Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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== Problem == | == Problem == | ||
<!-- [[Image:2006_AMC10A-16.png]] --> | <!-- [[Image:2006_AMC10A-16.png]] --> | ||
− | A | + | A circle of radius <math>1</math> is tangent to a circle of radius <math>2</math>. The sides of <math>\triangle ABC</math> are [[tangent]] to the circles as shown, and the sides <math>\overline{AB}</math> and <math>\overline{AC}</math> are congruent. What is the area of <math>\triangle ABC</math>? |
<asy> | <asy> | ||
Line 13: | Line 13: | ||
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);</asy> | MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);</asy> | ||
− | <math>\textbf{(A) | + | <math>\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad\textbf{(E) } 24\qquad</math> |
== Solution 1 == | == Solution 1 == |
Revision as of 12:03, 17 December 2021
Contents
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:
By the Pythagorean Theorem, we have .
Now using ,
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 | |
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.