Difference between revisions of "2006 AMC 12A Problems/Problem 19"
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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]], | 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} \left(\frac{5}{12}\right) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}</math> | + | :<math>\tan (2 \tan ^{-1} \left(\frac{5}{12}\right)) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}</math> |
:<math>= \frac{120}{119}</math> | :<math>= \frac{120}{119}</math> | ||
Revision as of 23:41, 5 February 2010
Problem
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
?
![AMC12 2006A 19.png](https://wiki-images.artofproblemsolving.com//8/8d/AMC12_2006A_19.png)
Solutions
Solution 1
- 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:
Solution 2
By skiron.
Let be the line that goes through
and
, and let
be the line
. If we let
be the measure of the acute angle formed by
and the x-axis, then
.
clearly bisects the angle formed by
and the x-axis, so
. We also know that
and
intersect at a point on the x-axis. The equation of
is
, so the coordinate of this point is
. Hence the equation of
is
, so
, and our answer choice is
.
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 |