2006 AIME II Problems/Problem 9
Problem
Circles and have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line is a common internal tangent to and and has a positive slope, and line is a common internal tangent to and and has a negative slope. Given that lines and intersect at and that where and are positive integers and is not divisible by the square of any prime, find
Solution
Call the centers , the points of tangency (with on and on , and on ), and the intersection of each common internal tangent to the X-axis . since both triangles have a right angle and have vertical angles, and the same goes for . By proportionality, we find that ; solving by the Pythagorean theorem yields . On , we can do the same thing to get and .
The vertical altitude of each of and can each by found by the formula (as both products equal twice of the area of the triangle). Thus, the respective heights are and . The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: , and by 30-60-90: .
From this information, the slope of each tangent can be uncovered. The slope of . The slope of .
The equation of can be found by substituting the point into , so . The equation of , found by substituting point , is . Putting these two equations together results in the desired . Thus, .
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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