<s>[INST] Finding the point inside a triangle to minimize the total distance to three vertices Let $\mathbf a, \mathbf b, \mathbf c$ be three vectors. We want to minimize the function
$$
f(\mathbf r)=|\mathbf r-\mathbf a|+|\mathbf r-\mathbf b|+|\mathbf r-\mathbf c|.
$$
Now as usual we evaluate the gradient:
$$
\nabla f=\frac{\mathbf r-\mathbf a}{|\mathbf r-\mathbf a|}+\frac{\mathbf r-\mathbf b}{|\mathbf r-\mathbf b|}+\frac{\mathbf r-\mathbf c}{|\mathbf r-\mathbf c|}=0.
$$
The sum of three unit vectors is zero, so the angle between any two of them is $\frac{2}{3}\pi$.
Question: can I find an expression of $\mathbf r$ in terms of $\mathbf a, \mathbf b, \mathbf c$? Please avoid writing the vectors into their components. I am looking forward to seeing a beautiful solution.
 [/INST] A beautiful solution can indeed be derived, with all the symmetry preserved. It is suitably more involved, as expected.
Start with the zero gradient condition
$$
\frac{\mathbf r-\mathbf A}{|\mathbf r-\mathbf A|}+\frac{\mathbf r-\mathbf B}{|\mathbf r-\mathbf B|}+\frac{\mathbf r-\mathbf C}{|\mathbf r-\mathbf C|}=0
$$
which leads to
$$\mathbf r =\frac{yz\mathbf A + zx\mathbf B + xy \mathbf C}{xy+yz+zx}
$$
where the scalers are $x=|\mathbf r-\mathbf A|$, $y=|\mathbf r-\mathbf B|$ and $z=|\mathbf r-\mathbf C|$. Because of the angles all being $2\pi/3$ between them, they satisfy, according to the cosine rule,
$$
a^2=y^2+z^2+yz\\
b^2=z^2+x^2+zx\\
c^2=x^2+y^2+xy
$$
where $a$, $b$ and $c$ are the side lengths of the triangle, opposite of vertices A, B and C, respectively. Specifically, $a=|\mathbf B-\mathbf C|$, $b=|\mathbf C-\mathbf A|$ and $c=|\mathbf A-\mathbf B|$.
The joint equations above can be worked out, albeit in lengthy and tedious steps. Nonetheless, it produces the following solution
$$
\mathbf r =\frac{k_a\mathbf A + k_b\mathbf B + k_c \mathbf C}{(8/\sqrt{3})(a^2+b^2+c^2)K+32K^2}
$$
with the coefficients
$$
k_a=\left(a^2+ 4K/\sqrt{3}\right)^2 -(b^2-c^2)^2\\
k_b=\left(b^2+ 4K/\sqrt{3}\right)^2 -(c^2-a^2)^2\\
k_c=\left(c^2+ 4K/\sqrt{3}\right)^2 -(a^2-b^2)^2
$$
and $K$ representing the area of the triangle, i.e.
$$
K=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}.
$$
</s>