Consider all quadrilaterals $ABCD$ such that $AB=14$, $BC=9$, $CD=7$, and $DA=12$. What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral? $\textbf{(A)}\ \sqrt{15} \qquad$ $ \textbf{(B)}\ \sqrt{21} \qquad$ $ \textbf{(C)}\ 2\sqrt{6} \qquad$ $ \textbf{(D)}\ 5 \qquad$ $ \textbf{(E)}\ 2\sqrt{7}$
Solution
Since $AB+CD=14+7=21=BC+DA$, the Pitot condition holds, so these sides can form a quadrilateral with an inscribed circle.
If a circle of radius $r$ lies inside a convex quadrilateral of perimeter $2s$, its center is at distance at least $r$ from each side, so $[ABCD]\ge\frac12r\cdot2s=rs$, i.e. $r\le\dfrac{[ABCD]}s$.
The area is largest for the cyclic quadrilateral; by Brahmagupta's formula, with $s=21$, it is $\sqrt{(21-14)(21-9)(21-7)(21-12)}=\sqrt{7\cdot12\cdot14\cdot9}=42\sqrt6$. A cyclic quadrilateral whose opposite sides have equal sums is also tangential, so $r=\dfrac{42\sqrt6}{21}=2\sqrt6$ is attained.
Consider all quadrilaterals $ABCD$ such that $AB=10$, $BC=7$, $CD=6$, and $DA=9$. What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?
$\textbf{(A)}\ \sqrt{15} \qquad$ $ \textbf{(B)}\ \sqrt{21} \qquad$ $ \textbf{(C)}\ 2\sqrt{6} \qquad$ $ \textbf{(D)}\ 5 \qquad$ $ \textbf{(E)}\ 2\sqrt{7}$ 
