Difference between revisions of "2006 AMC 12A Problems/Problem 19"
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<center>[[Image:AMC12_2006A_19.png]]</center> | <center>[[Image:AMC12_2006A_19.png]]</center> | ||
− | <math> \mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119} | + | <math> \mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}\qquad \mathrm{(E) \ } \frac{912}{119}</math> |
== Solutions == | == Solutions == |
Revision as of 11:51, 25 January 2008
Contents
[hide]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 ?
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 |