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Pascal's theorem

Pascal's 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.

Contents

1 Proof of Pascal's theorem

1.1 First step
1.2 Second step
1.3 Third step
1.4 Fourth step
1.5 Fifth step
1.6 Sixth step
1.7 Seventh step

2 Proof of Lemma One

3 Proof of Lemma Two

4 Proof of Lemma Three

5 External links

Proof of Pascal's theorem

The following proof will actually be just for a 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 x- and y-axis.

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 coördinates:

P 1 :[cosθ 1 :sinθ 1 :1]

P 2 :[cosθ 2 :sinθ 2 :1]

P 3 :[cosθ 3 :sinθ 3 :1]

P 4 :[cosθ 4 :sinθ 4 :1]

P 5 :[cosθ 5 :sinθ 5 :1]

P 6 :[cosθ 6 :sinθ 6 :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:

$((P_1 \times P_2)\times (P_4 \times P_5))\cdot ((P_2 \times P_3)\times (P_5 \times P_6))\times ((P_3 \times P_4)\times (P_6 \times P_1)) = 0,$

or using the notation <,,> for the scalar triple product:

$\langle (P_1 \times P_2)\times (P_4 \times P_5), (P_2 \times P_3)\times (P_5 \times P_6), (P_3 \times P_4) \times (P_6 \times P_1) \rangle = 0.$

Let

$\Gamma = \langle (P_1 \times P_2)\times (P_4 \times P_5), (P_2 \times P_3)\times (P_5 \times P_6), (P_3 \times P_4) \times (P_6 \times P_1) \rangle.$

Then the objective is to show that Γ = 0.

First step

Apply the following identity of vector calculus:

$(A \times B)\times (C \times D) = C \langle A,B,D\rangle - D \langle A,B,C\rangle$

to produce

$\Gamma = \langle \langle P_1, P_2, P_5\rangle P_4 - \langle P_1, P_2, P_4\rangle P_5,$

$\langle P_2, P_3, P_6\rangle P_5 - \langle P_2, P_3, P_5\rangle P_6,$

$\langle P_3, P_4, P_1\rangle P_6 - \langle P_3, P_4, P_6\rangle P_1\rangle.$

Second step

The scalar triple product satisfies the following linearity relations:

$\langle A+B,C,D\rangle = \langle A,C,D\rangle + \langle B,C,D\rangle$

$\langle A,B+C,D\rangle = \langle A,B,D\rangle + \langle A,C,D\rangle$

$\langle A,B,C+D\rangle = \langle A,B,C\rangle + \langle A,B,D\rangle$

$\langle a A, b B, c C\rangle = a b c \langle A,B,C\rangle$ .

Applying the linearity relations, and cancelling out some of the terms through the following "nullification identities":

$\langle A,A,B\rangle = \langle C,D,D\rangle = \langle E,F,E\rangle = 0,$

the target becomes

$\Gamma = \langle P_1,P_2,P_5\rangle \langle P_2,P_3,P_6\rangle \langle P_3,P_4,P_1\rangle \langle P_4,P_5,P_6\rangle$

$- \langle P_1,P_2,P_5\rangle \langle P_2,P_3,P_6\rangle \langle P_3,P_4,P_6\rangle \langle P_4,P_5,P_1\rangle$

$+ \langle P_1,P_2,P_5\rangle \langle P_2,P_3,P_5\rangle \langle P_3,P_4,P_6\rangle \langle P_4,P_6,P_1\rangle$

$- \langle P_1,P_2,P_4\rangle \langle P_2,P_3,P_5\rangle \langle P_3,P_4,P_6\rangle \langle P_5,P_6,P_1\rangle.$

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

$\langle P_i,P_j,P_k\rangle = 4 \sin \left( {\theta_i - \theta_j \over 2}\right) \sin \left( {\theta_j - \theta_k \over 2}\right) \sin \left( {\theta_k - \theta_i \over 2}\right).$

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

$S_{ij} = \sin \left( {\theta_i - \theta_j \over 2}\right)$

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

$\Gamma = 256 \, S_{12} S_{25} S_{51} S_{23} S_{36} S_{62} S_{34} S_{41} S_{13} S_{45} S_{56} S_{64}$

