Four disks are arranged in a plane such that each is externally tangent to two others. Prove that the four points of tangency lie on a circle.

Going either CW or CCW, let’s call the centers of the circles O1, O2, O3 and O4 and their respective radii R1, R2, R3 and R4.

If you take any two tangent circles and draw a line connecting their centers, it will always pass through the point of tangency (if someone needs that justified let me know). So you can connect all four pairs of tangent circles and get a quadrilateral with sides (R1+R2), (R2+R3), (R3+R4), and (R4+R1). I will also mark the tangency points on this quadrilateral as T12, T23, T34 and T41.

If the four points of tangency do indeed lie on a circle, then I should be able to do the following:
Connect two consecutive tangency points (ex: T12 and T23).
Find the midpoint and draw the perpendicular bisector.
Repeat for the other three pairs of consecutive tangency points.
Now I should see that all four perpendicular bisectors intersect at the same point if they do indeed all lie on a circle.

I can see it if I mock it up, but I’m having trouble proving it.

Another point I’d like to make is that the perpendicular bisector of the lines connecting two consecutive tangency points is also the angle bisector of the angle formed by the two tangency points and the vertex of the rectangle between them. Wow, I don’t know if I just made sense there. Here (I hope this won’t be more confusing):

Let’s consider one of the vertices of the rectangle and just call it O. Let’s call the two points of tangency on each leg of O, A and B. Remember, due to the construction of the quadrilateral, OA = R and OB = R. Therefore triangle OAB is an isosceles triangle.

Bisect AB and call the midpoint M. Now consider triangles OMA and OMB. Since we bisected AB, we know that MA = MB. Both triangles share OM, so those two sides are equal. And remember we said OA=R=OB. So triangles OMA and OMB are congruent.

Therefore angle MOA = angle MOB. So MO bisects angle AOB.

Also, angle OMA = OMB, plus OMA and OMB are a linear pair so they add to 180. Therefore OMA = OMB = 90. This means that OM is perpendicular to AB, and remember M is the midpoint of AB, so OM is the perpendicular bisector of AB.

Hopefully I’m on the right track and this will turn a light bulb on, letting me come back with a thorough answer later =)  Or maybe it will just help someone else out.

Posted by nikki on 2005-02-02 14:26:50