 Application Center - Maplesoft

# Permutation of vertices of pascal's hexagon

You can switch back to the summary page by clicking here.

pascal1.mws

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.!

> 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.

>        >  >  >  >