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Difference between revisions of "2011 AMC 12A Problems/Problem 24"

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=== Solution 2 ===
 
=== Solution 2 ===
  
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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 a of opposite sides is equal to the semiperimeter of the quadrilateral. So, <math>14+7=12+9</math>. 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.  
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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, <math>14+7=12+9</math>. 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 <math>\sqrt{(s-a)(s-b)(s-c)(s-d)}</math> where <math>s</math> is the semiperimeter of the cyclic quadrilateral and <math>a, b, c,</math> and <math>d</math> are the sides of the quadrilateral. Compute this area to get <math>42\sqrt{6}</math>. The area of a tangential quadrilateral is given by the <math>rs</math> formula, where <math>rs=A</math>. We can substitute <math>s</math>, the semiperimeter, and <math>A</math>, the area and solve for r to get <math>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
 
For cyclic quadrilaterals, the area is <math>\sqrt{(s-a)(s-b)(s-c)(s-d)}</math> where <math>s</math> is the semiperimeter of the cyclic quadrilateral and <math>a, b, c,</math> and <math>d</math> are the sides of the quadrilateral. Compute this area to get <math>42\sqrt{6}</math>. The area of a tangential quadrilateral is given by the <math>rs</math> formula, where <math>rs=A</math>. We can substitute <math>s</math>, the semiperimeter, and <math>A</math>, the area and solve for r to get <math>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.

Revision as of 20:04, 22 September 2013

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 $\boxed{\textbf{(C)}\ 2\sqrt{6}}$.

See also

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
All AMC 12 Problems and Solutions

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