When Circle Meets Hyperbola

Problem

Assume real, non-zero, numbers $a,b,c,d$ are pairwise distinct and that points $\displaystyle A(a,\frac{1}{a}),\;$ $\displaystyle B(b,\frac{1}{b}),\;$ $\displaystyle C(c,\frac{1}{c}),\;$ $\displaystyle D(d,\frac{1}{d}),\;$ are concyclic. Find $abcd.$

Solution 1

Assume the circle in question is defined by the equation

$x^2+y^2+2gx+2fy+c=0$

Then $a,b,c,d\;$ are the roots of the equation

$\displaystyle t^2+\frac{1}{t^2}+2gt+2f\frac{1}{t}+c=0$

which is equivalent to

$t^4+2gt^3+ct^2+2ft+1=0.$

By one of Viète's formulas then $abcd=1.$

Solution 2

In complex numbers, points $A,B,C,D$ are concyclic iff $\displaystyle\frac{A-C}{B-C}:\frac{A-D}{B-D}\;$ is a real number. This is equivalent to

$\displaystyle\frac{\displaystyle\frac{a^2+i}{a}-\frac{c^2+i}{c}}{\displaystyle\frac{b^2+i}{b}-\frac{c^2+i}{c}}\cdot \frac{\displaystyle\frac{a^2+i}{a}-\frac{d^2+i}{d}}{\displaystyle\frac{b^2+i}{b}-\frac{d^2+i}{d}} \in\mathbb{R}$

which, in turn, is equivalent to

$\displaystyle\frac{\displaystyle\frac{a-c}{ac}}{\displaystyle\frac{b-c}{bc}}\cdot\frac{ac-i}{bc-i}\cdot\frac{\displaystyle\frac{b-d}{bd}}{\displaystyle\frac{a-d}{ad}}\cdot\frac{bd-i}{ad-i} \in\mathbb{R},$

or,

$\displaystyle\frac{ac-i}{bc-i}\cdot\frac{bd-i}{ad-i} \in\mathbb{R},$

and, finally, to

$\displaystyle\frac{abcd-1-(ac+bd)i}{abcd-1-(ad+bc)i} \in\mathbb{R}.$

If $abcd \ne 1,$ then $ac+bd=ad+bc,\;$ i.e., $(a-b)(c-d)=0$ which would contradict the stipulations of the problem. Therefore, $abcd=1.$

Solution 3

Let the circle be $(x-u)^2+(y-v)^2=1.$ Then with $\displaystyle y=\frac{1}{x},\;$ this becomes

$x^4-2ux^3+(u^2+v^2-1)x^2-2vx+1=0,$

From which, by one of Viète's formulas then $abcd=1.$

Generalization

Assume real, non-zero, numbers $a,b,c,d$ are pairwise distinct and that points $\displaystyle A(a,\frac{1}{a}),\;$ $\displaystyle B(b,\frac{1}{b}),\;$ $\displaystyle C(c,\frac{1}{c}),\;$ $\displaystyle D(d,\frac{1}{d}),\;$ lie on an ellipse. Find $abcd.$

Solution

If $x^2+Mxy+Ny^2+Px+Qy+S=0\;$ is the equation of the ellipse, then, with $\displaystyle y=\frac{1}{x},\;$ it becomes

$x^4+Px^3+(M+S)x^2+Qx+N=0$

from which $abcd=N.$

Acknowledgment

The story began with Leo Giugiuc posting the problem and his and Dan Sitaru solution at the CutTheKnotMath facebook page. Leo also provided a link to the original post by Wahab Raiz who referred to Ritesh Sharma as the author of the problem. The discussion there included a solution by Ravi Prakash. Without letting up, Leo has served Daniel Dan's solution and hid and Dan Sitaru's generalization that extended the result to four points on an ellipse.

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