Desargues' TheoremDate: 07/03/98 at 13:27:44 From: Jos Verhulst Subject: Theorem of Desargues Is there any (fundamental) reason why Desargues' two-triangle theorem is so easy to prove in three dimensions, and seems impossible to prove when we restrict ourselves to two dimensions? Date: 07/04/98 at 13:57:22 From: Doctor Floor Subject: Re: Theorem of Desargues Hi Jos, Thank you for sending your question to Dr. Math! I will try to give you an impression of what is known on Desargues' theorem, but a lot of it is quite difficult. Proving Desargues' two-triangle theorem in two-dimensional geometry is usually done using Menelaus' theorem. Menelaus' theorem says: Given a triangle ABC, and points P on BC, Q on AC and R on AB, then P, Q and R are collinear if and only if: AR BP CQ -- * -- * -- = -1 RB PC QA This theorem can be used in Euclidean and affine geometry, so Desargues' theorem can be proved there without jumping to three dimensions. But Menelaus' theorem does not make any sense in projective geometry, because ratios of distances of 3 collinear points, like AR/RB, are not invariant under central projection. So there has been a big question whether Desargues' theorem could be proved without ratios and without third dimension, but just using the axioms on connection of points and intersection of lines. In Hilbert's investigations in 1899, this is proved to be impossible. I don't know the proof though. The same (it cannot be proved by the mentioned axioms alone) goes for Pappus' theorem: Given a hexagon A1A2A3A4A5A6, if A1, A3 and A5 are on a line l1, and A2, A4, A6 are one a second line l2, then the points of intersection of opposite sides of the hexagon are collinear. I tell you this about Pappus' theorem, because it was proved by Hessenberg in 1905 that Pappus' theorem implies Desargues' theorem. I don't know this proof either, but it is said to be very ingenious. So usually Pappus' theorem is added to the axioms in projective geometry. I hope to have given you a little idea of the answer to your question. - Doctor Floor, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/