# Find a heptagon with mirror symmetry that can tile a flat plane

A seven-sided flat shape of fixed size in which all angles are equal and all sides of the same length, called a regular heptagon, cannot tile a flat plane. The only regular shapes that can are the equilateral triangle, the square, and the regular hexagon.

But if we drop the condition of regularity, there is a nice tiling of the plane using a five-sided shape: a pentagon that has all of its sides the same length (it is equilateral), and which is also a mirror image of itself (it has reflectional symmetry), but which has angles of different sizes (it is not equiangular), namely the Cairo pattern: It is also possible to use an irregular hexagon (six sides) that is neither equilateral nor equiangular: The puzzle is to find a convex equilateral heptagon that permits such a tiling and has reflective symmetry; or to prove that none exist. In other words, we are looking for a shape that

• has seven straight sides of equal length
• is the mirror image of itself
• contains no pair of points that cannot be connected by a straight line inside the shape
• can tile a flat plane

This is not a trick question. Euler-style tilings where the shape is allowed to vary in size are not allowed, not even if you can find a way to distort the following so that the heptagons, while remaining of different sizes, all become similar in shape! (Note: originally I forgot to include the convexity condition, and M Oehm posted an excellent solution using a heptagon formed from a regular hexagon with a triangular bite taken out of it. This was a correct answer to the question as stated, and deserved the tick, but I have now added the convexity condition with his permission.)

(Further note: for the avoidance of doubt, I am using the word "side" to mean a polygon's edge that meets adjacent edges at each of its ends and does not form part of a single straight line with either of them. In other words it meets each of the two adjacent edges at a corner, where the interior angle at a corner is not 180 degrees. This is just ordinary English usage.)

• your second picture uses an irregular hexagon
– JMP
Oct 1 '17 at 11:08
• Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :)
– Rubio
Oct 11 '17 at 14:37

One possible way is to use ...

... a concave, symmatrical and equilateral heptagon that is created by incising a regular hexagon or by taking away one of the equilateral triangles that make up a regular hexagon. Two of these "pacmanised" shapes can form an elongated hexagon. These hexagons can tile a plane: • Ouch. This fine and completely correct solution is a "cook" of what I intended to be the problem, which was to find a convex shape. Shall I give you the fully deserved tick, or do you mind if I add the convexity condition?
– h34
Oct 1 '17 at 11:22
• @h34: If that's not what you intended, amend your question and give the tick to whoever solves the problem as intended. I'm easy. :) Oct 1 '17 at 11:25
• @h34 No such convex heptagon exists. At any point $P$ where 3 or more tiles meet, the mean of the angles at $P$ is at most $2\pi/3$. A heptagon's interior angles sum to $5\pi$ so its mean angle is $5\pi/7>2\pi/3$. Sustaining so high a mean entails points where the interior angles of only 2 tiles meet. This means that either the tile is concave, or you are counting 2 or more pieces of a single side as separate sides and your "heptagon" in fact has no more than 6 sides. Oct 1 '17 at 12:50
• @RosieF - Nicely shown. If you write this as answer I will accept it.
– h34
Oct 1 '17 at 14:06

(This question, as posted earlier, had not specified that the heptagon must be convex. My answer here addresses the later version of the question, which specifies that it must be convex.)

No such convex heptagonal tile $T$ exists. At any point $P$ where 3 or more tiles meet, the mean of the angles at $P$ is at most $2\pi/3$. $T$'s interior angles sum to $5\pi$ so $T$'s mean interior angle $M(T)=5\pi/7>2\pi/3$.

Consider the "mean angle of a tiling" defined as follows. Pick an arbitrary point $O$ in the tiling. For any radius $r>0$, let $C(r)$ be the circle centre $O$, radius $r$. Consider the set $S$ of all points $P$ where the interior angles of 2 or more tiles meet, where $|OP|<r$. What is the mean $M(r)$ of all these interior angles? It is a weighted average of the respective averages over all points in $S$; the weight for $P$ being proportional to the number of angles meeting at $P$. For large $r$, $M(r)\approx M(T)$. The two means might not be exactly equal, because of tiles that straddle $C$ and have at least one vertex within $C(r)$ and at least one outside $C(r)$. But such tiles can be made an arbitrarily small proportional of the total by taking $r$ large enough. Thus $$\lim_{r\to\infty} M(r)=M(T).$$

As shown above, $M(r)$ cannot be made that high using only points where 3 or more tiles meet. This entails points where the interior angles of only 2 tiles meet. At any such point $P$, one of two possibilities occurs.

