Find the area of square $ABCD$

Question

A square $ABCD$ has all its vertices on the curve $x^2y^2=1$. The mid points of its side also lie on the same curve. Find the area of square $ABCD$.

I drew the graph of $x^2y^2=1$. It is a discontinuous curve having two points of discontinuity.

Then I assumed coordinates of the vertices and put them in the equation of curve. But I got nothing.

Any help is greatly appreciated.

Answer 1

$x^2y^2=1 \iff xy=1,-1$. So I have two equilateral hyperbolas in the $4$ quadrants. Let be $A=(t_1,\frac1{t_1})$ with $t_1>0$ that lie in the vertex of the square in the first quadrant. $D=(-t_2,-\frac1{t_2})$ with $t_2>0$ that lie in the vertex of the square in the second quadrant. Let be $C=(-t_1,-\frac1{t_1})$ that lie in the vertex of the square in the third quadrant. At least let be $B=(t_2,-\frac1{t_2})$ that lie in the vertex of the square in the fourth quadrant. Then

$$\frac{t_1+t_2}{2}\cdot \frac{\frac1{t_1}-\frac1{t_2}}{2}=1 \implies t_1^2-t_2^2=4t_1t_2$$$$\frac{1}{t^2_1}\cdot \left(-\frac{1}{t^2_2}\right)=-1\implies t_1^2-t_2^2=4$$

$$t^2_1+t_2^2=\sqrt{4^2+4}=2\sqrt5$$

$$t_1^2=2+\sqrt 5 \quad \mapsto \quad \frac{1}{t^2_1}=\sqrt 5-2$$

Therefore

$$|AB|^2=(t_1-t_2)^2+\left(\frac{1}{t_1}+\frac{1}{t_2}\right)^2=2\left(t_1^2+\frac{1}{t^2_1}\right)=2\left(2+\sqrt{5}+\frac{1}{2+\sqrt{5}}\right)$$ $$2\cdot \left(2\sqrt{5}\right)=4\sqrt 5$$

Answer 2

The curve for $ \ x^2·y^2 \ = \ 1 \ \rightarrow \ xy \ = \ \pm 1 \ $ has "four-fold" symmetry about the origin: if $ \ (x \ , \ y) \ $ is a point on the curve, so are $ \ (x \ , \ -y) \ , \ (-x \ , \ y) \ , \ $ and $ \ (-x \ , \ -y) \ \ . \ $ The requirement that the square must have both its vertices and the midpoints of its sides lie on the curve means that two points of the square must be located on each "branch". This makes it impossible for the sides of the square to be parallel to the coordinate axes, since either the set of vertices or the set of midpoints would not be accommodated; this forces us to look for a solution in which the square is canted at some angle to the axes. (The aforementioned symmetry will also locate the center of the square at the origin, to respond to Ennar's comment.)

We then anticipate a solution with one vertex on each branch of this "double rectangular hyperbola" in such a way that the side connecting "consecutive" vertices intersects the curve as well. We then have one vertex and one midpoint on each branch. The change in sign for the product of coordinates in the first and third versus the second and fourth quadrants "breaks" the four-fold symmetry for the square, but we still have a symmetry about the origin , placing the vertices at $ \ \left(X \ , \ \frac{1}{X} \right) \ \ , \ \ \left(-X \ , \ -\frac{1}{X} \right) \ $ on $ \ xy \ = \ +1 \ $ and $ \ \left(\frac{1}{X} \ , \ -X \right) \ \ , \ \ \left(-\frac{1}{X} \ , \ X \right) \ $ on $ \ xy \ = \ -1 \ \ ; \ $ in this designation, we have chosen $ \ X \ > \ \frac{1}{X} \ > \ 0 \ \ . \ $

The midpoints of the sides connecting consecutive vertices also "obey" this "partial-broken" symmetry. In an earlier attempt, I found that making use of the geometry of a square only yielded relations that told nothing about its size. It turns out that the requirement that the midpoints of the sides lie on the branches on the curve "sets the scale" for the square. So saying, we will call these midpoints $ \ \left(a \ , \ \frac{1}{a} \right) \ \ , \ \ \left(\frac{1}{a} \ , \ -a \right) \ \ ,$ $ \left(-a \ , \ -\frac{1}{a} \right) \ \ , \ \ $ and $ \ \left(\frac{1}{a} \ , \ -a \right) \ . \ $ We will focus on the midpoint in the first quadrant and the vertices in the first and second, although a similar discussion would follow, with appropriate sign changes, for any of the other midpoints. [Sebastiano presents a fine argument based on the requirement that the product of the coordinates of these midpoints must equal either $ \ +1 \ $ or $ \ -1 \ \ . \ $ An alternative approach is offered here.]

