16. Are there tetrads with no common vertices?

More exactly: Are there tetrads where no pair of the four congruent polygons has common vertices?
But for tetrads with no hole all vertices of the inner polygon belong to at least one other polygon.

Say VP is the number of vertices of the polygonal tile then the number VA of all vertices of the Tetrad (including all inner vertices) is 4 times greater than VP.
VA = 4 · VP VA = 24

(B)    A tetrad made of order-22 polyiamonds (11-gons) with no common vertices. VA = 44

 Open problem: What is the smallest order of polyiamonds which form a tetrad with no common vertex?

(C)    A tetrad made of order-16 polyominoes (octagons) with no common vertices. VA = 32

 Open problem: What is the smallest order of polyominoes which form a tetrad with no common vertex?

(D)    For a tetrad without hole all vertices of the inner polygon also belong to at least one other polygon, therefore the maximum number of vertices is:   VA,max = 3 · VP - 3
The maximum number of vertices is reached when the shape of the whole tetrad is a n-gon with n = 2 · VP:   VT ≤ 2 · VP
This means a tetrad made of hexagons can be at most a 12-gon.
The following well-known tetrad made of hexagons has 12 corners, more corners are not possible. VP = 6, VT = 12, VA = 18

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