A set of points in a plane such that for each subset of 3 there is a circle surrounding them

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A set of $ngt3$ distinct points from plane $pi$ satisfies the following criteria: for every subset of 3 points there is a circle of radius $R$ surrounding them. Prove that you can surround all points with a circle of radius $R$.



I have tried to encompass all points in a rectangle or a polygon, then tried to prove that such shape can be "covered" by a circle of the given radius, but the idea did not work. Also my attempt to make a proof by induction fell short on the induction step.




euclidean-geometry plane-geometry




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  • Wouldn't any sufficiently large circle be able to surround a set of any number of coplanar points?
    – Karn Watcharasupat
    Jul 28 at 7:11










  • @KarnWatcharasupat I think the idea is to have a predefined radius $R$. You cannot pick points first and then select some big $R$ to cover them all.
    – Oldboy
    Jul 28 at 7:14











  • oooh I see, my bad
    – Karn Watcharasupat
    Jul 28 at 7:21










  • @KarnWatcharasupat I have edited the text, it should be more clear now.
    – Oldboy
    Jul 28 at 7:21










  • @Oldboy Also, Is this proposition true in, for example, Banach spaces? (Maybe this could be added to the question.)
    – Ma Joad
    Jul 28 at 8:20














up vote
4
down vote

favorite
1












A set of $ngt3$ distinct points from plane $pi$ satisfies the following criteria: for every subset of 3 points there is a circle of radius $R$ surrounding them. Prove that you can surround all points with a circle of radius $R$.



I have tried to encompass all points in a rectangle or a polygon, then tried to prove that such shape can be "covered" by a circle of the given radius, but the idea did not work. Also my attempt to make a proof by induction fell short on the induction step.




euclidean-geometry plane-geometry




share|cite|improve this question





















  • Wouldn't any sufficiently large circle be able to surround a set of any number of coplanar points?
    – Karn Watcharasupat
    Jul 28 at 7:11










  • @KarnWatcharasupat I think the idea is to have a predefined radius $R$. You cannot pick points first and then select some big $R$ to cover them all.
    – Oldboy
    Jul 28 at 7:14











  • oooh I see, my bad
    – Karn Watcharasupat
    Jul 28 at 7:21










  • @KarnWatcharasupat I have edited the text, it should be more clear now.
    – Oldboy
    Jul 28 at 7:21










  • @Oldboy Also, Is this proposition true in, for example, Banach spaces? (Maybe this could be added to the question.)
    – Ma Joad
    Jul 28 at 8:20












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





A set of $ngt3$ distinct points from plane $pi$ satisfies the following criteria: for every subset of 3 points there is a circle of radius $R$ surrounding them. Prove that you can surround all points with a circle of radius $R$.



I have tried to encompass all points in a rectangle or a polygon, then tried to prove that such shape can be "covered" by a circle of the given radius, but the idea did not work. Also my attempt to make a proof by induction fell short on the induction step.




euclidean-geometry plane-geometry




share|cite|improve this question













A set of $ngt3$ distinct points from plane $pi$ satisfies the following criteria: for every subset of 3 points there is a circle of radius $R$ surrounding them. Prove that you can surround all points with a circle of radius $R$.



I have tried to encompass all points in a rectangle or a polygon, then tried to prove that such shape can be "covered" by a circle of the given radius, but the idea did not work. Also my attempt to make a proof by induction fell short on the induction step.





euclidean-geometry plane-geometry





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 28 at 7:32
























asked Jul 28 at 7:07









Oldboy

2,5901316




2,5901316











  • Wouldn't any sufficiently large circle be able to surround a set of any number of coplanar points?
    – Karn Watcharasupat
    Jul 28 at 7:11










  • @KarnWatcharasupat I think the idea is to have a predefined radius $R$. You cannot pick points first and then select some big $R$ to cover them all.
    – Oldboy
    Jul 28 at 7:14











  • oooh I see, my bad
    – Karn Watcharasupat
    Jul 28 at 7:21










  • @KarnWatcharasupat I have edited the text, it should be more clear now.
    – Oldboy
    Jul 28 at 7:21










