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.

It can be found very easily if we can show that diagonal of this square meet at the origin.

TonysennY2cp

Beginner2022-03-19Added 2 answers

If $({x}_{0},{y}_{0})$ is the centre of the square and the vector $(2u,2v)$ corresponds to one of the edges (wlog. $u\ge v\ge 0$), then all the points

$({x}_{0}+\alpha u+\beta v,{y}_{0}+\alpha v-\beta v)$

with $\alpha ,\beta \in \{-1,0,1\}$, except $\alpha =\beta =0$, are on the curve. That is,

The only quadratic polynomials $A{x}^{2}+Bx+C$ that take values $\pm 1$ for all $x\in \{-1,0,1\}$ are given by

- $A=B=0,C=\pm 1$

- $A=\pm 2,B=0,C=\mp 1$

- $A=1,B=\pm 1,C=-1$

- $A=-1,B=\pm 1,C=1$

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 $uv\in \{1,2\}$, which leaves us only with the possibilities $-2{\beta}^{2}+1,-{\beta}^{2}+\beta +1,-{\beta}^{2}+\beta +1$. At any rate,

$({x}_{0}+u)({y}_{0}+v)=1.$

The same argument for $\alpha =-1$ leads to

$({x}_{0}-u)({y}_{0}-v)=1.$

By subtracting, we find $({x}_{0},{y}_{0})\mathrm{\perp}(v,u)$.

Swapping the roles of $\alpha$ and $\beta$, we find

$({x}_{0}+v)({y}_{0}-u)=-1{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}({x}_{0}-v)({y}_{0}+u)=-1.$

This time, by subtracting, we find $({x}_{0},{y}_{0})\mathrm{\perp}(-u,v)$. $As(v,u)\mathrm{\perp}(-u,v)$, one of the three vectors must be zero, and the only candidate is $({x}_{0},{y}_{0})$, as desired.

With that, $yv=1\text{}\text{and}\text{}{u}^{2}-{v}^{2}=(u+v)(u-v)=1$, hence $u}^{2$ is a root of ${x}^{2}-x-1$, etc.

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