Definitions

# Pascal's theorem

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 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:

$P_1 : \left[cos theta_1 : sin theta_1 : 1\right]$
$P_2 : \left[cos theta_2 : sin theta_2 : 1\right]$
$P_3 : \left[cos theta_3 : sin theta_3 : 1\right]$
$P_4 : \left[cos theta_4 : sin theta_4 : 1\right]$
$P_5 : \left[cos theta_5 : sin theta_5 : 1\right]$
$P_6 : \left[cos theta_6 : sin theta_6 : 1\right]$.
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:

$\left(\left(P_1 times P_2\right)times \left(P_4 times P_5\right)\right)cdot \left(\left(P_2 times P_3\right)times \left(P_5 times P_6\right)\right)times \left(\left(P_3 times P_4\right)times \left(P_6 times P_1\right)\right) = 0,$
or using the notation 〈,,〉 for the scalar triple product:
$langle \left(P_1 times P_2\right)times \left(P_4 times P_5\right), \left(P_2 times P_3\right)times \left(P_5 times P_6\right), \left(P_3 times P_4\right) times \left(P_6 times P_1\right) rangle = 0.$

Let

$Gamma = langle \left(P_1 times P_2\right)times \left(P_4 times P_5\right), \left(P_2 times P_3\right)times \left(P_5 times P_6\right), \left(P_3 times P_4\right) times \left(P_6 times P_1\right) rangle.$
Then the objective is to show that Γ = 0.

### First step

Apply the following identity of vector calculus:
$\left(A times B\right)times \left(C times D\right) = C langle A,B,Drangle - D langle A,B,Crangle$
to produce
$Gamma = langle langle P_1, P_2, P_5rangle P_4 - langle P_1, P_2, P_4rangle P_5,$
$langle P_2, P_3, P_6rangle P_5 - langle P_2, P_3, P_5rangle P_6,$
$langle P_3, P_4, P_1rangle P_6 - langle P_3, P_4, P_6rangle P_1rangle.$

### Third step

Lemma One. If Pi, Pj, Pk are points on a unit circle in a projective plane and are expressed in homogeneous coördinates like so:
$P_i : \left[cos theta_i : sin theta_i : 1\right]$
$P_j : \left[cos theta_j : sin theta_j : 1\right]$
$P_k : \left[cos theta_k : sin theta_k : 1\right],$
then
$langle P_i,P_j,P_krangle = 4 sin left\left(\left\{theta_i - theta_j over 2\right\}right\right) sin left\left(\left\{theta_j - theta_k over 2\right\}right\right) sin left\left(\left\{theta_k - theta_i over 2\right\}right\right).$

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

$S_\left\{ij\right\} = sin left\left(\left\{theta_i - theta_j over 2\right\}right\right)$
for {i,j} ⊂ {1,2,3,4,5,6}, the target becomes
$Gamma = 256 , S_\left\{12\right\} S_\left\{25\right\} S_\left\{51\right\} S_\left\{23\right\} S_\left\{36\right\} S_\left\{62\right\} S_\left\{34\right\} S_\left\{41\right\} S_\left\{13\right\} S_\left\{45\right\} S_\left\{56\right\} S_\left\{64\right\}$
$\left\{\right\} -256 , S_\left\{12\right\} S_\left\{25\right\} S_\left\{51\right\} S_\left\{23\right\} S_\left\{36\right\} S_\left\{62\right\} S_\left\{34\right\} S_\left\{46\right\} S_\left\{63\right\} S_\left\{45\right\} S_\left\{51\right\} S_\left\{14\right\}$
$\left\{\right\} + 256 , S_\left\{12\right\} S_\left\{25\right\} S_\left\{51\right\} S_\left\{23\right\} S_\left\{35\right\} S_\left\{52\right\} S_\left\{34\right\} S_\left\{46\right\} S_\left\{63\right\} S_\left\{46\right\} S_\left\{61\right\} S_\left\{14\right\}$
$\left\{\right\} - 256 , S_\left\{12\right\} S_\left\{24\right\} S_\left\{41\right\} S_\left\{23\right\} S_\left\{35\right\} S_\left\{52\right\} S_\left\{34\right\} S_\left\{46\right\} S_\left\{63\right\} S_\left\{56\right\} S_\left\{61\right\} S_\left\{15\right\}.$

