## Task 3 - 220923

For a square $A B C D$ and four points $P, Q, R, S$, the following is assumed:

(1) $P$ lies on the segment $A B$ between $A$ and $B$,

(2) $Q$ lies on the segment $B C$ between $B$ and $C$,

(3) $R$ lies on the segment $C D$ between $C$ and $D$,

(4) $S$ lies on the segment $D A$ between $D$ and $A$,

(5) it holds that $P R \perp Q S$.

Investigate whether for every position of the points for which conditions (1) to (5) are satisfied, one of the three statements $P RQ S$ always holds!

If this is the case, name this statement!
We place a coordinate system in the plane of the square such that $A$ is at the origin, $B$ is at the point $(1,0)$, and $D$ is at the point $(0,1)$. Then $C$ is at the point $(1,1)$. Furthermore, there exist real numbers $p, q, r, s$ such that $P$ has coordinates $(p, 0)$, $Q$ has coordinates $(1, q)$, $R$ has coordinates $(r, 1)$, and $S$ has coordinates $(0, s)$.

If $p=r$, then the line $P R$ is parallel to the $y$-axis and thus parallel to the $y$-axis. Therefore, according to (5), the segment $Q S$ must be parallel to the $x$-axis, so $q=s$ must hold. This results in the lengths of both segments being

$$
|P R|=\sqrt{(r-p)^{2}+(1-0)^{2}}=1=\sqrt{(1-0)^{2}+(s-q)^{2}}=|Q S|
$$

Otherwise, the line $P R$ has the slope $m_{1}=\frac{1-0}{r-p}=\frac{1}{r-p}$ and the line $Q S$ has the slope $m_{2}=\frac{s-q}{1-0}=s-q$. According to (5), the two lines are perpendicular to each other, so their slopes multiply to $-1$.

Thus, $\frac{1}{r-p} \cdot(s-q)=-1$ or $s-q=-(r-p)$, so $(s-q)^{2}=(r-p)^{2}$. For the lengths of the segments $P R$ and $Q S$, we then get

$$
|P R|=\sqrt{(r-p)^{2}+(1-0)^{2}}=\sqrt{(1-0)^{2}+(s-q)^{2}}=|Q S|
$$

so in any case $|P R|=|Q S|$ holds.

Remark: We have only used the conditions (1) to (4) that the points $P$ to $S$ lie on the specified lines. Therefore, the additional requirement that the points should lie in the interior of the respective sides of the square can be omitted.