In geometry, a configuration is a collection of points and lines so that every point is coincident with the same number of lines, and every line contains the same number of points. The Fano plane, Möbius-Kantor Configuration, and Pappus configuration are examples on 7, 8, and 9 points. Starting with 7 points, the number of 3-configurations is 1, 1, 3, 10, 31, 229, ... A001403. A023994 gives 4-configurations. A099999 enumerates geometrically realizable 3-configuration. A100001 counts self-dual 3-configuration. Below are some 4-configurations: every point has 4 lines, every line has 4 points (more 4-configurations).
For barycentric coordinates, consider a triangle with vertices A, B, C and an area of 1. Add an interior point D. The barycentric coordinates of D are the areas of triangles ABD, ADC, and DBC. The sum of the coordinates equals one since the original triangle had area 1. In the below. A =(1,0,0), B =(0,1,0), C =(0,0,1). For the interior points add the step of dividing by the total. The point marked 210 is actually (2/3, 1/3, 0). The point marked 123 is (1/6, 1/3, 1/2). Showing them as fractions would be a much messier picture.
For barycentric lines, a barycentric point a b c is on a barycentric line A B C if their dot product A a + B b + C c equals 0. A canonical barycentric line can have three forms:
- (a, b, c) with total 1.
- (a, -1-a, 1) with total 0, going through the center point (1/3, 1/3, 1/3).
- (1,-1,0), the vertical line through the center point.
Based on this, I wanted to try to make a self-dual configuration where the set of canonical barycentric points was the same as the set of canonical barycentric lines. Is that possible?
Here is one attempt. In the below, overbar means that number is negative. Divide each triple by their total to get the canonical fraction forms, then take all permutations to obtain a set of 111 barycentric points and 111 barycentric lines.
$$\bar{5}34 \enspace \bar{4}35 \enspace \bar{2}\bar{1}6 \enspace \bar{2}12 \enspace \bar{2}13 \enspace \bar{1}02 \enspace \bar{1}03 \enspace \bar{1}11 \enspace \bar{1}12 \enspace \bar{1}13 \enspace \bar{1}22$$ $$\bar{1}23 \enspace \bar{1}33 \enspace \bar{1}36 \enspace 001 \enspace 011 \enspace 012 \enspace 013 \enspace 112 \enspace 114 \enspace 122 \enspace 133 \enspace 157$$
Here's a picture of what that looks like. Each point is on six or more lines, and every line has six or more points. To get a configuration, the "or more" isn't allowed. Can a 3, 4, or 5 configuration be pulled out of this? Or a 6, but no geometric 6-configuration is currently known. Leah Berman found a 5-configuration of 48 points in 2007. The colors indicate the number of incident points/lines.
Code for all this is at Extreme Orchards for Gardner. Can a set of sum=1 triples represent the canonical points and lines of a self-dual barycentric configuration?
Yes. A self-dual $27_4$ barycentric configuration exists.
The set of lines AND the set of points are both as follows:
{{-2, 3, 1}, {-2, 4, -1}, {-1, -2, 4}, {-1, 0, 2}, {-1, 1, 1}, {-1, 1,2}, {-1, 2, 3}, {-1, 2, 4}, {0, 2, -1}, {0, 2, 1}, {1, -2, 3}, {1, -1, 1}, {1, 0, 2}, {1, 1, -1}, {1, 1, 2}, {1, 2, -1}, {1, 2,1}, {2, -1, 0}, {2, -1, 1}, {2, 1, 0}, {2, 1, 1}, {2, 3, -1}, {2, 4, -1}, {3, -1, 2}, {3, 1, -2}, {4, -1, -2}, {4, -1, 2}}
A point is on a line if the dot product is zero.