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:

$P\_1\; :\; [cos\; theta\_1\; :\; sin\; theta\_1\; :\; 1]$

$P\_2\; :\; [cos\; theta\_2\; :\; sin\; theta\_2\; :\; 1]$

$P\_3\; :\; [cos\; theta\_3\; :\; sin\; theta\_3\; :\; 1]$

$P\_4\; :\; [cos\; theta\_4\; :\; sin\; theta\_4\; :\; 1]$

$P\_5\; :\; [cos\; theta\_5\; :\; sin\; theta\_5\; :\; 1]$

$P\_6\; :\; [cos\; theta\_6\; :\; sin\; theta\_6\; :\; 1]$.

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:

$(A\; times\; B)times\; (C\; times\; D)\; =\; C\; langle\; A,B,Drangle\; -\; D\; langle\; A,B,Crangle$

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

$P\_i\; :\; [cos\; theta\_i\; :\; sin\; theta\_i\; :\; 1]$

$P\_j\; :\; [cos\; theta\_j\; :\; sin\; theta\_j\; :\; 1]$

$P\_k\; :\; [cos\; theta\_k\; :\; sin\; theta\_k\; :\; 1],$

then

$langle\; P\_i,P\_j,P\_krangle\; =\; 4\; sin\; left(\{theta\_i\; -\; theta\_j\; over\; 2\}right)\; sin\; left(\{theta\_j\; -\; theta\_k\; over\; 2\}right)\; sin\; left(\{theta\_k\; -\; theta\_i\; over\; 2\}right).$

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

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

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

$S\_\{ab\}\; S\_\{cd\}\; -\; S\_\{ac\}\; S\_\{bd\}\; =\; S\_\{ad\}\; S\_\{cb\},$

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

$Gamma\; =\; \{256,\; Tover\; D\}\; (-S\_\{16\}\; S\_\{26\}\; S\_\{35\}\; S\_\{45\}\; +\; S\_\{16\}\; S\_\{26\}\; S\_\{35\}\; S\_\{45\})$

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

quod erat demonstrandum.

Proof of Lemma One

$=\; cos\; theta\_i\; sin\; theta\_j\; +\; sin\; theta\_i\; cos\; theta\_k\; +\; cos\; theta\_j\; sin\; theta\_k,$

$\{\}\; -\; cos\; theta\_i\; sin\; theta\_k\; -\; sin\; theta\_i\; cos\; theta\_j\; -\; sin\; theta\_j\; cos\; theta\_k,$

$=\; (cos\; theta\_i\; sin\; theta\_j\; -\; sin\; theta\_i\; cos\; theta\_j)\; ,$

$\{\}\; +\; (sin\; theta\_i\; cos\; theta\_k\; -\; cos\; theta\_i\; sin\; theta\_k),$

$\{\}\; +\; (cos\; theta\_j\; sin\; theta\_k\; -\; sin\; theta\_j\; cos\; theta\_k).,$

Applying the trigonometric identity

$cos\; a\; sin\; b\; -\; sin\; a\; cos\; b\; =\; sin\; (b\; -\; a)\; ,$

results in

$langle\; P\_i,P\_j,P\_krangle\; =\; sin\; (theta\_j\; -\; theta\_i)\; +\; sin\; (theta\_i\; -\; theta\_k)\; +\; sin\; (theta\_k\; -\; theta\_j).\; ,$

Lemma Three:

$sin\; (theta\_a\; -\; theta\_b)\; +\; sin\; (theta\_b\; -\; theta\_c)\; +\; sin\; (theta\_c\; -\; theta\_a)\; ,$

$=\; 4\; sin\; left(\{theta\_a\; -\; theta\_b\; over\; 2\}\; right)\; sin\; left(\{theta\_b\; -\; theta\_c\; over\; 2\}\; right)\; sin\; left(\{theta\_a\; -\; theta\_c\; over\; 2\}\; right).,$

Applying Lemma Three yields

$langle\; P\_i,P\_j,P\_krangle\; =\; 4\; sin\; left(\{theta\_j\; -\; theta\_i\; over\; 2\}\; right)\; sin\; left(\{theta\_i\; -\; theta\_k\; over\; 2\}right)\; sin\; left(\{theta\_j\; -\; theta\_k\; over\; 2\}right),$

$\{\}\; =\; 4\; sin\; left(\{theta\_i\; -\; theta\_j\; over\; 2\}\; right)\; sin\; left(\{theta\_j\; -\; theta\_k\; over\; 2\}right)\; sin\; left(\{theta\_k\; -\; theta\_i\; over\; 2\}right),\; ,$

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

$S\_\{ab\}\; S\_\{cd\}\; =\; (S\_a\; C\_b\; -\; C\_a\; S\_b)\; (S\_c\; C\_d\; -\; C\_c\; S\_d)\; ,$

$\{\}\; =\; S\_a\; S\_c\; C\_b\; C\_d\; -\; S\_a\; S\_d\; C\_b\; C\_c\; ,$

$\{\}\; -\; S\_b\; S\_c\; C\_a\; C\_d\; +\; S\_b\; S\_d\; C\_a\; C\_c,\; ,$

$S\_\{ac\}\; S\_\{bd\}\; =\; (S\_a\; C\_c\; -\; C\_a\; S\_c)\; (S\_b\; C\_d\; -\; C\_b\; S\_d),$

$\{\}\; =\; S\_a\; S\_b\; C\_c\; C\_d\; -\; S\_a\; S\_d\; C\_b\; C\_c\; ,$

$\{\}\; -\; S\_b\; S\_c\; C\_a\; C\_d\; +\; S\_c\; S\_d\; C\_a\; C\_b,\; ,$

so that

$S\_\{ab\}\; S\_\{cd\}\; -\; S\_\{ac\}\; S\_\{bd\},$

$\{\}\; =\; S\_a\; S\_c\; C\_b\; C\_d\; -\; S\_a\; S\_b\; C\_c\; C\_d\; ,$

$\{\}\; -\; S\_c\; S\_d\; C\_a\; C\_b\; +\; S\_b\; S\_d\; C\_a\; C\_c\; ,$

$\{\}\; =\; (S\_a\; C\_d\; -\; C\_a\; S\_d)\; (S\_c\; C\_b\; -\; C\_c\; S\_b),$

$\{\}\; =\; S\_\{ad\}\; S\_\{cb\},\; ,$

quod erat demonstrandum.

Proof of Lemma Three

$sin\; (A\; -\; B)\; +\; sin\; (B\; -\; C)\; +\; sin\; (C\; -\; A),$

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

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

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

$\{\}\; +\; sin\; (C\; -\; A),$

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

$\{\}\; +\; sin\; (C\; -\; A)\; ,$

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

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

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

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

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

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

quod erat demonstrandum.

External links

• Interactive demo of Pascal's theorem (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