1996 AMC 12 Problem 30

Question

A hexagon inscribed in a circle has three consecutive sides each of length

3

and three consecutive sides each of length

5.

The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length

3

and the other with three sides each of length

5,

has length equal to

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m+n.

309

349

369

389

409

Let

\alpha

and

\beta

be the half-central angles subtended by sides

3

and

5

Use

\alpha+\beta=60^\circ

and the chord ratio to find

\sin^2\alpha

, then apply

\frac{\sin3\alpha}{\sin\alpha}

Let the circle have radius

R,

and let

\alpha,\beta

be the half-central angles for the sides

3,5.

Then

3=2R\sin\alpha,

5=2R\sin\beta,

and

\alpha+\beta=60^\circ.

Thus

\frac53=\frac{\sin(60^\circ-\alpha)}{\sin\alpha}

=\frac{\sqrt3}{2}\cot\alpha-\frac12,

so

\tan\alpha=\frac{3\sqrt3}{13}

and

\sin^2\alpha=\frac{27}{196}.

The dividing chord spans the three consecutive sides of length

3,

so

\frac L3=\frac{\sin3\alpha}{\sin\alpha}

=3-4\sin^2\alpha=3-\frac{27}{49}=\frac{120}{49}.

Therefore

L=\frac{360}{49},

so

m+n=409.

The correct answer is E.

Problem 30 in other years