Sequence of triangles from incircle tangency points

[veronica1999]

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) ?

The answer is 1509/28.

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

My answer 6036/4098 = 1509/1024

 

[CaptainBlack]

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) ?

The answer is 1509/128.

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

My answer 6036/4098 = 1509/1024

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

 

[veronica1999]

Thanks!

I see I forgot to consider the triangle inequality rule.

6036/512 = 11.787

11.787/3 = 3.92

2.92 3.92 4.926036/1024 = 5.89

5.89/3 = 2.96

0.96 1.96 2.96

 

[veronica1999]

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

 

[CaptainBlack]

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

Which is what I get.

CB