Difference between revisions of "2006 AMC 10A Problems/Problem 16"
m (→Solution) |
m (→Solution) |
||
(7 intermediate revisions by the same user not shown) | |||
Line 15: | Line 15: | ||
== Solution == | == Solution == | ||
+ | |||
+ | 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</math> be <math>D</math> and <math>E</math>, respectively. | ||
+ | We can then draw the following diagram: | ||
+ | |||
<!-- [[Image:2006_AMC10A-16a.png]] --> | <!-- [[Image:2006_AMC10A-16a.png]] --> | ||
<asy> | <asy> | ||
Line 27: | Line 32: | ||
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); </asy> | MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); </asy> | ||
− | + | We see that <math>\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the [[proportion]]: | |
<div style="text-align:center;"><math>\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3</math></div> | <div style="text-align:center;"><math>\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3</math></div> | ||
− | By the [[Pythagorean Theorem]] we have | + | By the [[Pythagorean Theorem]], we have <math>AD = \sqrt{3^2-1^2} = \sqrt{8}</math>. |
Now using <math>\triangle ADO_1 \sim \triangle AFC</math>, | Now using <math>\triangle ADO_1 \sim \triangle AFC</math>, | ||
Line 37: | Line 42: | ||
<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{16\sqrt{2}\ \mathrm{(D)}}</cmath> | |
− | |||
== See also == | == See also == |
Revision as of 17:51, 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 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
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.