Question

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 $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n?$

$129$

$247$

$353$

$391$

$421$

Answer: D

Concepts:Ptolemy’s Theorem chord polynomial

Difficulty rating: 2650

Solution:

Because arcs $AB, CD$ are equal and arcs $BC, DE$ are equal, the chords $AC, BD, CE$ are all equal; let $x = AC = BD = CE,$ $y = AD,$ and $z = BE.$

Ptolemy's theorem on $ABCD,$ $BCDE,$ and $ABDE$ gives $\begin{gathered} 10y + 9 = x^2, \\ \quad 100 + 3z = x^2, \\ \quad 30 + 14x = yz. \end{gathered}$ Solving the first two for $y$ and $z$ and substituting into the third yields $\begin{gathered} x^3 - 109x - 420 = 0 \\ = (x-12)(x+5)(x+7). \end{gathered}$

So $x = 12,$ $y = \tfrac{135}{10} = \tfrac{27}{2},$ and $z = \tfrac{44}{3}.$ The five diagonals are $AC, BD, CE, AD, BE,$ summing to $\begin{gathered} 3x + y + z = 36 \\ {}+ \tfrac{27}{2} + \tfrac{44}{3} \\ = \tfrac{385}{6}. \end{gathered}$

Thus $m + n = 385 + 6 = 391,$ and the correct answer is D.

2014 AMC 12B Problem 24

Answer: D

Difficulty rating: 2650

Solution:

Problem 24 in Other Years