2011 AMC 12A Problems/Problem 24

Problem

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

Solution 1

Note as above that ABCD must be cyclic to obtain the circle with maximal radius. Let $E$ , $F$ , $G$ , and $H$ be the points on $AB$ , $BC$ , $CD$ , and $DA$ respectively where the circle is tangent. Let $\theta=\angle BAD$ and $\alpha=\angle ADC$ . Since the quadrilateral is cyclic, $\angle ABC=180^{\circ}-\alpha$ and $\angle BCD=180^{\circ}-\theta$ . Let the circle have center $O$ and radius $r$ . Note that $OHD$ , $OGC$ , $OFB$ , and $OEA$ are right angles.

Hence $FOG=\theta$ , $GOH=180^{\circ}-\alpha$ , $EOH=180^{\circ}-\theta$ , and $FBE=\alpha$ .

Therefore, $AEOH\sim OFCG$ and $EBFO\sim HOGD$ .

Let $x=CG$ . Then $CF=x$ , $BF=BE=9-x$ , $GD=DH=7-x$ , and $AH=AE=x+5$ . Using $AEOH\sim OFCG$ and $EBFO\sim HOGD$ we have $r/(x+5)=x/r$ , and $(9-x)/r=r/(7-x)$ . By equating the value of $r^2$ from each, $x(x+5)=(7-x)(9-x)$ . Solving we obtain $x=3$ so that $\boxed{\textbf{(C)}\ 2\sqrt{6}}$ .

Solution 2

To maximize the radius of the circle, we also need to maximize the area. To maximize the area of the circle, the quadrilateral must be tangential (have an incircle). In a tangential quadrilateral, the sum of opposite sides is equal to the semiperimeter of the quadrilateral. So, $14+7=12+9$ . Therefore, it has an incircle. By definition, a cyclic quadrilateral has the maximum area for a quadrilateral with corresponding side lengths. Therefore, to maximize the area of the quadrilateral and thus the incircle, we assume that this quadrilateral is cyclic.

For cyclic quadrilaterals, the area is $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semiperimeter of the cyclic quadrilateral and $a, b, c,$ and $d$ are the sides of the quadrilateral. Compute this area to get $42\sqrt{6}$ . The area of a tangential quadrilateral is given by the $rs$ formula, where $rs=A$ . We can substitute $s$ , the semiperimeter, and $A$ , the area and solve for r to get $\boxed{\textbf{(C)}\ 2\sqrt{6}}$ .

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2011 AMC 12A Problems/Problem 24

Contents

Problem

Solution

Solution 1

Solution 2

See also