A sequence in a triangle Let $T_1$ be a triangle with sides $2011, 2012,$ 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 $( T_n )$?

Here is the diagram:
https://wiki-images.artofproblemsolving.com//thumb/e/e4/File2011AMC10B25.png/260px-File2011AMC10B25.png
Anyways So it appears to me that it is a infinite sequence. After all, for what I understand it, it seems to want me to draw a circle inscribed in the ABC triangle, take the 3 sides it gives, and construct another triangle, in which you repeat the process until no such triangle exists.  However, I believe that there always will be a triangle that can be constructed, there is just no way that the three smaller sides of a triangle can't be constructed into another. Where is the flaw in my logic?
Thanks!
Max0815
 A: If you start with:
$$AD+BE=c, \ BE+CF=a, \ CF+AD=b$$
you get the following:
$$AD=\frac12(b+c-a)\tag{1}$$
$$BE=\frac12(c+a-b)\tag{2}$$
$$CF=\frac12(a+b-c)\tag{3}$$
Let $a$ be the length of the shortest side:
$$a_1=2011,\ b_1=a_1+1, \ c_1=a_1+2$$
By applying (1),(2),(3) you get:
$$AD=\frac12(a_1+3)=\frac {a_1-1}2+2$$
$$BE=\frac12(a_1+1)=\frac {a_1-1}2+1$$
$$CF=\frac12(a_1-1)=\frac {a_1-1}2$$
So in the next iteration we'll have:
$$a_2=\frac{a_1-1}2, \ b_2=a_2+1, \ c_2=a_2+2$$
The point is: we still have an arithmetic progression with the same difference (+1). So you can quickly proceeed from one triangle to the next one just by calculating the shortest side in the next iteration with a fairly simple formula:
$$a_{n+1}=\frac{a_n-1}{2}\tag{4}$$
The other two sides are simply:
$$b_{n+1}=a_{n+1}+1, \ c_{n+1}=a_{n+1}+2$$
How far can we go down the road? As long as:
$$a_n+b_n>c_n\implies a_n+a_n+1>a_n+2\implies a_n>1$$
So we have to stop with iterations when $a_n$ becomes less than or equal to 1.
By using (4), we can now calculate the sequence which represents the length of the shortest triangle side $a_i$:
$$2011, \ 1005, \ 502.0, \ 250.5, \ 124.75, \\ 
61.875, \ 30.4375, \ 14.71875, \ 6.859375, \ 2.9296875$$
The next value would be less than 1 so we have to stop here. The last triangle has the following side lengths:
$$a_{10}=2.9296875, \ \ b_{10}=3.9296875, \ \ c_{10}=4.9296875$$
...with circumference equal to $11.7890625$.
