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

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==Solution==
 
==Solution==
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{{solution}}
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By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites. Hence <math>AD=AF</math> and <math>BD=BE</math> and <math>CE=CF</math>. Let <math>AD = x, BD = y</math> and <math>CE = z</math> gives three equations:
 +
 
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<math>x+y = a-1</math>
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<math>x+z = a</math>
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<math>y+z = a+1</math>
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(where <math>a = 2012</math> for the first triangle.)
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Solving gives:
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<math>x= \frac{a}{2} - 1</math>
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<math>y = \frac{a}{2}</math>
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<math>z = \frac{a}{2}+1</math>
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Subbing in gives that <math>T_2</math> has sides of <math>1005, 1006, 1007</math>.
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Repeating gives <math>T_3</math> with sides <math>502, 503, 504</math>.
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<math>T_4</math> has sides <math>\frac{501}{2}, \frac{503}{2}, \frac{505}{2}</math>.
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<math>T_5</math> has sides <math>\frac{499}{4}, \frac{503}{4}, \frac{507}{4}</math>.
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<math>T_6</math> has sides <math>\frac{495}{8}, \frac{503}{8}, \frac{511}{8}</math>.
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<math>T_7</math> has sides <math>\frac{487}{16}, \frac{503}{16}, \frac{519}{16}</math>.
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<math>T_8</math> has sides <math>\frac{471}{32}, \frac{503}{32}, \frac{535}{32}</math>.
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<math>T_9</math> has sides <math>\frac{439}{64}, \frac{503}{64}, \frac{567}{64}</math>.
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<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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triangle as <math>\frac{247}{256} + \frac{503}{256} < \frac{759}{256}</math>.
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Hence the perimeter is <math>\frac{375}{128}, \frac{503}{128}, \frac{631}{128} = \frac{1509}{128} \qquad\textbf{(D)}</math>
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2011|ab=B|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2011|ab=B|num-b=24|after=Last Problem}}

Revision as of 22:57, 9 June 2011

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. 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 $1005, 1006, 1007$.

Repeating 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 length do not make a triangle as $\frac{247}{256} + \frac{503}{256} < \frac{759}{256}$.

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

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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