Given $n$ points on $\mathbb{R}^{2}$, s.t no three are on the same straight line, and not all the points are on the same circle, prove that there exists a circle which contains only three of those points. We are considering only the circles that are defined by the above points, of course.

I am trying to find a direct geometrical proof. I was thinking of looking at the circle and point which have minimal positive distance between them (since I have seen a similar question considering straight lines and points which uses this strategy), or looking at all the circles containing a specific point, but I still can't seem to find the trick.

Thanks alot!

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    $\begingroup$ Do you mean a circle which passes through only 3 or a circle which contains three inside? $\endgroup$ – N. S. May 5 '12 at 20:46
  • $\begingroup$ Passes through 3. Sorry if it wasn't clear $\endgroup$ – IBS May 5 '12 at 21:24

The above result was proven by Motzkin (1951) on page 3 of this document. The proof follows from the Sylvester - Gallai Theorem (presumably the "similar problem" you mentioned) after inverting around a chosen point.

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  • $\begingroup$ I'm sorry, but he uses many terms over there which i am not familiar with... I am not able to follow. Any chance that someone could simplify this here? Thanks $\endgroup$ – IBS May 6 '12 at 13:45

This was too long to make it a comment to EuYu's post. Credit is to him (and Motzkin).

I am assuming this is the result you know already: if $n$ points in the plane are arranged such that no line contains exactly two of them, then they are all on the same line.

The strategy outlined in the document by Motzkin is the following: pick any of the points, say $P$. Call the other points $Q_i$. Apply an inversion with respect to $P$ (check wikipedia if you don't know what an inversion is, and how it acts on circles and lines). Let's say the $Q_i$ get mapped to $Q_i'$ by the inversion.

Because there was no circle containing all the original points, there is not line containing all the $Q_i'$'s. This means that some line $l$ contains exactly two such $Q_i'$, say $Q_1'$ and $Q_2'$. This line does not pass through $P$, because that would mean that $Q_1$, $Q_2$ and $P$ are on the same line. Apply an inversion for the second time (again with respect to $P$ and using the same radius as before). Now $Q_1'$ and $Q_2'$ are mapped to $Q_1$ and $Q_2$ respectively, and $l$ is mapped to a circle containing precisely $P$, $Q_1$ and $Q_2$.

Hope this helps. If it's not clear, try experimenting a little with inversions to get a feel for their properties.

Note that we could pick the point $P$ at random. This means we have actually proven a stronger result: through any of the $n$ points there is a circle containing exactly three of the given points.

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  • $\begingroup$ I gave the credits as you asked :) Thank you for the help. $\endgroup$ – IBS May 8 '12 at 13:01

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