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

 
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== Problem ==
 
== Problem ==
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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>?
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[[Image:2006amc10b24.gif]]
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<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>
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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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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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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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By the [[Pythagorean Theorem]]:
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<math>OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}</math>.
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The area of <math>OADX</math> is <math>2\cdot4\sqrt{2}=8\sqrt{2}</math>.
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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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Using similar steps, the area of quadrilateral <math>OBCP</math> is also <math>12\sqrt{2}</math>
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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 17:16, 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