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 ?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Points , and all lie on the perpendicular bisector of . Assume lies between and . Let and . Then and , so and . Subtracting the two equations gives , from which , and the only positive solution is . Thus , and .
Note that circles and are determined by the assumption that lies between and . Thus because the four circles are noncongruent, does not lie between and . Let and . Then and , so and . Subtracting the two equations gives , from which , and the only positive solution is . Thus and . Again, the uniqueness of the solution implies that must indeed lie between and .
The requested sum is .
The problems on this page are the property of the MAA's American Mathematics Competitions