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Monge from 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.)

Following is a proof derived from Desargues' theorem.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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Denote the centers of the three circles as A, B, C. Let A', B', C' be the intersections of the common tangents external to ABC, see the applet. In A'B'C' the lines AA', BB', CC' serve as bisectors of angles A', B', and C', respectively. Therefore, the three lines concur at the incenter (I') of A'B'C'. This shows that the two triangles ABC and A'B'C' are perspective from a point. By Desargues' theorem, they are also perspective from a line.

Remark

There is by far a more direct derivation of Monge's theorem from Desargues' theorem.

References

  1. J. McCleary, An Application of Desargues' Theorem, Mathematics Magazine, Vol. 55, No. 4 (Spet., 1982), 233-235.
  1. Three Circles and Common Tangents
  2. Monge from Desargue
  3. Monge via Desargue

Desargues' Theorem

Copyright © 1996-2008 Alexander Bogomolny

28678049Page copy protected against web site content infringement by Copyscape


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