Date: 09/22/1999 at 08:53:50 From: Rezvan Subject: Algebra, Desargues' Theorem I'd like to find out more about Desargues' theorem and the ways to prove it. I've searched the Internet and many books in my college's library. I found some materials on the Internet but none in the library. I'd be grateful if you could help me and also give me some questions related to Desargues' theorem. Thank you.
Date: 09/22/1999 at 10:07:29 From: Doctor Anthony Subject: Re: Algebra, Desargues' Theorem Three concurrent straight lines 0A, 0B and 0C are produced to A', B', and C', respectively. Prove that the points of intersection of AB and A'B', BC and B'C', and CA and C'A' are collinear. This is Desargues' theorem. If BC, B'C' meet at P; CA, C'A' meet at Q; and AB, A'B' meet at R; then we must prove P, Q and R are collinear. Use the well-known result that if P, Q, and R are collinear, then the two conditions that will be satisfied are: k1.p + k2.q + k3.r = 0 k1 + k2 + k3 = 0 where p, q and r are position vectors respectively of P, Q, and R, and k1, k2, and k3 are scalar constants. Since OAA' are collinear we have (taking v as position vector of O) v + k1.a + k1'.a' = 0 1 + k1 + k1' = 0 Since OBB' are collinear we have v + k2.b + k2'.b' = 0 1 + k2 + k2' = 0 Since OCC' are collinear we have v + k3.c + k3'.c' = 0 1 + k3 + k3' = 0 From the second and third pairs of equations k2.b - k3.c k2'b' - k3'c' ----------- = ------------- k2 - k3 k2' - k3' and these must both represent the point P where BC intersects B'C' = p and so (k2-k3) x p = k2.b - k3.c Similarly (k3-k1) x q = k3.c - k1.a (k1-k2) x r = k1.a - k2.b adding these three equations we get (k2-k3)p + (k3-k1)q + (k1-k2)r = 0 and of course (k2-k3) + (k3-k1) + (k1-k2) = 0 and this is the condition for P, Q, and R to be collinear. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
Date: 09/22/1999 at 11:54:43 From: Doctor Floor Subject: Re: Algebra, Desargues' Theorem Hi, Rezvan: In addition to what Dr. Anthony stated, the complete theorem says: Let ABC and A'B'C' be two triangles, with sides abc and a'b'c' respectively. Then the following conditions are equivalent: (1) AA', BB', CC' are concurrent (2) The points of intersection of a+a', b+b' and c+c' are collinear. Dr. Anthony proved, using calculations, (1) --> (2). One could prove in the same way that (2) --> (1), but this can also be seen "by duality." Let me explain: In geometry, there is some symmetry in points and lines: * two points define one line * two lines define one point This symmetry is found in all of Euclid's axioms (but not the parallel postulate). We say that points and lines are dual. Each geometric statement about lines and points has a dual version: * A triangle is formed out of three non-collinear points. * A trilateral is formed out of three non-concurrent lines. * Side a is the line through vertices B and C. * Vertex A is the intersection of sides b and c. There are some other, more vague dual forms: * In Cartesian coordinates a point (a,b) is given by two numbers a, b * and a line y = mx+n is given by two numbers m and n. The symmetry in Euclid's axioms has a consequence: when a certain statement is correct, then so is its dual. Duality does NOT mean that if a certain condition holds in a figure, then immediately the dual condition holds in the same figure. Three lines that intersect in a point do not imply three collinear points in the same figure (whatever meaning that could have). Normally the dual version needs a new figure. So three intersecting lines in one figure are dual to three collinear points in another figure. Back to Desargues' theorem. Note that conditions (1) and (2) of Desargues' theorem are dual: 1. The pairs AA', BB', and CC' of the vertices of two triangles define three concurrent lines, 2. The pairs of sides aa', bb' and cc' of the side lines of two triangles define three collinear points. Desargues' theorem says that these two duals hold together in one figure. If you prove (1) -> (2), then the dual statement (2) -> (1) must also be true. So Dr. Anthony did in fact fully prove Desargues' theorem. Another way to 'prove' Desargues' theorem is by seeing the two triangles as projection of a (double) pyramid. Triangles ABC and A'B'C' are formed by two planes that intersect the pyramid. The point of intersection of AA', BB', and CC' is the top of the pyramid. The line connecting the intersections aa', bb', and cc' is the line of intersection of the two planes. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 09/23/1999 at 10:02:14 From: Rezvan Moghimi Subject: Thank you! Dear Dr. Math: I'd like to thank both Dr. Anthony and Dr. Floor for answering my question. Their answers helped me to have a better understanding of the problem that I had. I have one more question about Desargues' theorem. Since I'm doing a project on this subject, I need to have a question as my thesis statement. I'd be grateful if you can help me to find a suitable question (a question that can be solved using Desargues' theorem.) Thank you.
Date: 09/26/1999 at 05:04:47 From: Doctor Floor Subject: Re: Thank you! Dear Rezvan, Thanks for your response. Your question is difficult, but I would suggest something like: "Given two triangles with sides a, b, c and a', b', c', under what conditions do the points of intersection X of a and a', Y of b and b', and Z of c and c' lie on one line?" I hope that is suitable. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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