## Task 5 - 340835

On the boundary of a square $A B C D$ with a given side length $a$, let $P_{1}, P_{2}, P_{3}$ be the following points: $P_{1}$ lies on $B C$ such that $\overline{B P_{1}}=\overline{P_{1} C}$, $P_{2}$ lies on $C D$ such that $\overline{P_{2} D}=3 \cdot \overline{C P_{2}}$, and $P_{3}$ lies on $D A$ such that $\overline{P_{3} A}=3 \cdot \overline{D P_{3}}$.

A point $X$ moves on the segments $P_{1} B, B A, A P_{3}$ from $P_{1}$ to $P_{3}$. We are looking for all points $X$ on this path (including its starting and ending points) for which the area of the quadrilateral $X P_{1} P_{2} P_{3}$

a) is as small as possible,

b) is as large as possible.

Find all these points and also calculate the area of the quadrilateral $X P_{1} P_{2} P_{3}$ for each of them!

Hint: In this task, a quadrilateral is also allowed to "degenerate into a triangle" if two of its vertices coincide.
The quadrilateral $X P_{1} P_{2} P_{3}$ consists of the triangle $X P_{1} P_{3}$ and the triangle $P_{1} P_{2} P_{3}$, or it is equal to the latter triangle in the cases $X=P_{1}, X=P_{3}$.

Therefore, its area is smallest in these two cases; and it is largest exactly when the area of the triangle $X P_{1} P_{3}$ is largest.

This is the case exactly when the distance of the point $X$ from the line $g$ through $P_{1}$ and $P_{3}$ is largest. This is true for $X=A$.

Proof (see figure):

The parallel line $h$ through $A$ to $g$ has, due to $P_{3} A > P_{1} B$, only the point $A$ in common with the square $A B C D$, and thus also with the entire path of the point $X$. Consequently, all points of this path except $A$ lie in the strip enclosed by $g$ and $h$, closer to $g$ than the point $A$. Thus, the points $X$ sought in (a) and (b) are found.

Since $C D = a$ and $P_{1} C = \frac{1}{2} a, P_{3} D = \frac{1}{4} a$, the trapezoid $P_{1} C D P_{3}$ has the area

$$
F\left(P_{1} C D P_{3}\right) = a \cdot \frac{1}{2}\left(\frac{1}{2} a + \frac{1}{4} a\right) = \frac{3}{8} a^{2}
$$

Since $P_{2} C = \frac{1}{4} a$ and $P_{2} D = \frac{3}{4} a$, we have further

$$
\begin{aligned}
& F\left(P_{1} C P_{2}\right) = \frac{1}{2} \cdot \frac{1}{2} a \cdot \frac{1}{4} a = \frac{1}{16} a^{2} \\
& F\left(P_{2} D P_{3}\right) = \frac{1}{2} \cdot \frac{3}{4} a \cdot \frac{1}{4} a = \frac{3}{32} a^{2}
\end{aligned}
$$

For the points $X = P_{1}$ and $X = P_{3}$ found in (a), the area of $X P_{1} P_{2} P_{3}$ is thus

$$
F\left(P_{1} P_{2} P_{3}\right) = F\left(P_{1} C D P_{3}\right) - F\left(P_{1} C P_{2}\right) - F\left(P_{2} D P_{3}\right) = \frac{7}{32} a^{2}
$$

With $A B = a$ and $B P_{1} = \frac{1}{2} a$, we get $F\left(A B P_{1}\right) = \frac{1}{4} a^{2}$, and thus for the point $X = A$ found in (b), the area of $X P_{1} P_{2} P_{3}$ is:

$$
F\left(P_{1} P_{2} P_{3}\right) = F(A B C D) - F\left(P_{1} C P_{2}\right) - F\left(P_{2} D P_{3}\right) - F\left(A B P_{1}\right) = \frac{19}{32} a^{2}
$$