Consider three consecutive vertices labeled $A_k,A_m,A_{\ell}$. Take the unique $1\leq n \leq 2012$ for which $n \equiv (k+\ell) - m \pmod{2012}$. Since then the chords $A_kA_{\ell}$ and $A_mA_n$ would meet, it means the only possibility is to have $n=m$; this forces $k+\ell$ to be even, and $\ell \equiv 2m-k \pmod{2012}$. This can also be interpreted as having the labels following an arithmetic progression, of ratio $m-k \pmod{2012}$ Applied to our case, with $k=1$ and $m=4$, this yields the $t$-th vertex to be labeled $A_{1 + (3(t - 1) \pmod{2012})}$, hence the tenth label will be $\boxed{A_{28}}$. Remark. It behooves a true mathematician (even if not asked in the problem) to assure oneself that this labeling actually obeys the conditions stated. First, let us notice that since $3\nmid 2012$, the values $1 + (3(t - 1) \pmod{2012})$ run over all possible indices $\{1,2,\ldots,2012\}$ of the labels, $1\leq t\leq 2012$. Now, for two vertices labeled with indices $1 + (3(k - 1) \pmod{2012})$ and $1 + (3(\ell - 1) \pmod{2012})$, for $1\leq k<\ell \leq 2012$, then to the vertex labeled with index $1 + (3(m - 1) \pmod{2012})$, for some $k<m<\ell$, corresponds a vertex labeled with index $1 + (3(n - 1) \pmod{2012})$ where $n = (k+\ell) - m$, thus also $k<n<\ell$, and clearly the corresponding chords do not meet. Therefore the labeling is consistent with the conditions of the problem; the only thing that matters is that $3\nmid 2012$.