1969 Canadian MO Problems/Problem 3

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.

Page

Toolbox

Search

1969 Canadian MO Problems/Problem 3

Problem

Solution