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

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== Problem ==
 
== Problem ==
  
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Let <math>T_1</math> be a triangle with sides <math>2011, 2012,</math> and <math>2013</math>. For <math>n \ge 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, 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>( T_n )</math>?
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Let <math>T_1</math> be a triangle with side lengths <math>2011, 2012,</math> and <math>2013</math>. For <math>n \ge 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, 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>( T_n )</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>
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
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==Solution==
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==Solution 1==
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By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.  
 
By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.  
  
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Solving gives:
 
Solving gives:
  
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<math>x= \frac{a}{2} - 1</math>
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<math>x= \frac{a}{2} -1</math>
  
 
<math>y = \frac{a}{2}</math>
 
<math>y = \frac{a}{2}</math>
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Subbing in gives that <math>T_2</math> has sides of <math>1005, 1006, 1007</math>.
 
Subbing in gives that <math>T_2</math> has sides of <math>1005, 1006, 1007</math>.
  
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<math>T_3</math> can easily be derivied from this as the sides still differ by 1 hence the above solutions still work (now with <math>a=1006</math>). 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.
+
<math>T_3</math> can easily be derived from this as the sides still differ by 1 hence the above solutions still work (now with <math>a=1006</math>). 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 <math>T_3</math> with sides <math>502, 503, 504</math>.
 
Subbing in gives <math>T_3</math> with sides <math>502, 503, 504</math>.
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Hence the perimeter is <math>\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}</math>
 
Hence the perimeter is <math>\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}</math>
  
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==Solution 2 ==
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==Solution 2==
  
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Proceeding similarly to the first solution, we have that sides of each triangle are of the form <math>a, a+1, a+2</math>. Also note that the perimeter of each triangle is half of the previous one. In order for the triangle to not exist, it must not satisfy the triangle inequality, meaning that <math>a + a + 1 < a+2 \Rightarrow a<1</math>. Then, the perimeter would be <math>a + a + 1 + a + 2 = 3a + 3 < 6</math>. So, we have <math>\frac{3018}{2^{k}} < 6 \Rightarrow 2^k > 503 \Rightarrow 2^{k} \geq 512</math>. The first triangle to not work would have perimeter <math>\frac{3018}{512} = \frac{1509}{256}</math>, thus the answer is <math>\boxed{\textbf{(D)} \frac{1509}{128}}</math>.
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Proceeding similarly to the first solution, we have that sides of each triangle are of the form <math>a, a+1, a+2</math> for some number <math>a</math>. Also, note that the perimeter of each triangle is half of the previous one. In order for the triangle to not exist, it must not satisfy the triangle inequality, meaning that <math>a + a + 1 < a+2 \Rightarrow a<1</math>. Then, the perimeter would be <math>a + a + 1 + a + 2 = 3a + 3 < 6</math>. So, to have a proper triangle, we have <math>\frac{3018}{2^{k}} > 6 \Rightarrow 2^k < 503 \Rightarrow 2^{k} \leq 512</math>. The first triangle to not work would have perimeter <math>\frac{3018}{512} = \frac{1509}{256}</math>, thus the answer is   <math>\boxed{\textbf{(D)} \frac{1509}{128}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 11:03, 27 December 2020

Problem

Let $T_1$ be a triangle with side lengths $2011, 2012,$ 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 1

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 derived 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 $\frac{247}{256} + \frac{503}{256} < \frac{759}{256}$.


Likewise, you could create an equation instead of listing all the triangles to $T_{11}$. The sides of a triangle $T_{k}$ would be $\frac{503}{2^{k-3}} - 1, \frac{503}{2^{k-3}}, \frac{503}{2^{k-3}} + 1$. We then have $503 - 2^{k-3} + 503 > 503 + 2^{k-3} \rightarrow 1006 - 2^{k-3} > 503 + 2^{k-3} \rightarrow 503 > 2^{k-2} \rightarrow 9 > k-2 \rightarrow k < 11$. Hence, the first triangle which does not exist in this sequence is $T_{11}$.


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

Solution 2

Proceeding similarly to the first solution, we have that sides of each triangle are of the form $a, a+1, a+2$ for some number $a$. Also, note that the perimeter of each triangle is half of the previous one. In order for the triangle to not exist, it must not satisfy the triangle inequality, meaning that $a + a + 1 < a+2 \Rightarrow a<1$. Then, the perimeter would be $a + a + 1 + a + 2 = 3a + 3 < 6$. So, to have a proper triangle, we have $\frac{3018}{2^{k}} > 6 \Rightarrow 2^k < 503 \Rightarrow 2^{k} \leq 512$. The first triangle to not work would have perimeter $\frac{3018}{512} = \frac{1509}{256}$, thus the answer is $\boxed{\textbf{(D)} \frac{1509}{128}}$.

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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