As the center of symmetry is the origin, we extend a segment to the midpoint at $ \ \left(a \ , \ \frac{1}{a} \right) \ \ ; \ $ this segment then has slope $ \ m' \ = \ \frac{1}{a^2} \ \ . \ $ The side of the square containing this point is perpendicular to the segment: its slope is then $ \ m \ = \ -a^2 \ \ $ and this side lies on the line $$ y \ - \ \frac{1}{a} \ \ = \ \ -a^2·(x \ - \ a) \ \ \rightarrow \ \ y \ \ = \ \ -a^2·x \ + \ a^3 \ + \ \frac{1}{a} \ \ . $$ We wish to find where this line intersects the branches of the "double hyperbola" in order to locate the vertices of this side of the square, so we need to solve the equations $ \ -a^2·x + a^3 + \frac{1}{a} \ = \ \pm \frac{1}{x} \ \ . \ $ The vertex in the first quadrant is on the same branch as our midpoint, for which we find $$ -a^2·x \ + \ a^3 \ + \ \frac{1}{a} \ \ = \ \ \frac{1}{x} \ \ \Rightarrow \ \ a^3·x^2 \ - \ (a^4 \ + \ 1)·x \ + \ a $$ $$ = \ \ (a \ - \ x)·(1 \ - \ a^3·x ) \ \ = \ \ 0 \ \ . $$ This locates the first quadrant vertex at $ \ \left(\frac{1}{a^3} \ , \ a^3 \right) \ \ . \ $ The equation to find the vertex on the second-quadrant branch of $ \ y \ = \ -\frac{1}{x} \ $ is far less pleasant to solve, but here the anti-symmetry across coordinate axes comes to our rescue(!), giving us the coordinates as $ \ \left(-a^3 \ , \ \frac{1}{a^3} \right) \ \ . \ $

The midpoint must be equidistant from these two vertices, so we can write $$ \left(a \ - \ \frac{1}{a^3} \right)^2 \ + \ \left(\frac{1}{a} \ - \ a^3 \right)^2 \ \ = \ \ ( \ a \ - \ [-a^3] \ )^2 \ + \ \left(\frac{1}{a} \ - \ \frac{1}{a^3} \right)^2 $$ $$ \Rightarrow \ \ a^2 \ - \ \frac{2}{a^2} \ + \ \frac{1}{a^6} \ + \ \frac{1}{a^2} \ - \ 2·a^2 \ + \ a^6 $$ $$ = \ \ a^2 \ + \ 2·a^4 \ + \ a^6 \ + \ \frac{1}{a^2} \ - \ \frac{2}{a^4} \ + \ \frac{1}{a^6} \quad \mathbf{[ \ 1 \ ]} $$ $$ \Rightarrow \ \ a^4 \ - \ \frac{1}{a^4} \ + \ a^2 \ + \ \frac{1}{a^2} \ \ = \ \ \frac{1}{a^4}·\left( a^4 \ + \ 1 \right)· \left( a^4 \ + \ a^2 \ - \ 1 \right) \ \ = \ \ 0 \ \ . $$

Since $ \ a \ $ is a non-zero real number, only the quadratic factor above yields a solution, given by $$ a^2 \ \ = \ \ -\frac12 \ + \ \frac{\sqrt{1^2 \ - \ 4·1·(-1)}}{2} \ \ = \ \ \frac{\sqrt5 \ - \ 1}{2} \ \ , $$ upon "discarding" the non-real result $ \ a^2 \ = \ -\frac{ \sqrt5 \ + \ 1}{2} \ \ . $ [One may recognize the acceptable value as $ \ a^2 \ = \ \frac{1}{\phi} \ \ , \ \ \phi \ $ being the "Golden Ratio".]

Either side of equation $ \ \mathbf{1} \ $ gives the distance-squared from the midpoint of the side of the square to one or the other of the vertices, so the area of the square is obtained from, say,

$$ \mathcal{A} \ \ = \ \ 4 · \left( \ a^6 \ + \ \frac{1}{a^6} \ - \ a^2 \ - \ \frac{1}{a^2} \ \right) $$ $$ = \ \ 4 · \left[ \ \left(\frac{\sqrt5 \ - \ 1}{2} \right)^3 \ + \ \left(\frac{\sqrt5 \ + \ 1}{2} \right)^3 \ - \ \left(\frac{\sqrt5 \ - \ 1}{2} \right) \ - \ \left(\frac{\sqrt5 \ + \ 1}{2} \right) \ \right] \ \ $$ [using $ \ \frac{1}{a^2} \ = \ \frac{\sqrt5 \ + \ 1}{2} \ ] $

$$ = \ \ \frac48 · [ \ (\sqrt5 \ - \ 1)^3 \ + \ (\sqrt5 \ + \ 1 )^3 \ - \ 4·(\sqrt5 \ - \ 1) \ - \ 4·(\sqrt5 \ + \ 1) \ ] \ \ $$ $$ = \ \ \frac12 · ( \ 5\sqrt5 \ + \ 3\sqrt5 \ + \ 5\sqrt5 \ + \ 3\sqrt5 \ - \ 4\sqrt5 \ - \ 4\sqrt5 \ ) \ \ = \ \ 4·\sqrt5 \ \ . $$

We also note that the slopes of the sides of the square are therefore $ \ m \ = \ -a^2 \ = \ -\frac{1}{\phi} \ $ and $ \ m' \ = \ \frac{1}{a^2} \ = \ \phi \ \ $ (another of many problems on this site in which the Golden Ratio makes a surprise appearance). A graph to scale of the geometrical situation is shown below.