1969 Canadian MO Problems/Problem 3: Difference between revisions

Revision as of 08:21, 28 July 2006

Problem

Let $\displaystyle c$ be the length of the hypotenuse of a right angle 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.

The equality condition is clearly that $(a-b)^2 = 0$ -- the isosceles right triangle.

@@ Line 3: / Line 3: @@
 == Solution ==
-Since <math>\displaystyle a,b,c</math> are all positive, squaring  preserves the inequality; <math>\displaystyle 2c^2\ge (a+b)^2.</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>.
-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.
+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]].
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1969 Canadian MO Problems/Problem 3: Difference between revisions

Revision as of 08:21, 28 July 2006

Problem

Solution