## Task B-2.5.

Complex numbers $z_{1}, z_{2}$, $z_{3}$ are associated with points $A, B, C$ in the complex plane that are 2016 units away from the origin. If for the complex numbers $z_{1}, z_{2}, z_{3}$ it holds that $z_{1}+z_{2}+z_{3}=0$, calculate the lengths of the sides of triangle $ABC$.
## Solution.

Let $z_{1}=x_{1}+y_{1} i, z_{2}=x_{2}+y_{2} i, z_{3}=x_{3}+y_{3} i$.

Then we have $\left|z_{1}\right|^{2}=x_{1}^{2}+y_{1}^{2}=2016^{2},\left|z_{2}\right|^{2}=x_{2}^{2}+y_{2}^{2}=2016^{2},\left|z_{3}\right|^{2}=x_{3}^{2}+y_{3}^{2}=2016^{2}$.

Furthermore, from $z_{1}+z_{2}+z_{3}=0$ it follows that

$$
x_{1}+x_{2}+x_{3}=0, \quad y_{1}+y_{2}+y_{3}=0 \text {. }
$$

The points $A, B, C$ have coordinates $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$.

The length of the side $\overline{A B}$ is calculated using the distance formula:

$$
|A B|^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}=x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}-2 x_{1} x_{2}-2 y_{1} y_{2}
$$

From (*) by squaring we get

$$
x_{1}^{2}+x_{2}^{2}-x_{3}^{2}=-2 x_{1} x_{2}, \quad y_{1}^{2}+y_{2}^{2}-y_{3}^{2}=-2 y_{1} y_{2}
$$

It follows that

$$
\begin{aligned}
|A B|^{2} & =\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}=x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{1}^{2}+x_{2}^{2}-x_{3}^{2}+y_{1}^{2}+y_{2}^{2}-y_{3}^{2} \\
& =2\left(x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}\right)-\left(x_{3}^{2}+y_{3}^{2}\right)=4 \cdot 2016^{2}-2016^{2}=3 \cdot 2016^{2}
\end{aligned}
$$

Similarly, we get

$$
\begin{aligned}
& |B C|^{2}=2\left(x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2}\right)-\left(x_{1}^{2}+y_{1}^{2}\right)=3 \cdot 2016^{2} \\
& |C A|^{2}=2\left(x_{3}^{2}+y_{3}^{2}+x_{1}^{2}+y_{1}^{2}\right)-\left(x_{2}^{2}+y_{2}^{2}\right)=3 \cdot 2016^{2}
\end{aligned}
$$

We conclude that $A B C$ is an equilateral triangle with the length of each side being $2016 \sqrt{3}$.

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - B variant

Primosten, April 5, 2016.