Call a point $(x, y)$ in the plane hyperbolic if it lies on one of the hyperbolas $y = 1/x$ or $y = -1/x$. Find a square such that its four vertices and the midpoints of its four sides are all hyperbolic.
I trying to prove it with distance formula not with vectors.
Let's say one of the vertices of the square is $(a,1/a)$. Then its $90^\circ$ rotations around the origin are all hyperbolic, and in particular $(-1/a,a)$, a neighbouring vertex, is also hyperbolic. Now suppose the midpoint of the two points discussed is hyperbolic and lies on $y=1/x$: $$\frac{1/a+a}2=\frac2{a-1/a}$$ This rearranges to $$a^2-\frac1{a^2}=4$$ $$a^4-4a^2-1=0$$ $$a^2=\frac{4+\sqrt{20}}2=2+\sqrt5$$ $$a=\sqrt{2+\sqrt5}$$ So for this particular value of $a$, the square formed by $(a,1/a)$ and its $90^\circ$ rotations satisfies the given conditions.
The square looks like this: