2013 AMC 12A Problems/Problem 12
Because the angles are in an arithmetic progression, and the angles add up to , the second largest angle in the triangle must be . Also, the side opposite of that 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 , therefore . By using the quadratic formula, , therefore .
When the second longest side is x, we get that , therefore .
When the second longest side is 4, we get that , therefore . Using the quadratic formula, . However, is not real, so the second longest side cannot equal 4.
Adding the two other possibilities gets , with , and . , which is answer choice .