## Special Angles are Golden

This is George Odom‘s elegant construction of the ever-fascinating golden ratio, $$\phi$$:

If $$a$$ measures a “midpoint segment” of an equilateral triangle, and if $$b$$ measures the extension of that segment meeting the triangle’s circumcircle, then $$\frac{a}{b} = \phi = 1.618\dots$$

As it turns out, Odom’s construction generalizes very nicely when we replace “midpoint segment” with “trisecting segment” or “quadrisecting segment”, and “equilateral triangle” with “square” or “regular hexagon”.

I became aware of the square result while contemplating this post to Math Stack Exchange by Dr. Elliot McGucken (aka, “Astrophysics Math”). This prompted my own (re?-)discovery of the hexagon result, which simply falls out of the straightforward analysis I provide below. It’s possible —even likely— that all of these observations already appear somewhere in the vast lore on the golden ratio, but I’m too lazy to do a literature search.
Update: The essence of the square construction appears in an isosceles-right-triangle construction
published in 2015 by Q. H. Tran (PDF). Curiously, despite noting the Odom-esque flavor of his result,
Tran doesn’t seem to realize that his figure features an obvious, and even-more-Odom-esque,
segment trisector. (The missed connection with the square is less surprising, given the semi-circular
context.) Nevertheless, it appears appropriate to assign priority to what I’ll call “Tran’s Square”.
Be that as it may …

To see the underlying connections here, consider an inscribed angle of half-measure $$\theta$$, and suppose the “$$(n+1)$$-secting segment” through the sides of this angle is such that the ratio of its length to that of its circle-meeting extension is $$r : 1$$, for some not-necessarily-golden ratio $$r$$.
By the chord-chord aspect of the Power of a Point theorem, we have
$$n\cdot 1 = x \cdot ( x + r x )\quad\to\quad x = \sqrt{\frac{n}{1+r}}$$
But, clearly, we also have $$r x = 2\sin\theta$$, so that
$$\sin\theta = \frac{rx}{2} = \frac{\sqrt{n}}{2}\cdot\frac{r}{\sqrt{1+r}}\qquad(\star)$$
Now, recall that what makes the particular ratio $$\phi$$ “golden” is that the square of its value is precisely $$1$$ unit greater than the value itself; that is, $$\phi^2 = 1 + \phi$$. Consequently, when $$r = \phi$$, the second factor in $$(\star)$$ vanishes, leaving …

$$\sin\theta = \frac{\sqrt{n}}{2}$$
… which should be familiar to you
from the common mnemonic
involving trig’s “special angles” …
\begin{align} \sin \phantom{0}0^\circ &= \sqrt{0}\,/2 \\[4pt] \sin \color{blue}{30^\circ} &= \sqrt{\color{blue}{1}}\,/2 \quad\to\quad\text{Odom’s Triangle}\\[4pt] \sin \color{red}{45^\circ} &= \sqrt{\color{red}{2}}\,/2 \quad\to\quad\text{Tran’s Square}\\[4pt] \sin \color{violet}{60^\circ} &= \sqrt{\color{violet}{3}}\,/2 \quad\to\quad\text{Someone’s Hexagon} \\[4pt] \sin 90^\circ &= \sqrt{4}\,/2 \end{align}
In this very peculiar sense, then, special angles are indeed golden.

Our sine relation establishes that $$n$$ cannot exceed four. Therefore, the triangle, square, and hexagon above represent all available (non-degenerate) Odom-like constructions for integer $$n$$. The composite view at right shows a bonus feature: All the dividing points lie on an ellipse. (No, the ellipse isn’t “golden”: neither its eccentricity, nor its aspect ratio, is $$\phi$$.)

Of course, no one requires $$n$$ to be an integer,
so there remains more to say. For instance, the Odom-like construction for the regular pentagon, which already admits so many connections with the golden ratio, has $$n = \phi^2 = \phi+1$$. Also, $$n = \phi$$ implies $$\theta \approx 39.49^\circ$$. (Is there anything (else) remarkable about that angle?) And so on.

I’ll leave further considerations to the reader.