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Difference between revisions of "2006 AMC 10B Problems/Problem 24"

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== Solution ==
 
== Solution ==
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Since a tangent line is perpendicular to the radius containing the tangent point, <math>\angle OAD = \angle PDA = 90^\circ</math>
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Since a tangent line is perpendicular to the radius containing the tangent point, <math>\angle OAD = \angle PDA = 90^\circ</math>.
  
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Construct a perpendicular to <math>DP</math> that goes through point <math>O</math>. Label the point of intersection <math>X</math>
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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.  
 
Clearly <math>OADX</math> is a rectangle.  
  
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Therefore <math>DX=2</math> and <math>PX=2</math>
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Therefore <math>DX=2</math> and <math>PX=2</math>.
  
 
By the [[Pythagorean Theorem]]:
 
By the [[Pythagorean Theorem]]:
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The area of <math>OXP</math> is <math>\frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}</math>.
 
The area of <math>OXP</math> is <math>\frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}</math>.
  
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So the area of quadrilateral <math>OADP</math> is <math>8\sqrt{2}+4\sqrt{2}=12\sqrt{2}</math>
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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>
 
Using similar steps, the area of quadrilateral <math>OBCP</math> is also <math>12\sqrt{2}</math>

Revision as of 13:21, 18 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

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