Difference between revisions of "1969 Canadian MO Problems/Problem 9"
m |
(box) |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | Show that for any quadrilateral inscribed in a [[circle]] of [[radius]] <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> | + | 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> | + | 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. |
| − | - | + | {{Old CanadaMO box|num-b=8|num-a=10|year=1969}} |
| − | |||
| − | |||
| − | |||
Revision as of 21:42, 17 November 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.
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 8 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 10 |
