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

 
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== Problem ==
 
== Problem ==
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>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 ==
Since <math>\displaystyle a,b,c</math> are all positive, squaring  preserves the inequality; <math>\displaystyle 2c^2\ge (a+b)^2.</math>
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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>.
  
By the Pythagorean Theorem, <math>\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0,</math> since the square of a real number is always positive.
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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.
  
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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.
* [[1969 Canadian MO Problems/Problem 2|Previous Problem]]
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* [[1969 Canadian MO Problems/Problem 4|Next Problem]]
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{{Old CanadaMO box|num-b=2|num-a=4|year=1969}}
* [[1969 Canadian MO Problems|Back to Exam]]
 

Latest revision as of 22:39, 17 November 2007

Problem

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

Solution

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

Thus $2c^2\ge (a+b)^2.$ Since $a,b,c$ 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.

1969 Canadian MO (Problems)
Preceded by
Problem 2
1 2 3 4 5 6 7 8 Followed by
Problem 4


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