2006 AMC 10A Problems/Problem 16
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[hide]Problem
A circle of radius is tangent to a circle of radius
. The sides of
are tangent to the circles as shown, and the sides
and
are congruent. What is the area of
?
Solution 1
Let the centers of the smaller and larger circles be and
, respectively.
Let their tangent points to
be
and
, respectively.
We can then draw the following diagram:
We see 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 .
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)
Hence, the area of the triangle is
Solution 2
Since we have that
.
Since we know that the total length of
We also know that , so
Also, since we have that
Since we know that and
we have that
This equation simplified gets us
Let
By the Pythagorean Theorem on we have that
We know that ,
and
so we have
Simplifying, we have that
Recall that .
Therefore,
Since the height is we have the area equal to
Thus our answer is .
~mathboy282
See also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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All AMC 10 Problems and Solutions |
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