2011 AMC 10B Problem 25

Question

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 )?

\dfrac{1509}{8}

\dfrac{1509}{32}

\dfrac{1509}{64}

\dfrac{1509}{128}

\dfrac{1509}{256}

Answer: D

Concepts:incircle, incenter, and inradius recursion triangle inequality

Difficulty rating: 2490

Small hint:

Tangent lengths from the same vertex are equal

Big hint:

For side lengths

s-1,s,s+1

, the next middle side is

\frac{s}{2}

Solution:

For a triangle with side lengths

a=BC

,

b=CA

, and

c=AB

, equal tangents from the same vertex give the next side lengths

\begin{gathered} \dfrac{b+c-a}{2}, \\ \quad \dfrac{a+c-b}{2}, \\ \quad \dfrac{a+b-c}{2}. \end{gathered}

If the current side lengths are

s-1,s,s+1

, then the next side lengths are

\dfrac{s}{2}-1,\dfrac{s}{2},\dfrac{s}{2}+1

. Thus the same form persists while the middle side halves each time.

For

T_n

, the middle side is

\frac{2012}{2^{n-1}}

. A triangle of the form

s-1,s,s+1

exists exactly when

s>2

.

The last valid triangle has

\frac{2012}{2^{n-1}}>2

, but the next one does not. This gives

n=10

, with middle side

\frac{2012}{2^9}=\frac{503}{128}

.

The perimeter is

3\cdot\dfrac{503}{128}=\dfrac{1509}{128}

.

Thus, D is the correct answer.