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