8. An equilateral triangle $\triangle A_{1} B_{1} C_{1}$ is inscribed in a circle of unit radius. A circle is inscribed in the triangle $\triangle A_{1} B_{1} C_{1}$, and the points of tangency are denoted as $A_{2}, B_{2}$, and $C_{2}$. A circle is inscribed in the triangle $\triangle A_{2} B_{2} C_{2}$, and the points of tangency are denoted as $A_{3}, B_{3}$, and $C_{3} \ldots$ This process is repeated 100 times, resulting in 100 triangles and 100 circles (the first three steps of this process are shown in the figure). Find the total area

![](https://cdn.mathpix.com/cropped/2024_05_06_94cb46a707e666e84118g-3.jpg?height=554&width=518&top_left_y=1425&top_left_x=1053)

of all 300 shaded parts.
Answer: $\left(\frac{4 \pi}{3}-\sqrt{3}\right) \cdot\left(1-4^{-100}\right)$. Solution. Note that at each step, the size of the circle and the triangle is halved. Therefore, the areas of the specified parts form a geometric progression with a common ratio of $\frac{1}{4}$. The area of the first one can be obtained by subtracting the area of the equilateral triangle $A_{1} B_{1} C_{1}$, which is $\frac{3 \sqrt{3}}{4}$, from the area of the circle $S_{\text {circle }}=\pi$. The sum of the geometric progression is

$$
b_{1} \cdot \frac{q^{N}-1}{q-1}=\left(\pi-\frac{3 \sqrt{3}}{4}\right) \frac{(1 / 4)^{100}-1}{-3 / 4}=\left(\frac{4 \pi}{3}-\sqrt{3}\right) \cdot\left(1-4^{-100}\right) .
$$