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It seems to me that if a tiling is tile-uniform, then it must be vertex-uniform as well. But is this the case? How would one go about devising a proof?

By 'tile-uniform', I mean a tiling whose tile-types are the same; and a tiling which is 'vertex-uniform' is one whose vertex-types are all the same. I suppose to prove the above one would first need to look at the properties of the tile itself?

By tiling, I refer to any kind of polygon.

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Can you define your terms? Most definitions I see for uniform tilings have vertex-uniformity as part of the definition. What does "tile-uniform" mean exactly? – mjqxxxx Oct 28 '13 at 18:43
Should regular tiles be assumed? – hardmath Oct 28 '13 at 18:56
If by "tile-uniform" you mean that all the tiles are congruent, then no, such a tiling need not be vertex-uniform. For instance, to tile with unit squares, you can place the lower left corners at $((2y-1)^2+x,y)$ where $x,y$ are integers. There will be vertices of multiple types (e.g., both degree three and degree four). – mjqxxxx Oct 28 '13 at 19:02
The standard definition of "vertex-uniform" (or "vertex-transitive" or "isogonal") is, more-specifically, having the property that, for any two distinct vertices, $A$ and $B$, there's an isometry of the tiling that carries $A$ to $B$. Similarly for "tile-uniform" (or "face-transitive" or "isohedral"). Note that a tiling can also be "edge-uniform" (or "edge-transitive" or "isotoxal"). – Blue Oct 28 '13 at 19:13
up vote 5 down vote accepted

No, standard example is the $60^\circ - 120^\circ$ rhombus. Three of those make up a regular hexagon, which then tiles. But some vertices, i.e. hexagon centers, have valence three, other vertices as high as six, depending on choices you make in rotating the hexagons or not. Regardless, this way there are vertices where at least two $60^\circ$ angles meet, three in each hexagon. Each such vertex has valence among $4,5,6.$

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