A problem with concyclic points on $\mathbb{R}^2$

I am thinking about the following problem:

If a collection $\{P_1,P_2,\ldots,P_n\}$ of $n$ points are given on the $\mathbb{R^2}$ plane, has the property that for every $3$ points $P_i,P_j,P_k$ in the collection there is a fourth point $P_l$ in the collection such that $P_l$ is con-cyclic with $P_i,P_j,P_k$, (i.e. $P_l$ lies on the circle passing through the points $P_i,P_j,P_k$), does it follow that all the points are necessarily con-cyclic ?

I would really appreciate if someone finds a proof with basic Euclidean Geometry.

I would call a class of Convex Geometric figure (upto Homothety) on $\mathbb{R}^2$, $k$-determined if exactly $k$ points are required to determine the figure uniquely. For example a circle is $3$-determined, one needs exactly $3$ points on the plane to determine a circle uniquely. An ellipse is $4$-determined.

From here I would like to ask the following question : If a collection $S$ of $n$ points on $\mathbb{R}^2$, has the property that every sub-collection $T_i=\{P_{i_1},\ldots,P_{i_k}\}$ of $k$ points of $S$ has the property that there is a $k+1^{th}$ point, $P_i \in S\setminus T_i$ (distinct from the sub-collection $T_i$) that lies on the $k$-determined convex figure, determined by $T_i$, then does it follow that all points of $S$ lie on the $k$-determined convex figure?

Inspired from The Sylvester-Gallai Theorem

Solution 1:

The conjecture is true, and can be recovered from the standard Sylverster-Gallai theorem in the plane.

We first prove:

Lemma 1 Suppose $C$ is a finite configuration of points in real 3-space ${\bf R}^3$ such that for every $Q_1,Q_2,Q_3 \in C$ there is a plane meeting $C$ in $Q_1,Q_2,Q_3$, and at least one more point of $C$. Then $C$ is contained in a plane.

(Of course the plane is unique unless $Q_1,Q_2,Q_3$ are collinear.)

Proof of Lemma 1: fix $Q_1 \in C$ and let $\Pi_1$ be a plane containing $Q_1$ but no other point of $C$. Let $\Pi \neq \Pi_1$ be any plane parallel to $\Pi$, and let $C'$ be the set consisting of the projections from $Q_1$ to $\Pi$ of all $Q \in C$ other than $Q_1$ (that is, the intersections with $\Pi$ of the lines $\overline{Q_1 Q}$). Applying the hypothesis only to triples that contain $Q_1$ shows that $C'$ satisfies the hypothesis of the Sylvester-Gallai theorem. Hence $C'$ is contained in a line, whence $C$ is contained in the plane spanned by this line and $Q_1$. $\ \diamondsuit$

We connect this with the problem at hand using the following observation:

Lemma 2 Points $(x_i,y_i)$ in the plane are concyclic or collinear iff the corresponding points $(x_i, y_i, x_i^2 + y_i^2)$ on the round paraboloid $z = x^2 + y^2$ are coplanar.

Proof of Lemma 2: the intersection of $z = x^2 + y^2$ with any plane $A_0 + A_1 x + A_2 y + A_3 z = 0$ projects to the locus of $A_0 + A_1 x + A_2 y + A_3 (x^2+y^2) = 0$, which is line if $A_3 = 0$ and a circle otherwise. $\ \diamondsuit$

Now assume that $S \subset {\bf R}^2$ is a finite set of points such that for every $P_1,P_2,P_3 \in S$ there is a circle meeting $S$ in $P_1,P_2,P_3$, and at least one more point of $S$. Then by Lemma 2 the associated configuration $$ C_S := \{ (x,y,x^2+y^2) \in {\bf R}^3 \mid (x,y) \in S \} $$ satisfies the hypotheses of Lemma 1. Hence $C_S$ is contained in a plane. Applying Lemma 2 in reverse, we conclude that $S$ is contained in a circle or line. The line does not satisfy the hypothesis, so $S$ is concyclic, as desired. QED

The same linearization trick applies to other such problems. Lemma 1 generalizes to ${\bf R}^d$ for any $d>2$: a finite configuration $C$ such that every $d$ points are on a hyperplane through a $(d+1)$st point must be contained in a hyperplane; this is proved by projection to a plane from any $d-2$ points of $C$ in general linear position. So, for instance, if $S \subset {\bf R}^2$ has the property that every five points are on a conic that contains a sixth point of $S$ then the associated configuration $$ \{ (x,y,x^2,xy,y^2) \in {\bf R}^5 \mid (x,y) \in S \} $$ (which lies on the affine version of the Veronese surface) is contained in a hyperplane, whence $S$ is contained in a conic.

