Subfield: Geometry


Question: Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.

The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.

Solution:
Let $H$ denote the orthocenter of $A B C$, and let $e$ denote its Euler line. Let $e_{1}, e_{2}, e_{3}$ denote the respective reflections of $e$ in $B C, C A, A B$. The proof naturally divides into two parts: we first show that pairwise intersections of the circles in question correspond to pairwise intersections of $e_{1}, e_{2}, e_{3}$, and then prove that $e_{1}, e_{2}, e_{3}$ intersect in a single point on the circumcircle of $A B C$.

<img_3343>



Now consider for example the circumcircles of $O_{1} O_{2} C$ and $G_{1} G_{2} C$. By construction, it is clear that $\angle O_{2} C O_{1}=\angle G_{2} C G_{1}=2 \angle A C B$. Let $G_{1} O_{1}$ and $G_{2} O_{2}$ meet at $X$, and let $e$ meet $e_{1}, e_{2}$ at $E_{1}, E_{2}$, respectively, as shown in the diagram below. Chasing angles,

$$
\begin{aligned}
\angle G_{2} X G_{1}=\angle O_{2} X O_{1} & =\angle E_{2} X E_{1}=180^{\circ}-\angle E_{1} E_{2} X-\angle X E_{1} E_{2} \\
& =180^{\circ}-2 \angle E_{1} E_{2} C-\left(\angle C E_{1} E_{2}-\angle C E_{1} X\right) .
\end{aligned}
$$

But $\angle C E_{1} X=\angle B E_{1} O_{1}=\angle B E_{1} O=180^{\circ}-\angle C E_{1} E_{2}$, and thus

$$
\angle G_{2} X G_{1}=\angle O_{2} X O_{1}=2\left(180^{\circ}-\angle E_{1} E_{2} C-\angle C E_{1} E_{2}\right)=2 \angle A C B .
$$

<img_3902>

It follows from this that $X$ lies on the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$. In other words, the second point of intersection of the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$ is the intersection of $e_{1}$ and $e_{2}$. Similarly, the circumcircles of $G_{1} G_{3} B$ and $O_{1} O_{3} B$ meet again at the intersection of $e_{1}$ and $e_{3}$, and those of $G_{2} G_{3} A$ and $O_{2} O_{3} A$ meet again at the intersection of $e_{2}$ and $e_{3}$.



<img_3869>

It thus remains to show that $e_{1}, e_{2}, e_{3}$ are concurrent, and intersect on the circumcircle of $A B C$. Let $e$ meet the circumcircles of the triangles $B C H, A C H, A B H$ at $X_{1}, X_{2}, X_{3}$, respectively. It is well known that the reflections of $H$ in the sides of $A B C$ lie on the circumcircle of $A B C$. For this reason, the circumcircles of $B C H, A C H, A B H$ have the same radius as the circumcircle of $A B C$, and hence the reflections $X_{1}^{\prime}, X_{2}^{\prime}, X_{3}^{\prime}$ of $X_{1}, X_{2}, X_{3}$ in the sides $[B C],[C A],[A B]$ lie on the circumcircle of $A B C$. By definition, $X_{1}^{\prime}, X_{2}^{\prime}, X_{3}^{\prime}$ lie on $e_{1}, e_{2}, e_{3}$, respectively. It thus remains to show that they coincide.

To show that, for example, $X_{1}^{\prime}=X_{2}^{\prime}$, it will be sufficient to show that $\angle X_{2} A C=\angle X_{1}^{\prime} A C$, since we have already shown that $X_{1}^{\prime}$ and $X_{2}^{\prime}$ lie on the circumcircle of $A B C$. But, chasing angles in the diagram above,

$$
\begin{aligned}
\angle X_{1}^{\prime} A C & =\angle X_{1}^{\prime} A B-\angle B A C=\left(180^{\circ}-\angle X_{1}^{\prime} C B\right)-\angle B A C \\
& =180^{\circ}-\angle X_{1} C B-\angle B A C=\left(180^{\circ}-\angle B A C\right)-\angle X_{1} H B \\
& =\angle B H C-\angle X_{1} H B=\angle X_{2} H C=\angle X_{2} A C,
\end{aligned}
$$

