1

This is a question from the IMO shortlist which I found here.

Given $n > 3$ points in the plane, no 3 collinear, show that there is a circle through 3 of the points such that none of the points lies inside the circle.

So I found this very similar question on stackexchange but it's the exact opposite. That question asks for a circle which contains all the points.

So what I have tried till now is to start with 3 points and draw a circle through them. After that I add one more point inside the circle. I make a new circle with this new point and the two closest points.
If a point lies inside the new circle again I repeat the process until we reach n points and the desired circle.

But the problem is: Sometimes if one of the previous points comes inside the new circle then if we use that point to make a new circle and another point comes inside that circle again then points may keep coming inside the circle infinitely.

So I think I can omit this strategy. So I'm gonna start from three points, make a circle and keep adding points like the previous strategy. And after adding the new point inside the circle I need to make a new circle passing through it. But I need to choose two other points to make the circle with.
So I'm looking for a strategy to decide which other 2 points to choose.

If I can choose the right two points to make the new circle with in each step of this process of adding points, then eventually I will get the desired circle. But this strategy is what I'm looking for.

  • Try new point ("N"), closest point ("A") and "whichever of the other two makes the smaller circle", which I suspect is "whichever has a bisector with A closest to N". – Joffan Mar 19 '21 at 08:28
  • Making the smallest possible circle seems like a good idea. I'll try it –  Mar 19 '21 at 09:32
  • @Joffan could you explain the "whichever has a bisector with A closest to N" part? I didn't get how you got that. –  Mar 19 '21 at 09:40
  • I was thinking about how to find the centre of the circle, using intersection of the bisectors of the lines joining the points, but it's just a guess. In fact there;s only three circles to try so I'd say forget about the "closest to N" being an automatic choice either. – Joffan Mar 19 '21 at 12:23

1 Answers1

5

Consider the closest two points in the set, $M$ and $N$. Obviously if we draw the circle that has those two points on a diameter, there will be no other points inside it (and no other points on its circumference wither, so this is not the circle we are looking for).

Now imagine shifting the centre of the circle away from the midpoint of $MN$, out along the bisector of $MN$, increasing in radius to keep $M$ and $N$ on the circumference. Assuming there are some points on this side of $MN$ (otherwise we move the circle the other way), we will eventually make the circle big enough to touch another point. This enlarged circle then fulfills the condition.

Joffan
  • 39,627