2013 AMC 12A Problems/Problem 12
Contents
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 .
Video Solution
https://www.youtube.com/watch?v=XQpQaomC2tA
(NOTE: Problem 12 starts at 8:50 in the youtube video ~jyang66)
~sugar_rush
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.