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 $p + q.$

答案

Solution:

Let $r$ be the radius, and let each chord of length $22$ subtend central angle $\alpha$ and each chord of length $20$ subtend $\beta.$ The six central angles fill the circle: $4\alpha + 2\beta = 360^\circ,$ so $\frac{\beta}{2} = 90^\circ - \alpha$ and $\sin\frac{\beta}{2} = \cos\alpha.$

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, $1 - \frac{242}{r^2} = \frac{10}{r} \quad\Longrightarrow\quad r^2 - 10r - 242 = 0,$ so $r = 5 + \sqrt{267}$ (taking the positive root).

Therefore $p + q = 5 + 267 = 272.$

2013 AIME II Problem 8

Solution:

Problem 8 in Other Years