Difference between revisions of "2006 AMC 10A Problems/Problem 16"
Hashtagmath (talk | contribs) |
m (→Solution) |
||
Line 15: | Line 15: | ||
<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> | <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 == | + | == Solution 1 == |
Let the centers of the smaller and larger circles be <math>O_1</math> and <math>O_2</math> , respectively. | Let the centers of the smaller and larger circles be <math>O_1</math> and <math>O_2</math> , respectively. | ||
Line 43: | Line 43: | ||
<div style="text-align:center;"><math>\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}</math></div> | <div style="text-align:center;"><math>\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}</math></div> | ||
− | Hence, the area of the triangle is <cmath>\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{ | + | Hence, the area of the triangle is <cmath>\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{\mathrm{(D)} 16\sqrt{2}}</cmath> |
== Solution 2 == | == Solution 2 == |
Revision as of 10:14, 13 February 2021
Contents
[hide]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 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.