Monge's theorem
en.wikipedia.org
en.wikipedia.org
Edit: Nope, nevermind, two of the points both see the medium and small circles!
q3 = (r1 * p2 - r2 * p1) / (r1 - r2)
(with similar expression for q1 and q2). Using homogenous coordinates we get the following list for q1, q2, q3: ( r2 p3 - r3 p2 ) ( r3 p1 - r1 p3 ) ( r1 p2 - r2 p1 )
( r2 - r3 ), ( r3 - r1 ), ( r1 - r2 )
now note that r1 q1 + r2 q2 + r3 q3 = 0, showing that they're not linearly independent and hence collinear.A somewhat neater but more advanced proof, follows by making the positions pi also homogeneous and noting that (using Einstein summation notation):
0 = ε_ijk rj rk
so we can determine the intersections to be qi = ε_ijk rj pk
hence ri qi = ε_ijk ri rj pk = ε_jik rj ri pk = -ε_ijk ri rj pk = 0.On the other end it was a Projective Geometry course and at the end we used a lot the General Linear Group, in particular 4x4 matrices for bilinear forms, so it was like warming up with Geometry to arrive to the Algebra tools.