 Jeffrey Ali

2022-03-18

A square ABCD has all it's vertices on ${x}^{2}{y}^{2}=1$. The midpoints of it's sides also lie on the same curve . What is the area of this square?
It can be found very easily if we can show that diagonal of this square meet at the origin. TonysennY2cp

If $\left({x}_{0},{y}_{0}\right)$ is the centre of the square and the vector $\left(2u,2v\right)$ corresponds to one of the edges (wlog. $u\ge v\ge 0$), then all the points
$\left({x}_{0}+\alpha u+\beta v,{y}_{0}+\alpha v-\beta v\right)$
with $\alpha ,\beta \in \left\{-1,0,1\right\}$, except $\alpha =\beta =0$, are on the curve. That is,

$\begin{array}{}\text{(1)}& \left({x}_{0}+\alpha u+\beta v\right)\left({y}_{0}+\alpha v-\beta u\right)=±1.\end{array}$

The only quadratic polynomials $A{x}^{2}+Bx+C$ that take values $±1$ for all $x\in \left\{-1,0,1\right\}$ are given by
$A=B=0,C=±1$
$A=±2,B=0,C=\mp 1$
$A=1,B=±1,C=-1$
$A=-1,B=±1,C=1$
For $\alpha =1$, (1) must turn into one of these in terms of $\beta$. As at least one of u, v is non-zero, at least one of the factors in (1) is non-constant and hence the first of the four options, degree 0, is excluded. It follows that the polynomial is of degree 2, i.e., both u and v are non-zero. In fact, we find that $uv\in \left\{1,2\right\}$, which leaves us only with the possibilities $-2{\beta }^{2}+1,-{\beta }^{2}+\beta +1,-{\beta }^{2}+\beta +1$. At any rate,
$\left({x}_{0}+u\right)\left({y}_{0}+v\right)=1.$
The same argument for $\alpha =-1$ leads to
$\left({x}_{0}-u\right)\left({y}_{0}-v\right)=1.$
By subtracting, we find $\left({x}_{0},{y}_{0}\right)\mathrm{\perp }\left(v,u\right)$.
Swapping the roles of $\alpha$ and $\beta$, we find

This time, by subtracting, we find $\left({x}_{0},{y}_{0}\right)\mathrm{\perp }\left(-u,v\right)$$As\left(v,u\right)\mathrm{\perp }\left(-u,v\right)$, one of the three vectors must be zero, and the only candidate is $\left({x}_{0},{y}_{0}\right)$, as desired.
With that, , hence ${u}^{2}$ is a root of ${x}^{2}-x-1$, etc.

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