Comments on: Phi and Fibonacci in Kepler and Golden Triangles

By: Gary B Meisner

Gary B Meisner — Fri, 21 Aug 2026 18:51:57 +0000

In reply to Mark.

The focus on Fibonacci-based triangles is not an attempt to approximate a true Kepler triangle. It’s simply the fact that this site is dedicated to appearances and applications of the golden ratio and the Fibonacci sequence.

You’re absolutely correct that there are many other ways to approximate a Kepler triangle. Actually there are an infinite number of ways. The focus here is simply to share interesting applications of Fibonacci numbers.

By: Mark

Mark — Fri, 21 Aug 2026 07:11:54 +0000

I still wonder why not just the roots of 3 consecutive Fibonacci-numbers are used to approximate a true Kepler “Phi-angle” and instead b²-a² = (b-a)(b+a) in particular, Mainly because the súm of 2 consecutive F-numbers is another F-number i.e. 3² + 5² = √34² for example with √(34/25), being √(9/25 +25/25) or √(0.618+1) phi-wise , which in contrast, also accounts, in a similar fashion. for two non-consecutive squares, like 25-4=21 or a √4 : √21 : √25. Both are being somewhat negated by pointing out solely the one based on the difference between consecutive Fibonacci-squares. Is there a special reason for that, which I am missing ? It also makes me wonder if √4+√25=√29 has any particular meaning as well. Or what kind of Phi-angle that represent. That one does not (in contrast) converge to a Kepler-triangle, hat’s for sure.
It is a 1 : phi : √phi × √√5 , probably making little to no sense. A √16 : √25 : √41 does however , I believe, with 40/25 being just ‘1’ off, like 25/16 (instead of 25/15) as well. Next would be a √39 : √64 : √103 , both 39 and 103 being 1 off a golden rectangle as well. Converging to a (true) Kepler Phiangle (?)

By: Just Cause

Just Cause — Wed, 29 Oct 2025 11:22:55 +0000

I have a bit of an issue with the ‘analogy’ of a Kepler triangle being “in contrast” in a geometric progression; just ‘like’ 3²,4²,5² is arithmatically ? Actually 1,2,3 is, as a arithmatic square sum of their roots. Yes, the roots of Phi are also powers of root phi, but that is sort of missing the point of the special (Fibonacci) case of a √2/3 : √1 : √5/3 rationally adding up as consecutive ‘powers’ , of phi. Their roots do not. Just like 9,16,25 is nowhere near an arithmatic progression. In my opinion you are comparing 2 (totally) different things, even thought 5² – 3² = 2×8 and therefore Phibonacci-alike. Fibonacci & Pythagoras are not the like. Be asured, Mathologer makes (exactly) the same mistake, as his bend(over) method accounts just as well for any (a+b) × (b-a). Not any (root phi) triple in particular. 🤣
https://youtu.be/94mV7Fmbx88?si=bZ8pYAeRkzP0bXtP

By: Mr. M

Mr. M — Fri, 09 Dec 2022 14:47:55 +0000

I have found a different approach
(a+b)²-(b-a)²=4(ab)
(a²+b²)²-(b²-a²)²=4(ab)²
(1²+2²)²-(2²-1²)²=4(1×2)²
5²=4²+3²
13²=12²+5²
34²=30²+16²
(3²+5²)²=4(3×5)²-(5²-3²)²

https://math-journal.blogspot.com/2012/02/fibonacci-meets-pythagoras.html?m=1

By: Mr M

Mr M — Tue, 08 Nov 2022 11:23:20 +0000

144 is the only Fib.square nr. that can be related to 3x 2,3,5,8 (6,9,15,24) or (15-9)(15+9) and a 9,12,15 Pythagoras triangle (9²+12²=15²) but also to a Pythagoras 5²+12²=13² which gives a nice equation:
15²-9²=13²-5²
225-81=169-25
250=160+90

By: Mr. M

Mr. M — Tue, 30 Aug 2022 11:56:40 +0000

In reply to Gary B Meisner.

Somewhat less mathemagical
(b-a)(a+b) = b²- a²
(5-3)(5+3) = 5²-3²=4²
5² = 4² + 3²

By: Gary B Meisner

Gary B Meisner — Tue, 30 Aug 2022 02:49:28 +0000

In reply to Mr. M. There are likely many combinations that will work. The article is just showing how Fibonacci sequence numbers consistently follow this pattern, which is not meant to imply that only those numbers work.

By: Mr. M

Mr. M — Sat, 27 Aug 2022 14:44:27 +0000

ps I notice(d) that it is not typically Fibonacci:
as 7, 8, 15, 23 works too, or 12² + 11 x 35 = 23².

By: Mr. M

Mr. M — Tue, 26 Jul 2022 09:25:09 +0000

I am sort of missing the link between the square sum of the 3 consecutive natural numbers i.e. 9 + 16 = 25 and the sum of the 3 consecutive Fibonacci numbers: 3 + 5 = 8 as 3 times those numbers gives: 9 + 15 = 24 which reveales the similarity a little bit more. as there is only 1 less on either side. Maybe it is all to obvious for math enthousiasts; it really was not to me.

By: Mr. M

Mr. M — Fri, 22 Jul 2022 14:17:53 +0000

I should have used the roots of Fibonacci-products: (2×13),(3×13),(5×13) for all sides, to make it an even better one.