One possibility is that one of those angles is greater than $\pi$, i.e. it is a reflex angle, which means that the tile containing it is concave.

The other possibility is that both of the angles there are equal to $\pi$. But in that case, $P$ is not a vertex of either of the 2 tiles meeting there. To call $P$ a point where two vertices meet would be to count 2 or more pieces of a single side as separate sides; the "heptagon" would in fact have no more than 6 sides.

Note that it is OK for 3 or more tiles to meet at $P$ and for one of the angles at $P$ to be $\pi$. This would mean that vertices of 2 or more tiles met at $P$ along with some point of another tile which is not a vertex.

• Nice argument. If you really want to make it rigorous, you can notice that the tiling of the plane is equivalent to a tiling of a torus. Using the Euler characteristic, you can then show that if at least m edges of an n-gon meet at each vertex, you must have $1 + n/m \geq n/2$. In particular, if $m = 3$, then we must have $n \leq 6$. Oct 1 '17 at 14:17
• A tiling of a torus is equivalent to a tiling, periodic in both dimensions, of the plane. But I don't assume that every tiling of the plane by congruent tiles is periodic. Oct 1 '17 at 14:21
• Hmm...that last sentence of the answer. Why must there be any points where at least three corners meet? It is possible to tile a plane using a 2x1 rectangle without having such points. And if a corner of one of our hypothetical heptagons meets a corner of another on the side of a third, the average size of the two relevant angles is only $\frac{\pi}{2}$.
– h34
Oct 1 '17 at 14:32
• I don't think this proof holds as it stands. We can have points at which three tiles meet but only two vertices meet, summing to $\pi$, partway along a side of the third tile. It's easy to tessellate squares so that at each vertex exactly three squares meet, but the mean interior angle does not become $\frac{2\pi}{3}$.
– h34
Oct 1 '17 at 17:25

I'd like to think that your first image is a hint:

Each set of 4 pentagons in that tiling forms a pretty obvious irregular hexagon. If you remove one of the pentagons that have three outside edges, you get a (non-convex) symmetric heptagon that tiles very nicely.

Here's a low-tech demonstration. The letters show heptagons with the same orientation; I suspect there are other tilings as well.

• A very nice tiling!
– h34
Oct 1 '17 at 14:38
• Do you mean that you remove one of the pentagons with two inside edges? Looking at shape C you have removed the left hand pentagon which has three external edges and two internal ones... Or is this one of those things where if I look at it in a different way what you say is true in a different way? Oct 1 '17 at 23:35
• Oops, yes, I meant to write "three outside edges! Thanks! Oct 2 '17 at 15:31

Edit: This answer is now excluded by yet another edit to the question. The revised question explains very wordily that the intended polygons are strictly convex. Starting at the lowest vertex and proceeding either clockwise or counterclockwise, the interior angles are $\pi/3$, $\pi$, $2\pi/3$, $2\pi/3$, $2\pi/3$, $2\pi/3$, and $\pi$. All sides have the same length.

These combine into a somewhat hexagonal unit cell. And these tile the plane. • @h34 : I disagree. These are equilateral heptagons. I see none of your criteria lead to a rejection of straight angles. Oct 1 '17 at 23:10
• OK, I take your point, but my excuse is I was trying to write something as close as possible to ordinary English. If we allow straight angles, then every square is a pentagon, a hexagon, a heptagon, etc.
– h34
Oct 1 '17 at 23:23
• @h34 : But a square is not an equilateral pentagon, hexagon, heptagon, nonagon, decagon, undecagon, tridecagon, et c. Equilaterality is rather strict. Oct 2 '17 at 0:29
• The degenerate case of allowing 180 degree angles would also allow trapezoids, which are boringly easy to tile. So no, even though your tiling is very pretty, it's not made of heptagons. (Mainly because convex 7+gons cannot tile a plane.)
– Bass
Oct 2 '17 at 3:39
• @zzzzBov : I see. When "M Oehm posted an excellent solution using a heptagon formed from a regular hexagon with a triangular bite taken out of it. This was a correct answer to the question as stated, and deserved the tick". But when I propose a solution to the question as written, I'm engaging in an "obvious attempt at bending the rules". Your double standard is not sufficiently interesting to merit further attention. Oct 2 '17 at 18:22

This is kind of cheap but, quoting Wolfram:

There are no tilings for identical convex n-gons for $n\geq7$, although non-identical convex heptagons can tile the plane (Steinhaus 1999, p. 77; Gardner 1984, pp. 248-249).

So no such tiling exists.

• That's not cheating, that's research, actually :)
– yo'
Jan 11 '18 at 23:20