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.

Figure 1. Tangled-up hexagon ABCDEF is inscribed in a circle. Its sides are extended so that pairs of opposite sides intersect on Pascal's line. Each pair of extended opposite sides has its own color: one red, one yellow, one blue. Pascal's line is shown in white.

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.

[edit] Proof of Pascal's theorem

The 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:

P₁:[cosθ₁:sinθ₁:1]

P₂:[cosθ₂:sinθ₂:1]

P₃:[cosθ₃:sinθ₃:1]

P₄:[cosθ₄:sinθ₄:1]

P₅:[cosθ₅:sinθ₅:1]

P₆:[cosθ₆:sinθ₆:1].

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 Γ = 0.

[edit] First step

Apply the following identity of vector calculus:

to produce

[edit] Third step

Lemma One. If P_i, P_j, P_k are points on a unit circle in a projective plane and are expressed in homogeneous coördinates like so:

P_i:[cosθ_i:sinθ_i:1]

P_j:[cosθ_j:sinθ_j:1]

P_k:[cosθ_k:sinθ_k:1],

then

This lemma will be proved below, later. Meanwhile, applying it to the target, and letting

for {i,j} ⊂ {1,2,3,4,5,6}, the target becomes

[edit] Fourth step

The target's sum has four terms, each one a product of twelve S_ij′s out of 15 possible ones.

For each S_ij, if i > j then replace it with its equivalent −S_ji. Then, for any pair of adjacent S_ij S_kl in each product, commute them if i > k or if i=k but j > l. The result is

The S_ij factors which are raised to the zeroth power denote factors which are actually missing.

[edit] Fifth step

Let

T = S₁₂S₁₃S₁₄S₁₅S₁₆S₂₃S₂₄S₂₅S₂₆S₃₄S₃₅S₃₆S₄₅S₄₆S₅₆.

Then the target becomes

[edit] Seventh step

Lemma Two:

Using S_ij notation, Lemma Two becomes

S_abS_cd − S_acS_bd = S_adS_cb,

which when applied to the target yields

Replace S₅₃ with −S₃₅, resulting in

quod erat demonstrandum.

[edit] Proof of Lemma One

= cosθ_isinθ_j + sinθ_icosθ_k + cosθ_jsinθ_k

− cosθ_isinθ_k − sinθ_icosθ_j − sinθ_jcosθ_k

= (cosθ_isinθ_j − sinθ_icosθ_j)

+ (sinθ_icosθ_k − cosθ_isinθ_k)

+ (cosθ_jsinθ_k − sinθ_jcosθ_k).

Applying the trigonometric identity

cosasinb − sinacosb = sin(b − a)

results in

Lemma Three:

sin(θ_a − θ_b) + sin(θ_b − θ_c) + sin(θ_c − θ_a)

Applying Lemma Three yields

quod erat demonstrandum.

[edit] Proof of Lemma Two

Since

and letting

then

S_abS_cd = (S_aC_b − C_aS_b)(S_cC_d − C_cS_d)

= S_aS_cC_bC_d − S_aS_dC_bC_c

− S_bS_cC_aC_d + S_bS_dC_aC_c,

S_acS_bd = (S_aC_c − C_aS_c)(S_bC_d − C_bS_d)

= S_aS_bC_cC_d − S_aS_dC_bC_c

− S_bS_cC_aC_d + S_cS_dC_aC_b,

so that

S_abS_cd − S_acS_bd

= S_aS_cC_bC_d − S_aS_bC_cC_d

− S_cS_dC_aC_b + S_bS_dC_aC_c

= (S_aC_d − C_aS_d)(S_cC_b − C_cS_b)

= S_adS_cb,

quod erat demonstrandum.

[edit] Proof of Lemma Three

sin(A − B) + sin(B − C) + sin(C − A)

= sinAcosB − cosAsinB + sinBcosC − cosBsinC + sin(C − A)

= cosB(sinA − sinC) + sinB(cosC − cosA) + sin(C − A)

+ sin(C − A)

+ sin(C − A)

quod erat demonstrandum.

[edit] External links

• Interactive demo of Pascal's theorem (Java required) at cut-the-knot
• 60 Pascal Lines (Java required) at cut-the-knot
• Another one
• Pascal's Mystic Hexagram Theorem Proof applying Menelaus’ Theorem