Eight identical spheres, each of diameter 20, fit tightly into a cube of side length 40 so that each sphere just touches three of the faces of the cube. The radius of the largest sphere that will fit in the central space, just touching all eight spheres, is closest to
(A) 7.0
(B) 7.3
(C) 7.6
(D) 7.9
(E) 8.2
Divide the large cube of side length 40 into 8 smaller cubes of side length 20 , by making three cuts of the large cube through its centre using planes parallel to the pairs of faces.

Each of these small cubes has the centre of the large cube as its vertex.

Each of these small cubes also just encloses one of the large spheres, in the sense that the sphere just touches each of the faces of the small cube.

We call the sphere that fits in the central space the inner sphere. To make this sphere as large possible, its centre will be at the centre of the large cube. (If this was not the case, the centre be outside one of the small cubes, and so would be farther away from one of the large spheres than from another.)

To find the radius of the inner sphere, we must find the shortest distance from the centre of the large cube (that is, the centre of the inner sphere) to one of the large spheres. (Think of starting the inner sphere as a point at the centre of the cube and inflating it until it just touches the large spheres.)

Consider one of these small cubes and the sphere inside it.

Join the centre of the small cube to one of its vertices.

Since the small cube has side length 20 , then this new segment has length $\sqrt{10^{2}+10^{2}+10^{2}}$ or $\sqrt{300}$, since to get from the centre to a vertex, we must go over 10, down 10 and across 10 . (See below for an explanation of why this distance is thus $\sqrt{300}$.)

The inner sphere will touch the large sphere along this segment.

Thus, the radius of the inner sphere will be this distance $(\sqrt{300})$ minus the radius of the large sphere $(10)$, and so is $\sqrt{300}-10 \approx 7.32$.

Of the given answers, this is closest to 7.3.

(We need to justify why the distance from the centre of the small cube to its vertex is $\sqrt{10^{2}+10^{2}+10^{2}}$.

Divide the small cube into 8 tiny cubes of side length 10 each. The distance from the centre of the small cube to its vertex is equal to the length of a diagonal of one of the tiny cubes.

Consider a rectangular prism with edge lengths $a, b$ and $c$. What is the length, $d$, of the diagonal inside the prism?

By the Pythagorean Theorem, a face with side lengths $a$ and $b$ has a diagonal of length $\sqrt{a^{2}+b^{2}}$. Consider the triangle formed by this diagonal, the diagonal of the prism and one of the vertical edges of the prism, of length $c$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-152.jpg?height=371&width=331&top_left_y=649&top_left_x=951)

This triangle is right-angled since the vertical edge is perpendicular to the top face. By the Pythagorean Theorem again, $d^{2}=\left(\sqrt{a^{2}+b^{2}}\right)^{2}+c^{2}$, so $d^{2}=a^{2}+b^{2}+c^{2}$ or $d=\sqrt{a^{2}+b^{2}+c^{2}}$. $)$