A simple trigonometry problem: Put eight coins around a central coin

[Charles Link]

Using an identity from the Math Stack exchange: $4 \cos^2(\pi/5)=2 \cos(\pi/5)+1$, I was able to show that the solution @Ibix would have for for $a$, (post 35), is identical to mine in post 20, ( with $b=1$). (It takes a little algebra to show that they are the same). Very good @Ibix :)

 

[Charles Link]

I appear to no longer understand what we are talking about, and have not since I saw this post.
Is 'a' the radius of the outer spheres?
In that case, I can only see 'a' as being equal to 1.

In the 2-D case, you can put 6 pennies around a penny in the center, and $a=b$. It perhaps would be a neat thing to have a 3-D case where you could put 12 balls around a center ball, all of the same size, but we find with some calculations (e.g. see post 20 and @Ibix post 35) that nature/mathematics didn't work this way, but instead $a \approx 1.108$ when $b=1$.

 

[OmCheeto]

In the 2-D case, you can put 6 pennies around a penny in the center, and $a=b$. It perhaps would be a neat thing to have a 3-D case where you could put 12 balls around a center ball, all of the same size, but we find with some calculations (e.g. see post 20 and @Ibix post 35) that nature/mathematics didn't work this way, but instead $a \approx 1.108$ when $b=1$.

I just realized that you can put 12 balls around a center ball two different ways, so.... , NEVER MIND!
(It was in your next post!)

Number of spheres in the outer layers going down the y axis
1
5
5
1

&

3
6
3

I swear, that even in my mid-sixties, I'm still as impatient as I was at 12.