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Mystic Hexagon and Octagon
- A Nice Application of Bézout’s Theorem
Djordje Baralić
Mathematical Institute SASA, Belgrade, Serbia
djbaralic@mi.sanu.ac.rs
Adviser : Rade Živaljević
Problems I am working on
My Research Area
Currently I am studying quasitoric manifolds and small covers, topological generalizations of toric and
real toric varieties, relations between their topology and combinatorics of their naturally associated
simple polytopes and its application in geometric combinatorics
toric topology, algebraic topology, geometric combinatorics, combinatorial algebraic geometry, projective geometry
New Generalizations in Hexagrammum Mysticum
Introduction
Pappos, IV century A. D.
Pascal line for cubics D1 and D2
Let A1, A2, A3 be three points on a straight
line and let B1, B2, B3 be three points on
another line. Let the lines A1B2 , A2B3, A3B1
intersect the lines B1A2 , B2A3, B3A1, in the
points C1, C2 and C3 respectively. Then the
three points C1, C2 and C3 are collinear.
Let the points A, B, C, D, E and F lie
on cubics D1 and D2. Then the remaining 3 intersection points of the cubics D1
and D2, P , Q and R lie on the same line
l(ABCDEF ; D1, D2).
Baralić & Spasojević, 2012
The
lines
l(ABCDEF ; D1, D2),
l(ABCDEF ; D2, D3)
and
l(ABCDEF ; D3, D1) intersect in the
Steiner-Kirkman point for the cubics D1, D2
and D3 and we denote it by N(D1)(D2)(D3).
Blaise Pascal, 1639.
Let the points A1, A2, A3, A4, A5 and A6 lie
on conics C. The lines A1A2 and A4A5 meet
at P , the lines A2A3 and A5A6 meet at Q, the
lines A3A4 and A6A1 meet at R. Then the
points P , Q and R are collinear.
The point N(A.BD.E.CF ; D) = N(D)(l(AB) · l(EC) · l(DF ))(l(AD) · l(EF ) · l(BC)) is called
generalized D Steiner point.
Baralić & Spasojević, 2012
Bézout
The four generalized D Steiner points N(A.BC.F.DE; D), N(A.BD.F.CE; D),
N(A.DE.F.BC; D) and N(A.CE.F.BD; D) lie on the Generalized Steiner Line.
If two projective curves C and D in CP 2 of
degree n intersect at exactly n2 points and if
n·m of these points lie on an irreducible curve E
of degree m < n, then the remaining n·(n−m)
points lie on a curve of degree at most n − m.
Octagrammum Mysticum
Hexagrammum
Mysticum
60 possible ways for joining six points on a conic
produce 60 Pascal lines. They are concurrent by
triples in 20 ’Steiner points’ and also by triples in another 60 points, known as ‘Kirkman’s points’ which
form a (60)3-type configuration. ’Veronese’s Decomposition Theorem’ states that (60)3-type configuration splits properly into six Desargues Configurations of the type (10)31’s. He also proved the
existence of infinitely many systems comprised of
sixty lines and points, explaining the complete Mystic Hexagon Configuration.
Wilkinson 1872.
Let ABCDEF GH be an octagon inscribed in a
conic C and let Q1 and Q2 be distinct quartics
that pass through the points A, B, C, D, E,
F , G and H. Let L, M, N, O, P , Q, R and
S be the eight other points of the intersection
of Q1 and Q2. Then these eight points lie on
the same conic D.
Baralić & Spasojević, 2012.
Plücker
Let ABCDEF GH be an octagon inscribed
in a conic C and assume that Q1, Q2 and
Q3 are distinct quartics that pass through
A, B, C, D, E, F , G and H. Then
the
conics
D1(ABCDEF GH; Q2, Q3),
D2(ABCDEF GH; Q3, Q1)
and
D3(ABCDEF GH; Q1, Q2)
belong
to
the same pencil of conics.
20 Steiner points lie in fours on 15 SteinerPlücker lines, three through each point.
Cayley and Salmon
The Kirkman points lie in threes on 20 CayleySalmon lines. Each Steiner point lie on the
Cayley-Salmon line. The Cayley-Salmon lines
meet in threes in 15 Salmon points.
References
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4 G. Verones
-703.
Lincei, Vol. I, 1877, 649
Baralić & Spasojević, 2012.
Let Q be a quartic passing through 8 vertices of
mystic octagon ABCDEF GH, let C1, C2 and
C3 be three distinct conics through the points
A, B, C and D and let D1, D2 and D3 be three
distinct conics through the points E, F , G and
H. Let X1 be the mystic conic arising from the
curves Q and C1 · D1 and let Y1 be the mystic
conic arising from the curves Q and C3 · D2.
The conics X2, X3, Y2 and Y3 are defined in
analogous way. Then 12 intersection points of
X1 ∩ Y1, X2 ∩ Y2 and X3 ∩ Y3 lie on the same
conic.