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}{ | + | <math> \mathrm{(A) \ } \frac{908}{119}\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> |
== Solution == | == Solution == |
Revision as of 16:51, 28 September 2014
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 ?
Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.