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 "2006 AMC 10B Problems/Problem 24"

 
(Added problem and solution)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Circles with centers <math>O</math> and <math>P</math> have radii <math>2</math> and <math>4</math>, respectively, and are externally tangent. Points <math>A</math> and <math>B</math> on the circle with center <math>O</math> and points <math>C</math> and <math>D</math> on the circle with center <math>P</math> are such that <math>AD</math> and <math>BC</math> are common external tangents to the circles. What is the area of the concave hexagon <math>AOBCPD</math>?
 +
 +
[[Image:2006amc10b24.gif]]
 +
 +
<math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2} </math>
 +
 
== Solution ==
 
== Solution ==
 +
Since a tangent line is perpendicular to the radius containing the tangent point, <math>\angle OAD = \angle PDA = 90^\circ</math>
 +
 +
Construct a perpendicular to <math>DP</math> that goes through point <math>O</math>. Label the point of intersection <math>X</math>
 +
 +
Clearly <math>OADX</math> is a rectangle.
 +
 +
Therefore <math>DX=2</math> and <math>PX=2</math>
 +
 +
By the [[Pythagorean Theorem]]:
 +
<math>OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}</math>.
 +
 +
The area of <math>OADX</math> is <math>2\cdot4\sqrt{2}=8\sqrt{2}</math>.
 +
 +
The area of <math>OXP</math> is <math>\frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}</math>.
 +
 +
So the area of quadrilateral <math>OADP</math> is <math>8\sqrt{2}+4\sqrt{2}=12\sqrt{2}</math>
 +
 +
Using similar steps, the area of quadrilateral <math>OBCP</math> is also <math>12\sqrt{2}</math>
 +
 +
Therefore, the area of hexagon <math>AOBCPD</math> is <math>2\cdot12\sqrt{2}= 24\sqrt{2} \Rightarrow B </math>
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 16:16, 16 July 2006

Problem

Circles with centers $O$ and $P$ have radii $2$ and $4$, respectively, and are externally tangent. Points $A$ and $B$ on the circle with center $O$ and points $C$ and $D$ on the circle with center $P$ are such that $AD$ and $BC$ are common external tangents to the circles. What is the area of the concave hexagon $AOBCPD$?

2006amc10b24.gif

$\mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2}$

Solution

Since a tangent line is perpendicular to the radius containing the tangent point, $\angle OAD = \angle PDA = 90^\circ$

Construct a perpendicular to $DP$ that goes through point $O$. Label the point of intersection $X$

Clearly $OADX$ is a rectangle.

Therefore $DX=2$ and $PX=2$

By the Pythagorean Theorem: $OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}$.

The area of $OADX$ is $2\cdot4\sqrt{2}=8\sqrt{2}$.

The area of $OXP$ is $\frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}$.

So the area of quadrilateral $OADP$ is $8\sqrt{2}+4\sqrt{2}=12\sqrt{2}$

Using similar steps, the area of quadrilateral $OBCP$ is also $12\sqrt{2}$

Therefore, the area of hexagon $AOBCPD$ is $2\cdot12\sqrt{2}= 24\sqrt{2} \Rightarrow B$

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_24&oldid=7809"