Problem 1293

Solution. In what follows, $x(p, q)$ 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 $A B C$ 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 $O H$.

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 $S T$. In particular, $D^{*}$ lies on the circle with diameter $S D$. 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 $O N^{2}-N D^{2}-O D \cdot O D^{*}-R^{2}$. By rearranging, we see that $N D^{2}$ is the power of $N$ with respect to $w$.

Second, from $T$ being the pole of $S D$ it follows that $O T$ is perpendicular to $S D$. Let $M$ and $M^{\prime}$ denote the midpoints of $B C$ and $B^{\prime} C^{\prime}$. Then since $O M \perp B C$ and $O M^{\prime} \perp B^{\prime} C^{\prime}$ it follows that $O M M^{\prime} T$ is cyclic and
$$x(S D, B C)-x(O T, O M)-x\left(B^{\prime} C^{\prime}, M M^{\prime}\right) .$$

From $B B^{\prime} C C^{\prime}$ being cyclic we also have $x\left(B C, B B^{\prime}\right)-x\left(C C^{\prime}, B^{\prime} C^{\prime}\right)$, hence we obtain
$$\begin{aligned} x\left(S D, B B^{\prime}\right) & =x(S D, B C)+x\left(B C, B B^{\prime}\right) \\ & =x\left(B^{\prime} C^{\prime}, M M^{\prime}\right)+x\left(C C^{\prime}, B^{\prime} C^{\prime}\right)-x\left(C C^{\prime}, M M^{\prime}\right) . \end{aligned}$$

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 $\Varangle\left(S D, B B^{\prime}\right)-\dot{*}\left(C C^{\prime}, A A^{\prime}\right)$, 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 $x\left(B B^{\prime}, D M^{\prime}\right)-x\left(D M, C C^{\prime}\right)$. This yields
$$\begin{aligned} x\left(S D, D M^{\prime}\right) & =x\left(S D, B B^{\prime}\right)+x\left(B B^{\prime}, D M^{\prime}\right) \\ & =x\left(C C^{\prime}, M M^{\prime}\right)+x\left(D M, C C^{\prime}\right)-x\left(D M, M M^{\prime}\right), \end{aligned}$$
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 $D M M^{\prime}$ and $S M M^{\prime}$, so it lies on their radical axis $M M^{\prime}$.

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
$$\Varangle(B C, \ell)=\Varangle\left(B C, M M^{\prime}\right)=\Varangle\left(L L^{\prime}, B^{\prime} C^{\prime}\right)$$
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
$$P A \cdot P A^{\prime}=P L \cdot P L^{\prime}=P E \cdot P F \text {. }$$

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.