Permutation of Vertices of Pascal's Hexagon
dr.Laczik Blint
Technical University of Budapest, Hungary
mail: laczik@goliat.eik.bme.hu
Pascal's theorem (1640) says: If a hexagon =(123456) is inscribed in a conic, then opposite sides intersect in three points
=(12.45), = (23.56), =(34.61) which are collinear.
The representation of the hexagon and intersection points of opposite sides its by symbol is following:
By the way of elementar combinatorics can be proof that the number of different hexagons is 60, and the number of different points , , the intersections opposite sides of all hexagons are 45.
Let's show this special projective configuration by Maple.!
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Warning, new definition for Chi
Warning, new definition for draw
Let's make coordinates 6 randpoints P.i of circle C and the lines l.i through of points recommanded sequence and . and are element of by permutations generated points.
For example one of hexagons (z = 11) and its sides
The intersections of opposite sides l.1, l.4 is P, l.2, l.5 is Q, l.3, l.6 is S.
Make the line T_1 through of the points P and S, T_2 through of the points Q and S, T_3 through of the points P and Q,
and show the Pascal's hexagon and line.
The line T_1 through of points P and S, the line T_2 through of points Q and S, or the line T_3 through Q and P are equal by accuracy of calculating method - look the original Pascal's theorem.
Finally make the all hexagons, points and lines of combinatorical structure cc[i]. By simple way of representation show the full combinatorical and geometrical structure.
