2014 AMC 12B Problems/Problem 24
Problem
Let be a pentagon inscribed in a circle such that , , and . The sum of the lengths of all diagonals of is equal to , where and are relatively prime positive integers. What is ?
Solutions
Solution 1
Let , , and . Let be on such that . In we have . We use the Law of Cosines on to get . Eliminating we get which factorizes as Discarding the negative roots we have . Thus . For , we use Ptolemy's theorem on cyclic quadrilateral to get . For , we use Ptolemy's theorem on cyclic quadrilateral to get .
The sum of the lengths of the diagonals is so the answer is
Solution 2
Let denote the length of a diagonal opposite adjacent sides of length and , for sides and , and for sides and . Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain:
Using equations and , we obtain:
and
Plugging into equation , we find that:
Or similarly into equation to check:
, being a length, must be positive, implying that . In fact, this is reasonable, since in the pentagon with apparently obtuse angles. Plugging this back into equations and we find that and .
We desire , so it follows that the answer is
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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