2009/10 AHSMC Part 1 Problem 10

Problem 10 of 16

Two sides of an obtuse triangle of positive area are of length $5$ and $11$ . The number of possible integer lengths of the third side is

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Solution

By the Triangle Inequality, the third side must be from $7$ to $15$ . Now $9^2 < 11^2 - 5^2 < 10^2$ , so that $(5,7,11)$ , $(5,8,11)$ and $(5,9,11)$ are obtuse triangles. Also, $12^2 < 11^2 + 5^2 < 13^2$ , so that $(5,11,13)$ , $(5,11,14)$ and $(5,11,15)$ are obtuse triangles. The other three are not. The answer is (C).

University of Alberta

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