According to the problem:

$$
\begin{gathered}
A B=B C=C D=D A=a \quad ; \quad B P_{1}=P_{1} C=\frac{a}{2} \\
C P_{2}=D P_{3}=\frac{a}{4} \quad ; \quad A P_{3}=P_{2} D=\frac{3 a}{4}
\end{gathered}
$$

The polygon $P P_{1} P_{2} P_{3}$ has the smallest area if and only if the triangle $\triangle P_{1} P P_{3}$ (degenerates to the segment $P_{1} P_{3}$ and thus) has the smallest possible area of 0.

This occurs if and only if $P=P_{1}$ or $P=P_{3}$. In this case, the area of the polygon $P P_{1} P_{2} P_{3}$ is equal to that of the triangle $\triangle P_{1} P_{2} P_{3}$.

The area of the polygon $P P_{1} P_{2} P_{3}$ is the largest if and only if the area of the triangle $\triangle P P_{1} P_{3}$ is the largest, because the triangles lie on different sides of the line $g$. This is the case if and only if the point $P$ has the greatest distance from $P_{1} P_{3}$.

If a line $h$ is drawn through $A$ parallel to the line $g$ through $P_{1}$ and $P_{3}$, it is clear that among all possible positions of the point $P$, this point has the greatest distance from the fixed side $P_{1} P_{3}$ exactly when $P=A$; because for all other positions of the point $P$, this point lies within the strip bounded by $g$ and $h$ or on $g$, since $P_{3} A$ is greater than $P_{1} B$ by assumption.

a) The area $A_{D}$ of the triangle $\triangle P_{1} P_{2} P_{3}$ is obtained by subtracting the areas of the triangles $\triangle P_{2} P_{1} C$ and $\triangle P_{3} P_{2} D$ as well as the area of the trapezoid $P_{1} P_{3} A B$ from the area $A_{Q}$ of the square $A B C D$. Therefore, we have

$$
A_{D}=a^{2}-\frac{1}{2} \cdot \frac{a}{2} \cdot \frac{a}{4}-\frac{1}{2} \cdot \frac{3 a}{4} \cdot \frac{a}{4}-\frac{1}{2}\left(\frac{a}{2}+\frac{3 a}{4}\right) \cdot a=a^{2}-\frac{a^{2}}{16}-\frac{3 a^{2}}{32}-\frac{5 a^{2}}{8}=\frac{7 a^{2}}{32}
$$

The desired ratio of the areas in this case, given $A_{Q}=a^{2}$ and $A_{V}=A_{D}$, is

$$
A_{D}: A_{Q}=\frac{7 a^{2}}{32}: a^{2}=7: 32
$$

b) Similarly, the area $A_{V}$ of the quadrilateral $P_{1} P_{2} P_{3} A$ can be determined:

$$
A_{v}=a^{2}-\frac{1}{2} \cdot \frac{a}{2} \cdot \frac{a}{4}-\frac{1}{2} \cdot \frac{3 a}{4} \cdot \frac{a}{4}-\frac{1}{2} \cdot \frac{a}{2} \cdot a=a^{2}-\frac{a^{2}}{16}-\frac{3 a^{2}}{32}-\frac{a^{2}}{4}=\frac{19 a^{2}}{32}
$$

The desired largest ratio of the areas is therefore $A_{V}: A_{Q}=19: 32$.
## Task 3 - 110733

Given a square $A B C D$ with side length $a$. On $B C$ lies a point $P_{1}$ such that $B P_{1}=P_{1} C$ holds, on $C D$ lies a point $P_{2}$ with $P_{2} D=3 C P_{2}$, and on $D A$ lies a point $P_{3}$ with $P_{3} A=3 D P_{3}$.

A point $P$ moves along the sides of the square from $P_{1}$ over $B$ and $A$ to $P_{3}$.

Let $A_{Q}$ be the area of the square $A B C D$ and $A_{V}$ be the area of the polygon $P P_{1} P_{2} P_{3}$.

Determine all positions of $P$ for which the ratio $A_{Q}: A_{V}$

a) is the largest,

b) is the smallest!

Calculate the ratio for each of these cases!

It is also allowed that $P$ coincides with $P_{1}$ or $P_{3}$ if one of the desired ratios occurs in these cases.