**Problem**: (2006 AMC 12A - #22) 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 $\frac{1}{2}$. What is $r$?

**Solution**: Consider the six arcs formed by extending two side to intersect the circle as shown in the diagram ($AB$ is one of these arcs). They represent the points from which one can see three entire sides of the hexagon (key step). Since there are six of them and must encompass half of the total arc length of the circle, or $180^{\circ}$, each must be $30^{\circ}$. Therefore we have $ \angle AOB = 30^{\circ} $.

Then we have $ \angle BAO = 75^{\circ} $ because $ \triangle AOB $ is isosceles. Note that $ \triangle ACB $ is equilateral because $ \angle ACB = 60^{\circ} $ and it is isosceles. Let $D$ be the foot of the perpendicular from $O$ to $AC$. We have $ \angle DAO = \angle CAO = \angle BAO-\angle BAC = 75^{\circ}-60^{\circ} = 15^{\circ} $. Note that $ DO = \sqrt{3} $ by the $ 30-60-90 $ triangles formed by the hexagon. Then the radius is $ AO = \frac{DO}{\sin{15^{\circ}}} = \frac{\sqrt{3}}{\frac{\sqrt{6}-\sqrt{2}}{4}} = 3\sqrt{2}+\sqrt{6} $. QED.

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Comment: I really liked this problem, because it was cool to translate the wording into a diagram and then if you drew the right line ($OD$) you were set with simple trigonometry. Other solutions use the Law of Cosines or the Law of Sines, but I personally favor this one which is quick and rather elegant (besides the unavoidable $\sin{15^{\circ}}$).

Hey, what do you think you got on the AMC A test?

ReplyDeleteI got a 144. I missed the last question, sadly enough :(

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