TITLE: Six spheres in a flying saucer?
QUESTION [6 upvotes]: I take a sphere of diameter d and remove two ends to create two bowls each having a depth of d/4. If I bring these two bowls together it forms a 3D, flying saucer shaped, Vesica Piscis whereby the saucer's diameter around (x), divided by its height/depth (d/2) equals Root3.
I drop three smaller spheres into one bowl & they bunch together around its central vertical axis...I sit another three into the first to form a six sphered octahedron...and then I place the second bowl over...the upper three spheres sit as perfectly in the top bowl as the lower sit in the bottom....a perfect fit of the six within the flying saucer..
How do I determine the diameter of each of the six smaller spheres please?
I say how but in all honesty I probably wouldn't understand any process offered...personally, with little knowledge of math, I just assume there to be some constant appertaining to the diameter of the original sphere and the diameter of the smaller?
Eg: Six regular 40mm ping-pong balls require a Vesica/Saucer formed from a sphere of diameter 40 x some Constant?
Thanks/Gill

REPLY [2 votes]: Let $1$ be the radius of each small sphere (and thus $2$ the edge of the octahedron), and $r$ the radius of the large spheres forming the vesica. The center $O$ of the octahedron is also the center of the vesica, and is at a distance $r/2$ from the center $C$ of one of the large spheres.
Point $O$ is at a distance $\sqrt{2/3}$ from the center $G$ of the upper face of the octahedron, while $G$ is at a distance $2/\sqrt3$ from a vertex $A$ of that face. But $AGC$ is a right triangle and $AC$ must be equal to $r-1$ if the large sphere has to be tangent to the sphere centred at $A$. Hence we have the equation:
$$
(r-1)^2=(r/2+\sqrt{2/3})^2+(2/\sqrt3)^2,
$$
which can be solved to:
$$
r=
\frac{2}{9} \left(6+\sqrt{6}+\sqrt{69+12 \sqrt{6}}\right)
\approx4.08197.
$$