From Euclid to Combinatorics
Antreas P. Hatzipolakis
xpolakis at otenet.gr
Thu Nov 4 21:19:37 CET 2004
AN OLD THEOREM OF EUCLIDEAN GEOMETRY:
Let ABC be a triangle.
1,2,3 := three parallel lines through A,B,C, resp.
1',2',3' := three perpendiculars to 1,2,3, through A,B,C, resp.
Denote: i/\ j' = (ij') := (j'i) [so we have A = (11'), B = (22'), C = (33')]
The lines (23')(32ã), (31')(13'), (21ã)(12'), are concurrent
at a point on the Nine Point Circle of the triangle (11')(22')(33') = ABC
2' 1' 3'
| | |
2 ---B-----21'-----23'---------
| | |
1 --12'-----A------13'---------
| | |
| | |
3 --32'----31'------C----------
| | |
PROBLEMS
Let 1,2,3,...., n be n horizontal lines, and 1',2',3',..., n'
n vertical lines.
Let i/\ j' = (ij') := (j'i)
Two points are "collinear" if they lie on the same horizontal
or vertical line.
Problem #1:
Which is the number f(n) of the non-ordered triangles whose
the vertices are not "collinear" ?
If n = 3, there are 6 triangles, namely:
(11')(22')(33')
(11')(23')(31')
(12')(23')(31')
(12')(21')(33')
(13')(22')(31')
(13')(21')(32')
Geometry Problem: Where are lying the respective Nine Point
Circles points?
Problem #2:
Let n = 2k + 1
Which is the number c(n) of the non-ordered triangles with
non "collinear" vertices, that have as vertex the central point
([k+1][k+1]') ?
If k = 1, there are two triangles with vertex the point (22'):
(11')(22')(33')
(13')(22')(31')
Antreas
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