<s>[INST] Viewing sides of a hexagon in a circle I need help with this problem:

Take a circle with radius r, and place a regular hexagon of side length 2 so that the circle and hexagon are concentric. The probability of picking a point on the circle where three entire sides of the hexagon can be seen is 0.5. What is r?

My approach would be like this:

The hexagon would be made up of six triangles congruent to triangle ACB. Therefore, radius r would be equal to two times the height of ACB plus the radius of sector CDE.
Now, consider this diagram:

Beautiful diagram indeed. In this diagram, the shaded areas are where three hexagon sides can be seen. Therefore, these six sectors should have an area that’s half the area of the entire circle. We also notice that, since each of the sectors have an angle of 60, they form one circle, with a radius of r minus two times the height of ACB.
First we find the height of ACB. With some Pythagoras, we find that the height is $\sqrt{3}$, so two times that height would be $2\sqrt{3}$. The radius of the smaller circle would therefore be $r-2\sqrt{3}$. Solve:
$$2\pi(r-2\sqrt{3})^2=\pi r^2$$
$$2(r-2\sqrt{3})^2=r^2$$
$$2(r^2-4r\sqrt{3}+12)=r^2$$
$$2r^2-8r\sqrt{3}+24=r^2$$
$$r^2-8r\sqrt{3}+24=0$$
$$r=\dfrac{8\sqrt{3}\pm\sqrt{192-96}}{2}$$
$$r=\dfrac{8\sqrt{3}\pm 4\sqrt{6}}{2}=\color{blue}{4\sqrt{3}\pm 2\sqrt{6}}$$
This was the answer I got (though I’m wary of $4\sqrt{3}-2\sqrt{6}$, seems weird, as it’s very close to 2). However, the textbook I’m using says it’s $\color{red}{3\sqrt{2}+\sqrt{6}}$. Does the textbook answer include the area inside the hexagon, where 6 sides can be seen? What am I doing wrong?
 [/INST] Here's one way . . .

The length of arc $DE$ must be equal to ${\large{\frac{1}{12}}}$ of the circumference, hence
$$
\angle{DOE}=\frac{\pi}{6}
$$
where $O$ is the center of the circle.

Let $x$ be the length of the line segment $DE$.

Applying the law of cosines to triangle $DOE$, we get
$$
x^2=r^2(2-\sqrt{3})\qquad(*)
$$
Let $F$ be the vertex of the hexagon adjacent to and counterclockwise from $A$.

Applying the law of cosines to triangle $EFO$, we get
$$
r^2=x^2+6x+12
$$
Using the above to replace $r^2$ in $(*)$, then solving the resulting quadratic equation for $x$, we get $x=2\sqrt{3}$ (the other root is negative).

Replacing $x$ by $2\sqrt{3}$ in $(*)$ yields
$$
r^2=24+12\sqrt{3}
$$
hence, noting that $24+12\sqrt{3}=(3\sqrt{2}+\sqrt{6})^2$, we get 
$$
r=3\sqrt{2}+\sqrt{6}
$$
</s>