Concurrent Lines and Collinear Points |

Let $n$ given points have the property that the line joining any two of them passes through a third point of the set. Must the $n$ points all lie on one line? This question was posed by Sylvester in his article “Mathematical question 11851” in 1893. The answer is no in general (complex projective plane), but on the ordinary plane the answer is yes. The paper by Coxeter, titled “A problem of Collinear points,” includes a geometric proof by L. M. Kelly. Additionally, the link provided features four sophisticated solutions, including Kelly’s own approach, and offers an intriguing historical perspective on the problem.

This post is about the corresponding dual problem that can be stated as follows: Consider a finite set $\mathcal{L}$ consisting of lines, none of which are entirely parallel, and arranged such that a third line from $\mathcal{L}$ passes through the intersection point of any two lines. Under these conditions, all the lines in $\mathcal{L}$ must intersect at a single point. The proof is as follows. Let $\ell_1, \ell_2$ be lines in the set $\mathcal{L}$ that intersect at point $C$ and let $\ell_3$ be another line that also passes through $C$ . For the sake of contradiction, let $\ell \in \mathcal{L}$ be a line that does not pass through $C$ . Since $\mathcal{L}$ is finite, the number of such line intersection points $C$ that are not in $\mathcal{L}$ is finite. Without any loss of generality, let’s assume that $C$ is the closest intersection point to $\ell$ . Let $A, D, B$ be the intersection points of the three lines with $\ell$ , and let $D$ be the middle point. By hypothesis another line $\ell'$ needs to go through $D$ (red dotted line) that results in a point $E$ that contradicts the fact that $C$ was the closest such point to $\ell$ . This is illustrated in the figure below.