Solution

In what follows, \(\sphericalangle(p, q)\) will denote the directed angle between lines \(p\) and \(q\), taken modulo \(180^{\circ}\). Denote by \(O\) the centre of \(\omega\). In any triangle, the homothety with ratio \(-\frac{1}{2}\) centred at the centroid of the triangle takes the vertices to the midpoints of the opposite sides and it takes the orthocentre to the circumcentre. Therefore the triangles \(ABC\) and \(A^{\prime}B^{\prime}C^{\prime}\) share the same centroid \(G\) and the midpoints of their sides lie on a circle \(\rho\) with centre on \(O H\). We will prove that \(\omega, \Omega\), and \(\rho\) are coaxial, so in particular it follows that their centres are collinear on \(OH\).

Let \(D=B B^{\prime} \cap C C^{\prime}\), \(E=C C^{\prime} \cap A A^{\prime}\), \(F=A A^{\prime} \cap B B^{\prime}\), \(S=B C^{\prime} \cap B^{\prime} C\), and \(T=B C \cap B^{\prime} C^{\prime}\). Since \(D, S\), and \(T\) are the intersections of opposite sides and of the diagonals in the quadrilateral \(B B^{\prime} C C^{\prime}\) inscribed in \(\omega\), by Brocard's theorem triangle \(DST\) is self-polar with respect to \(\omega\), i.e. each vertex is the pole of the opposite side. We apply this in two ways.

First, from \(D\) being the pole of \(S T\) it follows that the inverse \(D^{*}\) of \(D\) with respect to \(\omega\) is the projection of \(D\) onto \(ST\). In particular, \(D^{*}\) lies on the circle with diameter \(SD\). If \(N\) denotes the midpoint of \(S D\) and \(R\) the radius of \(\omega\), then the power of \(O\) with respect to this circle is \(ON^{2} - ND^{2}=O D \cdot OD^{*}=R^{2}\). By rearranging, we see that \(N D^{2}\) is the power of \(N\) with respect to \(\omega\).

Second, from \(T\) being the pole of \(SD\) it follows that \(OT\) is perpendicular to \(S D\). Let \(M\) and \(M^{\prime}\) denote the midpoints of \(B C\) and \(B^{\prime} C^{\prime}\). Then since \(OM \perp BC\) and \(OM^{\prime} \perp B^{\prime}C^{\prime}\) it follows that \(OMM^{\prime}T\) is cyclic and

\[\sphericalangle(S D, B C)=\sphericalangle(O T, O M)=\sphericalangle\left(B^{\prime} C^{\prime}, M M^{\prime}\right)\]

From \(B B^{\prime} C C^{\prime}\) being cyclic we also have \(\sphericalangle\left(B C, B B^{\prime}\right)=\sphericalangle\left(C C^{\prime}, B^{\prime} C^{\prime}\right)\), hence we obtain

\[\begin{aligned} \sphericalangle\left(S D, B B^{\prime}\right) & =\sphericalangle(S D, B C)+\sphericalangle\left(B C, B B^{\prime}\right) \\ & =\sphericalangle\left(B^{\prime} C^{\prime}, M M^{\prime}\right)+\sphericalangle\left(C C^{\prime}, B^{\prime} C^{\prime}\right)=\sphericalangle\left(C C^{\prime}, M M^{\prime}\right) . \end{aligned}\]

Now from the homothety mentioned in the beginning, we know that \(M M^{\prime}\) is parallel to \(A A^{\prime}\), hence the above implies that \(\sphericalangle\left(S D, B B^{\prime}\right)=\sphericalangle\left(C C^{\prime}, A A^{\prime}\right)\), which shows that \(\Omega\) is tangent to \(S D\) at \(D\). In particular, \(N D^{2}\) is also the power of \(N\) with respect to \(\Omega\). Additionally, from \(B B^{\prime} C C^{\prime}\) being cyclic it follows that triangles \(D B C\) and \(D C^{\prime} B^{\prime}\) are inversely similar, so \(\sphericalangle\left(B B^{\prime}, D M^{\prime}\right)=\sphericalangle\left(D M, C C^{\prime}\right)\). This yields

