Sylvester's Line

In 1893 Sylvester asked the mathematical community to prove that given a finite number of points in the plane, either, 1) A line exists that passes thru exactly two of the points or 2) All the points are collinear, that is they lie on the same straight line. His conjecture stumped mathematicians for nearly 50 years until it was correctly solved by Gallai in 1944. A simple and elegant book proof was given by Kelly four years later:

Proof Sketch:

Let $S$ be the given set of points (not all collinear). Consider the collection $C$ of pairs $(L,p)$, where $L$ is a line through at least two distinct points in $S$ and where $p \in S$ is a point not on $L$. Then $C$ is a nonempty finite set of pairs. Among those, pick $(L,p)$ such that the distance from $p$ to $L$ is minimal. We claim that $L$ harbors exactly two points from $S$.

Assume NOT, then $L$ contains 3 or more points. Let $q$ be the projection of $p$ onto $L$. In the figure below, on the horizontal line $L$ we label $q$ and the three points $a,b,c$ in $S$. Two of $a,b,c$ must be on either side of $q$, WLOG as in the figure. Consider $b$ and draw the line $L'$ through $p$ and $c$. In the figure, $L'$ is the slanted line. Then $(L',b)$ belongs to $C$ and the distance from $b$ to $L'$ is strictly less than the distance from $p$ to $L$, a CONTRADICTION.

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