Problem 1830

In what follows, $\Varangle(p, q)$ 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 $A'B'C'$ share the same centroid $G$ and the midpoints of their sides lie on a circle $\rho$ with centre on $OH$. We will prove that $\omega$, $\Omega$, and $\rho$ are coaxial, so in particular it follows that their centres are collinear on $OH$.

Let $D = BB' \cap CC'$, $E = CC' \cap AA'$, $F = AA' \cap BB'$, $S = BC' \cap B'C$, and $T = BC \cap B'C'$. Since $D$, $S$, and $T$ are the intersections of opposite sides and of the diagonals in the quadrilateral $BB'CC'$ 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 $ST$ 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 $SD$ and $R$ the radius of $\omega$, then the power of $O$ with respect to this circle is $ON^2 - ND^2 = OD \cdot OD^* = R^2$. By rearranging, we see that $ND^2$ is the power of $N$ with respect to $\omega$.

Second, from $T$ being the pole of $SD$ it follows that $OT$ is perpendicular to $SD$. Let $M$ and $M'$ denote the midpoints of $BC$ and $B'C'$. Then since $OM \perp BC$ and $OM' \perp B'C'$, it follows that $OMM'T$ is cyclic and
$$\Varangle(SD, BC) = \Varangle(OT, OM) = \Varangle(B'C', MM').$$
From $BB'CC'$ being cyclic we also have $\Varangle(BC, BB') = \Varangle(CC', B'C')$, hence we obtain
$$\begin{aligned} \Varangle(SD, BB') &= \Varangle(SD, BC) + \Varangle(BC, BB') \\ &= \Varangle(B'C', MM') + \Varangle(CC', B'C') = \Varangle(CC', MM'). \end{aligned}$$
Now from the homothety mentioned in the beginning, we know that $MM'$ is parallel to $AA'$, hence the above implies that $\Varangle(SD, BB') = \Varangle(CC', AA')$, which shows that $\Omega$ is tangent to $SD$ at $D$. In particular, $ND^2$ is also the power of $N$ with respect to $\Omega$.

Additionally, from $BB'CC'$ being cyclic it follows that triangles $DBC$ and $DC'B'$ are inversely similar, so $\Varangle(BB', DM') = \Varangle(DM, CC')$. This yields
$$\begin{aligned} \Varangle(SD, DM') &= \Varangle(SD, BB') + \Varangle(BB', DM') \\ &= \Varangle(CC', MM') + \Varangle(DM, CC') = \Varangle(DM, MM'), \end{aligned}$$
which shows that the circle $DMM'$ is also tangent to $SD$. Since $N$, $M$, and $M'$ are collinear on the Newton-Gauss line of the complete quadrilateral determined by the lines $BB'$, $CC'$, $BC'$, and $B'C$, it follows that $ND^2 = NM \cdot NM'$. 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.