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

Revision as of 01:54, 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

Since $\displaystyle a,b,c$ are all positive, squaring preserves the inequality; $\displaystyle 2c^2\ge (a+b)^2.$

By the Pythagorean Theorem, $\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0,$ since the square of a real number is always positive.

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

Revision as of 01:54, 28 July 2006

Problem

Solution