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 $\Phi=1.6180\cdots.$

New very simple Golden Ratio Phi construction with Circle and Two Segments (Circle Diameters).  Is there prior art?

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

  • $\begingroup$ Two tips to step-up your presentation game: (1) "PHI" looks like an acronym for something; use the LaTeX code for $\phi$: $\phi$. (2) GeoGebra allows you to hide elements to remove visual clutter; here, I'd hide the coordinate axes, and either point $E$ or $H$. (You can also hide labels; here, there's nothing I'd really hide, but in your triangle-square-pentagon question, I'd at least de-label "poly1", "poly2", "poly3". And "s". And I'd probably ditch the axes.) $\endgroup$ – Blue May 26 '16 at 3:51
  • $\begingroup$ Thanks so much @Blue ! Yes I will improve future illustrations as you suggest. Thanks again for introducing me to geogebra! And thanks for your advice on this construction: math.stackexchange.com/questions/1798843/… What's your general take on it--does it seem at all obvious? Thanks! $\endgroup$ – Astrophysics Math May 26 '16 at 4:03
  • $\begingroup$ Notice that if you also draw the diameter of the circle with endpoint at $I$ (that is draw the diameter perpendicular to point of tangency) then you get the figure in math.stackexchange.com/questions/1764736/… except plus the original circle. Since that construction was valid, so is this one. Also, compare cut-the-knot.org/do_you_know/GoldenJo.shtml --it arranges the last line segment differently, but knowing that $\triangle CDI$ is isoceles you can derive one construction from the other. $\endgroup$ – David K May 28 '16 at 16:13

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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. $\square$

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".)

  • $\begingroup$ Thanks so much @Blue ! Do you consider this construction to be obvious: math.stackexchange.com/questions/1798843/… ? Thanks so much! $\endgroup$ – Astrophysics Math May 26 '16 at 3:48
  • $\begingroup$ @AstrophysicsMath: The triangle-square-pentagon construction doesn't have a $1$-$2$-$\sqrt{5}$ triangle, so it passes my first "obvious" filter. :) My second filter, though, is "there's a pentagon": a pentagon has $\phi$ everywhere, so finding it is unsurprising. (In your construction, it's "obvious" that $a : (a+t) \;=\; \frac{1}{2}|ED| : \frac{1}{2}|HF| = 1 : \phi$.) But the construction is sneaky, in that it makes one think that the square and triangle matter, when they don't: $A$ could be anywhere on the perp bisector of $ED$, so long as $I$ is on the perpendicular through $E$.) $\endgroup$ – Blue May 26 '16 at 4:17
  • $\begingroup$ Thanks @Blue! Yes I see that and tested it by extending the lines! :) True! But at the same time, the equilateral triangle with sides the same size as the pentagon's sides sharing a side with a square with sides also of the same size which shares a side with the pentagon imply both the perpendicular bisector of the pentagon's side as well as the parallel line to the perpendicular bisector emanating from E. The three most basic polygons all with sides of the same size placed in a 3,4,5 progression, and the golden ratio is there. :) $\endgroup$ – Astrophysics Math May 26 '16 at 4:41
  • $\begingroup$ @AstrophysicsMath: You could replace the square with any "even-gon", and the triangle with any "odd-gon" ---you could even stack a zillion "even-gons" between the pentagon and the "odd-gon"--- and get the same result, provided the segment is divided by the perpendicular through $E$. Having a square in there somewhere has the advantage of making that perpendicular a "natural" part of the construction, but otherwise, there's just nothing special about the polygons you've used ... well, except for the sneaky factor. :) The $3$-$4$-$5$ progression looks cool, but is geometrically irrelevant. $\endgroup$ – Blue May 26 '16 at 5:01
  • $\begingroup$ could you replace the square with any "even-gon" that has the exact same length of side as the pentagram? If I replace the square with a hexagon or octogon with sides equal in length to the sides of the pentagon, the construction disappears. And too, the magic of the three polygons I used is that they are the three simplest polygons--the very first three polygons of an infinite number. Also, I'm not sure how many ways there are to arrange the three simplest polygons so that a) each one shares a side with another and b) the golden ratio naturally emerges. :) $\endgroup$ – Astrophysics Math May 26 '16 at 5:28

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 $r\sqrt{4 + \dfrac{1 - 3\sqrt{5}}{2}}$.

  • $\begingroup$ Thanks @StevenHarding ! So it is the golden ratio PHI! Feel free to be a bit more elaborate! :) $\endgroup$ – Astrophysics Math May 26 '16 at 1:35
  • $\begingroup$ I appreciate your very nice illustration! I did not notice at first, but the ratio is indeed the golden ratio since $(1 + \sqrt{5})^2 = 6 + 2\sqrt{5} = 2(3 + \sqrt{5}) = \dfrac{8}{3 - \sqrt{5}}$ $\endgroup$ – Steven Harding May 26 '16 at 1:52
  • $\begingroup$ Thanks @StevenHarding! Glad you enjoy the simple drawing! $\endgroup$ – Astrophysics Math May 26 '16 at 1:53

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

  • $\begingroup$ thanks @PhyicsTheory! will keep googling/researching/seeking! $\endgroup$ – Astrophysics Math May 26 '16 at 2:14
  • $\begingroup$ I should have just blatantly stated that for best answer... $\endgroup$ – Steven Harding May 26 '16 at 2:16
  • $\begingroup$ oops @StevenHarding I thought I could check two answers! I just rechecked yours! As it is more complete! :) $\endgroup$ – Astrophysics Math May 26 '16 at 2:21

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