Let the triangle $ABC$ be equilateral.

$1^{0}$. How does the perimeter and area of the quadrilateral $APMQ$ change as $M$ moves along the side $BC$? $BC=a$.

$2^{0}$. Let $M$ be the midpoint of $BC$ and the quadrilateral $APMQ$ be rotated around $BC$. The volume of the solid of revolution is to be calculated!
$1^{0}$. If $\triangle ABC$ is equilateral, then the triangles $\triangle BPM$ and $\triangle MQC$ are also equilateral, so

$$
\begin{aligned}
PM & = BM, \quad QM = QC \text{ and } \\
AP + PM + QM + AQ & = AP + PB + AQ + QC = AB + AC = 2a
\end{aligned}
$$

Thus, the perimeter of parallelogram $APMQ$ is constant.

![](https://cdn.mathpix.com/cropped/2024_05_02_dbb343038ca7ed45cd15g-1.jpg?height=406&width=423&top_left_y=426&top_left_x=840)

The area of this parallelogram is twice the area of $\triangle APQ$, that is,

$$
t = 2 \cdot \frac{1}{2} AP \cdot AQ \sin 60^{\circ} = AP \cdot AQ \frac{\sqrt{3}}{2}
$$

Let $AP = x$, then $AQ = a - x$ and

$$
t = x(a - x) \frac{\sqrt{3}}{2} = \left(-x^2 + ax\right) \frac{\sqrt{3}}{2}
$$

The value of $t$ is maximized when $x = AP = \frac{a}{2} = AQ$.

Accordingly, the value of $t$ varies from 0 to a maximum; this maximum is half the area of $\triangle ABC$ and occurs when $M$ bisects $BC$. After reaching the maximum, it decreases back to zero. $\left(t_{\max} = \frac{a^2 \sqrt{3}}{8}\right)$.

$2^{0}$. If $M'$ is the midpoint of $BC$, then $P'$ and $Q'$ also bisect $AB$ and $AC$, respectively. In this case, $AP'M'Q'$ is a rhombus.

The volume of the solid generated by rotating around $BC$ is obtained by subtracting the volumes of the solids generated by rotating $\triangle BP'M'$ and $\triangle M'Q'C$ from the volume of the solid generated by rotating $\triangle ABC$. However, the latter two are equal and each is $\frac{1}{8}$ of the volume of the solid generated by rotating $\triangle ABC$, because their corresponding dimensions are in the ratio $1:2$. Therefore, the desired volume is $\frac{3}{4}$ of the volume of the solid generated by rotating $\triangle ABC$. This is a double cone with base $AM' = \frac{1}{2} a \sqrt{3}$ and height $2 \cdot \frac{a}{2} = a$. Thus,

$$
\frac{1}{3} \pi \overline{AM'}^2 \cdot BC = \frac{1}{3} \pi \cdot \frac{3a^2}{4} \cdot a = \frac{\pi}{4} a^3
$$

and

$$
V = \frac{3}{4} \cdot \frac{\pi}{4} a^3 = \frac{3\pi}{16} a^3
$$