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.

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A square ABCD has all it&#039;s vertices on x2y2=1. The midpoints of it&#039;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 · Accepted Answer

If&amp;nbsp;(x0,y0)&amp;nbsp;is the centre of the square and the vector&amp;nbsp;(2u,2v)&amp;nbsp;corresponds to one of the edges (wlog.&amp;nbsp;u≥v≥0), then all the points(x0+αu+βv,y0+αv−βv)with&amp;nbsp;α,β∈{−1,0,1}, except&amp;nbsp;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&amp;nbsp;Ax2+Bx+C&amp;nbsp;that take values&amp;nbsp;±1&amp;nbsp;for all&amp;nbsp;x∈{−1,0,1}&amp;nbsp;are given by-&amp;nbsp;A=B=0,C=±1-&amp;nbsp;A=±2,B=0,C=∓1-&amp;nbsp;A=1,B=±1,C=−1-&amp;nbsp;A=−1,B=±1,C=1For&amp;nbsp;α=1, (1) must turn into one of these in terms of&amp;nbsp;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&amp;nbsp;uv∈{1,2}, which leaves us only with the possibilities&amp;nbsp;−2β2+1,−β2+β+1,−β2+β+1. At any rate,(x0+u)(y0+v)=1.The same argument for&amp;nbsp;α=−1&amp;nbsp;leads to(x0−u)(y0−v)=1.By subtracting, we find&amp;nbsp;(x0,y0)⊥(v,u).Swapping the roles of&amp;nbsp;α&amp;nbsp;and&amp;nbsp;β, we find(x0+v)(y0−u)=−1&amp;nbsp;and&amp;nbsp;(x0−v)(y0+u)=−1.This time, by subtracting, we find&amp;nbsp;(x0,y0)⊥(−u,v).&amp;nbsp;As(v,u)⊥(−u,v), one of the three vectors must be zero, and the only candidate is&amp;nbsp;(x0,y0), as desired.With that,&amp;nbsp;yv=1&amp;nbsp;and&amp;nbsp;u2−v2=(u+v)(u−v)=1, hence&amp;nbsp;u2&amp;nbsp;is a root of&amp;nbsp;x2−x−1, etc.

A square ABCD has all it's vertices on

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