Circle inscribed in Equilateral Triangles The circle inscribed in the triangle $ABC$ touches the sides $BC$ , $CA$ , and $AB$  in the points $A_1,B_1,C_1$ respectively. Similarly the circle inscribed in the triangle $A_1B_1C_1$ touches the sides in $A_2,B_2,C_2$ respectively, and so on. If $A_nB_nC_n$ be the $n^{th}$ $\triangle$ so formed, Prove its angles are: $$\frac{\pi}{3} + (-2)^{-n}(A - \frac{\pi}{3})\ ,\ \ \frac{\pi}{3} + (-2)^{-n}(B - \frac{\pi}{3})\ ,\ \ \frac{\pi}{3} + (-2)^{-n}(C - \frac{\pi}{3})$$
Hence, prove that triangle so formed is an equilateral triangle.
 A: Quang Hoang has already provided a good hint.
Let us prove that by induction on $n$.
Let $O$ be the incenter of $\triangle{ABC}$. 
Then, noting that $OB\perp A_1C_1,OC\perp A_1B_1$, we have
$$\begin{align}\angle{B_1A_1C_1}&=\pi -\angle{B_1A_1C}-\angle{C_1A_1B}\\&=\pi-\left(\pi-\frac{\pi}{2}-\frac C2\right)-\left(\pi-\frac{\pi}{2}-\frac B2\right)\\&=\frac{B+C}{2}\\&=\frac{\pi-A}{2}\\&=\frac{\pi}{3}-2^{-1}\left(A-\frac{\pi}{3}\right)\end{align}$$
Similarly, we have
$$\angle{A_1B_1C_1}=\frac{\pi}{3}-2^{-1}\left(B-\frac{\pi}{3}\right),\qquad \angle{A_1C_1B_1}=\frac{\pi}{3}-2^{-1}\left(C-\frac{\pi}{3}\right)$$
Here, suppose that $$\angle{A_nB_nC_n}=\frac{\pi}{3}+(-2)^{-n}\left(B-\frac{\pi}{3}\right),\quad \angle{B_nC_nA_n}=\frac{\pi}{3}+(-2)^{-n}\left(C-\frac{\pi}{3}\right).$$
Let $O_n$ be the incenter of $\triangle{A_nB_nC_n}$.
Then, noting that $O_nB_n\perp A_{n+1}C_{n+1},O_nC_n\perp A_{n+1}B_{n+1}$, we have
$$\begin{align}&\angle{B_{n+1}A_{n+1}C_{n+1}}\\&=\pi -\angle{B_{n+1}A_{n+1}C_n}-\angle{C_{n+1}A_{n+1}B_n}\\&=\pi-\left(\frac{\pi}{2}-\frac 12\left(\frac{\pi}{3}+(-2)^{-n}\left(C-\frac{\pi}{3}\right)\right)\right)-\left(\frac{\pi}{2}-\frac 12\left(\frac{\pi}{3}+(-2)^{-n}\left(B-\frac{\pi}{3}\right)\right)\right)\\&=\frac{\pi}{3}+\frac{(-2)^{-n}}{2}(B+C)-(-2)^{-n}\cdot\frac{\pi}{3}\\&=\frac{\pi}{3}+\frac{(-2)^{-n}}{2}(\pi -A)-(-2)^{-n}\cdot\frac{\pi}{3}\\&=\frac{\pi}{3}+(-2)^{-n-1}\left(A-\frac{\pi}{3}\right)\end{align}$$
Similarly, we have
$$\small\angle{A_{n+1}B_{n+1}C_{n+1}}=\frac{\pi}{3}+(-2)^{-n-1}\left(B-\frac{\pi}{3}\right),\qquad \angle{A_{n+1}C_{n+1}B_{n+1}}=\frac{\pi}{3}+(-2)^{-n-1}\left(C-\frac{\pi}{3}\right)$$
A: The triangles created in such a way are known as "Contact Triangles".
If I understand correctly, the result you are looking for has been published. to quote form the above MathWorld link:

Beginning with an arbitrary triangle $\Delta$, find the contact triangle $C$. Then find the contact triangle $C^1$ of that triangle, and so on. Then the resulting triangle $C^{\infty}$ approaches an equilateral triangle Goldoni 2003. 

I don't have access to the paper, but here's a proof of my own:
Since the lines from the vertices to the incircle center are bisectors, we have that:
$$A_{n+1} = \frac{A_n+B_n}{2},\quad B_{n+1}= \frac{B_n+C_n}{2},\quad C_{n+1} = \frac{C_n+A_n}{2}$$
Now it's clear that at each iteration, we have that for each iteration $n$, and ${\bf v}_n \equiv(A_n,B_n,C_n)$, there exists some matrix $M$, s.t.:
$${\bf v}_n  = M_n {\bf v}_0$$
And the sum of each row is $1$, since the angles must sum to $\pi$. Let $T$ be the transformation defined by $M_{n+1} = T(M_n)$. We now wish to prove that $T$ is a contraction mapping. If $T$ is indeed a contraction mapping, we use the Banach Fixed Point Theorem to prove that $M_n$ converges to the unique fixed point, which we claim is $M^* = \{1/3\}_{ij}$.
To show this, take a look at an example:
$$A_{n+1} = \frac{M_n^{11} A + M_n^{12} B + M_n^{13} C + M_n^{21} A + M_n^{22} B + M_n^{23} C}{2}$$
Thus: $$M_{n+1}^{11} = \frac{M_n^{11}+M_n^{21}}{2}, \quad
M_{n+1}^{12} = \frac{M_n^{12}+M_n^{22}}{2},\quad \text{etc..}$$
Looking at $M_n$ as a vector in $\mathbb{R}^9$, we can find a suitable metric in which it is comfortable to show $T$ is indeed a contraction mapping.
To be continued.. 
