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Difference between revisions of "2006 AMC 12B Problems/Problem 15"

(Another solution , PLEASE edit as I am not sure how to make this clear, and please change to LaTeX. I tried to do it in LaTeX and I'm still new to the software.)
(→‎Solution 2)
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ADOP and OPBC are congruent right trapezoids with legs 2 and 4 and with OP equal to 6. Draw an altitude from O to either DP or CP, creating a rectangle with width 2 and base x, and a right triangle with one leg 2, the hypotenuse 6, and the other x. Using
 
ADOP and OPBC are congruent right trapezoids with legs 2 and 4 and with OP equal to 6. Draw an altitude from O to either DP or CP, creating a rectangle with width 2 and base x, and a right triangle with one leg 2, the hypotenuse 6, and the other x. Using
−
the Pythagorean theorem, x is equal to <math>4\sqrt{2}</math>, and x is also equal to the height of the trapezoid. The area of the trapezoid is thus 1/2(4+2)*4sqrt(2) = 12sqrt(2). and the total area is two trapezoids, or 24sqrt2.  
+
the Pythagorean theorem, x is equal to <math>4\sqrt{2}</math>, and x is also equal to the height of the trapezoid. The area of the trapezoid is thus <math>\frac{1}{2}\cdot(4+2)\cdot4\sqrt{2} = 12</math>
 +
. and the total area is two trapezoids, or <math>24\sqrt{2}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2006|ab=B|num-b=14|num-a=16}}

Revision as of 20:33, 29 January 2012

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Problem

Circles with centers $O$ and $P$ have radii 2 and 4, respectively, and are externally tangent. Points $A$ and $B$ are on the circle centered at $O$, and points $C$ and $D$ are on the circle centered at $P$, such that $\overline{AD}$ and $\overline{BC}$ are common external tangents to the circles. What is the area of hexagon $AOBCPD$?

[asy] // from amc10 problem series unitsize(0.4 cm); defaultpen(linewidth(0.7) + fontsize(11)); pair A, B, C, D; pair[] O; O[1] = (6,0); O[2] = (12,0); A = (32/6,8*sqrt(2)/6); B = (32/6,-8*sqrt(2)/6); C = 2*B; D = 2*A; draw(Circle(O[1],2)); draw(Circle(O[2],4)); draw((0.7*A)--(1.2*D)); draw((0.7*B)--(1.2*C)); draw(O[1]--O[2]); draw(A--O[1]); draw(B--O[1]); draw(C--O[2]); draw(D--O[2]); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SW); label("$D$", D, NW); dot("$O$", O[1], SE); dot("$P$", O[2], SE); label("$2$", (A + O[1])/2, E); label("$4$", (D + O[2])/2, E);[/asy]

$\textbf{(A) } 18\sqrt {3} \qquad \textbf{(B) } 24\sqrt {2} \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 24\sqrt {3} \qquad \textbf{(E) } 32\sqrt {2}$

Solution

Draw the altitude from $O$ onto $DP$ and call the point $H$. Because $\angle OAD$ and $\angle ADP$ are right angles due to being tangent to the circles, and the altitude creates $\angle OHD$ as a right angle. $ADHO$ is a rectangle with $OH$ bisecting $DP$. The length $OP$ is $4+2$ and $HP$ has a length of $2$, so by pythagorean's, $OH$ is $\sqrt{32}$.

$2 \cdot \sqrt{32} + \frac{1}{2}\cdot2\cdot \sqrt{32} = 3\sqrt{32} = 12\sqrt{2}$, which is half the area of the hexagon, so the area of the entire hexagon is $2\cdot12\sqrt{2} = \boxed{(B)} \qquad24\sqrt{2}$

Solution 2

ADOP and OPBC are congruent right trapezoids with legs 2 and 4 and with OP equal to 6. Draw an altitude from O to either DP or CP, creating a rectangle with width 2 and base x, and a right triangle with one leg 2, the hypotenuse 6, and the other x. Using the Pythagorean theorem, x is equal to $4\sqrt{2}$, and x is also equal to the height of the trapezoid. The area of the trapezoid is thus $\frac{1}{2}\cdot(4+2)\cdot4\sqrt{2} = 12$ . and the total area is two trapezoids, or $24\sqrt{2}$.

See also

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 12 Problems and Solutions
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