Difference between revisions of "2006 AMC 10B Problems/Problem 24"
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== Problem == | == Problem == | ||
[[Circle]]s 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>? | [[Circle]]s 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]] --> | |
− | [[Image:2006amc10b24.gif]] | + | <asy>size(200);defaultpen(linewidth(0.8)); |
+ | pair X=(-6,0), O=origin, P=(6,0), B=tangent(X, O, 2, 1), A=tangent(X, O, 2, 2), C=tangent(X, P, 4, 1), D=tangent(X, P, 4, 2); | ||
+ | pair top=X+15*dir(X--A), bottom=X+15*dir(X--B); | ||
+ | draw(Circle(O, 2)^^Circle(P, 4)); | ||
+ | draw(bottom--X--top); | ||
+ | draw(A--O--B^^O--P^^D--P--C); | ||
+ | pair point=X; | ||
+ | label("$2$", midpoint(O--A), dir(point--midpoint(O--A))); | ||
+ | label("$4$", midpoint(P--D), dir(point--midpoint(P--D))); | ||
+ | label("$O$", O, SE); | ||
+ | label("$P$", P, dir(point--P)); | ||
+ | pair point=O; | ||
+ | label("$A$", A, dir(point--A)); | ||
+ | label("$B$", B, dir(point--B)); | ||
+ | pair point=P; | ||
+ | label("$C$", C, dir(point--C)); | ||
+ | label("$D$", D, dir(point--D)); | ||
+ | fill((-3,7)--(-3,-7)--(-7,-7)--(-7,7)--cycle, white);</asy> | ||
<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> | <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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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]], so <math>DX=2</math> and <math>PX=2</math>. By the [[Pythagorean Theorem]], <math>OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}</math>. |
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− | By the [[Pythagorean Theorem]] | ||
− | <math>OX = \sqrt{6^2 - 2^2} = 4 | ||
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− | Using similar steps, the area of quadrilateral <math>OBCP</math> is also <math>12\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} \Longrightarrow \boxed{\mathrm{(B)}}</math>. |
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== See also == | == See also == | ||
{{AMC10 box|year=2006|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2006|ab=B|num-b=23|num-a=25}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] | ||
+ | [[Category:Circle Problems]] |
Revision as of 00:30, 21 August 2011
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 point of tangency, .
Construct a perpendicular to that goes through point
. Label the point of intersection
.
Clearly is a rectangle, so
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
.
See also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |