Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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== Solution == | == Solution == | ||
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+ | Let the centers of the smaller and larger circles be <math>O_1</math> and <math>O_2</math>, respectively. | ||
+ | Let their tangent points to <math>\triangle ABC be </math>D<math> and </math>E$, respectively. | ||
+ | We can then draw the following diagram: | ||
+ | |||
<!-- [[Image:2006_AMC10A-16a.png]] --> | <!-- [[Image:2006_AMC10A-16a.png]] --> | ||
<asy> | <asy> |
Revision as of 17:41, 22 October 2020
Problem
A circle of radius 1 is tangent to a circle of radius 2. The sides of are tangent to the circles as shown, and the sides
and
are congruent. What is the area of
?
Solution
Let the centers of the smaller and larger circles be and
, respectively.
Let their tangent points to
D
E$, respectively.
We can then draw the following diagram:
Note that . Using the first pair of similar triangles, we write the proportion:
![$\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3$](http://latex.artofproblemsolving.com/1/0/a/10ada06cd47acf9909068e7654b0ddc0944a8f67.png)
By the Pythagorean Theorem we have that .
Now using ,
![$\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}$](http://latex.artofproblemsolving.com/0/9/1/091c4ac75473dc96bea1372a641fa96643f04862.png)
The area of the triangle is .
Random Person: Great explanation!
See also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.