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Difference between revisions of "1969 Canadian MO Problems/Problem 9"

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== Problem ==
 
== Problem ==
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Show that for any quadrilateral inscribed in a circle of radius <math>\displaystyle 1,</math> the length of the shortest side is less than or equal to <math>\displaystyle \sqrt{2}</math>.
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Show that for any quadrilateral inscribed in a [[circle]] of [[radius]] <math>\displaystyle 1,</math> the length of the shortest side is less than or equal to <math>\displaystyle \sqrt{2}</math>.
  
 
== Solution ==
 
== Solution ==
−
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>.  
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Let <math>\displaystyle a,b,c,d</math> be the [[edge]]-[[length]]s and <math>\displaystyle e,f</math> be the lengths of the [[diagonal]]s of the [[quadrilateral]]. By [[Ptolemy's Theorem]], <math>\displaystyle ab+cd = ef</math>. However, each diagonal is a [[chord]] of the circle and so must be shorter than the [[diameter]]: <math>\displaystyle e,f \le 2</math> and thus <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. Thus, at least one of the sides must have length less than <math>\sqrt 2</math>, so certainly the shortest side must.
 
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.

Revision as of 11:37, 6 July 2007

Problem

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 $\displaystyle \sqrt{2}$.

Solution

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

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

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