Are there tetrads with no common vertices?

Further Questions about Tetrads

16. Are there tetrads with no common vertices?

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

(A) A tetrad made of hexagons with no common vertices.
Say V_P is the number of vertices of the polygonal tile then the number V_A of all vertices of the Tetrad (including all inner vertices) is 4 times greater than V_P.
V_A = 4 · V_P

V_A = 24

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

V_A = 44

Open problem (until 2025-12-10): What is the smallest order of polyiamonds which form a tetrad with no common vertex?

George Sicherman (reference [7]) could answer this question in December 2025 by creating a tetrad with no shared vertex made of order-16 polyiamonds. This is the smallest possible polyiamond tetrad with no shared vertex.

V_A = 32

Open problem (until 2025-12-10): What is the smallest order of polyominoes which form a tetrad with no common vertex?

George Sicherman also found an answer for this question in December 2025. For his solution he used polyominoes of order 12. A smaller polyomino tetrad with no shared vertex is not possible.

(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:   V_A,max = 3 · V_P - 3
The maximum number of vertices is reached when the shape of the whole tetrad is a n-gon with n = 2 · V_P:   V_T ≤ 2 · V_P
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.

V_P = 6, V_T = 12, V_A = 18

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