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

Since Area = $r \cdot$ semi-perimeter, and perimeter is fixed, we can maximize the area. Let the angle between the 14 and 12 be $\alpha$ degree, and the one between the 9 and 7 be $\beta$ .

2(Area) = $(14)(12) \sin \alpha + (9)(7) \sin \beta$

$\frac{2}{21}$ (Area) = $8 \sin \alpha + 3 \sin \beta$

By the law of cosines, $14^2 + 12 ^2 - 2(14)(12) \cos \alpha = 9^2 + 7^2 - 2(9)(7) \cos \beta$

$8 \cos \alpha - 3 \cos \beta = 5$ (simple algebra left to the reader)

$\frac{4}{441}$ (Area) $^2 + 25$ = $64 \sin^2 \alpha + 9 \sin^2 \beta + 64 \cos^2 \alpha + 9 \cos^2 \beta - 48 \cos \alpha \cos \beta + 48 \sin \alpha \sin \beta$

$= 73 - 48 \cos (\alpha + \beta)$

$\frac{4}{441}$ (Area) $^2$ = $48 ( 1- \cos (\alpha + \beta))$ , which reaches maximum when $( 1- \cos (\alpha + \beta)) = 2$ .

(and since it is a quadrilateral, it is possible to have $\alpha + \beta = \pi$ (hence cyclic quadrilateral, that would be the best guess and the Brahmagupta's formula would work for area and the work is simple).

$\frac{4}{441}$ (Area) $^2 \le 96$

(Area) $^2 \le 24 (441)$

(Area) $\le 42 \sqrt{6}$ , Area = $r \cdot$ semi-perimeter.

Hence, $r=\boxed{\textbf{(C)}\ 2\sqrt{6}}$

Solution 2

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 $OHB$ , $OGC$ , $OFB$ , 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 3

To maximize the radius of the circle, we also maximize the area. To maximize the area of the circle, the quadrilateral must be tangential (have an incircle). A tangential quadrilateral has the property that the sum of a of opposite sides is equal to the semiperimeter of the quadrilateral. In this case, $14+7=12+9$ . 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 tangential quadrilateral and thus the incircle, we assume that this quadrilateral is cyclic. For cyclic quadrilaterals, the area is given by $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ 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 $A$ , and area, for their respective values 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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