Question

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 $( T_n )?$

$\dfrac{1509}{8}$

$\dfrac{1509}{32}$

$\dfrac{1509}{64}$

$\dfrac{1509}{128}$

$\dfrac{1509}{256}$

Answer: D

Concepts:incircle, incenter, and inradius recursion triangle inequality

Difficulty rating: 2490

Solution:

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 $2012/2^{n-1}$ . A triangle of the form $s-1,s,s+1$ exists exactly when $s>2$ .

The last valid triangle has $2012/2^{n-1}>2$ , but the next one does not. This gives $n=10$ , with middle side $2012/2^9=503/128$ .

The perimeter is $3\cdot\dfrac{503}{128}=\dfrac{1509}{128}$ .

Thus, D is the correct answer.

2011 AMC 10B Problem 25

Answer: D

Difficulty rating: 2490

Solution:

Problem 25 in Other Years