\[\begin{aligned} \sphericalangle\left(S D, D M^{\prime}\right) & =\sphericalangle\left(S D, B B^{\prime}\right)+\sphericalangle\left(B B^{\prime}, D M^{\prime}\right) \\ & =\sphericalangle\left(C C^{\prime}, M M^{\prime}\right)+\sphericalangle\left(D M, C C^{\prime}\right)=\sphericalangle\left(D M, M M^{\prime}\right) \end{aligned}\]

which shows that the circle \(D M M^{\prime}\) is also tangent to \(S D\). Since \(N, M\), and \(M^{\prime}\) are collinear on the Newton-Gauss line of the complete quadrilateral determined by the lines \(B B^{\prime}, C C^{\prime}, B C^{\prime}\), and \(B^{\prime} C\), it follows that \(N D^{2}=N M \cdot N M^{\prime}\). Hence \(N\) has the same power with respect to \(\omega\), \(\Omega\), and \(\rho\). By the same arguments there exist points on the tangents to \(\Omega\) at \(E\) and \(F\) which have the same power with respect to \(\omega, \Omega\), and \(\rho\). The tangents to a given circle at three distinct points cannot be concurrent, hence we obtain at least two distinct points with the same power with respect to \(\omega, \Omega\), and \(\rho\). Hence the three circles are coaxial, as desired. **Comment 1.** Instead of invoking the Newton-Gauss line, one can also use a nice symmetry argument: If from the beginning we swapped the labels of \(B^{\prime}\) and \(C^{\prime}\), then in the proof above the labels of \(D\) and \(S\) would be swapped while the labels of \(M\) and \(M^{\prime}\) do not change. The consequence is that the circle \(S M M^{\prime}\) is also tangent to \(S D\). Since \(N\) is the midpoint of \(S D\) it then has the same power with respect to circles \(DMM^{\prime}\) and \(SMM^{\prime}\), so it lies on their radical axis \(MM^{\prime}\). **Comment 2.** There exists another triple of points on the common radical axis of \(\omega, \Omega\), and \(\rho\) which can be used to solve the problem. We outline one such solution.  Let \(L\) and \(L^{\prime}\) denote the feet of the altitudes from \(A\) and \(A^{\prime}\) in triangle \(A B C\) and \(A^{\prime} B^{\prime} C^{\prime}\), respectively. Since \(\rho\) is the nine-point circle of the two triangles it contains both \(L\) and \(L^{\prime}\). Furthermore, \(H A \cdot H L\) and \(H A^{\prime} \cdot H L^{\prime}\) both equal twice the power of \(H\) with respect to \(\rho\), so \(A, A^{\prime}, L, L^{\prime}\) are concyclic as well. Now let \(\ell=A A^{\prime}\) and denote \(P=L L^{\prime} \cap \ell, K=B C \cap \ell\), and \(K^{\prime}=B^{\prime} C^{\prime} \cap \ell\). As \(M M^{\prime} | \ell\) (shown in the previous solution) and \(L L^{\prime} M M^{\prime}\) is cyclic

\[ \sphericalangle(B C, \ell)=\sphericalangle\left(B C, M M^{\prime}\right)=\sphericalangle\left(L L^{\prime}, B^{\prime} C^{\prime}\right) \]

so \(K, K^{\prime}, L\), and \(L^{\prime}\) are also concyclic. From the cyclic quadrilaterals \(A A^{\prime} L L^{\prime}\) and \(K K^{\prime} L L^{\prime}\) we get \(P A \cdot P A^{\prime}=P L \cdot P L^{\prime}=P K \cdot P K^{\prime}\). This implies that \(P\) is the centre of the (unique) involution \(\sigma\) on \(\ell\) that swaps \(A, A^{\prime}\) and \(K, K^{\prime}\). On the other hand, by Desargues' involution theorem applied to the line \(\ell\), the quadrilateral \(B B^{\prime} C C^{\prime}\), and its circumcircle \(\omega\), the involution \(\sigma\) also swaps \(E\) and \(F\). Hence

\[PA \cdot PA^{\prime} = PL \cdot PL^{\prime} = PE \cdot PF\]

However, this means that \(P\) has the same power with respect to \(\omega, \Omega\), and \(\rho\), and by the same arguments there exist points on \(B B^{\prime}\) and \(C C^{\prime}\) with this property.