where we have used the fact that $B H C X_{1}$ and $A H X_{2} C$ are cyclic by construction, and the fact that $\angle B H C=180^{\circ}-\angle B A C$. This shows that $X_{1}^{\prime}=X_{2}^{\prime}$. Similarly, $X_{1}^{\prime}=X_{3}^{\prime}$, which completes the proof.
The proof consists of two parts. First, we show that if $P$ is any point inside the triangle $A B C$ and $P_{1}, P_{2}, P_{3}$ are its reflections in the sides $B C, C A, A B$, then the circumcircles of the triangles $P_{1} P_{2} C, P_{1} P_{3} B, P_{2} P_{3} A$ intersect in a point $T_{P}$ on the circumcircle of the triangle $A B C$. In the second part, we show that $T_{G}$ coincides with $T_{O}$.

Now let $P$ be any point inside the triangle $A B C$ and let $P_{1}, P_{2}, P_{3}$ be the reflections in the sides as above. Let $T_{P}$ be the second intersection of the circumcircles of the triangles $P_{1} P_{2} C$ and $A B C$. We want to show that $T_{P}$ lies on the circumcircles of the triangles $P_{1} P_{3} B$ and $P_{2} P_{3} A$.

<img_3515>

By construction, we have $P_{1} C=P_{2} C$, hence

$$
\angle C P_{1} P_{2}=90^{\circ}-\frac{1}{2} \angle P_{2} C P_{1}=90^{\circ}-\angle A C B .
$$

Similarly, $\angle P_{2} P_{3} A=90^{\circ}-\angle B A C$. This gives us

$$
\begin{aligned}
\angle P_{2} T_{P} A & =\angle C T_{P} A-\angle C T_{P} P_{2}=\angle C B A-\angle C P_{1} P_{2} \\
& =\angle C B A-90^{\circ}+\angle A C B=90^{\circ}-\angle B A C=\angle P_{2} P_{3} A,
\end{aligned}
$$

so $T_{P}$ lies on the circumcircle of the triangle $P_{2} P_{3} A$. Similarly, $T_{P}$ lies on the circumcircle of the triangle $P_{1} P_{3} B$ which completes the first part.

Note that if $P_{2}$ is given, then $T_{P}$ is the unique point on the circumcircle of the triangle $A B C$ with $\angle C T_{P} P_{2}=90^{\circ}-\angle A C B$. In the second part, we will use this as follows: If we can find a point $T$ on the circumcircle of the triangle $A B C$ with $\angle C T G_{2}=\angle C T O_{2}=90^{\circ}-\angle A C B$, then $T=T_{G}=T_{O}$ and we are done.

Let $H$ be the orthocenter of the triangle $A B C$ and let $H_{2}$ be the reflection in the side $A C$. It is known that $\mathrm{H}_{2}$ lies on the circumcircle of the triangle $A B C . G, O, H$ lie on the Euler line, so $G_{2}, \mathrm{O}_{2}, \mathrm{H}_{2}$ are collinear as well. Let $T$ be the second intersection of $G_{2} H_{2}$ and the circumcircle of the triangle $A B C$. We can now complete the proof by seeing that

$$
\angle C T G_{2}=\angle C T O_{2}=\angle C T H_{2}=\angle C B H_{2}=90^{\circ}-\angle A C B .
$$

<img_3257>
For every point $P$, let $p$ denote the corresponding complex number. Set $O$ to be the origin, so $o=0$, and without loss of generality we can assume that $a, b$ and $c$ lie on the unit circle. Then the centroid can be expressed as $g=\frac{a+b+c}{3}$.

The segments $o o_{1}$ and $b c$ have a common midpoint, so $o_{1}+o=b+c$, and then $o_{1}=b+c$. Similarly $o_{2}=a+c$ and $o_{3}=a+b$. In order to compute $g_{1}$, define $y$ to be the projection of $g$ onto $b c$. Since $b$ and $c$ are on the unit circle, it is well known that $y$ can be expressed as

$$
y=\frac{1}{2}(b+c+g-b c \bar{g})
$$

By using $\bar{a}=\frac{1}{a}, \bar{b}=\frac{1}{b}$ and $\bar{c}=\frac{1}{c}$ (points on the unit circle), we obtain

$$
g_{1}=b+c-\frac{a b+b c+c a}{3 a}
$$

Similarly, we get $g_{2}=a+c-\frac{a b+b c+c a}{3 b}$ and $g_{3}=a+b-\frac{a b+b c+c a}{3 c}$.

