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Pascal's theorem
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In projective geometry, Pascal's theorem (aka Hexagrammum Mysticum Theorem) states that if an arbitrary hexagon is inscribed in any conic section, and opposite pairs of sides are extended until they meet, the three intersection points will lie on a straight line, the Pascal line of that configuration. In the Euclidean plane, the theorem has exceptions; its natural home is the projective plane.
This theorem is a generalization of Pappus's hexagon theorem, and the projective dual of Brianchon's theorem. It was discovered by Blaise Pascal when he was only 16 years old.
The theorem was generalized by Möbius in 1847, as follows: suppose a polygon with 4n + 2 sides is inscribed in a conic section, and opposite pairs of sides are extended until they meet in 2n + 1 points. Then if 2n of those points lie on a common line, the last point will be on that line, too.
The simplest proof for Pascal's theorem is via Menelaus' theorem.
Proof of Pascal's theoremThe following proof will actually be just for a single unit circle in the projective plane, but a conic section can be turned into a circle by application of a projective transformation, and since projective transformations preserve incidence properties, then the proof of the circular version should imply the truth of the theorem for ellipses, hyperbolas, and parabolas. Ellipses in particular can be turned into circles by a rescaling of the plane along either the major or minor axis, and a circle of any size can be turned into a unit circle by simultaneous and proportional rescaling of both the x- and y-axes.
Let P1, P2, P3, P4, P5, and P6 be a set of six points on a unit circle of a projective plane, with the following homogeneous coordinates:
.
Pascal's theorem then states that the three points which are the intersections of: (1) lines P1P2 and P4P5, (2) lines P2P3 and P5P6, and (3) lines P3P4 and P6P1, are collinear.
Symbolically, this can be stated as:
or using the notation <,,> for the scalar triple product:
Let
Then the objective is to show that G = 0.
First stepApply the following identity of vector calculus:
to produce
Third stepLemma One. If Pi, Pj, Pk are points on a unit circle in a projective plane and are expressed in homogeneous coördinates like so:
then
This lemma will be proved below, later. Meanwhile, applying it to the target, and letting
for ? , the target becomes
Fourth stepThe target's sum has four terms, each one a product of twelve Sij′s out of 15 possible ones.
For each Sij, if i > j then replace it with its equivalent −Sji. Then, for any pair of adjacent Sij Skl in each product, commute them if i > k or if i=k but j > l. The result is
The Sij factors which are raised to the zeroth power denote factors which are actually missing.
Fifth stepLet
Then the target becomes
Seventh stepLemma Two:
Using Sij notation, Lemma Two becomes
which when applied to the target yields
Replace S53 with −S35, resulting in
quod erat demonstrandum.
Proof of Lemma One
Applying the trigonometric identity
results in
Lemma Three:
Applying Lemma Three yields
quod erat demonstrandum.
Proof of Lemma TwoSince
and letting
then
so that
quod erat demonstrandum.
Proof of Lemma Three-
quod erat demonstrandum.
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