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 $385+6=\boxed{391}$ .

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

Page

Toolbox

Search

2014 AMC 12B Problems/Problem 24

Problem

Solution