A square ABCD has all it's vertices on $x^2y^2 = 1$. Midpoints of it's sides also lie on the same curve . Can I show diagonals of it meet at origin?

Question

A square ABCD has all it's vertices on $x^2y^2 = 1$. 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.

Can anyone please help me ?

Answer 1

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, $$\tag1 (x_0+\alpha u+\beta v)(y_0+\alpha v-\beta u)=\pm1.$$ The only quadratic polynomials $Ax^2+Bx+C$ that take values $\pm1$ for all $x\in\{-1,0,1\}$ are given by

• $A=B=0$, $C=\pm1$
• $A=\pm2$, $B=0$, $C=\mp1$
• $A=1$, $B=\pm1$, $C=-1$
• $A=-1$, $B=\pm1$, $C=1$

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 $(x_0,y_0)\perp(v,u)$.

Swapping the roles of $\alpha$ and $\beta$, we find $$(x_0+v)(y_0-u)=-1\quad\text{and}\quad (x_0-v)(y_0+u)=-1.$$ This time, by subtracting, we find $(x_0,y_0)\perp(-u,v)$. As $(v,u)\perp (-u,v)$, one of the three vectors must be zero, and the only candidate is $(x_0,y_0)$, as desired.

With that, $ uv=1$ and $u^2-v^2=(u+v)(u-v)=1$, hence $u^2$ is a root of $x^2-x-1$, etc. etc.

Answer 2

Here is a solution with

$$A(a,1/a) \ \ \text{with} \ \ a=\sqrt{2+\sqrt{5}}\tag{1}$$

the other points $B,C,D$ being obtained by rotations from $A$ with angles $k\pi/2$.

Fig. 1.

Indeed, as there is one (hyperbolic) branch per quadrant $Q_k$, let us look for a solution where $A \in Q_1$, $B \in Q_2$ and midpoint $E=(A+B)/2 \in Q_1$.

Using complex number representation, let

$$A(x_1,\frac{1}{x_1})\leftrightarrow x_1+i \frac{1}{x_1} \ \ \text{and} \ \ B(x_2,\frac{1}{x_2})\leftrightarrow x_2-i \frac{1}{x_2}.$$

Let us express that $\vec{OB}$ is the image of $\vec{OA}$ by the $\frac{\pi}{2}$ rotation with center $O$:

$$i \left(x_1+i \frac{1}{x_1} \right)=\left(x_2-i \frac{1}{x_2} \right) \ \implies \ x_2=- \frac{1}{x_1} \ \implies \ x_1x_2=-1\tag{2}$$

Midpoint $E$ of $[AB]$ has coordinates.

$$\frac12 \left( x_1+i \frac{1}{x_1} + x_2-i \frac{1}{x_2}\right)= \underbrace{\frac12(x_1+x_2)}_x+i \underbrace{\frac12\left(\frac{x_2-x_1}{x_1x_2}\right)}_y $$

Now let us take into account the fact that we want the midpoint $E$ of $[AB]$ to belong to the branch in $Q_1$ by expressing that $xy=1$:

$$x_2^2-x_1^2=4x_1x_2\tag{3}$$

From (2) and (3), one gets :

$$\dfrac{1}{x_1^2}-x_1^2=-4\tag{4}$$

Setting $X=x_1^2$, on gets the quadratic equation:

$$X^2-4X-1=0\tag{5}$$

whose unique positive root is $X=2+\sqrt{5}$ giving $x_1=\sqrt{X}=\sqrt{2+\sqrt{5}}$ as said in (1).

easily obtain (1) by solving a quadratic.

Fig. 2: Explanation for the global solution: We have been working (red color) on the first branch of hyperbola $xy=1$ + right isosceles triangle. This basic pattern" is then rotated 3 times giving the green, blue, and black parts. The desired square is generated by the union of the 4 right isosceles triangles.

Consequence for the area of the square:

This area is given by the square $L^2$ of the following distance

$$L=dist(A,B)=dist(a,\dfrac{1}{a}),(-\dfrac{1}{a},a))$$

As a consequence:

$$L^2=(a+\dfrac{1}{a})^2+(a-\dfrac{1}{a})^2$$

$$L^2=2(a^2+\dfrac{1}{a^2})=2+\sqrt{5}-(2-\sqrt{5})=4 \sqrt{5}$$

Remark 1: By using reflections with respect to axes Ox and Oy, three other solutions can be obtained.

Reamrk 2: Maybe, by a suitable similitude (rotation + enlargment or shrinking)

$$\begin{cases}x'&=&(r \cos \theta) x - (r \sin \theta) y\\y'&=&(r \sin \theta) x + (r \cos \theta) y \end{cases},$$

the problem can be settled with a fixed square with vertices $(1,1),(-1,1),(-1,-1),(1,-1)$ and a certain fourth degree polynomial under one of its canonical forms as for example given in Some new canonical forms for polynomials by B. REZNICK amenable to the form $x^2y^2=1$ by the reversed change of variable...