2011 AMC 12A Problem 24

Suppose a circle of radius $r$ centered at $X$ fits in one of the quadrilaterals. If $h_1,h_2,h_3,h_4$ are the distances from $X$ to the four side lines, then each $h_i\ge r.$ Splitting the quadrilateral into four triangles gives $$\begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r. \end{aligned}$$ Bretschneider’s inequality bounds the area of any quadrilateral with these sides by the cyclic case: $$\begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6. \end{aligned}$$ Therefore $r\le \frac{K}{21}\le2\sqrt6.$ Equality is attainable: the cyclic quadrilateral with these sides is also tangential because $14+7=9+12,$ and its incircle has radius $\frac{K}{21}=2\sqrt6.$ Thus, the correct answer is C.