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Difference between revisions of "1969 Canadian MO Problems/Problem 3"

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== Problem ==
 
== Problem ==
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Let <math>\displaystyle c</math> be the length of the hypotenuse of a right angle triangle whose two other sides have lengths <math>\displaystyle a</math> and <math>\displaystyle b</math>. Prove that <math>\displaystyle a+b\le c\sqrt{2}</math>. When does the equality hold?
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Let <math>\displaystyle c</math> be the length of the [[hypotenuse]] of a [[right triangle]] whose two other sides have lengths <math>\displaystyle a</math> and <math>\displaystyle b</math>. Prove that <math>\displaystyle a+b\le c\sqrt{2}</math>. When does the equality hold?
  
 
== Solution ==
 
== Solution ==
 
By the [[Pythagorean Theorem]] and the [[trivial inequality]], <math>\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0</math>.
 
By the [[Pythagorean Theorem]] and the [[trivial inequality]], <math>\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0</math>.
  
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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.
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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.
  
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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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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.
  
 
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Revision as of 15:02, 13 October 2006

Problem

Let $\displaystyle c$ be the length of the hypotenuse of a right triangle whose two other sides have lengths $\displaystyle a$ and $\displaystyle b$. Prove that $\displaystyle a+b\le c\sqrt{2}$. When does the equality hold?

Solution

By the Pythagorean Theorem and the trivial inequality, $\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0$.

Thus $\displaystyle 2c^2\ge (a+b)^2.$ Since $\displaystyle a,b,c$ are all positive, taking a square root preserves the inequality and we have our result.

For equality to hold we must have $(a-b)^2 = 0$. In this case, we have an isosceles right triangle, and equality certainly holds for all such triangles.

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