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

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?

## marked as duplicate by Rahul, Winther, Théophile, choco_addicted, MCTMay 24 '16 at 3:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Yes, the prior art is the question you asked just two hours ago. – Rahul May 23 '16 at 23:45
• Thanks @Rahul, I realized I could make it even simpler with only two circles instead of a square and a circle. Circles are technically easier to construct, and thus this counts as a "simpler" golden ratio construct. Thanks! – Astrophysics Math May 23 '16 at 23:47
• @AstrophysicsMath Since the shape on the left isn't even used in the construction, it doesn't matter whether it's a circle or a square. In that sense, this question is a duplicate of your previous one. It would be better if you edited your first question instead. – Théophile May 24 '16 at 0:08
• @selfawareuser i used geogebra which gave the golden number PHI to fifteeen decimal places. :) – Astrophysics Math May 24 '16 at 0:14
• OP, you should consider submitting this as a MathBit article (under 1 page) to the American Mathematical Monthly. You should make both circles unit circles for clarity. – vadim123 May 24 '16 at 0:16

## 3 Answers

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

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.

• Not to take anything away from the OP's nice construction, which I think could not be any simpler, but the written history of the golden ratio goes back about 25 centuries, maybe more, so it's likely that somebody found this construction already. – DanielWainfleet May 24 '16 at 1:47
• yes @user254665! i have been looking for the construct or something similar! please, if anyone sees it anywhere (or something similar), please do let me know! :) i have checked many major websites/books on the golden ratio, and have been performing image searches on "golden ratio construct" "Golden section construct" etc. thanks! – Astrophysics Math May 24 '16 at 2:38
• @AstrophysicsMath I have not seen before the way you constructed. Another variation I proposed [PythagorasCostrn4GR] : math.stackexchange.com/questions/1797612/… – Narasimham May 24 '16 at 11:41
• wikipedia---golden ratio--- gives 3 constructions with diagrams, one of which is this one. There may be more in the references. – DanielWainfleet May 24 '16 at 13:01
• @user254665 are you talking to Narasimham 2? Please share the exact diagram you are talking about. Thanks! – Astrophysics Math May 24 '16 at 14:41

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.

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).

• Thanks @DavidK ! In my construct and figure above, how would you go about dividing the hypotenuse in the golden ratio as done in the figure you present? How would you go about dividing the longer leg into the golden ratio in my figure and construct above, as done in your figure from wikipedia? I do not believe this is a trivial manner, and thus that the constructs, while both incorporating a right triangle/right angle, are very different. Many constructs incorporate right angles, but are very different. Thanks! :) – Astrophysics Math May 24 '16 at 3:56
• while both my construct and the wikipedia construct incorporate right angles, the constructs appear to be completely different. – Astrophysics Math May 24 '16 at 4:02
• whereas the golden ratio is constructed in the long leg of the triangle in the example you provide: jwilson.coe.uga.edu/MATH7200/Sect4.4.html, the golden ratio in my construct is given in neither the longer leg of the triangle nor the hypotenuse, but along the segment FD, where the point E is at the golden cut of the segment FD. I asked you to please show how you would establish the golden cut on the longer leg of the triangle in my construct, but you have not done this. This is not a trivial matter. Ergo they are very different constructs. :) – Astrophysics Math May 24 '16 at 4:16
• i'm studying it over and over. i have seen the common right-triangle construct. my approach is very different as it does not find the golden ratio in the longer leg of any triangle, but rather along the segment FD, where the point E is at the golden cut of the segment FD. if they were similar constructs, you would be able to easily find PHI in the longer leg of the triangle in my construct. but you refuse to do so, as it is not a trivial matter, as the constructs are very, very different. :) they both use lines and arcs, but many constructs do :) – Astrophysics Math May 24 '16 at 4:27
• the constructs are obviously very, very different.You saw a right angle in both constructs and were off to the races, declaring the constructs to be the "exact same thing." But then, after failing to explain how they were the same or even similar, you failed to provide any drawing or figure mapping one to the other, as they are very different constructs. In the end, after waving your hands around a bit, you found his hasty contention to be impossible in reality. So now you keep waving your hands, saying "trust me i cannot show you but i know they are the same as they both use lines." – Astrophysics Math May 24 '16 at 5:35