An easy way to modify a Voderberg Nonagon is to add an additional point in the path from T to R. When constructing the polygon this path is transformed by rotations and reflections into the paths RV, WS and SZ. Thus we obtain 4 additional points and the 9-gon is modified to a 13-gon.

The construction starts with the points Z, T, R and S known from the Voderberg nonagon. In order to obtain a noteworthy modification we construct A₁ and A₂ in a way that the lengths of TA₁ and A₁A₂ are equal to the length of ZT. Additionally A₂R shall have the same length as RS. Experimentally we found that ε=15° and angle A₁TZ=109.5305082352° are suitable values for achieving a perfect polygon. Theoretical considerations may be added later. Note that A = A₁ would not admit the construction of a Voderberg nonagon, A would lay outside the circle describt in the nonagon page.



ZUB₁B₂S is obtained by rotating ZTA₁A₂R around Z with ε=15°. Next reflect TA₁A₂R at point R and ZUB₁B₂S at point S. We call this 13-gon perfect because the distances from B₁ to A₁A₂ and from B₂ to A₂C₂ are equal.


The path from T to R can be modified nearly arbitrarily as long as there are no intersections with the roatated image of the path. Another restriction is that no point A_k is allowed with angle A_kTZ < 180°-ε. For A_kTZ = 180°-ε the point C_k is an element of VW.

Now we use again TZ = 4 units and ε=15°. The path from T to R shall be a part of a regular 24-gon. T is a vertex of the 24-gon and R is the midpoint of an edge. There are 6 further vertices of the 24-gon between T and R. The size of the 24-gon has to allow the described positions for our fixed points T and R. In a 24-gon the angles between two neighbor edges are equal to 165° = 180°-ε.
A formula for the radius of the circum circle around the regular 24-gon and a formula for the side length can be derived:

r² = (w²+9) / ((w+2)wu²+1)  ,    s = 2ur      with  w=cot(ε/2),  u=sin(ε/2)


Now the problem is that TZ is too longer than the other sides. We also have to modify TZ. In order to preserve the Voderberg properties, the path from Z to T has to be point symmetric with respect to the midpoint Q of TZ. We add the point E between T and Q with TE = s and EQ = s/2.
F is the reflection of E at Q. U, G, H, B₁, B₂, ..., B₆ are the rotations of T, E, F, A₁, A₂, ..., A₆ around Z with angle ε.

Finaly we reflect the path from T to R at R and the path from Z to S at S.

When adding a points between T and R, and e points between Q and T we obtain 4a + 6e + 5 vertices, because the 5 points T, Z, U, V, and W (known from the nonagon) are always necessary. In the shown sample we have a=6 and e=1. Therefore we obtain a 35-gon.
There are two different ways to construct a suitable head of the VT-tile. In the first tile E is above TZ and in the second tile E is below TZ.



Tiling patches will be added later.