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

Let $a$ denote the length of a diagonal opposite adjacent sides of length $14$ and $3$ , $b$ for sides $14$ and $10$ , and $c$ for sides $3$ and $10$ . Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain:

$\begin{align} c^2 &= 3a+100 \\ c^2 &= 10b+9 \\ ab &= 30+14c \\ ac &= 3c+140\\ bc &= 10c+42 \end{align}$

Using equations $(1)$ and $(2)$ , we obtain:

$a = \frac{c^2-100}{3}$

and

$b = \frac{c^2-9}{10}$

Plugging into equation $(4)$ , we find that:

$\begin{align*} \frac{c^2-100}{3}c &= 3c + 140\\ \frac{c^3-100c}{3} &= 3c + 140\\ c^3-100c &= 9c + 420\\ c^3-109c-420 &=0\\ (c-12)(c+7)(c+5)&=0 \end{align*}$

Or similarly (to check):

$\begin{align*} \frac{c^2-9}{10}c &= 10c+42\\ \frac{c^3-9c}{10} &= 10c + 42\\ c^3-9c &= 100c + 420\\ c^3-109c-420 &=0\\ (c-12)(c+7)(c+5)&=0 \end{align*}$

$c$ , being a length, must be positive, implying that $c=12$ . In fact, this is reasonable, since $10+3\approx 12$ in the pentagon with apparently obtuse angles. Plugging this back into equations $(1)$ and $(2)$ we find that $a = \frac{44}{3}$ and $b= \frac{135}{10}=\frac{27}{2}$ .

We desire $3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{216+88+81}{6}=\frac{385}{6}$ , so it follows that the answer is $385 + 6 = \fbox{\textbf{(D) }391}$

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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