In February 1640 a sheet printed on one side only, like a poster, circulated in Paris under the title Essay on conics. It had no proofs, only definitions, three lemmas and a list of results. The author was 16 years old and his name was Blaise Pascal.

When Father Mersenne showed the sheet to Descartes, the reaction was suspicion: Descartes thought the young man had taken everything from Girard Desargues, the architect and geometer Pascal admired. Desargues’s influence was real, and Pascal himself acknowledged it. But the central result on the sheet was new.

The conics

Cut a cone with a plane. Depending on the tilt, the edge of the cut is an ellipse, a parabola or a hyperbola; the circle is a special case of the ellipse. These curves, the conics, had been studied since Apollonius in the 3rd century BC.

Desargues had an idea that changed the subject: look at conics as shadows of a circle. If you shine a torch on a circle, its shadow on a wall can be an ellipse, a parabola or a hyperbola. So a property of the circle that talks only about points and lines, without measuring distances or angles, survives in the shadow and holds for every conic. This is the idea of projective geometry.

The hexagon theorem

Mark six points $A,B,C,D,E,F$ on a conic, in that order, and join them to form a hexagon. The opposite sides are $AB$ and $DE$, $BC$ and $EF$, $CD$ and $FA$. Extend each pair until they meet:

$$P=AB\cap DE,\qquad Q=BC\cap EF,\qquad R=CD\cap FA.$$

Pascal’s theorem says that the three points $P$, $Q$ and $R$ always lie on one straight line.

The six points can be in any order on the curve, and the hexagon can even cross itself. If two opposite sides are parallel, they meet at a “point at infinity”, and the line is then parallel to them. Pascal called the figure the mystic hexagram.

Checking it on a parabola

Let us test the theorem with calculations. We choose six points on the parabola $y=x^2$, with $x$-coordinates $0,1,2,3,4,7$:

$$A=(0,0),\ B=(1,1),\ C=(2,4),\ D=(3,9),\ E=(4,16),\ F=(7,49).$$

First, a shortcut. The line through the points of the parabola with $x$-coordinates $u$ and $v$ is

$$y=(u+v)x-uv.$$

Indeed, at $x=u$ it gives $u^2+uv-uv=u^2$, and at $x=v$ it gives $v^2$. Now we work out the three meeting points.

Point P. $AB$ has $u=0,v=1$: $y=x$. $DE$ has $u=3,v=4$: $y=7x-12$. Setting them equal, $x=7x-12$, so $x=2$ and $P=(2,2)$.

Point Q. $BC$ is $y=3x-2$ and $EF$ is $y=11x-28$. Setting them equal, $8x=26$, so $x=\frac{13}{4}$ and $Q=\left(\frac{13}{4},\frac{31}{4}\right)$.

Point R. $CD$ is $y=5x-6$ and $FA$ is $y=7x$. Setting them equal, $-2x=6$, so $x=-3$ and $R=(-3,-21)$.

Pascal’s line

The three points lie on the line

$$5y=23x-36.$$

At $P$: $5\cdot2=10$ and $23\cdot2-36=10$. At $Q$: $5\cdot\frac{31}{4}=\frac{155}{4}$ and $23\cdot\frac{13}{4}-36=\frac{299-144}{4}=\frac{155}{4}$. At $R$: $5\cdot(-21)=-105$ and $23\cdot(-3)-36=-105$. All three check out.

Six points chosen almost at random, three intersections full of fractions, and everything falls on the same line. One calculation does not prove the theorem for all conics, but it shows that it is not obvious: there is no visible symmetry forcing that line.

Before and after Pascal

Around the year 300, Pappus of Alexandria had already proved a special case: when the “conic” is a pair of straight lines, with three points on each. Pascal saw that the same holds for any conic. In 1806 the Frenchman Charles Brianchon proved the “reverse” version: in a hexagon whose sides touch a conic, the three main diagonals pass through a single point.

Exercise: find the line through the points (1, 1) and (3, 9) of the parabola

With $u=1$ and $v=3$, the formula gives $y=4x-3$. At $x=1$, $y=1$; at $x=3$, $y=9$. Both points lie on the line.

Exercise: where do the lines AB and CD of the example cross?

$AB$ is $y=x$ and $CD$ is $y=5x-6$. Setting them equal, $x=5x-6$, so $x=\frac32$ and the point is $\left(\frac32,\frac32\right)$.

Sources: Pascal biography, MacTutor · Desargues biography, MacTutor.

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