2011 AMC 10B Problem 25

For a triangle with side lengths $a=BC$, $b=CA$, and $c=AB$, equal tangents from the same vertex give the next side lengths $$\begin{gathered} \dfrac{b+c-a}{2}, \\ \quad \dfrac{a+c-b}{2}, \\ \quad \dfrac{a+b-c}{2}. \end{gathered}$$ If the current side lengths are $s-1,s,s+1$, then the next side lengths are $\dfrac{s}{2}-1,\dfrac{s}{2},\dfrac{s}{2}+1$. Thus the same form persists while the middle side halves each time. For $T_n$, the middle side is $\frac{2012}{2^{n-1}}$. A triangle of the form $s-1,s,s+1$ exists exactly when $s>2$. The last valid triangle has $\frac{2012}{2^{n-1}}>2$, but the next one does not. This gives $n=10$, with middle side $\frac{2012}{2^9}=\frac{503}{128}$. The perimeter is $3\cdot\dfrac{503}{128}=\dfrac{1509}{128}$. Thus, D is the correct answer.