10.010. Each side of an equilateral triangle is divided into three equal parts, and the corresponding division points, counted in one direction, are connected to each other. A circle with radius \( r = 6 \, \text{cm} \) is inscribed in the resulting equilateral triangle. Determine the sides of the triangles.
Solution.

By the condition $A S=C S=A C, S B=B D=D C=M C=K M=A K=A F=$ $=F E=E S, r=6$ cm (Fig.10.10). $S_{\triangle F B M}=\frac{F B+B M+F M}{2} \cdot r=\frac{3 F B}{2} \cdot r$. $S_{\triangle F B M}=F B^{2} \frac{\sqrt{3}}{4}$, i.e., $F B=12 \sqrt{3} \mathrm{~cm}$. In $\triangle A F M \quad \angle \triangle F A M=60^{\circ}, A F=\frac{A S}{3}$, $A M=\frac{2}{3} A C, F M^{2}=A F^{2}+A M^{2}-2 \cdot A F \cdot A M \cdot \cos \angle F A M$ (by the cosine theorem), from which $F M^{2}=\frac{A S^{2}}{9}+\frac{4}{9} A S^{2}-2 \cdot \frac{2}{9} A S^{2} \cdot \frac{1}{2}, \quad F M^{2}=\frac{3 A S^{2}}{9}$; $F M=\frac{A S}{\sqrt{3}}, A S=F M \cdot \sqrt{3}, F M=B F, A S=12 \sqrt{3} \cdot \sqrt{3}=36 \mathrm{~cm}$.

Answer: $12 \sqrt{3}$ cm and 36 cm.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-561.jpg?height=450&width=356&top_left_y=93&top_left_x=188)

Fig. 10.11

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-561.jpg?height=388&width=570&top_left_y=156&top_left_x=635)

Fig. 10.12