Hubert Shutrick

The theorem in A conic in a triangle is really a theorem in projective geometry because the line at infinity can be replaced by any line that doesn't intersect the sides of the triangle. It can therefore be proved using homogeneous coordinates. The obvious choice is to use the given triangle ABC for the axes and by scaling the coordinates either the point P can be taken as (1,1,1) or the line can be assumed x + y + z = 0 but not both so we assume the latter and let P=(a,b,c) . There is conic through the points C_{a}B_{a}B_{c}A_{c}A_{b}C_{b} even if the chords do not intersect at P by Pascal's criterion so we concentrate on the second part where P is necessary.

This proof was supplied by Gordon Walsh. Use a'=b+c, b'=c+a, c'=a+b to keep the expressions more compact. The coordinates of the vertices of the above hexagon are obtained by (0,b,b') = (a,b,c)-a(1,0,-1) etc. The point A_{a} has coordinates (-aa,a'b,a'c) = a'(0,c',c) - a(a,a',0) = a'(0,b,b') - a(a,0,a') and then we know B_{b} and C_{c} by symmetry. The lines BA_{a} and AB_{b} are

{x \over aa}+{z \over a'c} = 0, {y \over bb}+{z \over b'c} = 0

and therefore they intersect at (aab',bba',-a'b'c) . The intersections of the other pairs of opposite sides are similarly (aac',-a'bc',cca'),(-ab'c',bbc',-ccb'). To prove that there is a conic through AC_{c}BA_{a}CB_{b} , one shows that the determinant with the three points as rows is zero. We leave the calculation to the reader with the tip that abc can be taken out from the columns and it turns out that a'b'c' is also a factor.