Let there be three circles of different radii lying completely outside each other. To exclude a trivial case, assume also that their centers are not collinear, i.e. the three centers do not lie on the same straight line. Under these conditions, six external tangents to two of the three circles, taken pairwise, intersect at three points. Prove that the three points are collinear. This is known as Monge's Circle Theorem.

This problem is almost on a par with the 4 Travelers and 9 dots problems. It has a "3D solution" but can also be solved without leaving the plane. D.Wells tells of John Edson Sweet, an engineer, whose reaction to the above diagram was, "Yes, that is perfectly self-evident."

Copyright © 1996-2008 Alexander Bogomolny

The following is the argment put forth by John Edson Sweet [Wells, p 153-154].

Imagine having three different spheres lying on a plane, say plane #1. In one particular case where there exists a second plane, plane #2, tangent to the three spheres but covering them from above the proof is particularly intuitive. For then planes #1 and #2 are bound to have a common line l, the line of their intersection. The bisector plane of the solid angle formed by planes #1 and #2 passes through the centers of all three spheres. This will be the plane, plane #3, depicted at the top of the page. By definition, plane #3 passes through l.

On the other hand, any two of the spheres can be inscribed into a cone. Any plane (externally) tangent to two spheres will also be tangent to the enclosing cone. Therefore, planes #1 and #2 are tangent to each of the three cones. Now, a plane tangent to a cone necessarily contains its vertex. Finally, vertices of the cones belong to both planes #1 and #2, hence to the line l, hence to the plane #3.

Note now that plane #3 intersects the cones along the common tangent lines as on the diagram.

Another 3D solution has been suggested by Nathan Bowler:

Solution 2

Consider 3 right cones on the three circles as bases. Now consider their common tangent planes: these intersect in three lines A', B' and C' corresponding to the three points A, B and C in the theorem. Now, A' and B' intersect at the apex of one of the cones. Similarly, B' and C' intersect and so do C' and A'. Thus A', B' and C' are coplanar, and so A, B and C are collinear.

Solution 3

This solution is due to Stuart Anderson.

Name the three circles a, b, and c, so that the common tangents of a and b meet at C, those of a and c at B, and those of b and c at A. We assign each circle a mass (ma, mb, or mc) inversely proportional to its radius. If we use a positive mass for each circle, then the center of mass of each pair of circles lies at the intersection of their internal tangents, but if we make one mass positive and one negative, the center of mass will lie at the intersection of their external tangents. This is easy to prove using similar triangles and the definition of center of mass. (If you don't like the idea of negative mass, think of forces applied either pointing into the screen or out of it, and use center of force instead.)

Now it is a known fact that the coordinates of the center of mass of a combined system is just the weighted average of the coordinates of the centers of each system separately, and hence is collinear with them. If I choose systems {ma, -mb}, {mb, -mc}, and {ma, -mc}, then superimposing the first two systems gives the third, hence its center of mass is collinear with theirs. QED

By the way, this also shows why you can't prove the theorem for internal tangent pairs: if all the masses are positive, you can't combine two pairs of circles to get the remaining pair, because no cancellation is possible. The theorem does work again with one external tangent pair and two internal tangent pairs: {ma, mb}{mc, -mb} = {ma, mc}.

Another proof (and a generalization) can be obtained with the help of the Menelaus' theorem. Further, any two circles are similar and could be obtained from each other by a similarity transformation (homothety) centered at a point of intersection of their common tangents. It's known that, in general, if one figure is homothetic to another, and the second to the third, and the product of the coefficients of homothety is positive, but distinct from 1, then the product of two homotheties is a homothety and the three homothety centers are collinear, see Yaglom, p. 27. Additional proofs follow as easily from Desargues' theorem.

As applied to circles, two circles may have 0, 1, or 2 pairs of common tangents. In the latter case, one pair is called internal, the other external. (In the diagram, three pairs of external tangents are shown, but any two circles have also internal tangents that meet in-between the circles.) If the number of external tangencies is odd (1 or 3) then the centers of homotheties are collinear. To understand this rule, think of circles and straight lines as oriented - an arrow is attached to either pointing in the positive direction of traversal. The opposite direction is considered negative.

An oriented circle may touch another oriented circle or an oriented line in one of two ways. Either the traversal directions of the two figures at the point of tangency are compatible or "against the grain" - depending on whether the figures have the opposite orientations or the same.

The above rule becomes clear if applied to oriented lines. Of the three circles, either all have the same orientation or the orientation of one differs from that of the other two. In both cases, the theorem applies to oriented tangent lines whose orientation is compatible with the orientation of the circles.

1. D. Wells, You are a Mathematician, John Wiley & Sons, 1995
2. I. M. Yaglom, Geometric Transformations II, MAA, 1968

2D problems that benefit from a 3D outlook

1. 4 travellers
2. Desargues' Theorem
3. Soddy Circles and Eppstein's Points
4. Symmetries in a triangle
5. Three Circles and Common Chords
6. Three circles and Common Tangents
7. Three equal circles
8. Monge via Desargue
9. Monge via Desargue II

Copyright © 1996-2008 Alexander Bogomolny