2013 AMC 12A Problems/Problem 12

Because the angles are in an arithmetic progression, and the angles add up to 180 degrees, the second largest angle in the triangle must be 60 degrees. Also, the side opposite of the 60 degree angle must be the second longest because of the angle-side relationship. Any of the three sides, 4, 5, or x, could be the second longest side of the triangle.

The law of cosines can be applied to solve for x in all three cases.

When the second longest side is five, we get that 5^2 = 4^2 + x^2 - 2(4)(x)cos 60, therefore x^2 - 4x - 9 = 0. By using the quadratic formula, x = (4 +- root(16 + 36)) / 2). Because root(52) is greater than 4, it must be positive root(52), so x = 2 + root(13).

When the second longest side is x, we get that x^2 = 5^2 + 4^2 - 40*cos 60, and x = root(21).

When the second longest side is 4, we get that 4^2 = 5^2 + x^2 - 2(5)(x)cos 60, therefore x^2 - 5x + 9 = 0. Using the quadratic formula x = (5 +- root(25 - 36)) / 2. However, because root(-11) is imaginary, there is no possible way that the longest side can be 4.

Adding the two other possibilities gets 2 + root(13) + root(21), with a = 2, b=13, and c=21. a + b + c = 36, answer choice A.