Points on a plane

A few puzzles about sets of points, and a nice conjecture following up one of those questions:

Suppose \( S \) is a finite set of points on a plane, such that for any two points \( A,B \in S \), there is a third point \( D \in S \), collinear with \( A \) and \( B \) (and different from \( A \) and \( B \)). Show that all points of \( \boldsymbol{S} \) are collinear.

The nicest solution I know of this problem comes by reductio ad absurdum, and goes like this:

Suppose that not all points in \( S \) are collinear. Among all possible triangles that we can form using points from this set as vertices, chose the one with the smallest height, say \( \delta \); label the vertices at the base of the triangle \( A \) and \( B \), and the remaining vertex \( C. \) It will not be hard for the reader to realize that this implies, in particular, that the projection of \( C \) on the line from \( A \) and \( B \) is either one of the previous two vertices, or it is located inside of the segment that joins them.

By hypothesis, there is a third point \( D \) which is collinear with \( A \) and \( B \). Consider now the new triangles that arose: \( \triangle ADC, \triangle BDC. \) The trick now is to realize that the smallest of the heights of one of these two triangles, say \( \delta’, \) is necessarily smaller than \( \delta \). This is a contradiction!

The second puzzle was sent by Ralph Howard:

Let \( S \) be a infinite set of points on a plane, such that the distance between any two is an integer. Prove that all points in \( \boldsymbol{S} \) are collinear.

Ralph pointed up to a clever solution to this riddle by Paul Erdös. He presented it in a short article, titled Integral distances, published in Bull. Amer. Math. Soc. 51, (1945). 996. It goes as follows:

Suppose you have three points \( A \), \( B \) and \( C \) with integer distances between them and not all on the same line. Let us denote \( \mathop{d}(A,B) \) the distance between points \( A \) and \( B \). If \( \mathop{d}(A,Q) \) and \( \mathop{d}(B,Q) \) are both integers, note that \( \mathop{d}(A,Q) - \mathop{d}(B,Q) \) is one of the integers in the closed interval \( [-\mathop{d}(A,B), \mathop{d}(A,B)]. \) Now for any given integer \( k \), the points \( Q \) satisfying \( \mathop{d}(A,Q) - \mathop{d}(B,Q) = k \) all lie on a branch of a hyperbola (or the degenerate cases: a straight line parallel or perpendicular to the line through \( A \) and \( B \)). Every point of the set \( S \) is an intersection of one of these curves, and one of the analogous curves for \( A \) and \( C \), and one of the curves for \( B \) and \( C \). But any two of the curves intersect in only a finite number of points. Therefore there are only a finite number of points with integer distances from \( A \), \( B \) and \( C. \) This is a contradiction!

Ralph conjectures too that the same result holds if the points lie in any other higher-dimension space:

I conjecture that the same is true in three and higher dimensions. That is: if \( S \) is an infinite set of points in \( \mathbb{R}^d \) such that the distance between any two is an integer, then the points of \( S \) are colinear.

How would you prove this conjecture?