Subfield: Geometry


Question: Let $P$ be a point on the circumcircle of an acute-angled triangle $A B C$. Let $D$, $E$, and $F$ be the reflections of $P$ in the midlines of triangle $A B C$ parallel to $B C, C A$, and $A B$, respectively. Denote by $\omega_{A}, \omega_{B}$, and $\omega_{C}$ the circumcircles of triangles $A D P, B E P$, and $C F P$, respectively. Denote by $\omega$ the circumcircle of the triangle formed by the perpendicular bisectors of segments $A D, B E$ and $C F$.

Show that $\omega_{A}, \omega_{B}, \omega_{C}$, and $\omega$ have a common point.

Solution:
Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the altitudes in triangle $A B C$, and let $m_{A}, m_{B}$, and $m_{C}$ be the midlines parallel to $B C, C A$, and $A B$, respectively. We always denote by $\sphericalangle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^{\circ}$.

Step 1: Circles $\omega_{A}, \omega_{B}$, and $\omega_{C}$ share a common point $Q$ different from $P$.

Notice that $m_{A}$ is the perpendicular bisector of $P D$, so $\omega_{A}$ is symmetric with respect to $m_{A}$. Since $A$ and $A_{1}$ are also symmetric to each other in $m_{A}$, the point $A_{1}$ lies on $\omega_{A}$. Similarly, $B_{1}$ and $C_{1}$ lie on $\omega_{B}$ and $\omega_{C}$, respectively.

Let $H$ be the orthocenter of $\triangle A B C$. Quadrilaterals $A B A_{1} B_{1}$ and $B C B_{1} C_{1}$ are cyclic, so $A H \cdot H A_{1}=B H \cdot H B_{1}=C H \cdot H C_{1}$. This means that $H$ lies on pairwise radical axes of $\omega_{A}$, $\omega_{B}$, and $\omega_{C}$. Point $P$ also lies on those radical axes; hence the three circles have a common radical axis $\ell=P H$, and the second meeting point $Q$ of $\ell$ with $\omega_{A}$ is the second common point of the three circles. Notice here that $H$ lies inside all three circles, hence $Q \neq P$.
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Step 2: Point $Q$ lies on $\omega$.

Let $p_{A}, p_{B}$, and $p_{C}$ denote the perpendicular bisectors of $A D, B E$, and $C F$, respectively; denote by $\Delta$ the triangle formed by those perpendicular bisectors. By Simson's theorem, in order to show that $Q$ lies on the circumcircle $\omega$ of $\Delta$, it suffices to prove that the projections of $Q$ onto the sidelines $p_{A}, p_{B}$, and $p_{C}$ are collinear. Alternatively, but equivalently, it suffices to prove that the reflections $Q_{A}, Q_{B}$, and $Q_{C}$ of $Q$ in those lines, respectively, are collinear. In fact, we will show that four points $P, Q_{A}, Q_{B}$, and $Q_{C}$ are collinear.

Since $p_{A}$ is the common perpendicular bisector of $A D$ and $Q Q_{A}$, the point $Q_{A}$ lies on $\omega_{A}$, and, moreover, $\sphericalangle\left(D A, D Q_{A}\right)=\sphericalangle(A Q, A D)$. Therefore,

$$
\sphericalangle\left(P A, P Q_{A}\right)=\sphericalangle\left(D A, D Q_{A}\right)=\sphericalangle(A Q, A D)=\sphericalangle(P Q, P D)=\sphericalangle(P Q, B C)+90^{\circ}
$$



Similarly, we get $\sphericalangle\left(P B, P Q_{B}\right)=\sphericalangle(P Q, C A)+90^{\circ}$. Therefore,

$$
\begin{aligned}
\sphericalangle\left(P Q_{A}, P Q_{B}\right)=\sphericalangle\left(P Q_{A}, P A\right) & +\sphericalangle(P A, P B)+\sphericalangle\left(P B, P Q_{B}\right) \\
& =\sphericalangle(B C, P Q)+90^{\circ}+\sphericalangle(C A, C B)+\sphericalangle(P Q, C A)+90^{\circ}=0,
\end{aligned}
$$

which shows that $P, Q_{A}$, and $Q_{B}$ are collinear. Similarly, $Q_{C}$ also lies on $P Q_{A}$.