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?
Best Answer
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:
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.