Pascal'sMysticHexagonTheo rem

							Pascal’s Mystic Hexagon Theorem

In this note we will prove Pascal’s Mystic Hexagon Theorem, but first a convention about labelling the sides
of a hexagon, simple or nonsimple.

Given a hexagon ABCDEF (simple or nonsimple) inscribed in a circle, we label the sides with the positive
integers so that

          1 ↔ AB,        2 ↔ BC,             3 ↔ CD,                 4 ↔ DE,               5 ↔ EF,             and       6 ↔ F A,

as in the figure below.

                                 B                                                                             B

                                         2                                                         1            2
                                                     C                             E                                          D
                         1                                                                                 4

                                                 3                                                     5                  3
             A                                                                 A
                                                                                                           6                  F
                  6                                  D
                                             4
                             5
                   F                                                                                                 C
                                     E

We say that sides 1 and 4, sides 2 and 5, and sides 3 and 6 are opposite sides, whether the hexagon is
simple or nonsimple.

Pacsal’s Theorem. Given a hexagon (simple or nonsimple) inscribed in a circle, the points of intersection
of opposite sides are collinear and form the Pascal line.

Proof. Given an inscribed hexagon as shown, we want to show that P, Q, and R are collinear.

                                                                                               X




                                                                       B


                                                         1                 2

                                         E           P
                                                                 4
                                                                                           D
                                                 Y                                     3
                                                             5                 Q
                                     A                                                 Z
                                                                 6                 R       F



                                                                               C
Create    XY Z by taking every second side of the hexagon, so that

                X = side 1 ∩ side 3,         Y = side 1 ∩ side 5,         and     Z = side 3 ∩ side 5.


Note that
                  P is on side 1,      Q is on side 5,          and       R is on side 3 (extended),
so that P, Q, and R are Menelaus points of          XY Z.

In order to show that P, Q, and R are collinear, by Menelaus’ theorem we need only show that

                                               XP Y Q ZR
                                                  ·   ·   = −1.
                                               PY   QZ RX


Applying Menelaus’ theorem to the following labeled transversals of             XY Z, we have


      −→
     ←−             XP Y E ZD
     EP D :            ·   ·   = −1
                    PY   EZ DX
      −→
     ←−             XB Y Q ZC
     BQC :             ·   ·   = −1
                    BY   QZ CX
      −
     ←→            XA Y F    ZR
     ARF :            ·    ·    = −1,
                   AY   F Z RX

and multiplying these together we get

              XP Y Q ZR Y E ZD XB ZC XA Y F
                 ·   ·  ·  ·  ·   ·   ·   ·
              PY   QZ RX EZ DX BY   CX AY   FZ

                               XP Y Q ZR               XA · XB              YE · YF           ZC · ZD
                           =      ·   ·   ·                           ·                   ·
                               PY   QZ RX              XC · XD              YA · YB           ZE · ZF

                           = (−1)3 = −1.


Now, the power of the point X with respect to the circle is

                                              XA · XB = XC · XD ,

the power of the point Y with respect to the circle is

                                               YA · YB = YE · YF ,

while the power of the point z with respect to the circle is

                                               ZC · ZD = ZE · ZF ,

and therefore
                          XA · XB              YE · YF           ZC · ZD
                                         ·                  ·                   = 1 · 1 · 1 = 1,
                          XC · XD              YA · YB           ZE · ZF
so that
                                               XP Y Q ZR
                                                  ·   ·   = −1,
                                               PY   QZ RX
and the points P, Q, and R are collinear.