*I’ve been encouraged to apply the type of analysis from my previous post to another situation.I’ll follow the same format. (Did I mention that I’m lazy?)*

The figure shows one of Đào Thanh Oai’s constructions of the golden ratio, \(\phi = 1.618\dots\).

Paraphrasing …

The circle through a vertex of an equilateral triangle, and through points one-fifth of the way up the adjacent sides, separates the third side into segments whose lengths are in the proportion $$1 \;:\; \phi \;:\; 1$$

For the general analysis, we consider an isosceles triangle with legs of unit length and vertex angle of half-measure \(\theta\). And we suppose that the circle through the vertex, and through points at distance \(s\) up the legs, separates the base in proportion \(1:r:1\) (for not-necessarily-golden ratio \(r\)), as shown. Calculating the power of a base vertex with respect to the circle in two ways (along the leg, along the base) gives …

$$s \cdot 1 = 2 x \cdot (2 x + 2 x r ) = 4 x^2 (1 + r)$$

Then, since \(\sin\theta = 2 x + x r = x (2 +r)\),

we can write …

$$s = 4 \sin^2\theta\;\frac{1+r}{(2+r)^2} \qquad(\star)$$

In the particular case \(r = \phi\), the fraction in \((\star)\) doesn’t

*completely vanish*as it did before;

it reduces to \(1/5\):

\displaystyle \qquad\quad s = \frac{4}{5}\sin^2\theta\qquad\text{or}\qquad \sin\theta = \frac{\sqrt{s}}{2}\cdot\frac{\sqrt{5}}{5}

\end{array}$$

\theta &= \;\;0^\circ :\; s = 0/5 \\[4pt]

\theta &= \color{blue}{30^\circ} :\; s = \color{blue}{1}/5 \quad\to\quad\text{Đào’s Triangle}\\[4pt]

\theta &= \color{red}{45^\circ} :\; s = \color{red}{2}/5 \quad\to\quad\text{Đào’s Square}\\[4pt]

\theta &= \color{violet}{60^\circ} :\; s = \color{violet}{3}/5 \quad\to\quad\text{Someone’s Hexagon}\\[4pt]

\theta &= 90^\circ :\; s = 4/5

\end{align}$$

Along with the degenerate cases for \(0^\circ\) and \(90^\circ\), the triangle, square, and hexagon configurations exhaust all possibilities of Đào-like constructions involving an “\(n\)-fifths circle” for integer \(n\).

The diagram at right combines the corresponding figures within a common circumcircle, and with a companion ellipse through separation points on appropriate chords. In this context, the chord for the \(90^\circ\) figure collapses to a point, which actually obscures an important phenomenon.

Holding the edges to a constant length (and focusing on the isosceles triangles formed with the chords), we find that the \(90^\circ\) case, in and of itself, is problematic: the corresponding “four-fifths circle” is actually a *straight line* that coincides with the flat triangle’s legs and its base, making the locations of the separation points *undefined*. The problem resolves, however, when this case is viewed as the limiting form in the *continuous family* of triangles. (An analogous observation can/should be added to my previous article. *(Lazy!)* )