1) Proof that circumcircles of triangles $a b c, o_{1} o_{2} c, o_{1} o_{3} b$ and $o_{2} o_{3} a$ have common point. Let $x$ be the point of intersection of circumcircles of triangles $o_{1} o_{2} c$ and $a b c(x \neq c)$. We know that $x, o_{1}, o_{2}$ and $c$ are concyclic if and only if $\frac{x-c}{o_{1}-c}: \frac{x-o_{2}}{o_{1}-o_{2}}$ is real number, which is equivalent to

$$
\frac{x-c}{\bar{x}-\bar{c}} \cdot \frac{o_{1}-o_{2}}{\overline{o_{1}}-\overline{o_{2}}}=\frac{o_{1}-c}{\overline{o_{1}}-\bar{c}} \cdot \frac{x-o_{2}}{\bar{x}-\overline{o_{2}}}
\tag{1}
$$

Since $x$ and $c$ are on the unit circle $\frac{x-c}{\bar{x} - \bar{c}} = -xc$. Also, $\frac{o_1-o_2}{\bar{o_1} - \bar{o_2}} = \frac{b-a}{\bar{b} - \bar{a}}=-ab$ and  $\frac{o_{1}-c}{\overline{o_{1}}-\bar{c}}=\frac{b}{\bar{b}}=b^{2}$. Since $\bar{x}=\frac{1}{x}$, from (1) and previous relations, we have:

$$
x=\frac{a b+b c+c a}{a+b+c} \text {. }
$$

This formula is symmetric, so we conclude that $x$ also belongs to circumcircles of $o_{1} o_{3} b$ and $\mathrm{o}_{2} \mathrm{o}_{3} a$.

2) Proof that $x$ belongs to circumcircles of $g_{1} g_{2} c, g_{1} g_{3} b$ and $g_{2} g_{3} a$.

Because of symmetry, it is enough to prove that $x$ belongs to circumcircle of $g_{1} g_{2} c$, i.e. to prove the following:

$$
\frac{x-c}{\bar{x}-\bar{c}} \cdot \frac{g_{1}-g_{2}}{\overline{g_{1}}-\overline{g_{2}}}=\frac{g_{1}-c}{\overline{g_{1}}-\bar{c}} \cdot \frac{x-g_{2}}{\bar{x}-\overline{g_{2}}} \tag{2}
$$

Easy computations give that

$$
g_{1}-g_{2}=(b-a) \frac{2 a b-b c-a c}{3 a b}, \quad \overline{g_{1}}-\overline{g_{2}}=(\bar{b}-\bar{a}) \frac{2-\frac{a}{c}-\frac{b}{c}}{3}
$$

and then

$$
\frac{g_{1}-g_{2}}{\overline{g_{1}}-\overline{g_{2}}}=\frac{c(b c+a c-2 a b)}{2 c-a-b}
$$



On the other hand we have

$$
g_{1}-c=\frac{2 a b-b c-a c}{3 a}, \quad \bar{g}_{1}-\bar{c}=\frac{2 c-a-b}{3 b c}
$$

This implies

$$
\frac{g_{1}-c}{{\bar{g_{1}}-\bar{c}}}=\frac{2 a b-b c-a c}{2 c-a-b} \cdot \frac{b c}{a} .
$$

Then (2) is equivalent to

$$
\begin{aligned}
-x c \cdot \frac{c(b c+a c-2 a b)}{2 c-a-b}=\frac{2 a b-b c-a c}{2 c-a-b} \cdot \frac{b c}{a} \cdot \frac{x-g_{2}}{\bar{x}-\overline{g_{2}}} \\
\Longleftrightarrow x c a\left(\bar{x}-\overline{g_{2}}\right)=b\left(x-g_{2}\right),
\end{aligned}
$$

which is also equivalent to

$\frac{a b+b c+c a}{a+b+c} \cdot c a\left(\frac{a+b+c}{a b+b c+c a}-\frac{1}{a}-\frac{1}{c}+\frac{a+b+c}{3 a c}\right)=b \cdot\left(\frac{a b+b c+c a}{a+b+c}-a-c+\frac{a b+b c+c a}{3 b}\right)$.

The last equality can easily be verified, which implies that $x$ belongs to circumcircle of triangle $g_{1} g_{2} c$. This concludes our proof.