A circle is inscribed in a sector that is one sixth of a circle of radius 6 . (That is, the circle is tangent to both segments and the arc forming the sector.) Find, with proof, the radius of the small circle.

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Because the circles are tangent, we can draw the line $A T$, which passes through the center $O$ of the other circle. Note that triangles $A D O$ and $A E O$ are symmetric (this is HL congruence: $A O$ is shared, radii $O D$ and $O E$ are equal, and angles $A D O$ and $A E O$ are right). Therefore, since $\angle B A C$ is $60^{\circ}$, angles $B A T$ and $T A C$ are each $30^{\circ}$. Now $A D O$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ right triangle. If we let $r$ be the radius of the small circle, then $O D=r$ and $O A=2 r$, but $O T=r$ so $A T=3 r$. But $A T$ is the radius of the large circle, so $3 r=6$ and $r=2$.