Inside an equilateral triangle with a side length of $4 \sqrt{3} \mathrm{~cm}$, a circle with a diameter of $1 \mathrm{~cm}$ moves such that it always touches the perimeter of the triangle. Calculate the area of the part of the triangle that can never be covered by the moving circle.

(M. Dillingerová)

![](https://cdn.mathpix.com/cropped/2024_04_17_af26ba8a6d6d981b210ag-1.jpg?height=431&width=491&top_left_y=1481&top_left_x=817)
The uncovered part of the triangle is marked in gray in Fig. 1. It consists of three "corners" (we will denote this area as $S_{2}$) and an inner equilateral triangle (we will denote this area as $S_{1}$). If we put the corners together, we get Fig. 2. We will successively calculate that $v=6 \mathrm{~cm}$, $v^{\prime}=1.5 \mathrm{~cm}$, and $a^{\prime}=\sqrt{3} \mathrm{~cm}$. By substituting into the relevant formulas, we obtain $S_{1}=3 \sqrt{3} \mathrm{~cm}^{2}$ and $S_{2}=\left(\frac{3}{4} \sqrt{3}-\right.$ $\left.-\frac{1}{4} \pi\right) \mathrm{cm}^{2}$. The total uncovered area is therefore

$$
S=S_{1}+S_{2}=\left(\frac{15}{4} \sqrt{3}-\frac{1}{4} \pi\right) \mathrm{cm}^{2}
$$