TITLE: The geometric median of a triangle
QUESTION [18 upvotes]: Let $\Omega\subset \mathbb R^n$ be a compact domain of dimension $n$. Define the geometric median on $\Omega$ as the point $m_{\Omega}\in \mathbb R^n$ such that the integral $\int_{\Omega}|x-m_{\Omega}|dx$ attains its minimum.
Question Suppose that the domain $\Omega$ is a triangle $\Delta$ in $\mathbb R^2$. Is there a closed formula for the geometric median of $\Delta$? 
Disclaimer. The name geometric median is taken from the Wikipedia article 
https://en.wikipedia.org/wiki/Geometric_median . There is huge amount of articles, in particular in statistics, probability, location theory, ect, that use this notion. It is clear as well that this notion has a lot of different names (some of which are given in the Wikipedia article). This notion is mainly applied to the case when $\Omega$ is a finite set. However, after an extensive search on Google, MathSciNet, Google Scholar, etc. I was not able to find any reasonable source treating the above question.

REPLY [5 votes]: Here is some evidence that there is no formula.  I took a right isoceles triangle as the simplest non-trivial example.  For the triangle on $(0,0), (1,-1), (1,1)$, the integral of distances to $(h,0)$ is
$$
\frac{h^3 - h^2s + 2s}{6}
+\frac{\sqrt{2}\ h^3}{12}\log
\dfrac{2-h+\sqrt{2}s}{(\sqrt{2}-1)h}
+\frac{(1-h)^3}{3} \log\dfrac{1+s}{1-h}
$$
where $s=\sqrt{2-2h+h^2}$.  Numerically I find this minimized at $h \simeq 0.648863$.  But any explicit formula for this median would have to minimize this function -- and more complicated functions with three more parameters for the general case.