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?

$\sqrt{15}$

$\sqrt{21}$

$2\sqrt{6}$

$5$

$2\sqrt{7}$

Solution:

Because $AB + CD = 14 + 7 = 21 = 9 + 12 = BC + DA,$ a tangential quadrilateral (one with an inscribed circle) with these sides exists. For a tangential quadrilateral the area equals $r \cdot s$ with semiperimeter $s = 21,$ so maximizing $r$ means maximizing the area.

Among tangential quadrilaterals with given sides, the largest area is achieved by the cyclic (bicentric) one, whose area is $\sqrt{(s-a)(s-b)(s-c)(s-d)} = \sqrt{7 \cdot 12 \cdot 14 \cdot 9} = 42\sqrt6.$

Then $r = \dfrac{42\sqrt6}{21} = 2\sqrt6.$

Thus, the correct answer is C.

2011 AMC 12A Problem 24

Solution:

Problem 24 in Other Years