2011 AMC 10B Problems/Problem 25


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.


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 \[503 - 2^{k-3} + 503 > 503 + 2^{k-3}\] \[1006 - 2^{k-3} > 503 + 2^{k-3}\] \[503 > 2^{k-2}\] \[9 > k-2\] \[k < 11\] 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

Identical problem to the 2011 AMC 12B Problems/Problem 22.

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