# Sequence

#### veronica1999

##### Member
Could I get some help please?

Let T1 be a triangle with sides 2011, 2012, and 2013. For n > = 1, if Tn = 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 Tn+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 (Tn) ?

Please do not laugh at my solution.

6036/2 , 6036/4 ,6036/8 ......... 6036/4096.
I put 4096 as the last term because the next one is 8192

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#### CaptainBlack

##### Well-known member
Could I get some help please?

Let T1 be a triangle with sides 2011, 2012, and 2013. For n > = 1, if Tn = 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 Tn+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 (Tn) ?

Please do not laugh at my solution.

6036/2 , 6036/4 ,6036/8 ......... 6036/4096.
I put 4096 as the last term because the next one is 8192

The stopping condition is where the longest of $$AD$$, $$BE$$, $$CF$$ for $$T_n$$ is greater than the sum of the other two.

So do you know the relationship between the sides of consecutive triangles in the sequence?

CB

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#### veronica1999

##### Member
Thanks!!!

I see I forgot to consider the triangle inequality rule.

6036/512 = 11.787

11.787/3 = 3.92

2.92 3.92 4.92

6036/1024 = 5.89

5.89/3 = 2.96

0.96 1.96 2.96

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#### veronica1999

##### Member
Oops it was my mistake. I meant 1509/128. Sorry.

#### CaptainBlack

##### Well-known member
Oops it was my mistake. I meant 1509/128. Sorry.
Which is what I get.

CB

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