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"

m (→‎Solution 2)
(→‎Solution)
Line 12: Line 12:
  
 
=== Solution 1 ===
 
=== Solution 1 ===
−
 
−
Since Area = <math>r \cdot </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>.
 
−
 
−
2(Area) = <math>(14)(12) \sin \alpha + (9)(7) \sin \beta</math>
 
−
 
−
<math>\frac{2}{21}</math> (Area) = <math>8 \sin \alpha + 3 \sin \beta</math>
 
−
 
−
<br />
 
−
By the law of cosines, <math>14^2 + 12 ^2 - 2(14)(12) \cos \alpha = 9^2 + 7^2 - 2(9)(7) \cos \beta</math>
 
−
 
−
<math>8 \cos \alpha - 3 \cos \beta = 5</math> (simple algebra left to the reader)
 
−
 
−
<br />
 
−
 
−
<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>
 
−
 
−
<math>= 73 - 48 \cos (\alpha + \beta)</math>
 
−
 
−
<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>.
 
−
 
−
(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).
 
−
 
−
<math>\frac{4}{441}</math> (Area)<math>^2 \le 96</math>
 
−
 
−
(Area)<math>^2 \le 24 (441)</math>
 
−
 
−
(Area)<math> \le 42 \sqrt{6}</math>, Area = <math>r \cdot </math> semi-perimeter.
 
−
 
−
Hence, <math>r=\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>
 
−
 
−
===Solution 2===
 
  
 
Note as above that ABCD must be cyclic 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, <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.  
 
Note as above that ABCD must be cyclic 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, <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.  
Line 52: Line 21:
 
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>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
 
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>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
  
−
=== Solution 3 ===
+
=== Solution 2 ===
  
 
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, <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 tangential quadrilateral and thus the incircle, we assume that this quadrilateral is cyclic. For cyclic quadrilaterals, the area is given by <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>, and area, for their respective values and solve for r to get <math>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
 
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, <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 tangential quadrilateral and thus the incircle, we assume that this quadrilateral is cyclic. For cyclic quadrilaterals, the area is given by <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>, and area, for their respective values and solve for r to get <math>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.

Revision as of 20:00, 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 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 $\boxed{\textbf{(C)}\ 2\sqrt{6}}$.

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=57145"