Subfield: Geometry


Question: Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$. Prove that the circumcentre of triangle $X S Y$ is equidistant from $P$ and $M$.

Solution:
Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $Q$ be the midpoint of $A H$ and $N$ be the nine-point centre of triangle $A B C$. It is known that $Q$ lies on the nine-point circle of triangle $A B C, N$ is the midpoint of $Q M$ and that $Q M$ is parallel to $A O$.

Let the perpendicular from $S$ to $X Y$ meet line $Q M$ at $S^{\prime}$. Let $E$ be the foot of altitude from $B$ to side $A C$. Since $Q$ and $S$ lie on the perpendicular bisector of $A D$, using directed angles, we have

$$
\begin{aligned}
\measuredangle S D Q & =\measuredangle Q A S=\measuredangle X A S-\measuredangle X A Q=\left(\frac{\pi}{2}-\measuredangle A Y X\right)-\measuredangle B A P=\measuredangle C B A-\measuredangle A Y X \\
& =(\measuredangle C B A-\measuredangle A C B)-\measuredangle B C A-\measuredangle A Y X=\measuredangle P E M-(\measuredangle B C A+\measuredangle A Y X) \\
& =\measuredangle P Q M-\angle(B C, X Y)=\frac{\pi}{2}-\angle\left(S^{\prime} Q, B C\right)-\angle(B C, X Y)=\measuredangle S S^{\prime} Q .
\end{aligned}
$$

This shows $D, S^{\prime}, S, Q$ are concyclic.

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Let the perpendicular from $N$ to $B C$ intersect line $S S^{\prime}$ at $O_{1}$. (Note that the two lines coincide when $S$ is the midpoint of $A O$, in which case the result is true since the circumcentre of triangle $X S Y$ must lie on this line.) It suffices to show that $O_{1}$ is the circumcentre of triangle $X S Y$ since $N$ lies on the perpendicular bisector of $P M$. From

$$
\measuredangle D S^{\prime} O_{1}=\measuredangle D Q S=\measuredangle S Q A=\angle(S Q, Q A)=\angle\left(O D, O_{1} N\right)=\measuredangle D N O_{1}
$$



since $S Q / / O D$ and $Q A / / O_{1} N$, we know that $D, O_{1}, S^{\prime}, N$ are concyclic. Therefore, we get

$$
\measuredangle S D S^{\prime}=\measuredangle S Q S^{\prime}=\angle\left(S Q, Q S^{\prime}\right)=\angle\left(N D, N S^{\prime}\right)=\measuredangle D N S^{\prime}
$$

so that $S D$ is a tangent to the circle through $D, O_{1}, S^{\prime}, N$. Then we have

$$
S S^{\prime} \cdot S O_{1}=S D^{2}=S X^{2}
\tag{1}
$$

Next, we show that $S$ and $S^{\prime}$ are symmetric with respect to $X Y$. By the Sine Law, we have

$$
\frac{S S^{\prime}}{\sin \angle S Q S^{\prime}}=\frac{S Q}{\sin \angle S S^{\prime} Q}=\frac{S Q}{\sin \angle S D Q}=\frac{S Q}{\sin \angle S A Q}=\frac{S A}{\sin \angle S Q A} .
$$

It follows that

$$
S S^{\prime}=S A \cdot \frac{\sin \angle S Q S^{\prime}}{\sin \angle S Q A}=S A \cdot \frac{\sin \angle H O A}{\sin \angle O H A}=S A \cdot \frac{A H}{A O}=S A \cdot 2 \cos A,
$$

which is twice the distance from $S$ to $X Y$. Note that $S$ and $C$ lie on the same side of the perpendicular bisector of $P M$ if and only if $\angle S A C<\angle O A C$ if and only if $\angle Y X A>\angle C B A$. This shows $S$ and $O_{1}$ lie on different sides of $X Y$. As $S^{\prime}$ lies on ray $S O_{1}$, it follows that $S$ and $S^{\prime}$ cannot lie on the same side of $X Y$. Therefore, $S$ and $S^{\prime}$ are symmetric with respect to $X Y$.

Let $d$ be the diameter of the circumcircle of triangle $X S Y$. As $S S^{\prime}$ is twice the distance from $S$ to $X Y$ and $S X=S Y$, we have $S S^{\prime}=2 \frac{S X^{2}}{d}$. It follows from (1) that $d=2 S O_{1}$. As $S O_{1}$ is the perpendicular bisector of $X Y$, point $O_{1}$ is the circumcentre of triangle $X S Y$.
Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $O_{1}$ be the circumcentre of triangle $X S Y$. Consider two other possible positions of $S$. We name them $S^{\prime}$ and $S^{\prime \prime}$ and define the analogous points $X^{\prime}, Y^{\prime}, O_{1}^{\prime}, X^{\prime \prime}, Y^{\prime \prime} O_{1}^{\prime \prime}$ accordingly. Note that $S, S^{\prime}, S^{\prime \prime}$ lie on the perpendicular bisector of $A D$.

As $X X^{\prime}$ and $Y Y^{\prime}$ meet at $A$ and the circumcircles of triangles $A X Y$ and $A X^{\prime} Y^{\prime}$ meet at $D$, there is a spiral similarity with centre $D$ mapping $X Y$ to $X^{\prime} Y^{\prime}$. We find that

$$
\measuredangle S X Y=\frac{\pi}{2}-\measuredangle Y A X=\frac{\pi}{2}-\measuredangle Y^{\prime} A X^{\prime}=\measuredangle S^{\prime} X^{\prime} Y^{\prime}
$$

and similarly $\measuredangle S Y X=\measuredangle S^{\prime} Y^{\prime} X^{\prime}$. This shows triangles $S X Y$ and $S^{\prime} X^{\prime} Y^{\prime}$ are directly similar. Then the spiral similarity with centre $D$ takes points $S, X, Y, O_{1}$ to $S^{\prime}, X^{\prime}, Y^{\prime}, O_{1}^{\prime}$. Similarly, there is a spiral similarity with centre $D$ mapping $S, X, Y, O_{1}$ to $S^{\prime \prime}, X^{\prime \prime}, Y^{\prime \prime}, O_{1}^{\prime \prime}$. From these, we see that there is a spiral similarity taking the corresponding points $S, S^{\prime}, S^{\prime \prime}$ to points $O_{1}, O_{1}^{\prime}, O_{1}^{\prime \prime}$. In particular, $O_{1}, O_{1}^{\prime}, O_{1}^{\prime \prime}$ are collinear.



<img_3383>

It now suffices to show that $O_{1}$ lies on the perpendicular bisector of $P M$ for two special cases.

Firstly, we take $S$ to be the midpoint of $A H$. Then $X$ and $Y$ are the feet of altitudes from $C$ and $B$ respectively. It is well-known that the circumcircle of triangle $X S Y$ is the nine-point circle of triangle $A B C$. Then $O_{1}$ is the nine-point centre and $O_{1} P=O_{1} M$. Indeed, $P$ and $M$ also lies on the nine-point circle.

Secondly, we take $S^{\prime}$ to be the midpoint of $A O$. Then $X^{\prime}$ and $Y^{\prime}$ are the midpoints of $A B$ and $A C$ respectively. Then $X^{\prime} Y^{\prime} / / B C$. Clearly, $S^{\prime}$ lies on the perpendicular bisector of $P M$. This shows the perpendicular bisectors of $X^{\prime} Y^{\prime}$ and $P M$ coincide. Hence, we must have $O_{1}^{\prime} P=O_{1}^{\prime} M$.
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