  Let $f(a,b,c,d)=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}-\sqrt{(a+c)^2+(b+d)^2}$, let
$(a,b,c,d)$ stand for $a,b,c,d\in\mathbb Z_{\geq 0}$ such that $ad-bc=1$.
Define \begin{equation} \label{eq_main} F(s) = \sum_{(a,b,c,d)} f(a,b,c,d)^s.
\end{equation} In other words, we consider the sum of the powers of the
triangle inequality defects for the lattice parallelograms (in the first
quadrant) of area one.
  We prove that $F(s)$ converges when $s>1$ and diverges at $s=1/2$. (This
papers differs from its published version: Fedor Petrov showed us how to easily
prove that $F(s)$ converges for $s>2/3$ and diverges for $s\leq 2/3$, see
below.) We also prove $$\sum\limits_{\substack{(a,b,c,d), 1\leq a\leq b, 1\leq
c\leq d}} \frac{1}{(a+b)^2(c+d)^2(a+b+c+d)^2} = 1/24,$$ and show a general
method to obtain such formulae. The method comes from the consideration of the
tropical analogue of the caustic curves, whose moduli give a complete set of
continuous invariants on the space of convex domains.