### Fourth step

The 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

$Gamma = 256 \left(S_\left\{12\right\} S_\left\{13\right\} S_\left\{14\right\} S_\left\{15\right\} S_\left\{16\right\}^0 S_\left\{23\right\} S_\left\{24\right\}^0 S_\left\{25\right\} S_\left\{26\right\} S_\left\{34\right\} S_\left\{35\right\}^0 S_\left\{36\right\} S_\left\{45\right\} S_\left\{46\right\} S_\left\{56\right\}$
$\left\{\right\} - S_\left\{12\right\} S_\left\{13\right\}^0 S_\left\{14\right\} S_\left\{15\right\}^2 S_\left\{16\right\}^0 S_\left\{23\right\} S_\left\{24\right\}^0 S_\left\{25\right\} S_\left\{26\right\} S_\left\{34\right\} S_\left\{35\right\}^0 S_\left\{36\right\}^2 S_\left\{45\right\} S_\left\{46\right\} S_\left\{56\right\}^0$
$\left\{\right\} + S_\left\{12\right\} S_\left\{13\right\}^0 S_\left\{14\right\} S_\left\{15\right\} S_\left\{16\right\} S_\left\{23\right\} S_\left\{24\right\}^0 S_\left\{25\right\}^2 S_\left\{26\right\}^0 S_\left\{34\right\} S_\left\{35\right\} S_\left\{36\right\} S_\left\{45\right\}^0 S_\left\{46\right\}^2 S_\left\{56\right\}^0$
$\left\{\right\} - S_\left\{12\right\} S_\left\{13\right\}^0 S_\left\{14\right\} S_\left\{15\right\} S_\left\{16\right\} S_\left\{23\right\} S_\left\{24\right\} S_\left\{25\right\} S_\left\{26\right\}^0 S_\left\{34\right\} S_\left\{35\right\} S_\left\{36\right\} S_\left\{45\right\}^0 S_\left\{46\right\} S_\left\{56\right\}\right).$

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

### Fifth step

Let
$T = S_\left\{12\right\} S_\left\{13\right\} S_\left\{14\right\} S_\left\{15\right\} S_\left\{16\right\} S_\left\{23\right\} S_\left\{24\right\} S_\left\{25\right\} S_\left\{26\right\} S_\left\{34\right\} S_\left\{35\right\} S_\left\{36\right\} S_\left\{45\right\} S_\left\{46\right\} S_\left\{56\right\}.$

Then the target becomes

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

### Seventh step

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

Using Sij notation, Lemma Two becomes

$S_\left\{ab\right\} S_\left\{cd\right\} - S_\left\{ac\right\} S_\left\{bd\right\} = S_\left\{ad\right\} S_\left\{cb\right\},$

which when applied to the target yields

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

Replace S53 with −S35, resulting in

$Gamma = \left\{256, Tover D\right\} \left(-S_\left\{16\right\} S_\left\{26\right\} S_\left\{35\right\} S_\left\{45\right\} + S_\left\{16\right\} S_\left\{26\right\} S_\left\{35\right\} S_\left\{45\right\}\right)$

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

## Proof of Lemma One

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

$= cos theta_i sin theta_j + sin theta_i cos theta_k + cos theta_j sin theta_k,$
$\left\{\right\} - cos theta_i sin theta_k - sin theta_i cos theta_j - sin theta_j cos theta_k,$

$= \left(cos theta_i sin theta_j - sin theta_i cos theta_j\right) ,$
$\left\{\right\} + \left(sin theta_i cos theta_k - cos theta_i sin theta_k\right),$
$\left\{\right\} + \left(cos theta_j sin theta_k - sin theta_j cos theta_k\right).,$

Applying the trigonometric identity

$cos a sin b - sin a cos b = sin \left(b - a\right) ,$
results in
$langle P_i,P_j,P_krangle = sin \left(theta_j - theta_i\right) + sin \left(theta_i - theta_k\right) + sin \left(theta_k - theta_j\right). ,$

Lemma Three:

$sin \left(theta_a - theta_b\right) + sin \left(theta_b - theta_c\right) + sin \left(theta_c - theta_a\right) ,$

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

Applying Lemma Three yields

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

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

quod erat demonstrandum.

## Proof of Lemma Two

Since

$sin \left(a - b\right) = sin a cos b - cos a sin b,,$

and letting

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

then

$S_\left\{ab\right\} S_\left\{cd\right\} = \left(S_a C_b - C_a S_b\right) \left(S_c C_d - C_c S_d\right) ,$
$\left\{\right\} = S_a S_c C_b C_d - S_a S_d C_b C_c ,$
$\left\{\right\} - S_b S_c C_a C_d + S_b S_d C_a C_c, ,$

$S_\left\{ac\right\} S_\left\{bd\right\} = \left(S_a C_c - C_a S_c\right) \left(S_b C_d - C_b S_d\right),$
$\left\{\right\} = S_a S_b C_c C_d - S_a S_d C_b C_c ,$
$\left\{\right\} - S_b S_c C_a C_d + S_c S_d C_a C_b, ,$

so that

$S_\left\{ab\right\} S_\left\{cd\right\} - S_\left\{ac\right\} S_\left\{bd\right\},$
$\left\{\right\} = S_a S_c C_b C_d - S_a S_b C_c C_d ,$
$\left\{\right\} - S_c S_d C_a C_b + S_b S_d C_a C_c ,$
$\left\{\right\} = \left(S_a C_d - C_a S_d\right) \left(S_c C_b - C_c S_b\right),$
$\left\{\right\} = S_\left\{ad\right\} S_\left\{cb\right\}, ,$
quod erat demonstrandum.

## Proof of Lemma Three

$sin \left(A - B\right) + sin \left(B - C\right) + sin \left(C - A\right),$

$= sin A cos B - cos A sin B + sin B cos C - cos B sin C + sin \left(C - A\right) ,$

$= cos B \left(sin A - sin C\right) + sin B \left(cos C - cos A\right) + sin \left(C - A\right),$

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

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

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

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

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

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

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

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

quod erat demonstrandum.