2013 AIME II Problem 8

Question

A hexagon that is inscribed in a circle has side lengths

22,

22,

20,

22,

22,

and

20

in that order. The radius of the circle can be written as

p + \sqrt{q},

where

p

and

q

are positive integers. Find

p + q.

Answer

If each

22

-chord subtends central angle

\alpha

and each

20

-chord subtends

\beta,

then

4\alpha + 2\beta = 360^\circ,

so

\sin\frac{\beta}{2} = \cos\alpha

The law of cosines gives

\cos\alpha = 1 - \frac{242}{r^2},

while

\sin\frac{\beta}{2} = \frac{10}{r};

equating them yields a quadratic in

r

Half a

20

-chord gives

\sin\frac{\beta}{2} = \frac{10}{r},

and the law of cosines on the isosceles triangle with legs

r

and base

22

gives

22^2 = 2r^2(1 - \cos\alpha),

so

\cos\alpha = 1 - \frac{242}{r^2}.

Equating,

\begin{aligned} 1 - \frac{242}{r^2} &= \frac{10}{r} \\ &\quad\Longrightarrow\quad r^2 - 10r - 242 \\ &= 0, \end{aligned}

so

r = 5 + \sqrt{267}

(taking the positive root).

Therefore

p + q = 5 + 267 = 272.

Problem 8 in other years