52. a) If $O$ is the center of the sought centrally symmetric polygon $m$, then by symmetrically reflecting the given $\triangle ABC \equiv T$ together with $m$ (which will transform into itself), we can see that $m$ is also inscribed in $\triangle A_{1} B_{1} C_{1} \equiv T_{1}$, which is symmetric to $\triangle ABC$ with respect to $O$. Since the intersection ${ }^{1}$ ) $n=T \cap T_{1}$ of triangles $T$ and $T_{1}$ is clearly a centrally symmetric polygon with center $O$, the smallest centrally symmetric polygon inscribed in $T$ with center $O$ has an area of $\frac{1}{2} S_{ABC}$, because if the height $OP$ of triangle $AOB$ were less than half the height $CH$ of triangle $ABC$, then point $C'$ would lie outside $\triangle ABC$ (beyond side $AB$) and the intersection of $\triangle ABC$ and $\triangle A'B'C'$ would not be a parallelogram.

In the second case, the convex hull of the considered points is a centrally symmetric hexagon $A C' B A' C B'$ (Fig. $109, \sigma$), the area of which is exactly twice the area of $\triangle ABC$. Indeed,

\[
\begin{aligned}
& S_{A C' B A' C B'} = S_{O A B'} + S_{O A' B} + S_{O C A'} + S_{O C' A} + S_{O B C'} + S_{O B' C} = \\
&= 2 S_{O A B'} + 2 S_{O C A'} + 2 S_{O B C'}
\end{aligned}
\]

but

\[
S_{O A B'} = S_{O A B}
\]

since $AO$ is the median of triangle $AB'B$, and similarly

\[
S_{O C A'} = S_{O C A}, \quad S_{O B C'} = S_{O B C}
\]

thus,

\[
S_{A C' B A' C B'} = 2 S_{O A B} + 2 S_{O C A} + 2 S_{O B C} = 2 S_{ABC}
\]

The simplest example of a centrally symmetric polygon (of the smallest possible area) with an area of $2 S_{ABC}$ (i.e., an area of 2 if $S_{ABC} = 1$), containing $\triangle ABC$ inside, is the parallelogram $A B A' C$ (Fig. $109, b$).
52. Given a triangle $T$.

a) Place a centrally symmetric polygon $m$ of the largest possible area inside $T$.

What is the area of $m$ if the area of $T$ is 1?

b) Enclose $T$ in a convex centrally symmetric polygon $M$ of the smallest possible area.

What is the area of $M$ if the area of $T$ is 1?