Relation between $\phi$ and $\pi$

Question

I saw this $$\frac 65 \phi^2 \sim \pi$$ with $99.9985\%$ of accuracy. There are many more estimations like this between $\phi $ and $\pi$, I think?!. Now the question is about the origination of this(those) estimation. Is it a geometrical representation of this concept?
can someone get a clue?

Answer 1

There are exact equivalences of $\pi^3$ using $\phi$ derived in this short paper. It states that $$\pi^3=\frac{125}{4}\frac{(3-\phi)^{3/2}}{\phi}\sum_{n=-\infty}^\infty\frac{1}{(1-5n)^3}$$and $$\pi^3=\frac{250}{\phi^3\sqrt{2+\phi}}\sum_{n=-\infty}^\infty\frac{1}{(1-10n)^3}$$ Take a look at this integral $$\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}dx=\frac{\pi}{2\sqrt{\phi}}$$From this answer. Here is another identity: $$\frac1{12}\int_0^{2\pi}\frac{x\,dx}{\phi-\cos^2 x}=\frac{\pi^2}6$$

Answer 2

Maybe the golden triangle could allow for some geometric intuition.

You can use the cosine law on the golden triangle to show that $$ \frac{6}{5}\phi^2=\frac{3}{5(1-\cos(\frac{\pi}{5}))}\approx\pi. $$

As a side note, if you notice that $\cos(\frac{\pi}{5})=\frac{1+\sqrt{5}}{4}=\frac{1}{2}\phi$, we arrive at $$ \phi^2=\frac{1}{2-\phi}. $$

Answer 3

This can be interpreted by the fact that the area of an inscribed regular decagon is close to the area of the disk and that the side of the regular decagon $a$ and the radius of the circle are linked by the formula: $$R= a \phi.$$

Answer 4

There are a vast combination of real numbers $a,p,q$ such that $a\phi^p\pi^q\approx 1$. You have found the example $a=6/5~,~p=2~,~q=-1$. But there are others, like $a=111/250~,p=q=1/2$. So one might ask

I have seen that $\frac{111}{250}\sqrt{\phi}=\frac{1}{\sqrt{\pi}}$ to within less than a 1% error. Why???

Because.... numbers??