# Cospherical Points

Leo Giugiuc has kindly communicated to me a problem from the Romanian Mathematical Magazine due to Dan Sitaru, along with his solution to the problem:

### Solution

In the Euclidean 2D space any three distinct points are *concyclic*, unless they are *collinear*. In the Euclidean 3D space any four distinct points are *cospherical*, unless they are *coplanar* but not concyclic.

The three distinct points $A,B,C\;$ define a plane, say $\alpha\;$ and a circle, say $\omega.\;$ It is immediately verifiable that $A,B,C\in S,\;$ where $S\;$ denotes the unit sphere $x^2+y^2+z^2=1.\;$ In particular, $\omega=\alpha\cap S.$

The condition for the for the four points be coplanar is given by

$\displaystyle\left|\begin{array}{cccc} 2\sqrt{3} & \sqrt{3} & 1 & 1\\ 2\sqrt{3} & 1 & \sqrt{3} & 1\\ 2 & 3 & \sqrt{3} & 1\\ 0 & 2 & 4x & 1 \end{array}\right|=0.$

The equation implies $\displaystyle x=\frac{5}{4},\;$ i.e., $\displaystyle D=\left(0,\frac{1}{2},\frac{5}{4}\right),\;$ meaning $D\notin S\;$ and, therefore, $D\notin\omega.\;$ So, if $\displaystyle x=\frac{5}{4},\;$ $A,B,C,D\;$ are coplanar, but not cospherical. If $\displaystyle x\ne\frac{5}{4},\;$ then $A,B,C,D\;$ are not coplanar and, hence, are cospherical.

Thus the answer is $\displaystyle\mathbb{R}\setminus\left\{\frac{5}{4}\right\}.$

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