2014 AMC 12B Problems/Problem 24

Problem

Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$ , $BC = DE = 10$ , and $AE= 14$ . The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

$\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }353\qquad \textbf{(D) }391\qquad \textbf{(E) }421\qquad$

Solution

Note that $ABCD$ and $BCDE$ are isosceles trapezoids. They must be cyclic quadrilaterals, so we can apply Ptolemy's Theorem. Let $d=AC=BD=CE$ , $e=AD$ , and $f=EB$ . Then we have:

$d^2=10e+9$

$d^2=3f+100$

$de=10d+42$

According to the first equation, $e=\frac{d^2-9}{10}$ . Plugging this into the third equation results in $d^3-109d-420=0$ . The only positive root of this cubic is $d=12$ . Substituting into the first and second equations gives $e=\frac{27}{2}$ and $f=\frac{44}{3}$ and thus the sum of all diagonals is $3d+e+f=\frac{385}{6}$ . Our answer is therefore $385+6=\boxed{391}$ .

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?

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2014 AMC 12B Problems/Problem 24

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