Difference between revisions of "1969 Canadian MO Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | Let <math> | + | Let <math>c</math> be the length of the [[hypotenuse]] of a [[right triangle]] whose two other sides have lengths <math>a</math> and <math>b</math>. Prove that <math>a+b\le c\sqrt{2}</math>. When does the equality hold? |
== Solution == | == Solution == | ||
| − | By the [[Pythagorean Theorem]] and the [[trivial inequality]], <math> | + | By the [[Pythagorean Theorem]] and the [[trivial inequality]], <math>2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0</math>. |
| − | Thus <math> | + | Thus <math>2c^2\ge (a+b)^2.</math> Since <math>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]] | + | For [[equality condition | equality]] to hold we must have <math>(a-b)^2 = 0</math>. In this case, we have an [[isosceles triangle | isosceles]] right triangle, and equality certainly holds for all such triangles. |
| − | - | + | {{Old CanadaMO box|num-b=2|num-a=4|year=1969}} |
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Latest revision as of 21:39, 17 November 2007
Problem
Let 
be the length of the hypotenuse of a right 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.
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 4 |
