Pappus configuration

In projective geometry, the Pappus configuration consists of a pair (("A","B","C"), ("D","E","F")) of triplets of points, which pair is located either on a pair of lines or on two sides of a conic section, with a hexagon "AECDBF" defined on the points, and a collinear triplet of intersections of opposite sides (e.g "AE" and "DB") of the hexagon. When each triplet of points is collinear, it forms a projective configuration of nine points incident to nine lines, with the notation (9₃9₃).

The Pappus configuration is related to both Pappus's hexagon theorem and Pascal's theorem. In particular, the Pappus configuration can be thought of as the incidence structure of the Pappus hexagon theorem, consisting of nine points:"P" = {1, 2, 3, 4, 5, 6, 7, 8, 9}and nine lines:"L" = 1,4,7}, {3,6,9}, {2,5,8}, {1,6,8}, {3,5,7}, {2,4,9}, {1,5,9}, {3,4,8}, {2,6,7,with each line passing through three points, and each point being the intersection of three concurrent lines.

Each of the lines in the above set "L" can be seen to correspond either to a term of the determinant of the "telephone keypad" matrix:or to a column of this matrix.

The following matrix, known also as a "configuration table", describesthe combniatorial structure of the Pappus configuration.

:

The Levi graph of the Pappus configuration is the Pappus graph, a bipartite distance-regular graph with 18 vertices and 27 edges.

:

Used to construct conic sections

A set of five points in general position in a plane is enough to specify a unique conic section in that plane which passes through all five points of the set. Such unique conic section can be constructed from the five points by using the Pappus configuration.

The construction is as follows: given five points labeled 1, 2, 3, 4, and 5; draw line 15 passing through points 1 and 5, line 24 passing through points 2 and 4, line 34 passing through points 3 and 4, and line 35 passing through points 3 and 5. Let point 9 be the intersection of lines 15 and 24, then choose some point, any point, on line 34 and call it point (8). Draw line 9(8) passing through points 9 and (8). Let point (7) be the intersection of lines 35 and 9(8). Draw lines 1(8) and 2(7), and let point (6) be the intersection of lines 1(8) and 2(7). Points with parenthesized labels are variable, whereas points with non-parenthesized labels are fixed.

Sliding point (8) along line 34 will cause point (6) to move in the plane, and tracing the trajectory of (6) will yield a conic section: this is the conic section which is uniquely specified by the set of points {1, 2, 3, 4, 5}.

Here is a summary of the construction:
* 1, 2, 3, 4, 5
* 15, 24, 34, 35
* 15 ∩ 24 = 9
* (8) ∈ 34
* 9(8)
* 9(8) ∩ 35 = (7)
* 1(8), 2(7)
* 1(8) ∩ 2(7) = (6).

elf-duality

The incidence structure of the Pappus configuration may be represented by a matrix of points "P",:and a matrix of lines "L",:where the matrices may be thought of as special kinds of ordered sets, with the ordering imposed upon the sets of points (and lines) in order to highlight certain duality properties (described below).

The incidence structure of the Pappus configuration may be described as a triple:$C\; =\; (P,\; L,\; I)$where "I" is an incidence relation:$I\; :\; P^3\; imes\; L$,e.g.:$I\; :\; (1,\; 4,\; 7)\; mapsto\; 147$,and "P" and "L" are unordered sets.

The dual incidence structure is then given by:$C^*\; =\; (L,\; P,\; I^*)$where "I^*" is the dual incidence relation:$I^*\; :\; L^3\; imes\; P$,e.g.:$I^*\; :\; (147,\; 168,\; 159)\; mapsto\; 1$.

Now, applying the matrix ordering to the sets "P" and "L", a duality mapping "g" can be given by:$g\; :\; P\_\{ij\}\; mapsto\; L\_\{ij\}$which is an isomorphism because:$I\; :\; (a,\; b,\; c)\; mapsto\; abc$implies:$I^*\; :\; (g\; a,\; g\; b,\; g\; c)\; mapsto\; g^\{-1\}(a\; b\; c)$,therefore the Pappus configuration is self-dual.

Hamiltonian path

A Hamiltonian path is a path which passes once through each point and each line of the incidence structure of the Pappus configuration. The problem is in finding such circumnavigating paths.

A method will be given here whose main idea is that since lines and points correspond through self-duality, and since the requirement of non-repetition is the same for points and lines, then given the path pointer is currently at a given point, that the pointer should be forced to move only through a line which is the dual of the given point: so that the line-path would end up being dual to the point-path: then they would satisfy the non-repetition requirement simultaneously.

Initially, this appears to be only possible for points 1, 5, and 8, since:1 ∈ "g"1 = 147,:5 ∈ "g"5 = 357,:8 ∈ "g"8 = 348,but other points do not belong to their own duals. This can be fixed by forcing the pointer at point 2 to move along line "g"3, and at point 3 to move along line "g"2, and likewise to swap the dual lines of points 4 and 6 and of points 7 and 9. This "forced move" mapping "F" is:$F\; :\; 1\; mapsto\; g1,\; 2mapsto\; g3,\; 3mapsto\; g2,\; 4mapsto\; g6,\; 5mapsto\; g5,\; 5mapsto\; g4,$::$7mapsto\; g9,\; 8mapsto\; g8,\; 9mapsto\; g7.$

If the pointer is at point "x", then it can only move to the set of points:"Fx" - {"x"}.Let mapping "M" be defined by:"Mx" = "Fx" - {"x"}.If the pointer on a path is at a given point "x", it is now only allowed to move to two other points belonging to "M"("x"): the problem has been reduced to doing a search of a non-repeating point-path down a binary tree.

Here is the mapping "M",:$M\; :\; 1mapsto\; \{4,7\},\; 2mapsto\; \{5,8\},\; 3mapsto\; \{6,9\},$::$4mapsto\; \{2,9\},\; 5mapsto\; \{3,7\},\; 6mapsto\; \{1,8\},$::$7mapsto\; \{2,6\},\; 8mapsto\; \{3,4\},\; 9mapsto\; \{1,5\}.$

Here is a search through the binary tree:

1____7
4____9
2____8
5________________________7____6____8____4
| |
[3] ____________ [9] __ [5] [2] [1] 3____9___ [5]
| |
[6] __ [8] __ [4] [1] [6] 1

[1] [3]

Starting from the first node and moving down the binary tree, keep track of the nodes already gone through. At a given node there will be a pair of successor nodes: if a node is the same as a node which appears less than nine steps "above" in the binary tree, then this node is "closed" and marked with a square (square brackets in the above diagram). If both successors of a node are closed, then the node itself is also closed. If, while searching, the pointer arrives at a closed node, then the pointer must backtrack one step to the previous node, and if this node has an alternative "open" node then the pointer should search down this node. Eventually the pointer will reach a node which is not closed but which is the same as the first node: this last node will be nine steps "below" the first node. When this happens the search ends, and a Hamiltonian path has been found. In the above diagram, the Hamiltonian path is
1 → 4 → 2 → 5 → 7 → 6 → 8 → 3 → 9 → 1.
Getting from one point to another requires travel through a connecting line. The Hamiltonian path with connecting lines is shown below:
1→147→4→249→2→258→5→357→7→267→6→168→8→348→3→369→9→159→1.