## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7c0a9615fb5cd46ea1c5g-15.jpg?height=827&width=854&top_left_y=798&top_left_x=603)

Since $D$ is the midpoint of side $\overline{B D}$, its length is $|B C|=2 \cdot 1=2 \text{ cm}$.

Point $T$ is the centroid of triangle $A B C$, so $|T D|=\frac{1}{3}|A D|$, which means $|A D|=3 \text{ cm}$.

Triangle $B D T$ is equilateral, so $\angle B D T=60^{\circ}$, and $\angle C D T=120^{\circ}$.

Apply the Law of Cosines to triangle $B D A$.

$$
\begin{aligned}
|A B|^{2} & =|B D|^{2}+|A D|^{2}-2|B D| \cdot|A D| \cos 60^{\circ} \\
|A B|^{2} & =1+9-2 \cdot 1 \cdot 3 \cdot \frac{1}{2}=7 \\
|A B| & =\sqrt{7} \text{ cm}
\end{aligned}
$$

Apply the Law of Cosines to triangle $C D A$.

$$
\begin{aligned}
|A C|^{2} & =|C D|^{2}+|A D|^{2}-2|C D| \cdot|A D| \cos 120^{\circ} \\
|A C|^{2} & =1+9+2 \cdot 1 \cdot 3 \cdot \frac{1}{2}=13 \\
|A C| & =\sqrt{13} \text{ cm}
\end{aligned}
$$

The area of triangle $A B C$ is calculated as twice the area of triangle $A D C$ because triangles $A D C$ and $A D B$ have the same base $(|B D|=|D C|)$ and height to that base.

$$
P=2 \cdot \frac{1}{2}|C D| \cdot|A D| \sin 120^{\circ}=\frac{3 \sqrt{3}}{2} \text{ cm}^2
$$

The radius of the circumscribed circle is

$$
R=\frac{a b c}{4 P}=\frac{2 \cdot \sqrt{7} \cdot \sqrt{13}}{4 \cdot \frac{3 \sqrt{3}}{2}}=\frac{\sqrt{91}}{3 \sqrt{3}}=\frac{\sqrt{273}}{9} \text{ cm}
$$
Task B-3.7. Point $T$ is the centroid of triangle $A B C$, and point $D$ is the midpoint of its side $\overline{B C}$. If the length of the side of the equilateral triangle $B D T$ is $1 \mathrm{~cm}$, determine the lengths of the sides of triangle $A B C$ and the radius of the circumcircle of triangle $A B C$.