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. | ||
| − | + | For [[equality condition | equality]] to hold we must have <math>(a-b)^2 = 0</math>. In this case, we have an [[isosceles]] [[right triangle]], and equality certainly holds for all such triangles. | |
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Revision as of 08:27, 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.
For equality to hold we must have 
. In this case, we have an isosceles right triangle, and equality certainly holds for all such triangles.
