2014 AMC 12B Problem 24

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

Equal chords subtend equal arcs, so

AC = BD = CE;

call this

x

Apply Ptolemy’s theorem to

ABCD,

BCDE,

and

ABDE

to get equations in

x, y = AD, z = BE

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.

Problem 24 in other years