Is it possible to locate $4$(or more than $4$) points on a plane such that every point is at an equal distance from every other point?
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No. Assume that such points exists and call them $A$, $B$, $C$, $D$.
It is clear that $ \triangle ABC$ is an equilateral triangle, and point $D$ is at an equal distance from $A$, $B$ and $C$, which means that $D$ is the centroid since they are one a single plane.
However, if $\overline {AB}=a$, then this implies that $\overline {AD}=\frac{\sqrt{3}a}{3}$ since $D$ is the centroid. However, this is a contradiction, as $\overline{AB}=\overline {AD}$.
S.C.B.
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