral. By Ptolemy's Theorem, 
. However, the diameter is the longest possible diagonal, so 
and 
.
If 
, then 
which is impossible. Thus, at least one of the sides must have length less than 
, so certainly the shortest side must.
