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nL-n-P12:  nL-QL-P4 Par1/Par2-Homothetic Center

nL-n-P12 is the Par1/Par2-Homothetic Center (nL-n-Luc5e) of mL-n-P12, where m=(n-1).

This recursive construction can be rolled up to increasing larger values of n.

Starting value for n is 4, where 4L-n-P12=QL-P4.

General remarks

With reservations nL-n-P12, nL-n-P13, nL-n-P14 are mentioned as n-points in the neos-system (see nL-1). There are no contra-indications and all Mathematica calculations with numeric examples until reasonable depth confirm they are recursive n-points indeed, however there is no synthetic proof of it yet. So they are mentioned as n-points because their existence is a fact (to a certain n-level) and therefore they deserve registration.

nL-n-P12, nL-n-P13, nL-n-P14 have starting points at n=4, where they represent QL-P4, QL-P28, QL-P29, which are centers related to Hofstadter points, resp. H(2)=X(3), H(3)=X(186), H(-2)=X(265).

Probably there will be corresponding nL-n-points for centers related to other Hofstadter points H(i).

Construction

The construction of nL-n-P12 is as follows.

  1. Every n-Line contains n (n-1)-Lines leaving one line being called the omitted line.
  2. Construct lines through the n versions of (n-1)L-n-P12 parallel to the omitted line.
  3. These n lines form a new n-Line being called Par1.
  4. In the same way a 2nd generation n-Line Par2 can be constructed.
  5. Par1 is homothetic to Par2 and so there will be a Homothetic Center, which is nL-n-P12.
  6. This recursive construction can be rolled up to increasing larger values of n. Starting value for n is 4, where 4L-n-P12=QL-P4.

In a corresponding way nL-n-P13 and nL-n-P14 are constructed.

Example in a 5-Line
Infovisual nL-n-P12-infovisual-cvt-01.png

5L-n-P12 is also the Par1/Per2-Homothetic Center (see nL-n-Luc5g) of QL-P4 wrt the Reference 5-Line, because Par2 coincides with Per2.

Infovisual nL-n-P12-infovisual-cvt-02.png
Correspondence with ETC/EQF
Properties
  • In a 5-Line 5L-n-P12 = Midpoint (5L-n-P7.5L-n-P3).
  • In a 5-Line 5L-n-P14 = 5L-n-P7. 5L-n-P5 (2:-1)
  • It looks like that for all n Par2 will coincide with Per2.
  • It looks like that for all n the lengths of the line segments of Par1 are equal to the corresponding line segments of Par2 as well as Per2.


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