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

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\textbf{(E)}\ 2\sqrt{7} </math>
 
\textbf{(E)}\ 2\sqrt{7} </math>
  
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== Solution ==
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== Solution 1 ==
  
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Since Area = <math>r \times</math> semi-perimeter, and perimeter is fixed, we can maximize the area. Let the angle between the 14 and 12 be <math>\alpha</math> degree, and the one between the 9 and 7 be <math>\beta</math>.
+
Note as above that ABCD must be tangential to obtain the circle with maximal radius. Let <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math> be the points on <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math> respectively where the circle is tangent. Let <math>\theta=\angle BAD</math> and <math>\alpha=\angle ADC</math>. Since the quadrilateral is cyclic(because we want to maximize the circle, so we set the quadrilateral to be cyclic), <math>\angle ABC=180^{\circ}-\alpha</math> and <math>\angle BCD=180^{\circ}-\theta</math>. Let the circle have center <math>O</math> and radius <math>r</math>. Note that <math>OHD</math>, <math>OGC</math>, <math>OFB</math>, and <math>OEA</math> are right angles.  
  
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2(Area) = <math>(14)(12) \sin \alpha + (9)(7) \sin \beta</math>
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Hence <math>FOG=\theta</math>, <math>GOH=180^{\circ}-\alpha</math>, <math>EOH=180^{\circ}-\theta</math>, and <math>FOE=\alpha</math>.
  
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<math>\frac{2}{21}</math> (Area) = <math>8 \sin \alpha + 3 \sin \beta</math>
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Therefore, <math>AEOH\sim OFCG</math> and <math>EBFO\sim HOGD</math>.
  
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<br />
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Let <math>x=CG</math>. Then <math>CF=x</math>, <math>BF=BE=9-x</math>, <math>GD=DH=7-x</math>, and <math>AH=AE=x+5</math>. Using <math>AEOH\sim OFCG</math> and <math>EBFO\sim HOGD</math> we have <math>r/(x+5)=x/r</math>, and <math>(9-x)/r=r/(7-x)</math>. By equating the value of <math>r^2</math> from each, <math>x(x+5)=(7-x)(9-x)</math>. Solving we obtain <math>x=3</math> so that <math>r=\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
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By law of cosine, <math>14^2 + 12 ^2 - 2(14)(12) \cos \alpha = 9^2 + 7^2 - 2(9)(7) \cos \beta</math>
 
  
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<math>8 \cos \alpha - 3 \cos \beta = 5</math> (simple algebra left to the reader)
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== Solution 2 ==
  
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<br />
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To maximize the radius of the circle, we also need to maximize its 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. <math>14+7=12+9</math>, so this particular quadrilateral has an incircle. By definition, given <math>4</math> side lengths, a cyclic quadrilateral has the maximum area of any quadrilateral with those 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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<math>\frac{4}{441}</math> (Area)<math>^2 + 25</math> = <math>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</math>  
+
For cyclic quadrilaterals, [[Brahmagupta's formula]] gives the area as <math>\sqrt{(s-a)(s-b)(s-c)(s-d)}</math> where <math>s</math> is the semiperimeter and <math>a, b, c,</math> and <math>d</math> are the side lengths. Breaking it up into <math>4</math> triangles, we see the area of a tangential quadrilateral is also equal to <math>r*s</math>. Equate these two equations.
 +
Substituting <math>s</math>, the semiperimeter, and <math>A</math>, the area and solving for <math>r</math>,we get <math>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
  
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<math>= 73 - 48 \cos (\alpha + \beta)</math>
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== Solution 3 (Trigonometry) ==
  
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<math>\frac{4}{441}</math> (Area)<math>^2</math> = <math>48 ( 1- \cos (\alpha + \beta))</math>, which reaches maximum when <math>( 1- \cos (\alpha + \beta)) = 2</math>.  
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By Pitot's Theorem, since <math>AB+CD=AD+BC</math>, there exists a circle tangent to all four sides of quadrilateral <math>ABCD</math>. Thus, we need to find the radius of this circle.
  
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(and since it is a quadrilateral, it is possible to have <math>\alpha + \beta = \pi</math> (hence cyclic quadrilateral, that would be the best guess and the Brahmagupta's formula would work for area and the work is simple).
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Let the circle be tangent to <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>AD</math> at points <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math>, respectively. Also, let <math>AP=x</math>. Then <math>AS</math> also equals <math>x</math>. Let the center of the circle be <math>O</math>. Observe that <math>AO</math> bisects angle <math>\angle A</math>, so <math>\cot\frac{A}{2}=\frac{x}{r}</math>. Moreover, <math>\cot\frac{C}{2}=\frac{9-(14-x)}{r}=\frac{x-5}{r}</math>. But <math>\angle \frac{C}{2}=90^\circ-\angle \frac{A}{2}</math>, so <math>\cot\frac{C}{2}=\tan\frac{A}{2}</math>, and we find that <math>\frac{r}{x}=\frac{x-5}{r}</math>. Hence, <math>r^2=x(x-5)</math>.
  
