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Difference between revisions of "2011 AMC 10B Problems/Problem 25"

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<math>T_{10}</math> has sides <math>\frac{375}{128}, \frac{503}{128}, \frac{631}{128}</math>.
 
<math>T_{10}</math> has sides <math>\frac{375}{128}, \frac{503}{128}, \frac{631}{128}</math>.
  
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<math>T_{11}</math> would have sides <math>\frac{247}{256}, \frac{503}{256}, \frac{759}{256}</math> but these length do not make a  
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<math>T_{11}</math> would have sides <math>\frac{247}{256}, \frac{503}{256}, \frac{759}{256}</math> but these lengths do not make a  
 
triangle as <math>\frac{247}{256} + \frac{503}{256} < \frac{759}{256}</math>.
 
triangle as <math>\frac{247}{256} + \frac{503}{256} < \frac{759}{256}</math>.
  

Revision as of 19:35, 17 January 2016

Problem

Let $T_1$ be a triangle with sides $2011, 2012,$ 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 $( T_n )$?

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

Solution

By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.

File2011AMC10B25.png

Hence $AD=AF$ and $BD=BE$ and $CE=CF$. Let $AD = x, BD = y$ and $CE = z$ gives three equations:

$x+y = a-1$

$x+z = a$

$y+z = a+1$

(where $a = 2012$ for the first triangle.)

Solving gives:

$x= \frac{a}{2} - 1$

$y = \frac{a}{2}$

$z = \frac{a}{2}+1$

Subbing in gives that $T_2$ has sides of $1005, 1006, 1007$.

$T_3$ can easily be derivied from this as the sides still differ by 1 hence the above solutions still work (now with $a=1006$). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinitely until the side lengths no longer form a triangle.

Subbing in gives $T_3$ with sides $502, 503, 504$.

$T_4$ has sides $\frac{501}{2}, \frac{503}{2}, \frac{505}{2}$.

$T_5$ has sides $\frac{499}{4}, \frac{503}{4}, \frac{507}{4}$.

$T_6$ has sides $\frac{495}{8}, \frac{503}{8}, \frac{511}{8}$.

$T_7$ has sides $\frac{487}{16}, \frac{503}{16}, \frac{519}{16}$.

$T_8$ has sides $\frac{471}{32}, \frac{503}{32}, \frac{535}{32}$.

$T_9$ has sides $\frac{439}{64}, \frac{503}{64}, \frac{567}{64}$.

$T_{10}$ has sides $\frac{375}{128}, \frac{503}{128}, \frac{631}{128}$.

$T_{11}$ would have sides $\frac{247}{256}, \frac{503}{256}, \frac{759}{256}$ but these lengths do not make a triangle as $\frac{247}{256} + \frac{503}{256} < \frac{759}{256}$.

Hence the perimeter is $\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}$

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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 10 Problems and Solutions

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