Does any edge-to-edge tiling of the Euclidean plane by convex regular polygons have only demiregular vertex configurations?

Question

In the Euclidean plane, a vertex figure of an edge-to-edge tiling by convex regular polygons is called demiregular if and only if its vertex configuration is $3.3.4.12, 3.3.6.6, 3.4.3.12,$ or $3.4.4.6$ in one of the given orders. A vertex configuration lists the cyclic order of the number of edges around each face that it touches.

For example, $3.6.3.6$ or $3.4.6.4$ are not demiregular. All non-demiregular usable vertex figures are semiregular, meaning that there exists a vertex-transitive tiling of the plane using only that vertex figure.

There does not exist any tiling of the plane that uses only one type of demiregular vertex figure (this is part of the inspiration the definition of demiregular.) Can any combination of demiregular vertex figures create a tiling of the plane without including any semiregular vertex figures?

Answer 1

Completing the other answer, here is a proof that there is no Euclidean plane tiling with two equilateral triangles, a square, and a regular dodecagon at each vertex.

We look at any square $ABCD$ in the tiling. The other faces adjacent to its edges must be either triangles or dodecagons. We cannot have two consecutive dodecagons: then a vertex would have two dodecagons passing through it. So there are two cases:

1. Along three consecutive edges of $ABCD$, the opposite face is a triangle.
2. $ABCD$ is adjacent to two dodecagons (along opposite faces) and two triangles.

Case 1. Suppose $ABCD$ is adjacent to three triangles along $AB$, $BC$, and $CD$; let $E$ be the third vertex of the triangle with edge $BC$. Then the other side of $BE$ is the dodecagon through vertex $B$; the other side of $CE$ is the dodecagon through vertex $C$; therefore $E$ lies on two dodecagons, contradiction.

Case 2. Suppose $ABCD$ is adjacent to a dodecagon along $AB$ and along $CD$, but a triangle along $BC$; let $E$ be the third vertex of the triangle with edge $BC$. Then the other side of $BE$ is the second triangle through vertex $B$; the other side of $CE$ is the second triangle through vertex $C$; therefore $E$ lies on three triangles, contradiction.

Answer 2

I proved that the configurations $3.3.6.6$ and $3.4.4.6$ are impossible by the constraints of this question.

First: the $3.3.6.6.$

The two touching triangles in this vertex configuration share a different vertex. On that vertex, there must be 3.3.6.6 or 3.3.4.12. Then the vertex that has one of the 6's in the 3.3.6.6 case or the 12 in the 3.3.4.12 case cannot be completed with a demiregular vertex configuration. Thus proved.

Second: the $3.4.4.6.$

If $3.3.6.6$ is not allowed (which has been proven), then hexagons can only occur in a $3.4.4.6$ configuration. A triangle adjacent to a hexagon, which always exists in $3.4.4.6,$ has to be part of two $3.4.4.6$ configurations. That implies that the triangle must touch two squares on two of its edges. But that can only occur when the vertex of the triangle that is touching both squares has a vertex configuration of $3.4.6.4$ or $3.3.4.3.4$, both of which are semiregular. Thus proved.

Therefore, any tiling that meets the restrictions provided by the question must have two triangles, a square, and a dodecagon around each vertex.