Subfield: Geometry


Question: In an acute-angled triangle $A B C$, point $H$ is the orthocentre and $A_{0}, B_{0}, C_{0}$ are the midpoints of the sides $B C, C A, A B$, respectively. Consider three circles passing through $H: \omega_{a}$ around $A_{0}, \omega_{b}$ around $B_{0}$ and $\omega_{c}$ around $C_{0}$. The circle $\omega_{a}$ intersects the line $B C$ at $A_{1}$ and $A_{2} ; \omega_{b}$ intersects $C A$ at $B_{1}$ and $B_{2} ; \omega_{c}$ intersects $A B$ at $C_{1}$ and $C_{2}$. Show that the points $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ lie on a circle.

Solution:
The perpendicular bisectors of the segments $A_{1} A_{2}, B_{1} B_{2}, C_{1} C_{2}$ are also the perpendicular bisectors of $B C, C A, A B$. So they meet at $O$, the circumcentre of $A B C$. Thus $O$ is the only point that can possibly be the centre of the desired circle.

From the right triangle $O A_{0} A_{1}$ we get

$$
O A_{1}^{2}=O A_{0}^{2}+A_{0} A_{1}^{2}=O A_{0}^{2}+A_{0} H^{2} .\tag{1}
$$

Let $K$ be the midpoint of $A H$ and let $L$ be the midpoint of $C H$. Since $A_{0}$ and $B_{0}$ are the midpoints of $B C$ and $C A$, we see that $A_{0} L \| B H$ and $B_{0} L \| A H$. Thus the segments $A_{0} L$ and $B_{0} L$ are perpendicular to $A C$ and $B C$, hence parallel to $O B_{0}$ and $O A_{0}$, respectively. Consequently $O A_{0} L B_{0}$ is a parallelogram, so that $O A_{0}$ and $B_{0} L$ are equal and parallel. Also, the midline $B_{0} L$ of triangle $A H C$ is equal and parallel to $A K$ and $K H$.

It follows that $A K A_{0} O$ and $H A_{0} O K$ are parallelograms. The first one gives $A_{0} K=O A=R$, where $R$ is the circumradius of $A B C$. From the second one we obtain

$$
2\left(O A_{0}^{2}+A_{0} H^{2}\right)=O H^{2}+A_{0} K^{2}=O H^{2}+R^{2} .\tag{2}
$$

(In a parallelogram, the sum of squares of the diagonals equals the sum of squares of the sides).

From (1) and (2) we get $O A_{1}^{2}=\left(O H^{2}+R^{2}\right) / 2$. By symmetry, the same holds for the distances $O A_{2}, O B_{1}, O B_{2}, O C_{1}$ and $O C_{2}$. Thus $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ all lie on a circle with centre at $O$ and radius $\left(O H^{2}+R^{2}\right) / 2$.

<img_3768>
We are going to show again that the circumcentre $O$ is equidistant from the six points in question.

Let $A^{\prime}$ be the second intersection point of $\omega_{b}$ and $\omega_{c}$. The line $B_{0} C_{0}$, which is the line of centers of circles $\omega_{b}$ and $\omega_{c}$, is a midline in triangle $A B C$, parallel to $B C$ and perpendicular to the altitude $A H$. The points $A^{\prime}$ and $H$ are symmetric with respect to the line of centers. Therefore $A^{\prime}$ lies on the line $A H$.

From the two circles $\omega_{b}$ and $\omega_{c}$ we obtain $A C_{1} \cdot A C_{2}=A A^{\prime} \cdot A H=A B_{1} \cdot A B_{2}$. So the quadrilateral $B_{1} B_{2} C_{1} C_{2}$ is cyclic. The perpendicular bisectors of the sides $B_{1} B_{2}$ and $C_{1} C_{2}$ meet at $O$. Hence $O$ is the circumcentre of $B_{1} B_{2} C_{1} C_{2}$ and so $O B_{1}=O B_{2}=O C_{1}=O C_{2}$.

Analogous arguments yield $O A_{1}=O A_{2}=O B_{1}=O B_{2}$ and $O A_{1}=O A_{2}=O C_{1}=O C_{2}$. Thus $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ lie on a circle centred at $O$.

<img_3671>
Let again $O$ and $R$ be the circumcentre and circumradius. Consider the vectors

$$
\overrightarrow{O A}=\mathbf{a}, \quad \overrightarrow{O B}=\mathbf{b}, \quad \overrightarrow{O C}=\mathbf{c}, \quad \text { where } \quad \mathbf{a}^{2}=\mathbf{b}^{2}=\mathbf{c}^{2}=R^{2}
$$

It is well known that $\overrightarrow{O H}=\mathbf{a}+\mathbf{b}+\mathbf{c}$. Accordingly,

$$
\overrightarrow{A_{0} H}=\overrightarrow{O H}-\overrightarrow{O A_{0}}=(\mathbf{a}+\mathbf{b}+\mathbf{c})-\frac{\mathbf{b}+\mathbf{c}}{2}=\frac{2 \mathbf{a}+\mathbf{b}+\mathbf{c}}{2}
$$

and

$$
\begin{gathered}
O A_{1}^{2}=O A_{0}^{2}+A_{0} A_{1}^{2}=O A_{0}^{2}+A_{0} H^{2}=\left(\frac{\mathbf{b}+\mathbf{c}}{2}\right)^{2}+\left(\frac{2 \mathbf{a}+\mathbf{b}+\mathbf{c}}{2}\right)^{2} \\
=\frac{1}{4}\left(\mathbf{b}^{2}+2 \mathbf{b} \mathbf{c}+\mathbf{c}^{2}\right)+\frac{1}{4}\left(4 \mathbf{a}^{2}+4 \mathbf{a} \mathbf{b}+4 \mathbf{a} \mathbf{c}+\mathbf{b}^{2}+2 \mathbf{b} \mathbf{c}+\mathbf{c}^{2}\right)=2 R^{2}+(\mathbf{a b}+\mathbf{a c}+\mathbf{b c}) ;
\end{gathered}
$$

here $\mathbf{a b}, \mathbf{b c}$, etc. denote dot products of vectors. We get the same for the distances $O A_{2}, O B_{1}$, $O B_{2}, O C_{1}$ and $O C_{2}$.