Prove that three points A, B, C are collinear

Question

We research the configuration of problem 3845 in Crux Mathematicorum:

Let six points $A_1, A_2, ...A_6$ lie on a circe, and the six points $B_1, B_2,...,B_6$ lie on another circle. If the quadruples $A_i, A_{i+1}B_{i+1}B_i$ lie on a circles with centers $O_i$ for $i=1,2,...,5$ then $A_6$, $A_1$, $B_1$, $B_6$ lie on a circle (namely center of the new circle is $O_6$) and $O_1O_4, O_2O_5, O_3O_6$ are concurrent.

By Miquel six circles theorem we can show that $A_i, B_i, B_{i+3}, A_{i+3}$ lie on a circle, for $i=1,2,3$. Denote these circles are $(O_7), (O_8), (O_9)$ respectively.

By problem 3845 we have:

• $O_1O_4, O_2O_5, O_3O_6$ are concurrent, denote the point of concurrence is $A$

• $O_1O_7, O_2O_8, O_3O_9$ are concurrent, denote the point of concurrence is $B$

• $O_4O_7, O_5O_8, O_6O_9$ are concurrent, denote the point of concurrence is $C$

My question: I am looking for a proof that A, B, C are collinear

Answer 1

In this proof we will automatically show also that $O_1O_4, O_2O_5, O_3O_6$ are concurrent.

$\textbf{Proof :}$

Consider the stereographic projection $\pi$. So consider the planes $\mathcal{P}_{i, i+1}$ which goes through the image of circles $(A_iA_{i+1}B_{i}B_{i+1})$ under $\pi$.

Let $\mathcal{P}_{1, 2}\cap\mathcal{P}_{4, 5} = \ell_1$, $\mathcal{P}_{2, 3}\cap\mathcal{P}_{4, 5} = \ell_2$, $\mathcal{P}_{3, 4}\cap\mathcal{P}_{6, 1} = \ell_3$.

Note that $\pi$ also can be seen as a projection on plane from space. So $O_1O_4$ is a projection of the polar line of $\ell_1$ wrt sphere on the basic plane and to prove that $O_1O_4, O_2O_5, O_3O_6$ are concurrent it's enough to prove that the polars of $\ell_1, \ell_2, \ell_3$ wrt sphere are concurrent, or in another words that $\ell_1, \ell_2, \ell_3$ are coplanar. Also to prove another concurrences and collinearities in this problem we can use stereographic reformulation and get statement about lines and planes in $3D$ planes.

But now we can consider other stereographic projections $\pi_{\alpha}$ of this sphere with circles on it. This projections correspond to an inversions on the plane, so in fact we just need to prove the original problem for any three different inversion images. So just consider three inversions which send circles $(A_1A_2A_3), (B_1B_2B_3)$ to a lines (to a concentric circles) and prove it just in this case

$\textbf{Comment :}$

Note that there is a generalization of this facts : In my proof we only need that $\pi(A_1A_2\ldots A_6B_1B_2\ldots B_6)$ lie on some quadratic surface (and of course, on a sphere). So we also get a proof of https://mathoverflow.net/questions/234722/a-chain-of-six-circles-associated-with-a-conic (because $\pi(\textbf{Conic})$ lie on a quadratic cone).