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 $k$ .

1995 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1995 AHSME Problems/Problem 23

Problem

Solution

See also