rijstmeel7d4t

2022-03-26

How can you prove that the midpoint of the square will be Origin?
Vertices A,B,C,D and the midpoints of the sides of the square ABCD lie on ${x}^{2}{y}^{2}=1$.
No other points of the square lies on the curve.
How can you prove that the midpoint of the square will be Origin ?
My approach: If I can show if (x,y) are the coordinates of A then (y, -x) will be so of B , I will be done. I supposed A lies on first quadrant , B lies on second quadrant etc..
Can anyone tell please me how to show it ?

zalutaloj9a0f

You can try to argue geometrically using symmetries or you can try to find the coordinates of the vertices of the square.
Let (x,y) be a vertex of the square in the first quadrant with smallest x. Then consider the point (y,-x). This point lies on the curve (in the second quadrant) and asking that the midpont $\left(\frac{x+y}{2},\frac{y-x}{2}\right)$ lies on the curve yields the equation ${x}^{4}+{y}^{4}=18.$ Substitution of ${y}^{2}=\frac{1}{{x}^{2}}$ yield as solution (take the positive x with the smallest value) that $x=\sqrt{\sqrt{5}-2}$ and $y={\sqrt{\sqrt{5}-2}}^{-1}$. Now you can check that the vertices form a square satisfying the requirements, and its midpoint is the origin.

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