2011 AMC 12B Problems/Problem 22

Problem

Let $T_1$ be a triangle with side lengths $2011$ , $2012$ , and $2013$ . For $n \geq 1$ , 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 $\left(T_n\right)$ ?

$\textbf{(A)}\ \frac{1509}{8} \qquad \textbf{(B)}\ \frac{1509}{32} \qquad \textbf{(C)}\ \frac{1509}{64} \qquad \textbf{(D)}\ \frac{1509}{128} \qquad \textbf{(E)}\ \frac{1509}{256}$

Solution

Answer: (D)

Let $AB = c$ , $BC = a$ , and $AC = b$

Then $AD = AF$ , $BE = BD$ and $CF = CE$

Then $a = BE + CF$ , $b = AD + CF$ , $c = AD + BE$

Hence:

Note that $a + 1 = b$ and $a - 1 = c$ for $n = 1$ , I claim that it is true for all $n$ , assume for induction that it is true for some $n$ , then

Furthermore, the average for the sides is decreased by a factor of 2 each time.

So $T_n$ is a triangle with side length $\frac{2012}{2^{n- 1}} - 1$ , $\frac{2012}{2^{n-1}}$ , $\frac{2012}{2^{n-1}} + 1$

and the perimeter of such $T_n$ is $\frac{(3)(2012)}{2^{n-1}}$

Now we need to find when $T_n$ fails the triangle inequality. So we need to find the last $n$ such that $\frac{2012}{2^{n-1}} > 2$

For $n = 10$ , perimeter is $\frac{(3)(2012)}{2^{9}} = \frac{1509}{2^7} = \frac{1509}{128}$

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2011 AMC 12B Problems/Problem 22

Problem

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