Difference between revisions of "1969 Canadian MO Problems/Problem 9"
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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>\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 [[ | + | 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 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.