Subfield: Geometry


Question: Let $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ be two triangles having the same circumcircle $\omega$, and the same orthocentre $H$. Let $\Omega$ be the circumcircle of the triangle determined by the lines $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$. Prove that $H$, the centre of $\omega$, and the centre of $\Omega$ are collinear.

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 $A B C$ 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 $D S T$ is self-polar with respect to $\omega$, i.e. each vertex is the pole of the opposite side. We apply this in two ways.

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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 $\omega$.

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

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

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.