We have inscribed a regular hexagon $A B C D E F$ in a circle with center $S$ and radius $12 \mathrm{~cm}$, and we have inscribed a regular hexagon $T U V X Y Z$ such that point $T$ is the midpoint of side $B C$. Calculate the area and perimeter of quadrilateral $T C U S$.

(M. Krejčová)
Let's call the circle described in the problem $k$ and its radius $r$, where $r=12 \text{ cm}$. The relationship between the circle and the hexagons can be interpreted as the circle $k$ being inscribed in a regular hexagon $A B C D E F$ and circumscribed around a regular hexagon $T U V X Y Z$, as shown in the figure. Furthermore, when constructing the regular hexagon $T U V X Y Z$, we can satisfy the condition that the vertex $T$ lies at the midpoint of side $B C$ because the point of tangency between the hexagon $A B C D E F$ and the inscribed circle $k$ is precisely at the midpoint of side $B C$. Therefore, the other vertices of the hexagon $T U V X Y Z$ also lie at the midpoints of the sides of the hexagon $A B C D E F$.

![](https://cdn.mathpix.com/cropped/2024_04_17_541460088524f502ebbcg-5.jpg?height=931&width=842&top_left_y=234&top_left_x=641)

In solving the problem, we will use the known property of a regular hexagon: any two adjacent vertices and the center of the circumscribed (or inscribed) circle form the vertices of an equilateral triangle. Thus, triangles $C S B$ and $C S D$ are equilateral, and they are symmetric with respect to the line $C S$. The point $T$ is the midpoint of side $B C$ of triangle $C S B$, and therefore it is also the foot of the perpendicular from $S$ to side $B C$. Triangle $C S T$ is thus a right triangle. The point $U$, which is the midpoint of side $D C$, is symmetric to $T$ with respect to $C S$, so triangles $C S T$ and $C S U$ are congruent.

To solve the problem, it suffices to find the length $|T C|$, which we can calculate using the Pythagorean theorem in triangle $C S T$ (using $|C S|=2|T C|$):

$$
\begin{aligned}
|C S|^{2} & =|S T|^{2}+|T C|^{2}, \\
4|T C|^{2} & =r^{2}+|T C|^{2}, \\
3|T C|^{2} & =r^{2} \\
|T C| & =\frac{r}{\sqrt{3}}=\frac{r \sqrt{3}}{3} .
\end{aligned}
$$

The quadrilateral $T C U S$ is formed by two congruent triangles, and we can determine its area:

$$
S_{T C U S}=2 \cdot S_{C S T}=|T C| \cdot|S T|=\frac{r \sqrt{3}}{3} \cdot r=\frac{r^{2} \sqrt{3}}{3} .
$$

The perimeter of the quadrilateral $T C U S$ is

$$
o_{T C U S}=2(|T C|+|S T|)=2\left(\frac{r \sqrt{3}}{3}+r\right)=2 r\left(\frac{\sqrt{3}}{3}+1\right) .
$$

Substituting $r=12 \text{ cm}$, we get the results:

$$
\begin{aligned}
& S_{T C U S}=48 \sqrt{3} \approx 83.1 \left(\text{ cm}^2\right) \\
& o_{T C U S}=8 \sqrt{3}+24 \approx 37.9 \left(\text{ cm}\right) .
\end{aligned}
$$