The circle inscribed in the triangle ABC touches the side BC, CA and AB in the point `A_(1)B_(1) and C_(1)` respectively. Similarly the circle inscribed in the `Delta A_(1) B_(1) C_(1)` touches the sieds in `A_(2), B_(2), C_(2)` respectively and so on. If `A_(n) B_(n) C_(n)` be the nth `Delta` so formed, prove that its angle are
`pi/3-(2)^-n(A-(pi)/(3)), pi/3-(2)^-n(B-(pi)/(3)),pi/3-(2)^-n(C-(pi)/(3)).` Hence prove that the triangle so formed is ultimately equilateral.

Answer

To solve the problem, we need to prove that the angles of the triangle \( A_nB_nC_n \) formed by the inscribed circles converge to \( \frac{\pi}{3} \) as \( n \) approaches infinity. We will derive the angles step by step.

### Step 1: Define the angles of triangle \( ABC \)
Let the angles of triangle \( ABC \) be:
- \( A = \angle A \)
- \( B = \angle B \)
- \( C = \angle C \)

### Step 2: Find the angles of triangle \( A_1B_1C_1 \)
The inscribed circle of triangle \( ABC \) touches the sides at points \( A_1, B_1, C_1 \). The angles of triangle \( A_1B_1C_1 \) can be derived as follows:

- The angle at \( A_1 \) is given by:
\[ \angle A_1 = \frac{1}{2}(\pi - A) = \frac{\pi}{2} - \frac{A}{2} \]

- Similarly, we can derive:
\[ \angle B_1 = \frac{\pi}{2} - \frac{B}{2} \]
\[ \angle C_1 = \frac{\pi}{2} - \frac{C}{2} \]

### Step 3: Find the angles of triangle \( A_2B_2C_2 \)
Now, we apply the same logic to triangle \( A_1B_1C_1 \):

- The angle at \( A_2 \) is:
\[ \angle A_2 = \frac{1}{2}(\pi - \angle A_1) = \frac{1}{2}\left(\pi - \left(\frac{\pi}{2} - \frac{A}{2}\right)\right) = \frac{1}{2}\left(\frac{\pi}{2} + \frac{A}{2}\right) = \frac{\pi}{4} + \frac{A}{4} \]

- Similarly, we can derive:
\[ \angle B_2 = \frac{\pi}{4} + \frac{B}{4} \]
\[ \angle C_2 = \frac{\pi}{4} + \frac{C}{4} \]

### Step 4: Generalize for triangle \( A_nB_nC_n \)
Continuing this process, we can generalize the angles for triangle \( A_nB_nC_n \):

- The angle \( A_n \) can be expressed as:
\[ \angle A_n = \frac{\pi}{2} - \frac{A}{2^n} \]

- Similarly:
\[ \angle B_n = \frac{\pi}{2} - \frac{B}{2^n} \]
\[ \angle C_n = \frac{\pi}{2} - \frac{C}{2^n} \]

### Step 5: Prove convergence to \( \frac{\pi}{3} \)
As \( n \) approaches infinity, the terms \( \frac{A}{2^n}, \frac{B}{2^n}, \frac{C}{2^n} \) approach zero. Therefore:

- The angles converge to:
\[ \angle A_n \to \frac{\pi}{3}, \quad \angle B_n \to \frac{\pi}{3}, \quad \angle C_n \to \frac{\pi}{3} \]

### Conclusion
Thus, we have shown that:
\[ \angle A_n = \frac{\pi}{3} - 2^{-n}(A - \frac{\pi}{3}), \quad \angle B_n = \frac{\pi}{3} - 2^{-n}(B - \frac{\pi}{3}), \quad \angle C_n = \frac{\pi}{3} - 2^{-n}(C - \frac{\pi}{3}) \]
As \( n \to \infty \), all angles converge to \( \frac{\pi}{3} \), proving that the triangle \( A_nB_nC_n \) ultimately becomes equilateral.