2011 AMC 10B Problems/Problem 25: Difference between revisions

Revision as of 23:00, 9 June 2011

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 $1005, 1006, 1007$ .

Repeating 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 length do not make a triangle as $\frac{247}{256} + \frac{503}{256} < \frac{759}{256}$ .

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

@@ Line 46: / Line 46: @@
 triangle as <math>\frac{247}{256} + \frac{503}{256} < \frac{759}{256}</math>.
-Hence the perimeter is <math>\frac{375}{128}, \frac{503}{128}, \frac{631}{128} = \frac{1509}{128} \qquad\textbf{(D)}</math>
+Hence the perimeter is <math>\frac{375}{128} + \frac{503}{128} + \frac{631}{128} = \boxed{\textbf{(D)} \frac{1509}{128}}</math>
 ==See Also==
 {{AMC10 box|year=2011|ab=B|num-b=24|after=Last Problem}}

2011 AMC 10B (Problems • Answer Key • Resources)
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2011 AMC 10B Problems/Problem 25: Difference between revisions

Revision as of 23:00, 9 June 2011

Problem

Solution

See Also