2015 AMC 12B Problems/Problem 24
Problem
Four circles, no two of which are congruent, have centers at , , , and , and points and lie on all four circles. The radius of circle is times the radius of circle , and the radius of circle is times the radius of circle . Furthermore, and . Let be the midpoint of . What is ?
Solution
First, note that lies the radical axis of any of the pairs of circles. Suppose that and are the centers of two circles that intersect exactly at and , with and lying on the same side of , and . Let , , and suppose that the radius of circle is and the radius of circle is .
Then the power of point in is
and the power of point in is
Also, note that .
Subtract the above two equations to find that or . As , we find that . Plug this into an earlier equation to find that . This is a quadratic equation with solutions , and as is a length, it is positive, hence , and . This is the only possibility if the two centers lie on the same same of their radical axis.
On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that . Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is .
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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