# Can similar convex heptagons tile the plane?

It is a fairly straightforward matter to apply Euler's formula $$V-E+F=2$$ for planar graphs to see that congruent convex heptagons cannot tile the plane. The graph associated to the heptagons within distance $$R$$ of the origin, with edges between heptagons that overlap along an edge, has degree $$\ge7$$ at every "interior" vertex but (for sufficiently large $$R$$) insufficiently many exterior vertices to balance things out. This can be extended to any family of convex heptagons whose area and perimeter are bounded - see e.g. this thread on the /r/mathriddles subreddit for a more detailed proof.

I am curious whether it is possible to tile the plane with convex heptagons which are similar to each other; I suspect it is impossible, but the above sort of argument seems to fail, because we cannot assume boundedness.

Note that there do exist tilings of the plane by heptagons if we relax these boundedness constraints; for instance, one can take a tiling of the hyperbolic plane by heptagons in the disc model and apply a map $$(r,\theta)\mapsto(\frac{r}{1-r},\theta)$$ to yield a tiling of the whole plane (if we straighten out the edges after applying the transformation).

Is it possible to tile the plane with similar convex heptagons? If the answer is "no", I am interested in the case of finitely many heptagon shapes, which I suspect is also impossible.

Edit: Edward H.'s answer made me realize that there are some additional tiling conditions that seem relevant when allowing for arbitrarily-scaled pieces.

• Say that a tiling is lower-bounded if there is some $$\epsilon>0$$ such that every tile contains a ball of radius $$\epsilon$$.

• Say that a tiling is locally finite if the boundary of every tile is covered by finitely many other tiles.

• Say that a tiling is traversable if you can reach any tile from any other tile by crossing over shared edges.

• Say that a tiling is complete if every point in the plane is in the closure of a tile (as opposed to being in the closure of the union of all tiles).

I'm most interested in the case of tilings that are locally finite, traversable, and complete, but I'm also curious to hear about possibility/impossibility results for any of the above restrictions. (I'm also interested in any implications between them - I think that being lower-bounded implies the other three conditions, but I don't have a proof yet.)

• By "convex", do you mean that angles should be $<180^\circ$ or $\le 180^\circ$?
– user632577
Mar 2, 2021 at 21:49
• Strictly less than $180^\circ$; if the inequality is weak, one can just use squares with extra vertices at the midpoints. Mar 2, 2021 at 21:54

Here's a construction that tiles a half-plane with two similar heptagons (four if discounting reflections). Two half-planes tile a plane. I'm having trouble coming up with a tiling using one similar heptagon, but at least the finite case is settled :)

• Nicely done! Your solution makes me realize that one can further classify tilings with unbounded sizes as to whether they have upper or lower bounds on the scales used - this one is unbounded in both directions, and I wonder if there are impediments to being bounded either above or below. Mar 3, 2021 at 1:17
• @RavenclawPrefect Ah yes, but this one isn't really unbounded above, as you could imagine isolating a single tile and extending downwards to form a trapezoid, and any trapezoid tiles the plane. I agree that unboundedness towards the smaller end seems essential here, maybe one could use Euler's formula to show this.
– user632577
Mar 3, 2021 at 1:49
• Thinking about this more, I was trying to figure out how to distinguish it from an Apollonian gasket kind of "tiling", since it doesn't cover the line between the two half-planes. I edited the original post to add some more criteria a collection of tiles can satisfy (that don't really come up when everything is the same size) - under that terminology, this is a locally finite but incomplete tiling. Mar 3, 2021 at 19:11
• @RavenclawPrefect Oh yeah that was a good thing to have been said explicitly. When I was thinking about these tilings I don't think I subconciously was aware of this local finite-ness. What I had in mind was the property that "every heptagon should be adjacent to exactly 7 other heptagons". (So a domino tiling could easily become a "hexagonal" tiling.) This makes my later trapezoidal attempt invalid though.
– user632577
Mar 9, 2021 at 16:54

Here is a fractal sort of tiling I've found with a single heptagon (up to similarity) which is traversable, but not lower-bounded, locally finite, or complete.

It's not obvious from the picture that this really is traversable, but the idea is to note that after adding the red and blue heptagons (as well as the geometric series of tiles directly above the middle of the blue heptagons), we can subdivide all leftover spaces into $$45-45-90$$ triangles such that every new tile added touches a larger tile along the hypotenuse of the triangle, and that we can inductively add new tiles to the voids left by these tiles to preserve the connectedness of the adjacency graph.

This doesn't really capture the sort of thing I'm hoping to find in the post, but I thought I would post it since it's still on the Pareto frontier of satisfiable properties for a tiling thus far, and might prompt someone else to discover a better tiling.

We can actually tile the plane with differently sized regular heptagons -- and pentagons and octagons. Probably any number of sides will work, although the process becomes more complicated with more than eight sides.

The proposed tiling below is not bounded below (heptagons have to become arbitrarily small) or complete (the angles of the triangles mean the corners cannot be tiled in finitely many steps), but the placement of heptagon sides along the sides of the triangles does appear to render the tiling traversible.

Divide and conquer

We can certainly tile a plane with any triangle. Now suppose a regular polygon is inscribed in the triangle and the area of the triangle outside the polygon can be divided into smaller triangles in which smaller copies of the regular polygon can be subsequently inscribed. By iterating this process of inscription and division, the entire triangle is tiled with similar regular polygons, and then the tiled triangle can be used to tile the plane.

We illustrate this idea first with the regular pentagon. The pentagon is inscribed in a golden tringle as shown below, and the excluded area of the triangle is then seen to be divided into three smaller triangles similar to the first. The required inscription and division iteration follows.

The regular octagon can be similarly inscribed in a right isosceles triangle. In this case two of the excluded regions are concave quadrilaterals instead of triangles, but by extending the appropriate sides of the octagon these may be divided into right triangles similar to the large one. Again smaller copies of the regular polygon may be inscribed iteratively to tile the remaining area of the triangle and thence the full plane.

Tag team

For the regular heptagon the process is slightly more complicated. the angles do not combine in the right way to allow inscription in any single triangle followed by division of the remaining area to be iterated. Instead we use inscriptions into two different triangles: one (blue) with angles 1, 2, 4 in multiples of $$\pi/7$$ radians, the other (gold) with angles of 2, 2, 3. When the two sets of excluded areas are divides, we obtain smaller copies of both triangles and no others, thus allowing the iterative process to tile them both. the plane can then be tiled with either tiled triangle or even an appropriate combination of both. Note that the mirror-image portions of the 2-2-3 triangle inscription are divided differently, implying a choice of triangulations.

• This is a very nice construction! I kind of want to make an illustration of regular heptagons tiling the plane in this way now. Aug 29, 2022 at 16:34
• Although if we just want a tiling with regular heptagons, we can also get one greedily. Take the unit square, and for each $n=1,2,3,\dots$, fit as many regular heptagons with side length $\frac1n$ as you can inside the empty spaces left over from previous steps. The closure of this tiling must contain the entire region, because if a point $x$ has a ball $B_\epsilon(x)$ around it, eventually we will place a heptagon that overlaps that ball, because for large enough $n$ a heptagon of side length $\frac1n$ will fit inside the ball. Aug 29, 2022 at 16:42