Difference between revisions of "2006 AMC 10B Problems/Problem 24"
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== 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> | ||
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== 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 17:16, 16 July 2006
Problem
Circles with centers and
have radii
and
, respectively, and are externally tangent. Points
and
on the circle with center
and points
and
on the circle with center
are such that
and
are common external tangents to the circles. What is the area of the concave hexagon
?
Solution
Since a tangent line is perpendicular to the radius containing the tangent point,
Construct a perpendicular to that goes through point
. Label the point of intersection
Clearly is a rectangle.
Therefore and
By the Pythagorean Theorem:
.
The area of is
.
The area of is
.
So the area of quadrilateral is
Using similar steps, the area of quadrilateral is also
Therefore, the area of hexagon is