# Limitations on a Growing Shape

Here is a way to create a shape:

For convex regular shape S with n sides: add a convex regular shape with n+1 sides to the outer edge of each open side (I'll explain what an open side is in a moment) of S, with one of the sides of the shape being added having its midpoint on the midpoint of the side it's being added to, the sides of the two shapes being parallel, and the side of the added shape being no longer than the side of the shape you're adding to. Then, one could repeat the process for each shape of n+1 sides, adding shapes of n+2 sides. This could continue.

As an example, one could start with n=3, an equilateral triangle. On each side of the triangle, the 4-sided convex regular shape, a square, is added, with the midpoint of one side being on the midpoint of a side of the triangle. You can decide the length (above 0) of the sides of the square.

After this step, you could add pentagons to the sides of each square that do not rest on the triangle (these sides are open).

The shape will be considered complete if no separate convex regular shapes overlap.

My question is this: Given n=3, and the initial side length is 1, what is the maximum side length in a complete shape, iterated up to n=7, of the heptagons (7-sided shape)?

You will have: 1 triangle, 3 squares, 9 pentagons, 36 hexagons, and 180 heptagons.

How can you maximize the side length of the heptagons to keep the shape complete (ensure that nothing is overlapping)?

Thank you!

• Are the shapes created in stages. In stage 1, you are allowed to attach squares to triangle, In stage 2, you are allowed to attach pentagons to existing squares or triangles (or only to squares?). In stage 3, you are allowed to attach hexagons but not triangles/squares/pentagons to existing polygons... Another question is do the immediate shapes obtained after each stage need to be complete or not? – achille hui Dec 1 '16 at 15:02
• There should be no open faces on a shape with x sides when you are adding shapes with x+2 sides. When adding shapes with x sides, you must add to all open sides of each shape with x-1 sides. And the final product must be complete, which would require each stage to be complete. While you're looking for your answer, feel free to use incomplete answers as testing. Thanks for your question! – Dashiell Shulman Dec 1 '16 at 15:11
• I see. If I didn't make any mistake, after adding the squares, the configuration is unique. after adding the pentagons, there are 2 topologically distinct configurations. Same thing happens after adding the hexagons (both of them contains 1 triangle/3 squares/6 pentagons/9 hexagons ). In each configuration, we are leave with $21$? open edges for adding heptagons. the problem now seems tractable. – achille hui Dec 1 '16 at 16:05

Note - It turns out this doesn't answer what OP is asking. I failed to notice the new polygons added need not has same side lengths as the original triangle.

Oh well, c'est la vie.

After adding the heptagons, all complete figures consists of

$$1 \text{ triangle},\; 3 \text{ squares},\; 6 \text{ pentagons},\; 9 \text{ hexagons and } 9 \text{ heptagons}.$$

Since adding a $n$-gon introduces $n-1$ new edges and remove $1$ old edges, the resulting figures has $$1(3) + 3(4-2) + 6(5-2) + 9(6-2) + 9(7-2) = 108\text{ sides}.$$

Above picture is two possible complete figures superposed into one. If you ignore the three red polygons, you get one complete figure. If you ignore the three green polygons, you get another one.

We will build the complete figures in stages.

• At stage $0$, we start from a single triangle.

• At stage $1$, we add the squares.

Since we can add squares to any sides of the triangles without intersecting each other. The resulting complete figure is unique, consists of $1$ triangles and $3$ squares.

• At stage $2$, we add the pentagons.

The figure from stage $1$ has $9$ edges. Adding a pentagon to any of them will not intersect the existing triangles and squares. $3$ of them is "non-interfering". You can add pentagon there without worrying it intersect with other new pentagons. For the remaining $6$ edges, they falls into $3$ pair. In each pair, you can add at most one pentagon to an edge belongs to the pair.

All resulting complete figures consists of $1$ triangles, $3$ squares and $6$ pentagons. There are essentially two topological distinct configures. To see what they look like, ignore either the red or the green polygons (but not both) and all hexagons/heptagons from above picture.

• At stage $3$, we add the hexagons.

For both complete figures from stage $2$, there are $9$ edges one can add hexagons without intersecting with the old polygons. Furthermore, adding any one of them will not interfere with another. All resulting complete figures has $9$ new hexagons. Once again, there are essentially two topological configurations.

• At final stage $4$, we add the heptagons.

For both complete figures from stage $2$, there are $15$ edges one can add hexagons without intersecting with the old polygons. $3$ of the edges are non-interfering, one can always add heptagons there. For the remaining $12$ edges, it falls into $3$ groups (the blue edges in above figure). In each group, you cannot add heptagon next to each other. This means you can add at most $2$ heptagons to each other. As a result, all complete figures after this stage has $9$ new heptagons.

• Wow! I appreciate the depth and clarity of this response, but I feel that it answers a slightly different question than the one I had hoped to propose. The thing I feel I must stress is that a shape MUST be added to each open side, which, to maintain completeness, requires the shortening of side lengths of pentagons/hexagons/heptagons. Though I stated this in my original post, I suppose my image made it slightly unclear. The shapes you add need not fill the whole edge, just fulfill the criteria above. At Stage 2, you would have to shorten your pentagon sides to allow for placement of each. – Dashiell Shulman Dec 1 '16 at 18:28
• Thank you, though--I really do appreciate the work you put into that answer. – Dashiell Shulman Dec 1 '16 at 18:29
• Hmm... in that case, the problem becomes significantly more complex – achille hui Dec 1 '16 at 19:03
• Yes. The problem involves following 3-7 variables. A significant simplification arises when you realize you only need look at a small portion of the growing shape (generally, one shape of each size and two of your current size). If you go too small early, you limit your later size because side length can only decrease (you can't add a shape with longer side length). If you go too big, you're left with little space in the corners in which to fit heptagons. – Dashiell Shulman Dec 1 '16 at 19:17