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<math>\frac{4}{441}</math> (Area)<math>^2 \le 96</math>
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Similarly, <math>\cot\frac{B}{2}=\frac{14-x}{r}</math>, and <math>\cot\frac{D}{2}=\frac{12-x}{r}=\tan\frac{B}{2}=\frac{r}{14-x}</math>. Therefore, <math>r^2=\frac{12-x}{14-x}</math>. Putting this together with the above equation yields <cmath>(12-x)(14-x)=x(x-5)</cmath><cmath>\Longrightarrow x^2-26x+168=x^2-5x</cmath><cmath>\Longrightarrow 21x=168</cmath><cmath>\Longrightarrow x=8.</cmath> Thus, <math>r^2=8(8-5)=8(3)=24</math>, and <math>r=\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
  
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(Area)<math>^2 \le 24 (441)</math>
 
  
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(Area)<math> \le 42 \sqrt{6}</math>, Area = <math>r \times</math> semi-perimeter.
+
== Solution 4 (wowe) ==
 +
Let <math>r</math> be the inradius. Then, the area of the quadrilateral can be expressed as <math>21r</math>. Clearly, the larger the area of the quadrilateral, the larger the value of the inradius.
  
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Hence, <math>r = 2 \sqrt{6}</math>, choice <math>(C)</math>
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Know that the area of a quadrilateral is maximized when it is cyclic. Applying Brahmagupta's, the area will be <math>\sqrt{(9)(7)(12)(14)} = 42\sqrt{6}</math>. Since <math>21r = 42\sqrt{6}</math>, <math>r = \boxed{\textbf{(C) } 2\sqrt6}</math>.
 +
 
 +
-skibbysiggy
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=23|num-a=25|ab=A}}
 
{{AMC12 box|year=2011|num-b=23|num-a=25|ab=A}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:49, 4 September 2024

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 1

Note as above that ABCD must be tangential 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(because we want to maximize the circle, so we set the quadrilateral to be 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 $FOE=\alpha$.

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

Solution 2

To maximize the radius of the circle, we also need to maximize its 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. $14+7=12+9$, so this particular quadrilateral has an incircle. By definition, given $4$ side lengths, a cyclic quadrilateral has the maximum area of any quadrilateral with those side lengths. Therefore, to maximize the area of the quadrilateral and thus the incircle, we assume that this quadrilateral is cyclic.

For cyclic quadrilaterals, Brahmagupta's formula gives the area as $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semiperimeter and $a, b, c,$ and $d$ are the side lengths. Breaking it up into $4$ triangles, we see the area of a tangential quadrilateral is also equal to $r*s$. Equate these two equations. Substituting $s$, the semiperimeter, and $A$, the area and solving for $r$,we get $\boxed{\textbf{(C)}\ 2\sqrt{6}}$.

Solution 3 (Trigonometry)

By Pitot's Theorem, since $AB+CD=AD+BC$, there exists a circle tangent to all four sides of quadrilateral $ABCD$. Thus, we need to find the radius of this circle.

Let the circle be tangent to $AB$, $BC$, $CD$, and $AD$ at points $P$, $Q$, $R$, and $S$, respectively. Also, let $AP=x$. Then $AS$ also equals $x$. Let the center of the circle be $O$. Observe that $AO$ bisects angle $\angle A$, so $\cot\frac{A}{2}=\frac{x}{r}$. Moreover, $\cot\frac{C}{2}=\frac{9-(14-x)}{r}=\frac{x-5}{r}$. But $\angle \frac{C}{2}=90^\circ-\angle \frac{A}{2}$, so $\cot\frac{C}{2}=\tan\frac{A}{2}$, and we find that $\frac{r}{x}=\frac{x-5}{r}$. Hence, $r^2=x(x-5)$.

Similarly, $\cot\frac{B}{2}=\frac{14-x}{r}$, and $\cot\frac{D}{2}=\frac{12-x}{r}=\tan\frac{B}{2}=\frac{r}{14-x}$. Therefore, $r^2=\frac{12-x}{14-x}$. Putting this together with the above equation yields \[(12-x)(14-x)=x(x-5)\]\[\Longrightarrow x^2-26x+168=x^2-5x\]\[\Longrightarrow 21x=168\]\[\Longrightarrow x=8.\] Thus, $r^2=8(8-5)=8(3)=24$, and $r=\boxed{\textbf{(C)}\ 2\sqrt{6}}$.


Solution 4 (wowe)

Let $r$ be the inradius. Then, the area of the quadrilateral can be expressed as $21r$. Clearly, the larger the area of the quadrilateral, the larger the value of the inradius.


Know that the area of a quadrilateral is maximized when it is cyclic. Applying Brahmagupta's, the area will be $\sqrt{(9)(7)(12)(14)} = 42\sqrt{6}$. Since $21r = 42\sqrt{6}$, $r = \boxed{\textbf{(C) } 2\sqrt6}$.

-skibbysiggy

See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 12 Problems and Solutions

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