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 ?
Solution
Note that and are isosceles trapezoids. They must be cyclic quadrilaterals, so we can apply Ptolemy's Theorem. Let , , and . Then we have:
According to the first equation, . Plugging this into the third equation results in . The only positive root of this cubic is . Substituting into the first and second equations gives and and thus the sum of all diagonals is . Our answer is therefore .
- This solution requires solving a cubic - however, I thought that it was in the rules that the AMC 12 cannot ask for one to solve a cubic - perhaps I am mistaken?
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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All AMC 12 Problems and Solutions |
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