  • @Oldboy Also, Is this proposition true in, for example, Banach spaces? (Maybe this could be added to the question.)
    – Ma Joad
    Jul 28 at 8:20
















  • Wouldn't any sufficiently large circle be able to surround a set of any number of coplanar points?
    – Karn Watcharasupat
    Jul 28 at 7:11










  • @KarnWatcharasupat I think the idea is to have a predefined radius $R$. You cannot pick points first and then select some big $R$ to cover them all.
    – Oldboy
    Jul 28 at 7:14











  • oooh I see, my bad
    – Karn Watcharasupat
    Jul 28 at 7:21










  • @KarnWatcharasupat I have edited the text, it should be more clear now.
    – Oldboy
    Jul 28 at 7:21










  • @Oldboy Also, Is this proposition true in, for example, Banach spaces? (Maybe this could be added to the question.)
    – Ma Joad
    Jul 28 at 8:20















Wouldn't any sufficiently large circle be able to surround a set of any number of coplanar points?
– Karn Watcharasupat
Jul 28 at 7:11




Wouldn't any sufficiently large circle be able to surround a set of any number of coplanar points?
– Karn Watcharasupat
Jul 28 at 7:11












@KarnWatcharasupat I think the idea is to have a predefined radius $R$. You cannot pick points first and then select some big $R$ to cover them all.
– Oldboy
Jul 28 at 7:14





@KarnWatcharasupat I think the idea is to have a predefined radius $R$. You cannot pick points first and then select some big $R$ to cover them all.
– Oldboy
Jul 28 at 7:14













oooh I see, my bad
– Karn Watcharasupat
Jul 28 at 7:21




oooh I see, my bad
– Karn Watcharasupat
Jul 28 at 7:21












@KarnWatcharasupat I have edited the text, it should be more clear now.
– Oldboy
Jul 28 at 7:21




@KarnWatcharasupat I have edited the text, it should be more clear now.
– Oldboy
Jul 28 at 7:21












@Oldboy Also, Is this proposition true in, for example, Banach spaces? (Maybe this could be added to the question.)
– Ma Joad
Jul 28 at 8:20




@Oldboy Also, Is this proposition true in, for example, Banach spaces? (Maybe this could be added to the question.)
– Ma Joad
Jul 28 at 8:20










1 Answer
1






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1
down vote



accepted










Consider a circle with minimal radius $r$ that contains all the points. At least three points lie on this circle (else we could shrink it), and at least three points that lie on this circle are not on the same semicircle (else we could shrink it). These three points are not contained in any circle of radius less than $r$, but by the premise they're contained in a circle of radius $R$. Hence $rle R$.






share|cite|improve this answer



















  • 1




    I have a problem with: "at least three points that lie on this circle are not in the same semicircle (else we could shrink it)"
    – Oldboy
    Jul 28 at 8:05










  • @Oldboy: Sorry, it should have said "on", not "in"; I've fixed that. Does that make it clearer?
    – joriki
    Jul 28 at 8:12










  • I still have a question: why is it possible to shrink the circle further if the three points are on the same semi-circle? It's something obvious but I'm missing that :)
    – Oldboy
    Jul 28 at 8:32






  • 2




    @Oldboy Because if all points on the circle are on the same semicircle, we can first move the circle a little so that the points are inside the circle, then we can shrink the circle. If not all the points are on the same semicircle, then there is a triple of points not on the same semicircle (which is a neat little geometry / combinatorics problem in itself). (Note, in this case we consider diametrically opposite points to not be on the same semicircle.)
    – Arthur
    Jul 28 at 8:51











  • @Arthur Thank you for explaining this in detail. My son just told me that there was a completely different proof using Helly's theorem but I think that Joriki's solution is a very simple and ellegant one, requiring no special, hard-to-prove theorems!
    – Oldboy
    Jul 28 at 10:22











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
1
down vote



accepted










Consider a circle with minimal radius $r$ that contains all the points. At least three points lie on this circle (else we could shrink it), and at least three points that lie on this circle are not on the same semicircle (else we could shrink it). These three points are not contained in any circle of radius less than $r$, but by the premise they're contained in a circle of radius $R$. Hence $rle R$.






