4B. In triangle $A B C$, the coordinates of points $A_{1}(-1,0)$, the midpoint of side $B C$, $B_{1}(2,1)$, the midpoint of side $A C$, and $C_{1}(0,3)$, the midpoint of side $A B$ are known. Determine the coordinates of the vertices of this triangle.
Solution. Let the coordinates of the vertices be $A\left(a^{\prime}, a^{\prime \prime}\right), B\left(b^{\prime}, b^{\prime \prime}\right)$, and $C\left(c^{\prime}, c^{\prime \prime}\right)$. Since $A_{1}(-1,0)$ is the midpoint of side $BC$, we get $\frac{b^{\prime}+c^{\prime}}{2}=-1, \frac{b^{\prime \prime}+c^{\prime \prime}}{2}=0$. Similarly, we obtain the equations

$$
\frac{a^{\prime}+b^{\prime}}{2}=0, \frac{a^{\prime \prime}+b^{\prime \prime}}{2}=3, \frac{a^{\prime}+c^{\prime}}{2}=2, \frac{a^{\prime \prime}+c^{\prime \prime}}{2}=1
$$

Now, solving the systems

$$
\begin{aligned}
& \frac{a^{\prime}+b^{\prime}}{2}=0, \frac{b^{\prime}+c^{\prime}}{2}=-1, \frac{a^{\prime}+c^{\prime}}{2}=2 \\
& \frac{a^{\prime \prime}+c^{\prime \prime}}{2}=3, \frac{b^{\prime \prime}+c^{\prime \prime}}{2}=0, \frac{a^{\prime \prime}+c^{\prime \prime}}{2}=1
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_a5e9539585a8c87968b2g-13.jpg?height=481&width=467&top_left_y=997&top_left_x=997)

for the coordinates of the vertices of the triangle, we get $A(3,4), B(-3,2)$, and $C(1,-2)$.