Difference between revisions of "2006 AMC 12A Problems/Problem 19"
m (→See also: +box) |
(→Solution: solution) |
||
| Line 9: | Line 9: | ||
== Solution == | == Solution == | ||
{{solution}} | {{solution}} | ||
| + | :''This solution needs a clearer explanation and a diagram.'' | ||
| + | Notice that both [[circle]]s are [[tangent]] to the [[x-axis]] and each other. Call the circles (respectively) A and B; the [[distance]] between the two centers is <math>4 + 9 = 13</math>. If we draw the parallel radii that lead to the common [[external tangent]], a line can be extended parallel to the tangent from A to the radius of circle B. This creates a 5-12-13 triangle. To find the slope of that line (which is parallel to the tangent), note that another 5-12-13 triangle can be drawn below the first one such that the side with length 12 is parallel to the x-axis. The slope can be found by using the [[double tangent identity]], | ||
| + | |||
| + | :<math>\tan (2 \tan ^{-1} (\frac{5}{12}) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}</math> | ||
| + | :<math>= \frac{120}{119}</math> | ||
| + | |||
| + | To find the x and y coordinates of the point of tangency of circle A, we can set up a ratio (the slope will be –119/120 because it is the negative reciprocal): | ||
| + | |||
| + | :<math>\frac{119}{\sqrt{119^2 + 120^2}}</math> <math>=</math> <math>\frac{119}{169} = \frac{y - 4}{4}</math> | ||
| + | |||
| + | :<math>\frac{-120}{\sqrt{119^2 + 120^2}} = \frac{-120}{169} = \frac{x - 2}{4}</math> | ||
| + | |||
| + | :<math>x = \frac{-142}{169}, y = \frac{1152}{169}</math> | ||
| + | |||
| + | We can plug this into the equation of the line for the tangent to get: | ||
| + | |||
| + | :<math>\frac{1152}{169} = \frac{120}{119}\frac{-142}{169} + b</math> | ||
| + | :<math>b = \frac{912}{119}</math> <math>\Rightarrow \mathrm{E}</math> | ||
| + | |||
== See also == | == See also == | ||
* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] | ||
Revision as of 19:51, 2 February 2007
Problem
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Circles with centers 
and 
have radii 
and 
, respectively. The equation of a common external tangent to the circles can be written in the form 
with 
. What is 
?
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
- This solution needs a clearer explanation and a diagram.
Notice that both circles are tangent to the x-axis and each other. Call the circles (respectively) A and B; the distance between the two centers is 
. If we draw the parallel radii that lead to the common external tangent, a line can be extended parallel to the tangent from A to the radius of circle B. This creates a 5-12-13 triangle. To find the slope of that line (which is parallel to the tangent), note that another 5-12-13 triangle can be drawn below the first one such that the side with length 12 is parallel to the x-axis. The slope can be found by using the double tangent identity,
To find the x and y coordinates of the point of tangency of circle A, we can set up a ratio (the slope will be –119/120 because it is the negative reciprocal):
We can plug this into the equation of the line for the tangent to get:
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 12 Problems and Solutions | |
