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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: 

$AD = AF = \frac{b + c - a}{2}$
$BE = BD = \frac{a + c - b}{2}$
$CF = CE = \frac{a + b - c}{2}$

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 

$AD = AF = \frac{a}{2}$
$BE = BD = \frac{a - 2}{2} = AD - 1$
$CF = CE = \frac{a + 2}{2} = AD + 1$

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 $\frac{2012}{2^{n-1}} > 2$ 

$\frac{2012}{2^{n-1}} > 2$
$2012 > 2^n$
$n \le 10$

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

Solution 2

This problem is a PoP(Power of Point) Bash exercise. What I did to solve this problem was look at recursive perimeters. The first ever perimeter is $P_1 = 2011 + 2012 + 2013$. Next, by PoP, inscribing the circle gives us three new lengths, namely $AD, BE, CF$. Denote $AD$ as $x_1$ and homogeneously the others. Then, $x_1 + x_2 = 2011, x_2 + x_3 = 2012$, and $x_1 + x_3 = 2013$. If we add all these equations up and divide by $2$, we get $x_1 + x_2 + x_3 = \frac{2011 + 2012 + 2013}{2}$. Writing this with $P_1 = 2011 + 2012 + 2013$, we get that our new perimeter, $P_2$, is indeed equal to $\frac{P_1}{2}$. Similarly, by the same concept, we get that $P_3 = \frac{P_1}{4}$ and the pattern keeps going. In general, I found that for each new perimeter $P_n$, $P_n = \frac{P_1}{2^{n-1}}$. Now, substituting in the numerical value of $P_1$, we get that $P_n = \frac{2011 + 2012 + 2013}{2^{n-1}}$. If you keep dividing the numerator and denominator by 2, I got that: $P_n = \frac{1509}{2^{n-3}}$. This representation of a new perimeter in terms of $n$ looks very similar to the option choices, so we're on the right path. Now, all we need to do is find out the last value of $n$ when the sequence stops working. By the Triangular Inequality, we may be able to finish this off. Now, I won't go into depths here, but listing out terms and using the Triangular Inequality gives us: First Sequence: $x_1 < 2012$

$x_2 < 2013$

$x_3 < 2011$. The ordered triple for this set of $(x_1, x_2, x_3)$ is $(1005, 1006, 1007)$ if you solve the PoP system of equations we got earlier. If you keep listing out terms, we can come up with the general form, and that is: $x_1 < \frac{503}{2^{n-4}}$

$x_2 < \frac{503}{2^{n-4}} + 1$

$x_3 < \frac{503}{2^{n-4}} - 1$. The ordered triple for this set of $(x_1, x_2, x_3)$ is $(\frac{503}{2^{n-4}}, \frac{503}{2^{n-4}} - 1, \frac{503}{2^{n-4}} + 1)$. Now, if we plug in these values of $x_1, x_2$, and $x_3$ into the inequalities, we see that the first two are always satisfied, but the last one is only satisfied when: $2^{n-2} < 503$(You will get this if you simplify the last inequality). The last $n$ for this to be satisfied is when $n = 10$. If we go up to our general representation of $P_n$, we see that $P_n = \frac{1509}{2^{n-3}}$. Plugging in $10$ because this is the last $n$(and also the last triangle), we our final answer of $\frac{1509}{2^{7}} = \boxed{\frac{1509}{128}}$ or $\boxed{D}$.

~ilikemath247365

See also

Identical problem to the 2011 AMC 10B Problems/Problem 25.

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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