New very simple Golden Ratio $\Phi$ construction with Circle and Two Equal Segments of Circle Diameter. Is there prior art? Proofs?

Question

Geogebra gives me the golden ratio $\Phi$ to fifteen decimal places for this simple construction illustrated below wherein the ratio of the blue line i to the red line h is $\Phi$ or 1.6180....

The golden ratio construction is made in the following manner:

1. Draw a circle resting on a line.
2. Draw a segment (segment f) equal to the diameter of the circle from the center of the circle to the line at point C.
3. Draw a second segment of the same length as the diameter of the circle (segment g) from point D where segment f intersects with the circle so that it also touches the line at point F.

The ratio of the blue segment i to the red segment h will then be the golden ratio $\Phi=1.6180\cdots.$

Has anyone seen any prior art relating to this construct? And again, both geometric and trigonometric proofs are welcome! :)

Answer 1

Computing the power of point $A$ with respect to $\bigcirc R$ in two ways, and the power of point $B$ with respect to $\bigcirc O$ in two ways: $$\begin{align} |\overline{AP}|^2 = |\overline{AO}| |\overline{AT}| &\qquad\to\qquad r \cdot 3 r = a^2 \\ |\overline{BP}| |\overline{BA}|\; = |\overline{BQ}| |\overline{BS}| &\qquad\to\qquad r \cdot 3 r = b\;( a + b ) \end{align}$$ Therefore, $$a^2 = b\;(a+b) \tag{$\star$}\qquad\to\qquad \frac{a}{b} = \frac{a+b}{a} \qquad\to\qquad \frac{a}{b} = \phi = 1.618\dots$$ by the definition of $\phi$, the golden ratio. $\square$

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As the literature on the golden ratio is vast, no one can definitively declare that there is "no prior art". I'll only say that this construction is not as "obvious" as some of your recent ones.

(For context, I'd say: If a construction involves a $1$-$2$-$\sqrt{5}$ triangle from the get-go, then the appearance of the golden ratio is likely to be "obvious", or, at least, "unsurprising".)

Answer 2

The ratio is $\sqrt{\dfrac{2}{3 - \sqrt{5}}}$.

Consider the right triangle HIC. We may calculate the length of the blue segment, which is $r\sqrt{3}$, and the angle ICH, which is $\pi/6$.

Consider the triangle FDC. From the Law of Sines, we may find angle CFD, which is $\sin^{-1}(1/4)$. Since the sum of the interior angles is $\pi$, we find angle FDC, which is $5\pi/6 - \sin^{-1}(1/4)$.

Consider the triangle FHD. Angles FDC and FDH are supplementary, so the measure of angle FDH is $\sin^{-1}(1/4) + \pi/6$. From the Law of Cosines, we may find the length of the segment FH, which is $r\sqrt{5 + \dfrac{1-3\sqrt{5}}{2}}$.

Consider the right triangle FIH. We may find the length of the red segment, which is $r\sqrt{4 + \dfrac{1 - 3\sqrt{5}}{2}}$.

Answer 3

I don't believe there exists prior art for this construction of the golden section.

Answer 4

This construction is effectively the same thing as the "3 lines" construction which I believe is first published by Jo Niemeyer in Forum Geometricorum in 2011. See this PDF file. The difference is that your original circle corresponds to Niemeyer's vertical line, your segment f is equal to the Niemeyer's second line, and your segment g is equal to Niemeyer's third line but drawn in the opposite direction.