Are the lengths from this recursive construction a geometric sequence?

Question

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!

Answer 1

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.

Answer 2

I realize this is an older question, with an accepted answer, but upon stumbling across it recently and reading Casselman's article from the reference, I would like to point out that the argument does not depend on convergence to 0 of the side and diagonal ratios for the successive pentagons as this question states, but upon convergence to a positive value less than the positive common measure $\epsilon$ assumed for the contradiction, and also that because the procedure is based upon subtraction (not multiplication) proceeding in the same way that the incommensurability argument appearing in proposition II of Euclid book X does. That is, the new diagonal is $d - s$, and the new side is $2s - d$, where $\epsilon$ measures both $s$ and $d$ as well as the new, smaller side diagonal pair obtained by subtracting $s$ and $d$ (both containing $\epsilon$). The 2 multiplier might seem problematic except for the observation that the new side is always smaller than the old side (by a number $s - (2s - d) = d - s$ measured by $\epsilon$). $\epsilon$ cannot be the common measure of side and diagonal when eventually this $\textbf{subtractive}$ process must produce a positive common measure smaller than $\epsilon$. Just as in the Euclidean algorithm, the $\textbf{ same } \epsilon$ must measure $\textbf{ all }$ the side/diagonal pairs of successively decreasing polygons. In other words $\epsilon$ cannot measure a lesser magnitude. None of this means that there is not a ratio involved as well (e.g. the appearance of the golden mean), but Casselman's argument depends on subtractions.

Answer 3

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 $t_1$.}