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2006 AMC 12A Problems/Problem 22

Revision as of 18:24, 5 January 2008 by 1=2 (talk | contribs) (→‎Problem)

Problem

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

$\mathrm{(A) \ } 2\sqrt{2}+2\sqrt{3}\qquad \mathrm{(B) \ } 3\sqrt{3}+\sqrt{2}\qquad \rm{(C) \ } 2\sqrt{6}+\sqrt{3}\qquad \rm{(D) \ } 3\sqrt{2}+\sqrt{6}$

$\mathrm{(E) \ } 6\sqrt{2}-\sqrt{3}$

Solution

Project any two non-adjacent and non-opposite sides of the hexagon to the circle; the arc between these two points is the locations where all three sides of the hexagon can be fully viewed. Since there are six such pairs of sides, there are six arcs. The probability of choosing a point is $1 / 2$, or if the total arc degree measures add up to $\frac{1}{2} * 360\deg = 180^o$. Divide that by six, and each arc must equal $30^\mathrm{o}$.

Call the center $O$, and the two endpoints of the arc $A$ and $B$. Angle AOB is equal to the central arc, so it is also 30 degrees. If we extend a line from the center to the midpoint between AB, and we project one of the sides of the hexagon to A and/or B, an isosceles triangle AXO is formed, with X being the intersection of the extension to the midpoint of AB and the extended side of the hexagon. It is isosceles because angle AOX is 15 degrees, and $OAX = OAB - 60 = \frac{180-30}{2} - 60 = 15$.

If we create the perpendicular bisector of AXO from X, then we get a right triangle:

$\cos 15 = \frac{\frac{r}{2}}{2\sqrt{3}} = \frac{r}{4\sqrt{3}}$

Since $\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4}$, we get:

$\frac{\sqrt{6} + \sqrt{2}}{4} 4\sqrt{3} = r$
$r = 3\sqrt{2} + \sqrt{6} \Rightarrow \mathrm{D}$

See also

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 12 Problems and Solutions
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