Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 12A Problems/Problem 24"
(→‎Solution)
Line 36: Line 36:
 
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>.
 
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>.
  
 +
=== Solution 4 (Areas) ===
  
 +
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>. 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 the inscribed circle have center <math>O</math> and radius <math>r</math>. Note that <math>OE</math>, <math>OF</math>, <math>OG</math>, and <math>OH</math> are perpendicular to sides <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively. Thus <math>\angle AOH=\angle AOE</math>, <math>\angle BOE=\angle BOF</math>, <math>\angle COF=\angle COG</math>, and <math>\angle DOG=\angle DOH</math>.
 +
 +
From <math>\angle AOH+\angle AOE+\angle BOE+\angle BOF+\angle COF+\angle COG+\angle DOG+\angle DOH=360^{\circ}</math>, we get the two equations <math>\angle AOE+\angle BOE+\angle COG+\angle DOG=180^{\circ}</math> and <math>\angle AOH+\angle DOH+\angle BOF+\angle COF=180^{\circ}</math>. These are equivalent to <math>\angle AOB+\angle COD=180^{\circ}</math> and <math>\angle AOD+\angle BOC=180^{\circ}</math>.
 +
 +
The ratio of the area of <math>AOB</math> to <math>COD</math> is <math>2</math>, since the heights are both <math>r</math>. However, we can also express this ratio as <math>\frac{\frac{1}{2}\times AO\times BO\times \sin (AOB)}{\frac{1}{2}\times CO\times DO\times \sin (COD)}</math>. Since <math>\angle AOB=180^{\circ}-\angle COD</math>, <math>\sin (AOB)=\sin (COD)</math>. Thus <math>\frac{AO\times BO}{CO\times DO}=2</math>. Similarly, <math>\frac{AO\times DO}{BO\times CO}=\frac{12}{9}=\frac{4}{3}</math>. From these two equations we have <math>BO=\frac{\sqrt{6}}{2}\times DO</math>.
 +
 +
Now let <math>AH=AE=a</math>, <math>BE=BF=b</math>, <math>CF=CG=c</math>, and <math>DG=DH=d</math>. We thus have the equations <math>a+b=14</math>, <math>b+c=9</math>, <math>c+d=7</math>, and <math>d+a=12</math>. Solving gives us <math>a=8</math>, <math>b=6</math>, <math>c=3</math>, and <math>d=4</math>. By the Pythagorean theorem, we have <math>BO^2-b^2=DO^2-d^2</math>. This is <math>DO^2\times\frac{3}{2}-36=DO^2-16</math>, which gives us <math>DO^2=40</math>. Thus, <math>r=\sqrt{DO^2-d^2}=\sqrt{24}</math>, and <math>r=\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:28, 28 December 2019

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 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, $\angle ABC=180^{\circ}-\alpha$ 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 $\boxed{\textbf{(C)}\ 2\sqrt{6}}$.

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 $\angle \frac{D}{2}=90^\circ-\angle \frac{A}{2}$, 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 (Areas)

By Pitot's Theorem, since $AB+CD=AD+BC$, there exists a circle tangent to all four sides of quadrilateral $ABCD$. Let $E$, $F$, $G$, and $H$ be the points on $AB$, $BC$, $CD$, and $DA$ respectively where the circle is tangent. Let the inscribed circle have center $O$ and radius $r$. Note that $OE$, $OF$, $OG$, and $OH$ are perpendicular to sides $AB$, $BC$, $CD$, and $DA$, respectively. Thus $\angle AOH=\angle AOE$, $\angle BOE=\angle BOF$, $\angle COF=\angle COG$, and $\angle DOG=\angle DOH$.

From $\angle AOH+\angle AOE+\angle BOE+\angle BOF+\angle COF+\angle COG+\angle DOG+\angle DOH=360^{\circ}$, we get the two equations $\angle AOE+\angle BOE+\angle COG+\angle DOG=180^{\circ}$ and $\angle AOH+\angle DOH+\angle BOF+\angle COF=180^{\circ}$. These are equivalent to $\angle AOB+\angle COD=180^{\circ}$ and $\angle AOD+\angle BOC=180^{\circ}$.

The ratio of the area of $AOB$ to $COD$ is $2$, since the heights are both $r$. However, we can also express this ratio as $\frac{\frac{1}{2}\times AO\times BO\times \sin (AOB)}{\frac{1}{2}\times CO\times DO\times \sin (COD)}$. Since $\angle AOB=180^{\circ}-\angle COD$, $\sin (AOB)=\sin (COD)$. Thus $\frac{AO\times BO}{CO\times DO}=2$. Similarly, $\frac{AO\times DO}{BO\times CO}=\frac{12}{9}=\frac{4}{3}$. From these two equations we have $BO=\frac{\sqrt{6}}{2}\times DO$.

Now let $AH=AE=a$, $BE=BF=b$, $CF=CG=c$, and $DG=DH=d$. We thus have the equations $a+b=14$, $b+c=9$, $c+d=7$, and $d+a=12$. Solving gives us $a=8$, $b=6$, $c=3$, and $d=4$. By the Pythagorean theorem, we have $BO^2-b^2=DO^2-d^2$. This is $DO^2\times\frac{3}{2}-36=DO^2-16$, which gives us $DO^2=40$. Thus, $r=\sqrt{DO^2-d^2}=\sqrt{24}$, and $r=\boxed{\textbf{(C)}\ 2\sqrt{6}}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_24&oldid=113592"