2011 AMC 10B Problems/Problem 25

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 $1005, 1006, 1007$ .

$T_3$ 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 $\frac{247}{256} + \frac{503}{256} < \frac{759}{256}$ .

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 $a, a+1, a+2$ . 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, we have $\frac{3018}{2^{k}} < 6 \Rightarrow 2^k > 503 \Rightarrow 2^{k} \geq 512$ . The first triangle to not work would have perimeter $\frac{3018}{512} = \frac{1509}{256}$ , thus the answer is $\boxed{\textbf{(D)} \frac{1509}{128}}$ .

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

Page

Toolbox

Search

2011 AMC 10B Problems/Problem 25

Contents

Problem

Solution

Solution 2

See Also