2011 AMC 12B Problem 22

Question

Let

T_1

be a triangle with sides

2011,

2012,

and

2013.

For

n\ge1,

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: 2350

Small hint:

The tangent lengths satisfy

AD=\tfrac12(b+c-a),

and similarly for

BE

and

CF

Big hint:

Each triangle keeps the form

(y-1,y,y+1)

with

y

halving; it fails the triangle inequality once

y\le2

Solution:

For a triangle with sides

a,b,c,

the tangent lengths are

AD=\tfrac12(b+c-a),

BE=\tfrac12(a+c-b),

and

CF=\tfrac12(a+b-c).

If

T_n

has sides

(y-1,y,y+1),

then

T_{n+1}

has sides

\left(\tfrac{y}{2}-1,\tfrac{y}{2},\tfrac{y}{2}+1\right).

Starting from

T_1

with middle side

2012,

the middle side halves each step and the perimeter of

T_{n+1}

is

\tfrac12

the perimeter of

T_n.

A triangle of this form exists only while its middle side exceeds

2.

The middle side of

T_n

is

\dfrac{2012}{2^{n-1}}.

This first drops to

2

or below at

n=11,

so the last valid triangle is

T_{10},

whose middle side is

\dfrac{2012}{2^9}

and whose perimeter is

3\cdot\dfrac{2012}{2^9}=\dfrac{6036}{512}=\dfrac{1509}{128}.

Thus, the correct answer is D.