62nd Ukrainian National Mathematical Olympiad

First, we will show that for points that are at a distance of $1$ from each other, there are no points $(X,Y)$ that satisfy the condition. Indeed, let us assume that such points exist. Then, $\angle AXB+\angle AYB=180^{\circ }$, which means that at least one of these angles is not less than $90^{\circ }$. Therefore, at least one of the points $X,Y$ must lie inside or on the circle with diameter $AB$, but this circle does not contain any other integer points except for $A$ and $B$.

Now, let's show that for all other pairs of points, such a pair $(X,Y)$ can be found. Let $A=(a_1,a_2)$ and $B=(b_1,b_2)$. If $a_1\ne b_1$ and $a_2\ne b_2$, then we can take $X=(a_1,b_2)$ and $Y=(b_1,a_2)$, and $AXBY$

will be a rectangle, which means that it is inscribed. Otherwise, without loss of generality, we can assume that $a_2=b_2=t$, and $A=(a_1,t)$ and $B=(b_1,t)$, where $|a_1-b_1|>1$. Without loss of generality, we can also assume that $a_1<b_1$. Then, it is sufficient to take the following points: $X=(a_1+1,t-1)$ and $Y=(a_1+1,t+(b_1-a_1-1))$. It is easy to see that the segments $AB$ and $XY$ intersect at the point $K=(a_1+1,t)$, and $AK\cdot BK=XK\cdot YK=1\cdot (b_1-a_1-1)$, so the quadrilateral $AXBY$ is indeed inscribed.