Question

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?

$\sqrt{15}$

$\sqrt{21}$

$2\sqrt{6}$

$5$

$2\sqrt{7}$

Answer: C

Concepts:incircle, incenter, and inradius Brahmagupta’s Formula cyclic quadrilateral optimization

Difficulty rating: 2460

Solution:

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 $r\le K/21\le2\sqrt6.$ Equality is attainable: the cyclic quadrilateral with these sides is also tangential because $14+7=9+12,$ and its incircle has radius $K/21=2\sqrt6.$

Thus, the correct answer is C.

2011 AMC 12A Problem 24

Answer: C

Difficulty rating: 2460

Solution:

Problem 24 in Other Years