The coordinates of the centroid of a triangle are $\left(5 ;-\frac{5}{3}\right)$, the orthocenter's coordinates are $(3 ;-1)$, and one of the vertices is $(7 ; 3)$. Find the coordinates of the other two vertices.
Solution. Let the vertices of the triangle be $A(7 ; 3), B\left(b_{1} ; b_{2}\right), C\left(c_{1} ; c_{2}\right)$, and let $S$ denote the centroid, and $M$ the orthocenter.

It is known that the $x$ and $y$ coordinates of the centroid of a triangle are the arithmetic means of the $x$ and $y$ coordinates of the vertices of the triangle. Therefore:

$$
\frac{7+b_{1}+c_{1}}{3}=5, \quad \text { and } \quad \frac{3+b_{2}+c_{2}}{3}=-\frac{5}{3}
$$

From these two equations, we get: $b_{1}+c_{1}=8$ and $b_{2}+c_{2}=-8$.

Since $M$ is the orthocenter, $\overrightarrow{M A}(4 ; 4)$ is perpendicular to $\overrightarrow{B C}\left(c_{1}-b_{1} ; c_{2}-b_{2}\right)$. Considering the dot product of the two vectors, which is 0 due to the perpendicularity, we can write the dot product of the vectors using their coordinates as: $4 \cdot\left(c_{1}-\right.$ $\left.b_{1}\right)+4 \cdot\left(c_{2}-b_{2}\right)=0$. This implies that $\left(c_{1}+c_{2}\right)-\left(b_{1}+b_{2}\right)=0$.

Adding the two equations derived from the coordinates of the centroid, we get $c_{1}+c_{2}=0$, which implies $b_{1}+b_{2}=0$. Therefore, $b_{1}=-b_{2}$ and $c_{1}=-c_{2}$. From this point on, we will use the following notation: $b_{1}=b$ and $c_{1}=c$.

Similarly to $\overrightarrow{M A}$ and $\overrightarrow{B C}$, $\overrightarrow{C A}(7-c ; 3+c)$ and $\overrightarrow{B M}(3-b ; b-1)$ are also perpendicular to each other. Let's write the dot product of these two vectors (which is also 0) using the coordinates of the vectors:

$$
\begin{aligned}
(7-c)(3-b)+(3+c)(b-1) & =0 \\
21-7 b-3 c+b c+3 b-3+b c-c & =0 \\
b c-2 b-2 c+9 & =0
\end{aligned}
$$

From the equation derived from the coordinates of the centroid, solve for $c$: $c=8-b$. Substitute this into the equation we just obtained:

$$
\begin{array}{r}
b(8-b)-2 b-2(8-b)+9=0 \\
8 b-b^{2}-2 b-16+2 b+9=0 \\
b^{2}-8 b+7=0
\end{array}
$$

This is a quadratic equation in $b$. Using the quadratic formula: $b=1$ or $b=7$.

If $b=1$, then the coordinates of the other two vertices are: $(1 ;-1)$ and $(7 ;-7)$.

If $b=7$, then we get the same two points, with $b$ and $c$ swapped.

Note. Most people followed the thought process of the solution found on the website, which involves writing the equation of line $BC$ using the midpoint of $BC$ and the normal vector of $\overrightarrow{M A}$, then determining the coordinates of the circumcenter using the Euler line, and finally calculating the coordinates of the intersection points of the circumcircle and line $BC$ by writing the equation of the circumcircle.