Monge via Desargues

Monge's theorem (Monge's Circle Theorem, Three Circles Theorem) claims that, given three circles of distinct radii situated entirely in each other's exterior, then the three points of intersection of the pairs of external common tangents are collinear. (The theorem is valid even if the circles intersect and has a sensible interpretation when two or three circles have the same radius.)

One of the proofs of the theorem has been derived from Desargues' theorem and has exploited the fact that, in a triangle, the angle bisectors concur in a point - the incenter of the triangle. As a matter of fact the relation between Desargues' and Monge's theorems is more direct as becomes clear from a demonstration below.

As was observed on several occasions, any two circles are similar. If, in addition, their radii differ, they are also homothetic with a finite center of homothety. This means that there exists a point O - the center of homothety (dilation) - such that

1. Any two corresponding points P and P' on the two circles are collinear with O.
2. The ratio OP/OP' does not depend on P.

If such circles have external common tangents, the latter intersect at the center of homothety, and the point of tangency of a common tangent correspond to each other.

Choose points A₁ and B₁ on one of the circles, and let A₂, B₂, A₃, B₃ be the points on the other two circles corresponding to A₁ and, respectively, B₁ via the naturally defined dilations. Denote the centers of the circles O₁, O₂, O₃. Then the lines O₁A₁ and O₂A₂ are equally inclined to the line of centers O₁O₂ and are, therefore, parallel. The same applies to other lines, so that

From here it follows that A₁B₁||A₂B₂||A₃B₃, which suggests a special case of Desargues' theorem: the lines joining the corresponding vertices of the two triangles A₁A₂A₃ and B₁B₂B₃ are parallel. In other words, the triangles A₁A₂A₃ and B₁B₂B₃ are perspective from a point, albeit, in this case, a point at infinity. It follows that the triangles are also perspective from a line, which completes the proof.

(Since homotheties and similarities were the main instrument in the above proof, it may not be entirely out of place to mention that the triangles A₁A₂A₃ and B₁B₂B₃, although perspective, need not be similar, let alone homothetic.)

1. Three Circles and Common Tangents
2. Monge via Desargue
3. Monge via Desargue II

Desargues' Theorem

• Desargues' Theorem
• 2N-Wing Butterfly Problem
• Cevian Triangle
• Do You Speak Mathematics?
• Desargues in the Bride's Chair (with Pythagoras)
• Menelaus From Ceva
• Monge from Desargues
• Monge via Desargue
• Nobbs' Points, Gergonne Line
• Soddy Circles and David Eppstein's Centers
• Pascal Lines: Steiner and Kirkman Theorems II
• Pole and Polar with Respect to a Triangle
• The Lepidoptera of the Circles

Copyright © 1996-2008 Alexander Bogomolny