# 1995 AHSME Problems/Problem 23

## Problem

The sides of a triangle have lengths $11,15,$ and $k$, where $k$ is an integer. For how many values of $k$ is the triangle obtuse? $\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 13 } \qquad \mathrm{(E) \ 14 }$

## Solution

By the Law of Cosines, a triangle is obtuse if the sum of the squares of two of the sides of the triangles is less than the square of the third. The largest angle is either opposite side $15$ or side $k$. If $15$ is the largest side, $$15^2 >11^2 + k^2 \Longrightarrow k < \sqrt{104}$$

By the Triangle Inequality we also have that $k > 4$, so $k$ can be $5, 6, 7, \ldots , 10$, or $6$ values.

If $k$ is the largest side, $$k^2 >11^2 + 15^2 \Longrightarrow k > \sqrt{346}$$

Combining with the Triangle Inequality $19 \le k < 26$, or $7$ values. These total $13\ \mathrm{(D)}$ values of $k$.

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