Difference between revisions of "2006 AMC 12B Problems/Problem 15"
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== Problem == | == Problem == | ||
| + | Circles with centers <math> O</math> and <math> P</math> have radii 2 and 4, respectively, and are externally tangent. Points <math> A</math> and <math> B</math> are on the circle centered at <math> O</math>, and points <math> C</math> and <math> D</math> are on the circle centered at <math> P</math>, such that <math> \overline{AD}</math> and <math> \overline{BC}</math> are common external tangents to the circles. What is the area of hexagon <math> AOBCPD</math>? | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <math> \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}</math> | ||
== Solution == | == Solution == | ||
| + | |||
| + | 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)24\sqrt{2}}</math> | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | <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\sqrt{2}</math>, and the total area is two trapezoids, or <math>\boxed{24\sqrt{2}}</math>. | ||
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2006|ab=B|num-b=14|num-a=16}} | |
| + | {{AMC10 box|year=2006|ab=B|num-b=23|num-a=25}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:40, 6 October 2019
Contents
Problem
Circles with centers 
and 
have radii 2 and 4, respectively, and are externally tangent. Points 
and 
are on the circle centered at , and points 
and 
are on the circle centered at , such that 
and 
are common external tangents to the circles. What is the area of hexagon 
?
![[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]](http://latex.artofproblemsolving.com/9/8/5/985d18b9bd1bc04f16562ee9a3eabc5afe6efcbd.png)
Solution
Draw the altitude from onto 
and call the point 
. Because 
and 
are right angles due to being tangent to the circles, and the altitude creates 
as a right angle. 
is a rectangle with 
bisecting . The length 
is 
and 
has a length of 
, so by pythagorean's, is 
.
, which is half the area of the hexagon, so the area of the entire hexagon is 
Solution 2
and 
are congruent right trapezoids with legs and 
and with equal to 
. Draw an altitude from to either or 
, creating a rectangle with width and base 
, and a right triangle with one leg , the hypotenuse , and the other . Using
the Pythagorean theorem, is equal to 
, and is also equal to the height of the trapezoid. The area of the trapezoid is thus 
, and the total area is two trapezoids, or 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
