Difference between revisions of "2006 AMC 10B Problems/Problem 24"
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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> | + | 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> | + | 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. | ||
− | Therefore <math>DX=2</math> and <math>PX=2</math> | + | 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>. | ||
− | So the area of quadrilateral <math>OADP</math> is <math>8\sqrt{2}+4\sqrt{2}=12\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> | 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 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