A square ABCD has all it's vertices on

Jeffrey Ali

Jeffrey Ali

Answered question


A square ABCD has all it's vertices on x2y2=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.

Answer & Explanation



Beginner2022-03-19Added 2 answers

If (x0,y0) is the centre of the square and the vector (2u,2v) corresponds to one of the edges (wlog. uv0), then all the points
with α,β{1,0,1}, except α=β=0, are on the curve. That is,

(1) ( x 0 + α u + β v ) ( y 0 + α v β u ) = ± 1.

The only quadratic polynomials Ax2+Bx+C that take values ±1 for all x{1,0,1} are given by
For α=1, (1) must turn into one of these in terms of β. 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{1,2}, which leaves us only with the possibilities 2β2+1,β2+β+1,β2+β+1. At any rate,
The same argument for α=1 leads to
By subtracting, we find (x0,y0)(v,u).
Swapping the roles of α and β, we find
(x0+v)(y0u)=1 and (x0v)(y0+u)=1.
This time, by subtracting, we find (x0,y0)(u,v)As(v,u)(u,v), one of the three vectors must be zero, and the only candidate is (x0,y0), as desired.
With that, yv=1 and u2v2=(u+v)(uv)=1, hence u2 is a root of x2x1, etc.

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