I have three points $a_1, a_2, a_3 \in \mathbb{R}^2$ which form a triangle. The triangle is defined as follows:
$$ \Delta = \{\lambda_1 a_1 + \lambda_2 a_2 + \lambda_3 a_3 \mid 0 \leq \lambda_i,\; \sum_{i=1}^3 \lambda_i = 1\} $$
Now let $x \in \mathbb{R}^2 \setminus \{a_1, a_2, a_3\}$ be a point such that the sum of the distances between $x$ and $a_i$ is minimal, i.e. $$ \sum_{i=1}^3 |x - a_i| = c $$
Forgive me if this little sketch is inaccurate. So intuitively I assumed that $x$ will always be the point where all "height lines" of the $a_i$ intersect.
Thus we get
$$ \begin{align*} |x - a_i| &= \lvert\begin{pmatrix} x_1 - a_{i_x} \\ x_2 - a_{i_y} \end{pmatrix}\rvert \\ &= \sqrt{(x_1 - a_{i_x})^2 + (x_2 - a_{i_y})^2} \end{align*} $$
where $x$ needs to be in the set $\Delta$.
Now the question is,
- How can I prove $x$ will always be this point and
- How can I prove that the angle between neighboring vectors $a_i - x$ is always $\frac{2 \pi}{3}$