Question

A circle of radius $r$ is concentric with and outside a regular hexagon of side length $2.$ The probability that three entire sides of the hexagon are visible from a randomly chosen point on the circle is $1/2.$ What is $r?$

$2\sqrt{2} + 2\sqrt{3}$

$3\sqrt{3} + \sqrt{2}$

$2\sqrt{6} + \sqrt{3}$

$3\sqrt{2} + \sqrt{6}$

$6\sqrt{2} - \sqrt{3}$

Solution:

Place the hexagon at the center of the circle. There are six congruent arcs from which three whole sides are visible; since the total probability is $\tfrac{1}{2},$ each arc measures $30^\circ.$

Take the arc centered at $(r, 0)$ with upper endpoint $P,$ so $\angle POA = 15^\circ.$ Then $P$ lies on the line containing a side whose distance from the center is the apothem $\sqrt{3}.$

Hence $\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4},$ giving $r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}.$

Thus, the correct answer is D.

2006 AMC 12A Problem 22

Solution:

Problem 22 in Other Years