# Are the lengths from this recursive construction a geometric sequence?

In his 1999 review of Edward Tufte's Visual Explanations in the Notices of the AMS (third page), Bill Casselman gives a very pretty proof of the irrationality of the golden mean. More precisely, Casselman shows that the side and diagonal of a regular pentagon cannot be commensurable. The proof assumes the contrary, and thereby constructs a vanishing sequence of commensurable terms, which is a contradiction.

Upon closer inspection, however, I realized that there's a step in the proof that I cannot justify to my satisfaction. Casselman does show that the commensurability of the side and the diagonal of a regular pentagon implies the existence of a strictly decreasing sequence of commensurable lengths (e.g. the sequence consisting of the lengths of the sides of successively constructed nested regular pentagons). It is not clear to me, however, that he showed that this sequence converges to 0 (and not to some positive number).

Of course, convergence to 0 would follow immediately if we could establish that the sequence of side lengths is a geometric one, i.e. that successive terms in it are in a constant ratio to each other (which has to be $< 1$, since the sequence is strictly decreasing). That this ratio is constant seems to me intuitively obvious somehow, but cannot I prove it rigorously.

Any help would be appreciated!

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The inner regular pentagon is similar* to the outer one, with similarity factor $\rho\lt 1$. So if we iterate $n$ times, we get a pentagon of side $\rho^n$ times the side of the original. – André Nicolas Apr 16 '13 at 2:03
I know, but the crux is showing that there is indeed a constant similarity factor between successively constructed pentagons. – kjo Apr 16 '13 at 2:06
It is the same construction. – André Nicolas Apr 16 '13 at 2:10
I know, but we're missing a theorem, or axiom, here to justify going from "the same construction" to "constant ratio". After all, one can devise recursive constructions that result in varying ratios from one stage to the next (Ford circles, for example). – kjo Apr 16 '13 at 2:15
Geometric (say straightedge/compass) constructions are invariant under similarity. – André Nicolas Apr 16 '13 at 2:17

It definitely is a constant ratio. If $P$ and $P^\prime$ are regular pentagons with side lengths $l$ and $l^\prime$ respectively, then any distance $d$ between points $A$ and $B$ in $P$ obeys $d^\prime = d \frac{l^\prime}{l}$, where $d^\prime$ is the distance between points $A^\prime$ and $B^\prime$ in $P^\prime$ isomorphic to $A$ an $B$ in $P$, respectively. One consequence of this is $a^\prime = \left( \frac{l^\prime}{l} \right)^2$, where $a$ and $a^\prime$ are the areas of $P$ and $P^\prime$, respectively.

Another way to look at it: Draw a schematic with only one iteration of the construction, and define the ratio of side-lengths $l^\prime/l = \rho$. For repeated constructions, redefine the points on your schematic adding a "prime" to everything. Then $l^{\prime\prime}/l^\prime = \rho^\prime$. Is it possible that $\rho^\prime \neq \rho$? The schematics are exactly the same.

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As far as I can tell, you have just reiterated the fact that all regular pentagons are similar, which is not in question. What is in question is that the sides of the first and second pentagon, for example, are in the same proportion as the sides of the second and third pentagon, and, more generally, of the $n$-th and ($n+1$)-th pentagon, for all $n$. – kjo Apr 16 '13 at 2:24
What I am saying is that, in the repeated constructions, all ratios between the first pentagon and the second are equal to the equivalent ratios between the second and the third, because the first pentagon is similar to the second one. Therefore, the side length of the $k^{\textrm{th}}$ pentagon is $\rho^k$ times the side length of the first, as stated by André. – Douglas B. Staple Apr 16 '13 at 2:35

OK, I think I have a an argument that formalizes my intuition. (It is, admittedly, pretty tedious, but I guess this is inevitable, since it is formalizing a reasoning that is intuitively obvious to most people.)

Let $p_0$ be the construction's initial regular pentagon, let $s_0$ be the "base" of $p_0$ (namely, the side of $p_0$ that runs horizontally along its bottom), and let $t_0$ be the triangle formed by $s_0$ and the two diagonals of $p_0$ incident on it. Now, for every $i \in \{1, 2, 3, \dots\}$, let $p_i$ be the first regular pentagon produced by the construction inside of $p_{i - 1}$. Let $s_i$ be the (unique) side of $p_i$ that is parallel to $s_{i - 1}$, and let $t_i$ be the triangle formed by $s_i$ and the two diagonals of $p_i$ that are incident on it.

Now, define $F$ as the (composite) figure consisting of all the triangles $t_0, t_1, t_2, \dots$. Likewise, define $G$ as the (composite) figure consisting of all the triangles $t_1, t_2, t_3, \dots$. It is easy to see that the $i$-th triangle of $F$ (namely $t_{i-1}$) is similar to the $i$-th triangle of $G$ (namely $t_i$). From this correspondence, and the details of the construction, it follows that $F$ is similar to $G$.1

Therefore, there is a "scaling factor" $\rho$ such that, for every length $l_F$ in $F$ and every corresponding length $l_G$ in $G$, the relation $l_G = \rho l_F$ holds. In particular, for every side length $s_i$ in $F$ and corresponding side length $s_{i+1}$ in $G$, we have $s_{i+1} = \rho s_i$.

From this it follows immediately that $s_n = \rho^n s_0$, for all $n \in \{0, 1, 2, \dots\}$.

1This implication, admittedly, is not entirely watertight, since one could make a figure $G^\prime$ consisting of triangles $t_1, t_2, t_3, \dots$ that make up $G$, but independently repositioned (translated and/or rotated) such that $G^\prime$ would not be similar to $F$ even though the same pairwise similarity between their component triangles exists. A more detailed accounting of all the angles involved would have to be made to establish the similarity of $F$ and $G$. It is clear, however, that the construction fixes the position of each triangle $t_i$ precisely within $t_{i-1}$, and that the vertices of $t_i$ and $t_{i-1}$ together defines 4 other triangles, which, together with $t_i$ fully tile $t_{i-1}$. The recursive construction preserves this tiling of each triangle, and it is easy to show that, for each $i$, the triangles that tile $t_i$ are similar to the corresponding triangles that tile $t_{i+1}$. In this way the figures $F$ and $G$ may be construed as similar tilings of the similar triangles $t_0$ and $t_1$.

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