Mind blowing question........

angle dear...........

Now way thats wrong

We have the formula

A_n = Ï€/2 (1 - 1/2 + 1/4 - 1/8+....+(-1)ⁿ 1/2ⁿ) + (-1)ⁿ A/2ⁿ

From this its obvious Lim _{n→∞}A_n = Ï€/3 = 60⁰

So, where is it wrong???

pankaj bro can u please post ur solution

yup pankaj plzzzz explain.......................and sorry prophet if u felt i was rude..........

incenter is formed by internal angle bisector lines.
if consider the first triangle A₁B₁C₁ formed by foot of angle bisector lines which intersect corresponding sides of triangle. we also say that I₁(incenter of ABC) will be cirumcenter of A₁B₁C₁.

from here we get angle BI₁C = Ï€ - B/2-C/2

therefore angle C₁A₁B₁ = 1/2(BI₁C) = 1/2(Ï€ - B/2-C/2) = Ï€/4+A/4

from the similar way u can get angle B₁I₂C₁ =Ï€ - B₁/2-C₁/2
C₂A₂B₂ = 1/2(BI₂C) = 5Ï€/4² +A/4²

and same way

C₃A₃B₃ = 21Ï€/4³ +A/4³

now by solving seq 1, 5, 21,......
u get t_n = (4ⁿ -1)/3

so

A_n = (4ⁿ -1)Ï€/(3.4ⁿ) + A/4ⁿ

lim n→∞ A_n = π/3

now given R= abc/4A = 13/2 ........1
where A represent area of triangle

r₁ = A/(s-a) = 3 ...........2

r = A/s = 2 ..........3

from 2 and 3 u get s = 3a ......4

from 1 and 4 u get bc = 12.13

so u can say that b = 12 and c = 13 or b = 13 and c = 12

from 4th u get a = 5

hence triangle will be right angle triangle...

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