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Can we prove or disprove that : There exists for any given closed figure, a point which is equidistant from all of its vertices?

Any closed figure means literally any closed figure?

I am gonna instinctively say no, but How!?

J.Gudal
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Adam Karlson
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  • If there is a point which is equidistant from all other points, you've defined a circle. – Batman Dec 27 '15 at 16:03
  • The circumcenter of a regular polygon is equidistant from every vertex of the polygon only, not from all points. – SchrodingersCat Dec 27 '15 at 16:14
  • @Batman or a cone or a sphere, the point could be anywhere – Adam Karlson Dec 27 '15 at 16:53
  • @Sky while referring to a polygon, I had a.discrete set of points (the vertices) only.in mind... I removed it to prevent confusion – Adam Karlson Dec 27 '15 at 16:55
  • @AdamKarlson Just to get some specific example going on ,if you have a cyclic quadrilatral (a quadrilateral inscribed in a circle )then you can find the point which is equidistant from the four vertices of the figure by finding the center of the circle circumscribed about the quadrilateral,now to prove your claim just take a point which is outside the circle.this example can easily be generalized (altough you don't need a cyclic quadrilateral in the first place,this was just to have a concrete example for you ) – Mr. Y Dec 27 '15 at 17:01
  • This question hasn't to be downvoted.It's a genuine question which the OP wants to know more about. – Mr. Y Dec 27 '15 at 17:07
  • The #1 issue here is that "any closed figure" means nothing, without further specification, and it may get technical to define it properly. Replace it with "any polygon", for instance, and it becomes a legitimate question (to which the answer is "no, there is not always an equidistant point"). – Federico Poloni Dec 27 '15 at 20:57

1 Answers1

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This is clearly false. First choose three random non-collinear points. Then there is a unique circle that goes through these three points, and hence there is a unique point (the circle's center) that is equidistant from the three points. Now add any other point which does not lie on the perimeter of the circle.

sbares
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    I think you could apply the same argument to a closed figure in any dimension. – J.Gudal Dec 27 '15 at 17:29
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    Choosing three collinear points would be quicker. – Wojowu Dec 27 '15 at 17:32
  • Agree. Two distinct vertices lie on a circle in many ways. Three distinct vertices will not lie on any circle if they are colinear (they "line up"), but if they are not colinear, there exists one and only one circle which passes through these three points. With four points, then, you have to be very lucky to have them on one circle, for the reason given in the above answer. With five, six, etc., even more so. A polygon whose vertices are on a circle is a cyclic polygon. Therefore any triangle is cyclic. – Jeppe Stig Nielsen Dec 27 '15 at 17:58
  • ... and for those "lucky" four-vertex configurations, see cyclic quadrilateral. – Jeppe Stig Nielsen Dec 27 '15 at 18:00
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    @Wojowu I thought about that, but I don't know if you could call that a "figure". It would at least be a sort of degenerate case. – sbares Dec 27 '15 at 18:53
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    @SBareS If you define your figure using only the 3 colinear points, then yes, the case is degenerate. However, since the question does not require the figure to be entirely convex, for any set of 3 colinear points, it's quite easy to define a non-degenerate figure that has an actual vertex (where the line segments coming into it are not colinear to each other) at each of those 3 points, by including extraneous points as also being vertices of the figure. Since the question requires a centre equidistant to all vertices, these extraneous ones are irrelevant so long as the 3 are shown colinear. – Matthew Najmon Dec 28 '15 at 01:30