Selected Problems from Open Contests

Assume without loss of generality that $A$ is located on the $x$-axis and $C$ is located on the $y$-axis, let these points have coordinates of $A(a,0)$ and $C(0,c)$. As the diagonals of a square bisect each other, we know that the intersection point $P$ of diagonal is also the mid-point of $AC$, i.e. $P(\frac{a}{2},\frac{c}{2})$ and $\overrightarrow{PC}=(-\frac{a}{2},\frac{c}{2})$.

As the diagonals of a square are perpendicular to each other and of the same length, the vectors $\overrightarrow{PB}$ and $\overrightarrow{PD}$ have the same length as the vector $\overrightarrow{PC}$ and are perpendicular to it. But for a given vector $\vec{u}=(s,t)$, there are exactly two vectors perpendicular to and having the same length as it: $\vec{v}=(-t,s)$ and $-\vec{v}=(t,-s)$. For the vector $\vec{u}=\overrightarrow{PC}=(-\frac{a}{2},\frac{c}{2})$ we get $\vec{v}=(\frac{c}{2},-\frac{a}{2})$ and w.l.o.g. we can assume that $\overrightarrow{PB}=\vec{v}$ and $\overrightarrow{PD}=-\vec{v}$. Now from here $B(\frac{a+c}{2},\frac{a+c}{2})$ and $D(\frac{a-c}{2},\frac{c-a}{2})$. Thus, we see that the point $B$ is located on the line $y=x$ and point $D$ is located on the line $y=-x$.

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Alternative solution.

W.l.o.g., assume that the vertices of the square are labelled counter-clockwise with $A(a,0)$, $C(0,c)$, where $a,c\ge 0$ (other cases are similar). Let $O$ be the origin, then $\angle AOC=90^{\circ }$, i.e. the circumcircle (with diameter $AC$) of the square $ABCD$ passes through the origin $O$. Based on the assumptions made, $B$ definitely lies in the first quadrant and $D$ has to lie in the second quadrant (Fig. 4) or in the fourth quadrant (Fig. 5), otherwise the circle with the diameter $BD$ cannot pass the origin.

Fig. 4

Fig. 5

diameter AC) of the square ABCD passes through the origin O. Now note that the vertices of the square divide its circumcircle into four equal arcs of $90^{\circ }$, each having an inscribed angle of $45^{\circ }$ subtending on it. Thus, $\angle AOB=\angle BOC=45^{\circ }$, i.e., $B$ lies on the line with equation $y=x$ (if $A=O$ or $C=O$, then one of those angles will lose its meaning, however, the other one is still $45^{\circ }$ and that is sufficient). Similarly, $\angle COD=45^{\circ }$, if $D$ lies in the second quadrant, or $\angle AOD=45^{\circ }$, if $D$ lies in the fourth quadrant. In both cases, $D$ lies on the line $y=-x$; this condition is also met in the special case $D=O$.