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 ?

Vertices A,B,C,D and the midpoints of the sides of the square ABCD lie on

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

Beginner2022-03-27Added 17 answers

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$(\frac{x+y}{2},\frac{y-x}{2})$ 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 $(x,y),(y,-x),(-x,-y)\text{}\text{and}\text{}(-y,x)$ form a square satisfying the requirements, and its midpoint is the origin.

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

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