AMC10 2011 B

Answer (D): Let $T_n=\triangle ABC$. Suppose $a=BC$, $b=AC$, and $c=AB$. Because $BD$ and $BE$ are both tangent to the incircle of $\triangle ABC$, it follows that $BD=BE$. Similarly, $AD=AF$ and $CE=CF$. Then $$ 2BE=BE+BD=BE+CE+BD+AD-(AF+CF)=a+c-b, $$ that is, $BE=\frac12(a+c-b)$. Similarly $AD=\frac12(b+c-a)$ and $CF=\frac12(a+b-c)$. In the given $\triangle ABC$, suppose that $AB=x+1$, $BC=x-1$, and $AC=x$. Using the formulas for $BE$, $AD$, and $CF$ derived before, it must be true that $$ BE=\frac12\big((x-1)+(x+1)-x\big)=\frac12x, $$ $$ AD=\frac12\big(x+(x+1)-(x-1)\big)=\frac12x+1,\ \text{and} $$ $$ CF=\frac12\big((x-1)+x-(x+1)\big)=\frac12x-1. $$ Hence both $(BC,CA,AB)$ and $(CF,BE,AD)$ are of the form $(y-1,y,y+1)$. This is independent of the values of $a$, $b$, and $c$, so it holds for all $T_n$. Furthermore, adding the formulas for $BE$, $AD$, and $CF$ shows that the perimeter of $T_{n+1}$ equals $\frac12(a+b+c)$, and consequently the perimeter of the last triangle $T_N$ in the sequence is $$ \frac{1}{2^{N-1}}(2011+2012+2013)=\frac{1509}{2^{N-3}}. $$ The last member $T_N$ of the sequence will fail to define a successor if for the first time the new lengths fail the Triangle Inequality, that is, if $$ -1+\frac{2012}{2^N}+\frac{2012}{2^N}\le 1+\frac{2012}{2^N}. $$ Equivalently, $2012\le 2^{N+1}$ which happens for the first time when $N=10$. Thus the required perimeter of $T_N$ is $\frac{1509}{2^7}=\frac{1509}{128}$.