$-256 \, S_{12} S_{25} S_{51} S_{23} S_{36} S_{62} S_{34} S_{46} S_{63} S_{45} S_{51} S_{14}$

$+256 \, S_{12} S_{25} S_{51} S_{23} S_{35} S_{52} S_{34} S_{46} S_{63} S_{46} S_{61} S_{14}$

$-256 \, S_{12} S_{24} S_{41} S_{23} S_{35} S_{52} S_{34} S_{46} S_{63} S_{56} S_{61} S_{15}.$

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

$\Gamma = 256 (S_{12} S_{13} S_{14} S_{15} S_{16}^0 S_{23} S_{24}^0 S_{25} S_{26} S_{34} S_{35}^0 S_{36} S_{45} S_{46} S_{56}$

$- S_{12} S_{13}^0 S_{14} S_{15}^2 S_{16}^0 S_{23} S_{24}^0 S_{25} S_{26} S_{34} S_{35}^0 S_{36}^2 S_{45} S_{46} S_{56}^0$

$+ S_{12} S_{13}^0 S_{14} S_{15} S_{16} S_{23} S_{24}^0 S_{25}^2 S_{26}^0 S_{34} S_{35} S_{36} S_{45}^0 S_{46}^2 S_{56}^0$

$- S_{12} S_{13}^0 S_{14} S_{15} S_{16} S_{23} S_{24} S_{25} S_{26}^0 S_{34} S_{35} S_{36} S_{45}^0 S_{46} S_{56}).$

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

Fifth step

Let

T = S 12 S 13 S 14 S 15 S 16 S 23 S 24 S 25 S 26 S 34 S 35 S 36 S 45 S 46 S 56 .

Then the target becomes

$\Gamma = 256 \left( {T \over S_{16} S_{24} S_{35}} - {T S_{15} S_{36} \over S_{13} S_{16} S_{24} S_{35} S_{56}} + {T S_{25} S_{46} \over S_{13} S_{24} S_{26} S_{45} S_{56}} - {T \over S_{13} S_{26} S_{45}} \right).$

Sixth step

Let

D = S 13 S 16 S 24 S 26 S 35 S 45 S 56

be the common denominator of all four fractions of the target in the fifth step. Then the target becomes

$\Gamma = {256\, T \over D} (S_{13} S_{26} S_{45} S_{56} - S_{15} S_{36} S_{26} S_{45} + S_{25} S_{46} S_{16} S_{35} - S_{16} S_{24} S_{35} S_{56}).$

Moreover, two pairs of summands have a pair of common S_ij factors, which when factored out yield

$\Gamma = {256\, T \over D} (S_{26} S_{45} (S_{13} S_{56} - S_{15} S_{36}) + S_{16} S_{35} (S_{25} S_{46} - S_{24} S_{56})).$

Seventh step

Lemma Two:

$\sin \left( {\theta_a - \theta_b \over 2}\right) \sin \left( {\theta_c - \theta_d \over 2} \right) - \sin \left( {\theta_a - \theta_c \over 2}\right) \sin \left( {\theta_b - \theta_d \over 2}\right)$

$= \sin \left( {\theta_a - \theta_d \over 2}\right) - \sin \left( {\theta_c - \theta_b \over 2}\right).$

Using S_ij notation, Lemma Two becomes

S a b S c d - S a c S b d = S a d S c b,

which when applied to the target yields

$\Gamma = {256 \, T \over D} (S_{26} S_{45} S_{16} S_{53} + S_{16} S_{35} S_{26} S_{45}).$

Replace S₅₃ with −S₃₅, resulting in

$\Gamma = {256\, T\over D} (-S_{16} S_{26} S_{35} S_{45} + S_{16} S_{26} S_{35} S_{45})$

$= {256\, T \over D} (0) = 0,$

quod erat demonstrandum. ö

Proof of Lemma One

$\langle P_i,P_j,P_k\rangle = \left| \begin{matrix} \cos \theta_i & \sin \theta_i & 1 \\ \cos \theta_j & \sin \theta_j & 1 \\ \cos \theta_k & \sin \theta_k & 1 \end{matrix} \right|$

= cosθ i sinθ j + sinθ i cosθ k + cosθ j sinθ k

- cosθ i sinθ k - sinθ i cosθ j - sinθ j cosθ k

= (cosθ i sinθ j - sinθ i cosθ j)

+ (sinθ i cosθ k - cosθ i sinθ k)

+ (cosθ j sinθ k - sinθ j cosθ k).

