From: ikastan@alumni.caltech.edu (Ilias Kastanas)
Newsgroups: sci.math
Subject: Re: Geometric problem ... ruler and compass
Date: 14 Dec 1997 12:41:06 GMT

In article <3492A3F5.3DD9@iprolink.ch>,
Laurent BADOUD  <lbadoud@iprolink.ch> wrote:
>Consider three points _inside_ a circle. Using ruler and compass, how to
>construct the inscribed triangle (to the circle) having each of its
>edges passing through one of the points ?
>Laurent Badoud - Geneva 



	Such "incidence" problems can be solved using just Projective
Geometry; in fact this one is no harder if K is an ellipse or other conic
section.   Given P1, P2, P3  let  A  be any point of K;  draw  AP1, find the
point where it meets K, do the same from there with P2 and then with P3,
calling the result A'.  The mapping  A -> A'  is a projectivity, and the
problem will be solved if you find an A with  A = A', that is, a fixed point.
This is standard:  for any  B, C, D  on K the points  BC'*B'C (intersection of
BC', B'C),  BD'*B'D, CD'*C'D  lie on a straight line  L; a familiar theorem
of Pascal/Brianchon (or, if K is a pair of straight lines, of Pappus).  So,
for any  E on K if  EB'*L = X  then  BX*K = E'.  But this means that  L*K = P
satisfies  P = P', and you are done.

	Note that the extra apparatus of Euclidean Geometry is not needed
and only clutters the issue.  For this construction, your compass stays in
its box!


						Ilias