AMC10 2011 10B, 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 $( T_n )$?

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 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 \[\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \fbox{\frac{1509}{128}}\].

Answer: \frac{1509}{128}