New, extremely simple golden ratio construction with two identical circles and line. Is there any prior art?

Question

This question is different from a previously asked question (linked above) as this golden ratio construction only utilizes two circles and a line, and is thus far simpler than the golden ratio construct in a previously asked question, which uses two squares, a circle, and a line. Thanks!

Illustrated below, please find a new, extremely simple golden ratio construction with just two identical adjacent circles and a line, wherein the ratio of the red line to the blue line is the golden ratio PHI (1.6180....)

Is there any prior art? I have been searching long and hard, but cannot find a similar golden ratio construction.

The simple construction is created as follows.

1. draw two adjacent circles with the same diameter.

2. draw a line from the top of one circle through the center of the second circle.

3. the ratio of line segment h to line segment g (the red segment to the blue segment) will then be exactly PHI or 1.6180....

I've been searching numerous books/online websites/resources for any previous similar constructions. If you know of any, please do share! Thanks!

P.S. User @Peter Woolfitt provides a seemingly very nice proof here of a slightly different construction, and any more proofs, either trigonometric or geometric would be weclome! New Golden Ratio Construction with Two Adjacent Squares and Circle. Have you seen anything similar?

Answer 1

Slightly more general answer.

Let $R$ and $B$ be the lengths of the red and blue lines respectively. If the radius of the circles is $r$, then we have the equations $$R=2r$$ since $R$ is the diameter of one of the circles, and $$B+r=\sqrt{r^2+(2r)^2}=r\sqrt5$$ since $B+r$ is the hypotenuse of a right triangle with legs of length $r$ and $2r$.

Hence $$\frac{R}{B}=\frac{2r}{r\sqrt5-r}=\frac{2}{\sqrt5-1}=\frac{2\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{2\left(\sqrt{5}+1\right)}{4}=\frac{\sqrt{5}+1}{2}=\varphi$$

Answer 2

Here's the proof: $$6^2+3^2=36+9=45$$ $$\frac{6}{\sqrt{45}-3}=\frac{2}{\sqrt{5}-1}$$ This was simple enough to just write down. For the last step we have: $$\frac{\sqrt{45}}{3}=\sqrt{x}$$ $$\frac{45}{9}=5$$ NB That was relatively simple so I can't claim any credit, the OP had the idea which may be original.

Answer 3

A common way to construct the golden ratio is to construct two perpendicular segments joined at endpoints, whose lengths are in the ratio $2:1$ like the segments $AC$ and $AD$ in the figure in the question. One then constructs the segment connecting the other two ends of these segments to form a right triangle (whose hypotenuse is $CD$ in the figure in the question).

Finally, one marks off the length of the shorter leg of the triangle ($AD$ in the question, but labeled $BC$ in the figure below) on the hypotenuse (at point $E$ in the question, $D$ in the figure below). The remaining portion of the hypotenuse ($DE$ in the question, $AD$ in the figure below) has length $\sqrt5 - 1$ times the length of the shorter leg of the triangle, and therefore is in the ratio $\phi : 1$ (the golden ratio) with the longer leg.

There are examples of this construction at http://www.goldenmuseum.com/0202geometry_engl.html and http://jwilson.coe.uga.edu/MATH7200/Sect4.4.html.
A typical diagram of the construction looks like the following figure, taken from https://commons.wikimedia.org/wiki/File:Construction_of_a_golden_ratio.svg:

There are a number of ways you can construct the two perpendicular segments, of which putting two congruent squares side by side (as in New Golden Ratio Construction with Two Adjacent Squares and Circle. Have you seen anything similar?) or by setting two congruent tangent circles' centers on a line to mark the longer leg, and use one of the circles to mark the distance on the shorter perpendicular leg.

The triangles in the figure in the question and the figure from Wikimedia are differently oriented (reflected left-right) and labeled with different letters, but those are unimportant differences. The main difference between the figures is that in the Wikimedia figure, the length of the shorter leg is marked off from the end of the hypotenuse on the shorter leg, whereas in the question that length is marked off from the end on the longer leg. If the two legs of the triangle are formed from the edges of two adjacent squares, as they are in New Golden Ratio Construction with Two Adjacent Squares and Circle. Have you seen anything similar?, then even this difference is not as significant, since we have two copies of the triangle.

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Note: At this point in the prior-art construction we have already constructed segment $AD$ so that $AB : AD$ is the golden ratio. If $BC = 3$ (the radius of the circles in the question), then $AD$ in the prior-art construction is exactly congruent to $DE$ in the question. If we must have both segments in the ratio lie end-to-end along the hypotenuse of the triangle, we could simply complete the circle with center at $C$ and extend segment $AC$ to meet the circle at $F$, at which point $DF : AD$ is the golden ratio, and the three-point figure $ADF$ in this construction would be congruent to the figure $DEF$ in the question.

Additional Note: This paragraph is not particularly relevant to the question, except that it shows a slight advantage to doing the construction in the "Wikimedia" fashion: namely, after having already constructed the golden ratio once, we get a "bonus" of constructing a second pair of segments in the golden ratio with merely one action of a collapsible compass. Namely, we strike an arc with center $A$ from $D$ to $E$, thereby copying the length of $AD$ onto the long leg of the triangle dividing the segment $AB$ into the two parts $AE$ and $EB$, which also are in the ratio $\phi : 1$ (the golden ratio). To divide the segment $AC$ in the question in this fashion, we have to construct a point $P$ between $A$ and $C$ such that $CP \cong DE$. This is easy enough with a non-collapsing compass (use segment $DE$ to set the compass, then put one end of the compass at $C$ and strike an arc across the segment $AC$) but rather tedious to do with a collapsible compass (although that too is a standard classical geometric construction).