Let $T_1$ be a triangle with side lengths $2011, 2012,$ and $2013$. For $n \ge 1$, if $T_n = \triangle ABC$ and $D, E,$ and $F$ are the points of tangency of the incircle of $\triangle ABC$ to the sides $AB, BC$, and $AC,$ respectively, then $T_{n+1}$ is a triangle with side lengths $AD, BE,$ and $CF,$ if it exists. What is the perimeter of the last triangle in the sequence $( T_n )$?
$\textbf{(A)}\ \frac{1509}{8} \qquad\textbf{(B)}\ \frac{1509}{32} \qquad\textbf{(C)}\ \frac{1509}{64} \qquad\textbf{(D)}\ \frac{1509}{128} \qquad\textbf{(E)}\ \frac{1509}{256}$
By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites. 

[File2011AMC10B25.png](https://artofproblemsolving.com/wiki/index.php/File:File2011AMC10B25.png)
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 
\[\frac{247}{256} + \frac{503}{256}  503 + 2^{k-3}\]
\[1006 - 2^{k-3} > 503 + 2^{k-3}\]
\[503 > 2^{k-2}\]
\[9 > k-2\]
\[k < 11\]
Hence, the first triangle which does not exist in this sequence is $T_{11}$.
Hence the perimeter is 
\[\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}\].