International Mathematical Olympiad

In what follows, $\text{Varangle}(p,q)$ 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 $A^{\prime }{B}^{\prime }{C}^{\prime }$ 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 $D=B{B}^{\prime }\cap C{C}^{\prime }$, $E=C{C}^{\prime }\cap A{A}^{\prime }$, $F=A{A}^{\prime }\cap B{B}^{\prime }$, $S=B{C}^{\prime }\cap B^{\prime }C$, and $T=BC\cap B^{\prime }{C}^{\prime }$. 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 $ST$ it follows that the inverse $D^{\ast }$ of $D$ with respect to $\omega$ is the projection of $D$ onto $ST$. In particular, $D^{\ast }$ 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 $O{N}^2-N{D}^2=OD\cdot O{D}^{\ast }=R^2$. 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 $SD$. Let $M$ and $M^{\prime }$ denote the midpoints of $BC$ and $B^{\prime }{C}^{\prime }$. Then since $OM\perp BC$ and $O{M}^{\prime }\perp B^{\prime }{C}^{\prime }$, it follows that $OM{M}^{\prime }T$ is cyclic and $$\text{Varangle}(SD,BC)=\text{Varangle}(OT,OM)=\text{Varangle}(B^{\prime }{C}^{\prime },M{M}^{\prime }).$$ From $B{B}^{\prime }C{C}^{\prime }$ being cyclic we also have $\text{Varangle}(BC,B{B}^{\prime })=\text{Varangle}(C{C}^{\prime },B^{\prime }{C}^{\prime })$, hence we obtain $$\begin{array}{rl}\text{Varangle}(SD,B{B}^{\prime })& =\text{Varangle}(SD,BC)+\text{Varangle}(BC,B{B}^{\prime })\\ & =\text{Varangle}(B^{\prime }{C}^{\prime },M{M}^{\prime })+\text{Varangle}(C{C}^{\prime },B^{\prime }{C}^{\prime })=\text{Varangle}(C{C}^{\prime },M{M}^{\prime }).\end{array}$$ 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 $\text{Varangle}(SD,B{B}^{\prime })=\text{Varangle}(C{C}^{\prime },A{A}^{\prime })$, which shows that $\Omega$ is tangent to $SD$ 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 $DBC$ and $D{C}^{\prime }{B}^{\prime }$ are inversely similar, so $\text{Varangle}(B{B}^{\prime },D{M}^{\prime })=\text{Varangle}(DM,C{C}^{\prime })$. This yields $$\begin{array}{rl}\text{Varangle}(SD,D{M}^{\prime })& =\text{Varangle}(SD,B{B}^{\prime })+\text{Varangle}(B{B}^{\prime },D{M}^{\prime })\\ & =\text{Varangle}(C{C}^{\prime },M{M}^{\prime })+\text{Varangle}(DM,C{C}^{\prime })=\text{Varangle}(DM,M{M}^{\prime }),\end{array}$$ which shows that the circle $DM{M}^{\prime }$ is also tangent to $SD$. 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 $N{D}^2=NM\cdot N{M}^{\prime }$. 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.