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1969 Canadian MO Problems/Problem 9: Difference between revisions

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== Problem ==
== Problem ==
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>.
Show that for any quadrilateral inscribed in a [[circle]] of [[radius]] <math>1,</math> the length of the shortest side is less than or equal to <math>\sqrt{2}</math>.


== Solution ==
== Solution ==
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>.  
Let <math>a,b,c,d</math> be the [[edge]]-[[length]]s and <math>e,f</math> be the lengths of the [[diagonal]]s of the [[quadrilateral]]. By [[Ptolemy's Theorem]], <math>ab+cd = ef</math>. However, each diagonal is a [[chord]] of the circle and so must be shorter than the [[diameter]]: <math>e,f \le 2</math> and thus <math>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>a,b,c,d > \sqrt{2}</math>, then <math>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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{{Old CanadaMO box|num-b=8|num-a=10|year=1969}}
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Revision as of 21:42, 17 November 2007

Problem

Show that for any quadrilateral inscribed in a circle of radius $1,$ the length of the shortest side is less than or equal to $\sqrt{2}$.

Solution

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

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

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


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