Packing Nine Balls in a Cube

Relevant wiki: 3D Coordinate Geometry - Problem Solving

If we let $r$ be the radius, the length of the line passing the opposing vertices of the cube is

$\sqrt{3}r+r+2r+r+\sqrt{3}r=100\sqrt{3},$ leading to:

$r=\dfrac{50\sqrt{3}}{2+\sqrt{3}}=50\sqrt{3}(2-\sqrt{3}) = 23.205080 \cdots \approx \boxed{23.2}.$

As requested, this is the illustration of how the above expression works.

Relevant wiki: 3D Coordinate Geometry - Problem Solving

Let r be the radius of each blue sphere. Construct a new, smaller cube whose vertices are the centers of the eight outer cubes. The side of this smaller cube A = 100 - 2r. The diagonal D of the smaller cube is: D = (100-2r) x √3. Since this same diagonal D starts at the center of an outer sphere passes through the center of the inner sphere and ends at the center of the opposite outer sphere, D = 4r. Setting the two expressions for D into one equations yields D = (100-2r) x √3 = 4r. So r = (100√3)/(4 + 2√3) = 23.206

Relevant wiki: 3D Coordinate Geometry - Problem Solving

We have 4 in the bottom layer and 4 in the top layer and 1 in the center.

From symmetry , the transition from a ball at a bottom vertex to the opposing top vertex will be along the major diagonals, one of which is along the unit vector $u = \dfrac{1}{\sqrt{3}}(1 , 1, 1) = ( \dfrac{1}{\sqrt{3}} , \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}} )$ .

Moving along the tangency points from a corner ball to the ball in the opposite corner, and adding the z-displacements, we get

$100 = r + 4 r ( \dfrac{1}{\sqrt{3}} ) + r = r (2 + \dfrac{4}{\sqrt{3}})$

Hence,

$r = \dfrac{100}{(2 + \dfrac{4}{\sqrt{3}})} = 23.2$

This cube's diagonals measure $100\sqrt{3}$ . Taking into account the three spheres whose diameters lay on a diagonal we have

$6r + 2x = 100\sqrt{3}$

where $r$ is the spheres' radius and $x$ is the small portion of diagonal between the vertex and a sphere surface. Now this is determined by focusing on a single sphere on the vertex. Since it is tangent to three faces of the cube, its centre is the vertex of a small cube of edge $r$ . The diagonal of this small cube is

$r + x = r\sqrt{3}$

and combining with the previous equation we get

$r = \frac{50\sqrt{3}}{2 + \sqrt{3}}$

This gives $23.2$ when truncated to the first decimal place.

The diagonal of a square is (side*√2) according to Pythagorean theorem. I just wonder how √3 came from. So (√2 r + 2r + r + √2 r) = 100 √2 That gives the answer as 20.71068 I don't think I am wrong. Am I?!

The centers of the outer spheres form the corners of a virtual cube with edges 100-2r, where r is the radius of one sphere.
Opposite corners of this virtual cube are separated by a distance of 4r.
Applying the Pythagorean Theorem twice (the 3D distance formula) we get the equation (100-2x)^2+(100-2x)^2+(100-2x)^2=(4x)^2.
This can be simplified and solved using the quadratic formula but since we are only asked for a one decimal approximation, why not just use a calculator to solve the equation?
(For example, we could graph the two sides of the equation on a TI84 and find an intersection point or use Solver.)
Either way, the one decimal approximation is 23.2.

Length of the longest diagonal of the cube is equal to 100 * root(2). this is equal to 6 times the radius of each sphere. (imagine a straight line passing through the two opposite corner sphere and the middle one).
100* root(2) = 6r