2000 AMC 12 Problem 24

Each arc has radius $AB,$ and $C$ is at distance $AB$ from both $A$ and $B,$ so $\triangle ABC$ is equilateral. Thus arc $BC$ subtends $60^\circ$ of a circle of radius $AB,$ whose full circumference is $6 \cdot 12 = 72.$ Let the small circle have radius $r$ and touch $\overline{AB}$ at its midpoint $D,$ where $AD = \tfrac12 AB.$ By Power of a Point, $AD^2 = AB(AB - 2r),$ so $$\frac{AB^2}{4} = AB^2 - 2r\,AB,$$ giving $2r = \tfrac34 AB,$ hence $r = \tfrac38 AB.$ The circumferences are in the ratio of the radii, so the small circle’s circumference is $\tfrac38 \cdot 72 = 27.$ Thus, the correct answer is D.