This problem is a doozy:

A circle is in the plane with center at O_a and some radius r_a. Another circle, not touching the first circle anywhere, has center at O_b and some radius r_b. Points A and B are both (arbitrary) points on circle A and circle B respectively. Finally, Point C is in such a location that ABC is an equilateral triangle. Suddenly, points A and B begin rotating in the same direction (say, counterclockwise) around their respective circles with the same angular speed (say, w) about their centers. During this process, point C moves so that ABC remains an equilateral triangle.

Prove that point C is moving in a circle with same direction (counterclockwise) and angular speed (w) about some center O_c somewhere in the plane.

My attempt at a solution:

At first, I thought that this problem might be solved by placing the whole thing on a coordinate axis... Not going to happen. There are so many arbitrary variables that even if I arrived at a solution that way, it would be an ugly algebraic mess. After some considerable time spent staring at my diagram I decided on one thing: if we take points O_a, O_b, and O_c, I believe they will always form an equilateral triangle. Thankfully, this fact need not be proven first--if we prove that point C moves in a circle around this O_c, then nothing else need be said.

A friend has suggested that I use vectors to solve this. I don't see the solution. The following is what I understand about the solution: in our diagram, we will place points A and B at their arbitrary locations on their circles. Then, I will rotate them to positions A' and B', an arbitrary angle $\displaystyle \phi$ counterclockwise. Then, I will want to show that point C', the third point on the equilateral triangle A'B'C' is the same C' as that point achieved by rotating point C about O_c by the angle $\displaystyle \phi$. This will suffice to solve the problem. My issue is precisely that I don't know how to show that above fact.

Help!