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.
Categories: math.NT math.AG math.CO math.SG