# Task 3.

Maximum 15 points

In an equilateral triangle with area $S_{1}$, a circle is inscribed, and in this circle, an equilateral triangle with area $S_{2}$ is inscribed. In the resulting new triangle, another circle is inscribed, and in this circle, another equilateral triangle with area $S_{3}$ is inscribed. The procedure is repeated $n$ times. The area of the triangle $S_{n}$ turned out to be 1. Find the function $S_{1}(n)$.
Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_c9c52a82f3617446c440g-05.jpg?height=474&width=580&top_left_y=525&top_left_x=801)

Step 1: $\Delta A_{1} B_{1} C_{1} \rightarrow \Delta A_{2} B_{2} C_{2}$.

$\Delta A_{1} B_{1} C_{1}$ is similar to $\Delta A_{2} B_{2} C_{2}$ with a similarity coefficient (ratio of side lengths) of 2. Therefore, the ratio of the areas of these triangles is 4.

If the area of $\Delta A_{1} B_{1} C_{1}$ is $S_{1}$, then the area of $\Delta A_{2} B_{2} C_{2}$ is $\frac{1}{4} S_{1}$, i.e.,

$S_{2}=\frac{1}{4} S_{1}$.

Step 2: $\Delta A_{2} B_{2} C_{2} \rightarrow \Delta A_{3} B_{3} C_{3}$.

The procedure is repeated, so the transition formula is the same, i.e.,

$S_{3}=\frac{1}{4} S_{2}=\frac{1}{4^{2}} S_{1}$.

Step n: $\Delta A_{n} B_{n} C_{n} \rightarrow \Delta A_{n+1} B_{n+1} C_{n+1}$.

$S_{n+1}=\frac{1}{4} S_{n}=\frac{1}{4^{n}} S_{1}$.

From the obtained formula $\frac{1}{4} S_{n}=\frac{1}{4^{n}} S_{1}$, considering the condition $S_{n}=1$, we get:

$\frac{1}{4}=\frac{1}{4^{n}} S_{1}$, from which we find $S_{1}(n)=4^{n-1}$.

Answer: $S_{1}(n)=4^{n-1}$.

## Criteria

If the correct transition for the first step is made and the formula

$S_{2}=\frac{1}{4} S_{1}$, connecting the areas of the original and the inscribed triangle, is obtained, then $25 \%$ - $\mathbf{4}$ points

If several steps are performed, the relationship between $S_{2}$ and $S_{3}, S_{3}$ and $S_{4}, \ldots$ is found, but the final formula is not obtained, then $50 \%$. - 8 points

If the formula $S_{n+1}=\frac{1}{4^{n}} S_{1}$ is obtained, and instead of $S_{n}=1$, $S_{n+1}=1$ is substituted, resulting in $S_{1}(n)=4^{n}$, then $80 \%$ - $\mathbf{1 2}$ points

If the formula $S_{1}(n)=4^{n-1}$ is obtained, then with justification $100 \%$ - 15 points