jmc

Question

The sides of a triangle have lengths 11, 15, and k, where k is a positive integer. For how many values of k is the triangle obtuse?

Accepted Answer

The longest side of the triangle either has length

15

or has length

k .

Take cases:

If the longest side has length

15,

then

k \leq 15.

The triangle must be nondegenerate, which happens if and only if

15 < 11 + k,

or

4 < k,

by the triangle inequality. Now, for the triangle to be obtuse, we must have

1 5^{2} > 1 1^{2} + k^{2},

or

1 5^{2} - 1 1^{2} = 104 > k^{2},

which gives

k \leq 10

(since

k

is an integer). Therefore, the possible values of

k

in this case are

k = 5, 6, \dots, 10.

If the longest side has length

k,

then

k \geq 15.

In this case, the triangle inequality gives

k < 15 + 11,

or

k < 26.

For the triangle to be obtuse, we must have

k^{2} > 1 1^{2} + 1 5^{2} = 346,

or

k \geq 19

(since

k

is an integer). Therefore, the possible values of

k

in this case are

k = 19, 20, \dots, 25.

In total, the number of possible values of

k

is

(10 - 5 + 1) + (25 - 19 + 1) = 13 .