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Difference between revisions of "2006 AMC 12B Problems/Problem 15"
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Draw the altitude from <math>O</math> onto <math>DP</math> and call the point <math>H</math>. Because <math>\angle OAD</math> and <math>\angle ADP</math> are right angles due to being tangent to the circles, and the altitude creates <math>\angle OHD</math> as a right angle. <math>ADHO</math> is a rectangle with <math>OH</math> bisecting <math>DP</math>. The length <math>OP</math> is <math>4+2</math> and <math>HP</math> has a length of <math>2</math>, so by pythagorean's, <math>OH</math> is <math>\sqrt{32}</math>.
 
Draw the altitude from <math>O</math> onto <math>DP</math> and call the point <math>H</math>. Because <math>\angle OAD</math> and <math>\angle ADP</math> are right angles due to being tangent to the circles, and the altitude creates <math>\angle OHD</math> as a right angle. <math>ADHO</math> is a rectangle with <math>OH</math> bisecting <math>DP</math>. The length <math>OP</math> is <math>4+2</math> and <math>HP</math> has a length of <math>2</math>, so by pythagorean's, <math>OH</math> is <math>\sqrt{32}</math>.
  
−
<math>2 \cdot \sqrt{32} + \frac{1}{2}\cdot2\cdot \sqrt{32} = 3\sqrt{32} = 12\sqrt{2}</math>, which is half the area of the hexagon, so the area of the entire hexagon is <math>2\cdot12\sqrt{2} = \boxed{(B)} \qquad24\sqrt{2}</math>
+
<math>2 \cdot \sqrt{32} + \frac{1}{2}\cdot2\cdot \sqrt{32} = 3\sqrt{32} = 12\sqrt{2}</math>, which is half the area of the hexagon, so the area of the entire hexagon is <math>2\cdot12\sqrt{2} = \boxed{(B)24\sqrt{2}}</math>
  
 
== Solution 2 ==  
 
== Solution 2 ==  
  
−
<math>ADOP</math> and <math>OPBC</math> are congruent right trapezoids with legs <math>2</math> and <math>4</math> and with <math>OP</math> equal to <math>6</math>. Draw an altitude from <math>O</math> to either <math>DP</math> or <math>CP</math>, creating a rectangle with width <math>2</math> and base <math>x</math>, and a right triangle with one leg <math>2</math>, the hypotenuse <math>6</math>, and the other <math>x</math>. Using
+
<math>ADPO</math> and <math>OPBC</math> are congruent right trapezoids with legs <math>2</math> and <math>4</math> and with <math>OP</math> equal to <math>6</math>. Draw an altitude from <math>O</math> to either <math>DP</math> or <math>CP</math>, creating a rectangle with width <math>2</math> and base <math>x</math>, and a right triangle with one leg <math>2</math>, the hypotenuse <math>6</math>, and the other <math>x</math>. Using
−
the Pythagorean theorem, <math>x</math> is equal to <math>4\sqrt{2}</math>, and <math>x</math> 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>\boxed{24\sqrt{2}}</math>.
+
the [[Pythagorean theorem]], <math>x</math> is equal to <math>4\sqrt{2}</math>, and <math>x</math> 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\sqrt{2}</math>, and the total area is two trapezoids, or <math>\boxed{24\sqrt{2}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 15:40, 6 October 2019

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 $2\cdot12\sqrt{2} = \boxed{(B)24\sqrt{2}}$

Solution 2

$ADPO$ 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 $\frac{1}{2}\cdot(4+2)\cdot4\sqrt{2} = 12\sqrt{2}$, and the total area is two trapezoids, or $\boxed{24\sqrt{2}}$.

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
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

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