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Source: 14-th iranian mathematical olympiad 1996/1997

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#1 h Oct 17, 2005, 3:15 PM • 2 Y

Y by Adventure10, Mango247

Suppose that $10$ points are given in the plane, such that among any five of them there are four lying on a circle. Find the minimum number of these points which must lie on a circle.

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#2 h Nov 3, 2005, 1:10 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

If we replace $10$ with $n\ge 5$ , then I believe the answer is $n-1$ for $n\ge 7$ and $4$ for $n=5,6$ . The case $n=5$ is obvious, and the fact that $4$ works when $n=6$ can be seen by considering the $6$ vertices of two unit squares in the plane which share a side. From now on we assume $n\ge 7$ .

First of all, we notice that if we can find a circle with $5$ points, then that circle must contain $n-1$ points.

Suppose the contrary, i.e. that we can find a circle with at least five points, $A,B,C,D,E$ , and two other points, $X,Y$ , not belonging to that circle. In the group $X,Y,A,B,C$ we must have four concyclic points, and one of $A,B,C$ must not belong to this group, so we may assume WLOG the group is $X,Y,A,B$ . Now look at the group $X,Y,A,C,D$ . The group of $4$ concyclic points that we can extract from must contain $X,Y$ , so it cannot contain $A$ . This means that it's $X,Y,C,D$ . Finally, look at the group $X,Y,A,C,E$ . The group of four concyclic points extracted from here must contain $X,Y$ , but cannot contain either $A$ or $C$ , so we get a contradiction.

What we need to do now is find those $n$ for which there is a configuration satisfying the hypothesis for which every circle passing through at least four points passes through no other points except for those four. We will show that no $n\ge 7$ works. Obviously, in order to do this, it suffices to prove it for $n=7$ .

Let $A,B,C,D$ be four concyclic points in our configuration. Just as above, if $X,Y,Z$ are the other three points, then we can find, say, circles through the quadruples $(X,Y,A,B),(X,Y,C,D)$ and $(X,Z,A,C),(X,Z,B,D)$ . Now perform an inversion of pole $X$ . For a general point $P$ , we denote its image by $P'$ . $A',B',C',D'$ are still concyclic, and $Y'=A'B'\cap C'D',\ Z'=A'C'\cap B'D'$ . The six points $A',B',C',D',Y',Z'$ must still satisfy the hypothesis, but it's clear that no four of the points $Y',Z',A',C',D'$ are concyclic.

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