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I have a question, but not sure how to prove this.

We are given $n$ points in the Euclidean plane such that there exists no disk of radius $a$ which contains all of the points.

Conjecture: There must exist three of these points which are not contained in a disk of radius $a$.

Any idea about how to prove this?


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closed as off-topic by alvarezpaiva, Stefan Kohl, Andrey Rekalo, Ryan Budney, David Roberts Apr 16 '14 at 1:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – alvarezpaiva, Stefan Kohl, Andrey Rekalo, Ryan Budney, David Roberts
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When you say that points are contained in a circle, do you mean that the points lie on the circle or that the points are within the closed disc bounded by the circle? – Ricardo Andrade Apr 10 '14 at 13:26
Sorry for the confusion. I actually mean "disk". I changed the statement of the question. – XiMS Apr 11 '14 at 4:22
I think the answer is given by Lev Soukhanov. See the post below. – XiMS Apr 11 '14 at 4:50
Thanks! A more concrete explanation can be found at…. This can help us have a better understanding of Helly's theorem and its applications. – XiMS Apr 11 '14 at 5:34
up vote 11 down vote accepted

There is a classical result of convex geometry, called Helly's theorem ( It states that if you have $n$ convex subsets of $R^d$ and any $d+1$ of these convex subsets have nontrivial intersection, then all of them have nontrivial intersection.

For your question you just apply this theorem to the balls of radius $a$ around your points.

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Thanks.This quite helpful. – XiMS Apr 11 '14 at 4:47

Start by choosing two of the points $A$ and $B$ such that all other points lie on the same side of the line $AB$. (This can be done by moving a line from the distance to the points until it meets a point, and then turning the line until you meet another point.) It there are other points on the line AB, you should choose $A$ and $B$ such that the other points lie between $A$ and $B$. Now consider circles through these two points with various radius where the larger part of the circle lies on the side of the other points. Then a circle with larger radius covers all the area of a circle of smaller radius (inside the half space where the points lie). Hence, the biggest circle you need to cover $A$, $B$ and a third point already covers all points.

So, $A$, $B$ and the third point for which one needed this biggest circle are the three points in your conjecture.

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