share|cite|improve this answer



















  • 1




    I have a problem with: "at least three points that lie on this circle are not in the same semicircle (else we could shrink it)"
    – Oldboy
    Jul 28 at 8:05










  • @Oldboy: Sorry, it should have said "on", not "in"; I've fixed that. Does that make it clearer?
    – joriki
    Jul 28 at 8:12










  • I still have a question: why is it possible to shrink the circle further if the three points are on the same semi-circle? It's something obvious but I'm missing that :)
    – Oldboy
    Jul 28 at 8:32






  • 2




    @Oldboy Because if all points on the circle are on the same semicircle, we can first move the circle a little so that the points are inside the circle, then we can shrink the circle. If not all the points are on the same semicircle, then there is a triple of points not on the same semicircle (which is a neat little geometry / combinatorics problem in itself). (Note, in this case we consider diametrically opposite points to not be on the same semicircle.)
    – Arthur
    Jul 28 at 8:51











  • @Arthur Thank you for explaining this in detail. My son just told me that there was a completely different proof using Helly's theorem but I think that Joriki's solution is a very simple and ellegant one, requiring no special, hard-to-prove theorems!
    – Oldboy
    Jul 28 at 10:22















up vote
1
down vote



accepted










Consider a circle with minimal radius $r$ that contains all the points. At least three points lie on this circle (else we could shrink it), and at least three points that lie on this circle are not on the same semicircle (else we could shrink it). These three points are not contained in any circle of radius less than $r$, but by the premise they're contained in a circle of radius $R$. Hence $rle R$.






share|cite|improve this answer



















  • 1




    I have a problem with: "at least three points that lie on this circle are not in the same semicircle (else we could shrink it)"
    – Oldboy
    Jul 28 at 8:05










  • @Oldboy: Sorry, it should have said "on", not "in"; I've fixed that. Does that make it clearer?
    – joriki
    Jul 28 at 8:12










  • I still have a question: why is it possible to shrink the circle further if the three points are on the same semi-circle? It's something obvious but I'm missing that :)
    – Oldboy
    Jul 28 at 8:32






  • 2




    @Oldboy Because if all points on the circle are on the same semicircle, we can first move the circle a little so that the points are inside the circle, then we can shrink the circle. If not all the points are on the same semicircle, then there is a triple of points not on the same semicircle (which is a neat little geometry / combinatorics problem in itself). (Note, in this case we consider diametrically opposite points to not be on the same semicircle.)
    – Arthur
    Jul 28 at 8:51











  • @Arthur Thank you for explaining this in detail. My son just told me that there was a completely different proof using Helly's theorem but I think that Joriki's solution is a very simple and ellegant one, requiring no special, hard-to-prove theorems!
    – Oldboy
    Jul 28 at 10:22













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Consider a circle with minimal radius $r$ that contains all the points. At least three points lie on this circle (else we could shrink it), and at least three points that lie on this circle are not on the same semicircle (else we could shrink it). These three points are not contained in any circle of radius less than $r$, but by the premise they're contained in a circle of radius $R$. Hence $rle R$.






share|cite|improve this answer















Consider a circle with minimal radius $r$ that contains all the points. At least three points lie on this circle (else we could shrink it), and at least three points that lie on this circle are not on the same semicircle (else we could shrink it). These three points are not contained in any circle of radius less than $r$, but by the premise they're contained in a circle of radius $R$. Hence $rle R$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 28 at 8:12


























answered Jul 28 at 7:47









joriki

164k10179328




164k10179328







  • 1




    I have a problem with: "at least three points that lie on this circle are not in the same semicircle (else we could shrink it)"
    – Oldboy
    Jul 28 at 8:05










  • @Oldboy: Sorry, it should have said "on", not "in"; I've fixed that. Does that make it clearer?
    – joriki
    Jul 28 at 8:12










  • I still have a question: why is it possible to shrink the circle further if the three points are on the same semi-circle? It's something obvious but I'm missing that :)
    – Oldboy
    Jul 28 at 8:32






