Question

The angles in a particular triangle are in arithmetic progression, and the side lengths are $4, 5,$ and $x.$ The sum of the possible values of $x$ equals $a + \sqrt{b} + \sqrt{c},$ where $a, b,$ and $c$ are positive integers. What is $a + b + c?$

$36$

$38$

$40$

$42$

$44$

Solution:

If the angles are $\alpha - \delta, \alpha, \alpha + \delta,$ their sum $3\alpha = 180^\circ$ gives $\alpha = 60^\circ,$ so one angle is $60^\circ.$

If $x$ is opposite the $60^\circ$ angle, the Law of Cosines gives $x^2 = 4^2 + 5^2 - 2\cdot 4\cdot 5\cos 60^\circ = 21,$ so $x = \sqrt{21}.$

If $5$ is opposite the $60^\circ$ angle, then $25 = x^2 - 4x + 16,$ whose positive solution is $x = 2 + \sqrt{13}.$ If $4$ is opposite, then $16 = x^2 - 5x + 25$ has no real solution.

The sum of the possible values is $2 + \sqrt{13} + \sqrt{21},$ so $a + b + c = 2 + 13 + 21 = 36.$

Thus, the correct answer is A.

2013 AMC 12A Problem 12

Solution:

Problem 12 in Other Years