Solution 2:

This follows directly by applying Sylvester-Gallai Theorem and inversion.

Consider any collection of $n+1$ points $\{ P_1, P_2, \ldots P_{n+1}\}$. Fix $P_{n+1}$, and apply inversion (with respect to a unit circle) to the remaining $n$ points to obtain $\{Q_1, Q_2, \ldots Q_{n}\}$. Note that $Q_{n+1}$ is the point at infinity, which lies on every line.

Then, Sylvester-Gallai tells us that for the points $\{Q_1, Q_2, \ldots Q_{n}\}$, either
1) all points are collinear or
2) there is a line with exactly 2 points.

Applying the inversion again, we get that
1) all points are concyclic or
2) there is a circle with exactly 3 points.

Hence, if there is no circle which contains exactly 3 points, then all points are concyclic.

Solution 3:

This is a very partial answer, just for a few cases ($n=4,5,6,7,9$). Consider all the possible circles that arise in the problem, and let $k$ be the largest number of points from the set that are on one circle. In other words, no circle has more than $k$ points from the set.

• Case $k=4$. Say $P_j$ for $j=1,2,3,4$ are on one of the maximal circles (actually, in this case, all circles will have exactly $4$ points), call it $\mathcal{C}$. Let $P_5$ be outside this circle. Then $P_5P_1P_2P_l$ must be on one circle. Moreover, $l$ has to be different than $3$ and $4$, as otherwise $P_5$ would be on $\mathcal{C}$ as well, contradicting the maximality of $k=4$. This solves the problem for $n=5$.

  If we have an additional point $P_6$, the previous considerations show $P_5P_1P_2P_6$ must be on the same circle. Similarly, $P_5P_1P_3P_6$ must be on the same circle. That is again impossible, since the previous two circles have $3$ points in common (so they are actually the same) and $5$ points on the circle, again contradicting the maximality of $k=4$. This solves the problem for $n=6$. (Well, almost so far, but we'll see later that the cases $k> 4$ easily work, or rather don't work, in this case as well.)

  Similar argument works for $n=7$. In a sense, there are two few points outside, and this is the argument for $k>4$. Essentially you must have at least $k$ points not on the $\mathcal{C}$, in order not to contradict the maximality of $k$. For $n=8$ the argument no longer works.

  For $n=9$, and in fact for any odd number, there is a slightly different argument. If $P_a$, $P_b$, and $P_c$ are three points not on the maximal circle $\mathcal {C}$, we cannot have $P_a P_1 P_2 P_b$ and $P_aP_1P_2P_c$ at the same time (contradicts maximality of $k=4$). Therefore, the circle containing $P_1$, $P_2$ and a point not on $\mathcal{C}$, must contain exactly one other point not on $\mathcal{C}$. But this is not possible if there are an odd number of points not on $\mathcal{C}$, since they "match" with $P_1$ and $P_2$ on pairs of two. This does not solve the problem for any odd number of points, since we still are in the restrictive case $k=4$. However, the case $n=9$ will not work for $k>4$ as well.

• Cases $k>4$. By an argument similar to the first part of the case $k=4$, we need at least $5$ points not on $\mathcal{C}$ in order not to contradict the maximality of $k=5$. This solves, as claimed, the cases $n=6$, $n=7$, and $n=9$. However, not $n=8$ since this remained unsolved in the case $k=4$.

That's about it. Note this argument is entirely combinatorial. I couldn't crack $n=8$, all I can say is that IF this case is possible in a non-trivial way (that is, all the points are indeed con-cyclic) it must happen in the case $k=4$. I also tried to build such an example , but I could not. Perhaps there is a geometric argument, rather than combinatorial, why this is not possible. Hope this makes sense.