Difference between revisions of "1969 Canadian MO Problems/Problem 3"
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Thus <math>\displaystyle 2c^2\ge (a+b)^2.</math> Since <math>\displaystyle a,b,c</math> are all positive, taking a square root preserves the inequality and we have our result. | Thus <math>\displaystyle 2c^2\ge (a+b)^2.</math> Since <math>\displaystyle a,b,c</math> are all positive, taking a square root preserves the inequality and we have our result. | ||
| − | The equality condition is clearly that <math>(a-b)^2 = 0</math> -- the [[isosceles]] [[right triangle]]. | + | The [[equality condition]] is clearly that <math>(a-b)^2 = 0</math> -- the [[isosceles]] [[right triangle]]. |
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Revision as of 08:22, 28 July 2006
Problem
Let 
be the length of the hypotenuse of a right angle triangle whose two other sides have lengths 
and 
. Prove that 
. When does the equality hold?
Solution
By the Pythagorean Theorem and the trivial inequality, 
.
Thus 
Since 
are all positive, taking a square root preserves the inequality and we have our result.
The equality condition is clearly that 
-- the isosceles right triangle.
