Difference between revisions of "Talk:2013 AMC 12A Problems/Problem 12"
Ukulifeguard (talk | contribs) (Created page with "2013 AMC 12A Problems/Problem 12Vt Problem The angles in a particular triangle are in arithmetic progression, and the side lengths are <math>4,5,x</math>. The sum of the possi...") |
(No difference)
|
Revision as of 20:55, 30 June 2015
2013 AMC 12A Problems/Problem 12Vt Problem The angles in a particular triangle are in arithmetic progression, and the side lengths are . The sum of the possible values of x equals where , and are positive integers. What is ?
Solution 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, , , or , could be the second longest side of the triangle.
The law of cosines can be applied to solve for in all three cases.
When the second longest side is , we get that , therefore . By using the quadratic formula, , therefore .
When the second longest side is , we get that , therefore .
When the second longest side is , we get that , therefore . Using the quadratic formula, . However, is not real, therefore the second longest side cannot equal .
Adding the two other possibilities gets , with , and . , which is answer choice .
What about a plain jane 3-4-5 triangle, where x would equal 3? Is there something obvious about this that I'm missing? Help!