# Unreflected infinitely simple polygon reflexivity

(This was retrofitted to more tightly match a surprise solution and to allow for another puzzle with the original intent.)

Reflexivity — When the self refers to itself. Above is a simple polygonal region divided into infinitely many different-sized copies of itself.   Each copy is √2 = 1.414... times as large as the next smaller one (in terms of linear scale, not area).   If 2 copies are removed, the remaining polygonal region is a scaled-down version of the original.

Can you find another simple polygon that has 4 or more sides and can be divided into infinitely many different-sized copies of itself, where the original polygonal region is geometrically similar, without reflection, to what remains if 2 or more component copies are removed?

The open-ended goal is a maximum successive-size ratio as close as possible to 1.

Reflection is not in play.   Each copy size occurs only once.   Polygons in this puzzle have finitely many vertices.   Note that the goal is to minimize the maximum, not average or smallest, ratio between any two successively sized copies.   The large composite polygon is not included in these ratios.

• Are you allowed to have multiple same-sized polygons in your dissection? – boboquack Feb 15 '17 at 21:37
• Each size is one-of-a-kind, @boboquack. I'll clarify. – humn Feb 15 '17 at 21:39
• I might be crazy, but these polygons appear to have a ratio of 2? – Sconibulus Feb 15 '17 at 21:55
• @Sconibulus I think maybe you're talking about the ratio of areas and humn is talking about the ratio of linear dimensions. – Gareth McCaughan Feb 15 '17 at 23:46
• Reminds me of a quote: "The square root of I is I" (Nabokov, Bend Sinister) – GoldenGremlin Feb 16 '17 at 0:20

I have a solution with area ratio

the golden ratio $\frac{1+\sqrt5}2$

and hence side ratio

its square root, or about 1.272.

Here's how it works.

Write $u$ for the reciprocal of our side ratio. Our polygon -- call it $A$ -- is simply a rectangle, 1 unit by $u$ units. We will chop it up into smaller rectangles whose longer sides are $u^2$, $u^3$, $u^4$, etc. (It is an easy exercise to verify that the total area checks out, so this isn't a crazy idea.)

So. First of all, observe that $1=u^2+u^4$ and $u=u^3+u^5$, leading to an obvious decomposition of $A$ into four smaller rectangles of sizes $u^2$-by-$u^3$, $u^4$-by-$u^3$, $u^2$-by-$u^5$, and $u^4$-by-$u^5$. The first, second and fourth of these are copies of $A$ scaled by $u^2$, $u^3$ and $u^4$. So far, so good.

Now

what remains is a $u^2$-by-$u^5$ rectangle. Since $u^2=u^4+u^6$ we can chop this into rectangles of sizes $u^4$-by-$u^5$ and $u^6$-by-$u^5$. The second of these is a copy of $A$ scaled by $u^5$. And the first -- which is all that remains after taking out the four copies of $A$ we have observed so far -- is a copy of $A$ scaled by $u^4$. We can now apply the same decomposition to this as we did to $A$, yielding four rectangles whose scales are the 6,7,8,9th powers of $u$ and a copy of $A$ scaled by $u^8$. And so on recursively ad infinitum.

Here's a diagram: Numbers indicate exponents of $u$.

• I couldn't resist putting the top left corner in curlers, Gareth, see what you think: : https://i.stack.imgur.com/ijegS.png (preview) – humn Feb 16 '17 at 3:18
• There's probably a way of drawing a purty spiral through that one. On the other hand, I think mine makes it clearer what the overall pattern is. – Gareth McCaughan Feb 16 '17 at 3:49
• I might be misunderstanding the problem, but don't you need to remove 3 copies to get something similar, where the original problem asked about removing 2? – TusanHomichi Feb 19 '17 at 21:59
• No, the original problem didn't have that clause at all. I think it didn't have it when I posted my answer, so it looks as if @humn changed the question after my answer was posted so as to invalidate the answer I gave. Did I get the timings wrong, @humn? – Gareth McCaughan Feb 19 '17 at 23:23
• (The comments on humn's edits suggest that the intention was to make the revised question match my answer...) – Gareth McCaughan Feb 19 '17 at 23:24

Gareth McCaughan’s surprise solution is laid out with the clear purpose of clarity, but a spiral form of it is too interesting to leave unnoticed. How it nestles two levels horizontally for each level vertically can tickle anyone familiar with the same old spirals of these two classics. OK, so I think humn needs to show me what's wrong with the following argument.

[EDIT: no, I don't think he does; see the end. I'm leaving this here in case others are inclined to make the same mistake as me, or merely enjoy seeing me make a fool of myself.]

Theorem: In such a configuration, the maximum ratio of areas is at least 2.

Before proving this, here's an

Observation: the area of the whole polygon equals the sum of the areas of the smaller scaled-down ones. (Proof: the larger is dissected into the smaller.) And hence a

Lemma: In such a configuration, the areas of the smaller polygons form a discrete set; they don't "accumulate" anywhere other than at zero. Proof: otherwise there would be infinitely many of 'em with area $\geq t$ for some $t>0$, which would mean infinite total area, contradiction. $\square$

Back to the theorem. Proof: Scale so that the area of the whole thing is 1. Because the areas are discrete, we can arrange them into a sequence in descending order of area; let the areas be $a_1>a_2>\cdots$. If all the area ratios $a_j/a_{j+1}$ are $\leq\rho$ then we have $a_1\geq 1/\rho$, $a_2\geq a_1/\rho\geq1/\rho^2$, $a_3\geq a_2/\rho\geq 1/\rho^3$, etc.; in general $a_j\geq\rho^{-j}$. Hence the total area of the smaller polygons is at least $\rho^{-1}+\rho^{-2}+\cdots=\frac{1}{\rho-1}$, so $\frac{1}{\rho-1}\leq1$ and hence $\rho\geq2$. $\square$

Hence the following Corollary: the maximum ratio of linear areas is at least $\sqrt{2}$. Proof: it's the square root of the maximum area ratio. $\square$

So, humn, would you care to clarify what careless assumption I've made? E.g., are you actually permitting the smaller polygons to overlap? Or are you considering "polygons" of zero or infinite area (but intermediate side-lengths, by some sort of fractal trickery)?

(If you prefer simply to indicate that you have fully understood what I wrote above but that you continue to maintain that there is a solution achieving a smaller maximum area-ratio, leaving it up to me to figure out what the hell I've done wrong, or indeed to ignore this completely, that is of course within your rights.)

[EDITED to add:] Oh, wait, I think I see what I missed. You say "the large composite polygon is not included in these ratios" so I am not entitled to assume $a_1\geq1/\rho$. D'oh.

• Whew, Gareth, I was enjoying the rare pleasure of reading a proof that matters to me, confident that it would be consistent (coming from you) while hoping that it would hinge on limiting the total area. Now that I see your postscript edit, I'm enjoying going through the body even more. – humn Feb 16 '17 at 0:23
• It's not very interesting. I just made the obvious mistake. Incidentally, I'm pretty sure I know what ratio you've got in your example, though I haven't found the right construction yet. – Gareth McCaughan Feb 16 '17 at 0:26
• Once again, you've presaged a potential follow-up puzzle, Gareth, one that allows fractals. I veered away from that here because it seemed to allow a ratio as close to 1 as desired, as triangles do. Interesting coincidence between the simplest and most complex "polygons," with no common ground inbetween, if true. – humn Feb 16 '17 at 0:27