  • 2




    @Oldboy Because if all points on the circle are on the same semicircle, we can first move the circle a little so that the points are inside the circle, then we can shrink the circle. If not all the points are on the same semicircle, then there is a triple of points not on the same semicircle (which is a neat little geometry / combinatorics problem in itself). (Note, in this case we consider diametrically opposite points to not be on the same semicircle.)
    – Arthur
    Jul 28 at 8:51











  • @Arthur Thank you for explaining this in detail. My son just told me that there was a completely different proof using Helly's theorem but I think that Joriki's solution is a very simple and ellegant one, requiring no special, hard-to-prove theorems!
    – Oldboy
    Jul 28 at 10:22













  • 1




    I have a problem with: "at least three points that lie on this circle are not in the same semicircle (else we could shrink it)"
    – Oldboy
    Jul 28 at 8:05










  • @Oldboy: Sorry, it should have said "on", not "in"; I've fixed that. Does that make it clearer?
    – joriki
    Jul 28 at 8:12










  • I still have a question: why is it possible to shrink the circle further if the three points are on the same semi-circle? It's something obvious but I'm missing that :)
    – Oldboy
    Jul 28 at 8:32






  • 2




    @Oldboy Because if all points on the circle are on the same semicircle, we can first move the circle a little so that the points are inside the circle, then we can shrink the circle. If not all the points are on the same semicircle, then there is a triple of points not on the same semicircle (which is a neat little geometry / combinatorics problem in itself). (Note, in this case we consider diametrically opposite points to not be on the same semicircle.)
    – Arthur
    Jul 28 at 8:51











  • @Arthur Thank you for explaining this in detail. My son just told me that there was a completely different proof using Helly's theorem but I think that Joriki's solution is a very simple and ellegant one, requiring no special, hard-to-prove theorems!
    – Oldboy
    Jul 28 at 10:22








1




1




I have a problem with: "at least three points that lie on this circle are not in the same semicircle (else we could shrink it)"
– Oldboy
Jul 28 at 8:05




I have a problem with: "at least three points that lie on this circle are not in the same semicircle (else we could shrink it)"
– Oldboy
Jul 28 at 8:05












@Oldboy: Sorry, it should have said "on", not "in"; I've fixed that. Does that make it clearer?
– joriki
Jul 28 at 8:12




@Oldboy: Sorry, it should have said "on", not "in"; I've fixed that. Does that make it clearer?
– joriki
Jul 28 at 8:12












I still have a question: why is it possible to shrink the circle further if the three points are on the same semi-circle? It's something obvious but I'm missing that :)
– Oldboy
Jul 28 at 8:32




I still have a question: why is it possible to shrink the circle further if the three points are on the same semi-circle? It's something obvious but I'm missing that :)
– Oldboy
Jul 28 at 8:32




2




2




@Oldboy Because if all points on the circle are on the same semicircle, we can first move the circle a little so that the points are inside the circle, then we can shrink the circle. If not all the points are on the same semicircle, then there is a triple of points not on the same semicircle (which is a neat little geometry / combinatorics problem in itself). (Note, in this case we consider diametrically opposite points to not be on the same semicircle.)
– Arthur
Jul 28 at 8:51





@Oldboy Because if all points on the circle are on the same semicircle, we can first move the circle a little so that the points are inside the circle, then we can shrink the circle. If not all the points are on the same semicircle, then there is a triple of points not on the same semicircle (which is a neat little geometry / combinatorics problem in itself). (Note, in this case we consider diametrically opposite points to not be on the same semicircle.)
– Arthur
Jul 28 at 8:51













@Arthur Thank you for explaining this in detail. My son just told me that there was a completely different proof using Helly's theorem but I think that Joriki's solution is a very simple and ellegant one, requiring no special, hard-to-prove theorems!
– Oldboy
Jul 28 at 10:22





@Arthur Thank you for explaining this in detail. My son just told me that there was a completely different proof using Helly's theorem but I think that Joriki's solution is a very simple and ellegant one, requiring no special, hard-to-prove theorems!
– Oldboy
Jul 28 at 10:22













 

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