Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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Since we know that <math>O_1 O_2 = 1 + 2 = 3,</math> the total length of <math>A O_2 = 2 \cdot 3 = 6.</math> | Since we know that <math>O_1 O_2 = 1 + 2 = 3,</math> the total length of <math>A O_2 = 2 \cdot 3 = 6.</math> | ||
− | We also know that <math>O_2 | + | We also know that <math>O_2 F = 2</math>, so <math>A F = A O_2 + O_2 F = 6 + 2 = 8.</math> |
− | Also, since <math>\triangle{ | + | Also, since <math>\triangle{ABF} \sim \triangle{A E O_2},</math> we have that <math>\frac{AC}{A O_2} = \frac{FC}{O_2 E}.</math> |
+ | |||
+ | Since we know that <math>A O_2 = 6</math> and <math>O_2 E = 2,</math> we have that <math>\frac{AC}{6} = \frac{FC}{2}.</math> | ||
+ | |||
+ | This equation simplified gets us <math>AC = 3 \cdot FC.</math> | ||
+ | |||
+ | Let <math>FC = a</math> | ||
+ | |||
+ | By the Pythagorean Theorem on <math>\triangle{AFC},</math> we have that <math>AF^2 + FC^2 = AC^2.</math> | ||
+ | |||
+ | We know that <math>AF = 8</math>, <math>FC = a</math> and <math>AC = 3a</math> so we have <math>8^2 + a^2 = (3a)^2.</math> | ||
+ | |||
+ | Simplifying, we have that <math>64 = 8a^2 \Rightarrow a^2 = 8 \Rightarrow a = \sqrt{8} = 2 \sqrt{2}.</math> | ||
+ | |||
+ | Recall that <math>FC=a</math>. | ||
+ | |||
+ | Therefore, <math>BC = 2 \cdot FC = 2 \cdot 2 \sqrt{2} = 4 \sqrt{2}.</math> | ||
+ | |||
+ | Since the height is <math>AF = 8,</math> we have the area equal to <math>\frac{4 \sqrt{2} \cdot 8}{2}=16\sqrt{2}.</math> | ||
+ | |||
+ | Thus our answer is <math>\boxed{\mathrm{(D) 16 \sqrt{2}.}}</math> | ||
== See also == | == See also == |
Revision as of 15:29, 25 December 2020
Contents
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 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
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.