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 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