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By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites. Hence $AD=AF$ and $BD=BE$ and $CE=CF$. Let $AD=x,BD=y$ and $CE=z$ gives three equations: $x+y=a-1$ $x+z=a$ $y+z=a+1$ (where $a=2012$ for the first triangle.) Solving gives: $x=\frac{a}{2}-1$ $y=\frac{a}{2}$ $z=\frac{a}{2}+1$ Subbing in gives that $T_2$ has sides of $1005,1006,1007$. $T_3$ can easily be derived from this as the sides still differ by 1 hence the above solutions still work (now with $a=1006$). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinitely until the side lengths no longer form a triangle. Subbing in gives $T_3$ with sides $502,503,504$. $T_4$ has sides $\frac{501}{2},\frac{503}{2},\frac{505}{2}$. $T_5$ has sides $\frac{499}{4},\frac{503}{4},\frac{507}{4}$. $T_6$ has sides $\frac{495}{8},\frac{503}{8},\frac{511}{8}$. $T_7$ has sides $\frac{487}{16},\frac{503}{16},\frac{519}{16}$. $T_8$ has sides $\frac{471}{32},\frac{503}{32},\frac{535}{32}$. $T_9$ has sides $\frac{439}{64},\frac{503}{64},\frac{567}{64}$. $T_{10}$ has sides $\frac{375}{128},\frac{503}{128},\frac{631}{128}$. $T_{11}$ would have sides $\frac{247}{256},\frac{503}{256},\frac{759}{256}$ but these lengths do not make a triangle as $$\frac{247}{256}+\frac{503}{256}<\frac{759}{256}$$ Likewise, you could create an equation instead of listing all the triangles to $T_{11}$. The sides of a triangle $T_k$ would be $$\frac{503}{2^{k-3}}-1,\frac{503}{2^{k-3}},\frac{503}{2^{k-3}}+1$$ We then have $$503-2^{k-3}+503>503+2^{k-3}$$ $$1006-2^{k-3}>503+2^{k-3}$$ $$503>2^{k-2}$$ $$9>k-2$$ $$k<11$$ Hence, the first triangle which does not exist in this sequence is $T_{11}$. Hence the perimeter is $$\frac{375}{128}+\frac{503}{128}+\frac{631}{128}=\boxed{\frac{1509}{128}}$$.