2006 AIME II Problems/Problem 9
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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 • Resources) | ||
| Preceded by Problem 8 | Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
