Cubes 'n Faces

Let's take a regular cube with side length 2. Now, cut this cube into 8 equal parts. Each smaller cube has a side length of 1 now. But, three of the corners of such a smaller cube was the face-centers of the bigger cube. Also, the corners of the bigger cube remained corners in the smaller cubes.

For the rest of this solution, we'll only deal with a smaller cube (in this case, a unit cube). The task now becomes to find the portion of volume that is closer to three red-colored corners than the green-colored corner. As all the 8 smaller cubes are identical, the answer will also be the same in the case of the bigger cube. (The red-colored corners represent the face-centers in the bigger cube and the green-colored one represents a corner in the bigger cube.)

Let's just position such a cube in a rectangular coordinate system such that $(0, 0, 1)$ , $(1, 0, 0)$ and $(0, 1, 0)$ are the red-colored corners and $(1, 1, 1)$ is the green-colored corner.

At first, think about one red-colored corner and the green-colored corner. The equidistant plane from this two points bisects the cube. Considering all three red points, there are 3 such planes. These are $x+y=1$ , $y+z=1$ and $z+x=1$ . These planes also outline the region we intend to find.

The portion of the cube that is closer to the green-point is shaped like a double pyramid. The base of which is shown in the first image in shaded color. One apex of that pyramid is the green-point itself and another one is the midpoint of the cube.

For finding out the volume of the double pyramid, we can use the following formula.

$V_{pyramid} = \frac{1}{3} \times base \times height$

Now, the base is an equilateral triangle whose side length is $\sqrt{2}$ . Hence, the area of the base is $\frac{\sqrt{3}}{4} \cdot (\sqrt{2})^2 = \frac{\sqrt{3}}{2}$ . And, the peak-to-peak height is half of the length of the diagonal. So, this height is $\frac{\sqrt{3}}{2}$ .

Hence, $V_{pyramid} = \frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}$ $\implies V_{pyramid} = \frac{1}{4}$

Therefore, the volume of our intended region is $1 - \frac{1}{4} = \boxed{\frac{3}{4}}$ .