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
Best Answer
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.