Desargues' Theorem

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/