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Difference between revisions of "2011 AMC 12B Problems/Problem 22"

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==Problem==
 
==Problem==
  
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Let <math>T_1</math> be a triangle with side lengths <math>2011</math>, <math>2012</math>, and <math>2013</math>. For <math>n \geq 1</math>, if <math>T_n = \Delta ABC</math> and <math>D, E</math>, and <math>F</math> are the points of tangency of the incircle of <math>\Delta ABC</math> to the sides <math>AB</math>, <math>BC</math>, and <math>AC</math>, respectively, then <math>T_{n+1}</math> is a triangle with side lengths <math>AD, BE</math>, and <math>CF</math>, if it exists. What is the perimeter of the last triangle in the sequence <math>\left(T_n\right)</math>?
+
Let <math>T_1</math> be a triangle with side lengths <math>2011</math>, <math>2012</math>, and <math>2013</math>. For <math>n \geq 1</math>, if <math>T_n = \triangle ABC</math> and <math>D, E</math>, and <math>F</math> are the points of tangency of the incircle of <math>\triangle ABC</math> to the sides <math>AB</math>, <math>BC</math>, and <math>AC</math>, respectively, then <math>T_{n+1}</math> is a triangle with side lengths <math>AD, BE</math>, and <math>CF</math>, if it exists. What is the perimeter of the last triangle in the sequence <math>\left(T_n\right)</math>?
  
 
<math>\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}</math>
 
<math>\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}</math>

Latest revision as of 02:10, 30 September 2024

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 $\frac{(3)(2012)}{2^{9}} = \frac{1509}{2^7} = \frac{1509}{128}$

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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