Applying the trigonometric identity

cos a sin b - sin a cos b = sin(b - a)

results in

$\langle P_i,P_j,P_k\rangle = \sin (\theta_j - \theta_i) + \sin (\theta_i - \theta_k) + \sin (\theta_k - \theta_j).$

Lemma Three:

sin(θ a - θ b) + sin(θ b - θ c) + sin(θ c - θ a)

$= 4 \sin \left( {\theta_a - \theta_b \over 2} \right) \sin \left( {\theta_b - \theta_c \over 2} \right) \sin \left( {\theta_a - \theta_c \over 2} \right).$

Applying Lemma Three yields

$\langle P_i,P_j,P_k\rangle = 4 \sin \left( {\theta_j - \theta_i \over 2} \right) \sin \left( {\theta_i - \theta_k \over 2}\right) \sin \left( {\theta_j - \theta_k \over 2}\right)$

$= 4 \sin \left( {\theta_i - \theta_j \over 2} \right) \sin \left( {\theta_j - \theta_k \over 2}\right) \sin \left( {\theta_k - \theta_i \over 2}\right),$

quod erat demonstrandum.

Proof of Lemma Two

Since

sin(a - b) = sin a cos b - cos a sin b,

and letting

$S_i = \sin \left( {\theta_i \over 2} \right), \qquad C_i = \cos \left( {\theta_i \over 2} \right),$

then

S a b S c d = (S a C b - C a S b)(S c C d - C c S d)

= S a S c C b C d - S a S d C b C c

- S b S c C a C d + S b S d C a C c,

S a c S b d = (S a C c - C a S c)(S b C d - C b S d)

= S a S b C c C d - S a S d C b C c

- S b S c C a C d + S c S d C a C b,

so that

S a b S c d - S a c S b d

= S a S c C b C d - S a S b C c C d

- S c S d C a C b + S b S d C a C c

= (S a C d - C a S d)(S c C b - C c S b)

= S a d S c b,

quod erat demonstrandum.

Proof of Lemma Three

$sin(A - B) + sin(B - C) + sin(C - A)$

$= sin A cos B - cos A sin B + sin B cos C - cos B sin C + sin(C - A)$

$= cos B (sin A - sin C) + sin B (cos C - cos A) + sin(C - A)$

$= 2 \cos B \sin \left( {A - C \over 2} \right) \cos \left( {A + C \over 2} \right) + 2 \sin B \sin \left( {A + C \over 2} \right) \sin \left( {A - C \over 2} \right)$

+ sin(C - A)

$= 2 \sin \left( {A - C \over 2} \right) \Bigg( \cos B \cos \left( {A + C \over 2}\right) + \sin B \sin \left( {A + C \over 2}\right) \Bigg)$

+ sin(C - A)

$= 2 \sin \left( {A - C \over 2}\right) \cos \left({A + C \over 2} - B\right) + \sin (C - A)$

$= 2 \sin \left( {A - C \over 2}\right) \cos \left( {A + C \over 2} - B\right) + 2\sin \left({C - A \over 2}\right) \cos \left({C - A \over 2}\right)$

$= 2 \sin \left({A - C\over 2}\right) \Bigg( \cos \left({A + C \over 2} - B\right) - \cos \left({C - A\over 2}\right) \Bigg)$

$= 2 \sin \left({A - C \over 2}\right) 2 \sin \left( {{C - A \over 2} + {A + C \over 2} - B \over 2} \right) \sin \left({{C - A \over 2} - {A + C \over 2} + B\over 2}\right)$

$= 4 \sin \left({A - C \over 2}\right) \sin \left({C - B \over 2}\right) \sin \left({B - A \over 2}\right)$

$= 4 \sin \left({A - B \over 2}\right) \sin \left({B - C \over 2}\right) \sin \left({A - C \over 2}\right),$
quod erat demonstrandum.

External links

Interactive demo of Pascal's theorem (requires a Java-enabled browser)
60 Pascal Lines (requires a Java-enabled browser)
Another one

Categories: Projective geometry | Theorems

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01-04-2007 01:21:04

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