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

[Charles Link]

Going back to post 20 for the 3-D problem, I'm going to attempt to post much of the solution here. We have 2 balls at the poles, surround by 5 balls just above the equator, and then 5 balls just below the equator. We can give as coordinates for the centers of one of the balls above the equator as $(x_1,y_1, z_1)$, and one just below the equator $(x_2,y_2, z_2)$. Let me go to the next post to write those equations, because if I hit the wrong button on the computer, it may delete everything.

 

[Charles Link]

Continuing from above:
$(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2=4 a^2$.
$x_1^2+y_1^2+z_1^2=(a+1)^2$.
$x_2^2+y_2^2+z_2^2=(a+1)^2$.
$x_1^2+y_1^2+(z_1-(a+1))^2=4a^2$.
$x_2^2+y_2^2+(z_2-(-a-1))^2=4a^2$.
$z_1=-z_2$.

 

[Charles Link]

Continuing:
$z_1=(a+1) \cos{\theta_o}$.
$x_2=(a+1) \sin{\theta_o} \cos(\pi/5)$.
$x_1=(a+1) \sin{\theta_o}$.

I think I've listed them all now. The rest was just to solve this set of equations, which we did. You eliminate $\theta_o$, and all of the other variables, and solve for $a$. (The last two in post 32 really count as one).

 

[OmCheeto]

...

Edit 2: and if it is fairly straightforward I do welcome anyone who will make the effort to compute it for the case of a dozen outer spheres, and see if they agree with the result that $a \approx 1.10$.

...

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.

 

[Ibix]

A slightly indirect way to do it is to observe that each corner of an icosohedron is surrounded by five equilateral triangles. The outer five edges of those five triangles form a pentagon of side $2a$, where $a$ is the radius of the outer balls. The center-to-corner distance of the pentagon is $P$ and satisfies $a/P=\sin 36$. But the distance from a corner to the center of the ball above the center of the pentagon must be $2a$, so the ball's center is a distance $z$ above the plane of the pentagon, where $z^2+P^2=(2a)^2$. So $z=\sqrt{4-\mathrm{cosec}^2 36}a$.

We can then place the center of the inner ball (radius $b$) a distance $z'$ directly below the center of the pentagon. We require $z+z'=a+b$ and $z'^2+P^2=(a+b)^2$ so that the outer balls all touch the inner one. If I've set that up correctly, I get that $$\begin{eqnarray*}
b&=&a\left[\frac{2}{\sqrt{4-\mathrm{cosec}^236}}-1\right]\\
&\approx&0.902a
\end{eqnarray*}$$

 

[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.