# Pascal'sMysticHexagonTheo rem

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```					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.

```
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