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Perimeter of last triangle in a sequence

Source: AMC 10 2011 b Problem 25

February 24, 2011
geometryperimeterratioinequalitiesgeometric seriestriangle inequalityAMC

Problem Statement

Let T1T_1 be a triangle with sides 2011,2012,2011, 2012, and 20132013. For n≥1n \ge 1, if Tn=△ABCT_n=\triangle ABC and D,E,D,E, and FF are the points of tangency of the incircle of △ABC\triangle ABC to the sides AB,BCAB,BC and ACAC, respectively, then Tn+1T_{n+1} is a triangle with side lengths AD,BE,AD,BE, and CFCF, if it exists. What is the perimeter of the last triangle in the sequence (Tn)(T_n)?
<spanclass=′latex−bold′>(A)</span> 15098<spanclass=′latex−bold′>(B)</span> 150932<spanclass=′latex−bold′>(C)</span> 150964<spanclass=′latex−bold′>(D)</span> 1509128<spanclass=′latex−bold′>(E)</span> 1509256 <span class='latex-bold'>(A)</span>\ \frac{1509}{8} \qquad <span class='latex-bold'>(B)</span>\ \frac{1509}{32} \qquad <span class='latex-bold'>(C)</span>\ \frac{1509}{64} \qquad <span class='latex-bold'>(D)</span>\ \frac{1509}{128} \qquad <span class='latex-bold'>(E)</span>\ \frac{1509}{256}
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