Imagine the world

This problem is actually simpler than it looks. Consider the cubes as repeating indefinitely in all directions. The vertices of the cubes form a simple cubic lattice, as do the centres of the cubes. You can interchange centres and vertices, and it will be the same. As such, the proportion of the volume that's nearer a centre than a vertex must be the same as vice versa.

For each point in the cube, we need only consider the vertex $V$ closest to it. Due to symmetry, the points nearest each of the 8 vertices comprise one eighth of the cube, more specifically a cube with a side $\frac{1}{2}$ , one vertex of which is the center, with $V$ on the opposite side of the cube. This means the problem is equivalent to finding the volume of the region of points closer to one vertex of a unit cube than the opposing one, which by symmetry is exactly $\frac{1}{2}$ .

Relevant wiki: Volume - Problem Solving - Medium

The border of the mystery shape would consist of points that are exactly the same distance from the center of the cube to one of the edges. If the cube is lying flat on one of its faces, then the highest point of the border of the mystery shape would be directly above and centered over the cube at a distance $x$ from the center of the cube and the same distance $x$ away from all $4$ top vertices of the cube. This point is the vertex of a right triangle made where one leg as half the diagonal of one of the faces of the cube ( $\frac{\sqrt{2}}{2}$ ), the other leg is the section above the center of the top face ( $x - \frac{1}{2}$ ), and the hypotenuse is the distance between the point and the one of the $4$ vertices of the cube ( $x$ ).

By Pythagorean's Theorem, $(\frac{\sqrt{2}}{2})^2 + (x - \frac{1}{2})^2 = x^2$ , which solves to $x = \frac{3}{4}$ . By symmetry, there are $6$ of these points by each face, all $\frac{3}{4}$ away from the center of the cube, which can all be joined together to make an octahedron with sides of $\frac{3\sqrt{2}}{4}$ . Since the mystery shape must be inside the cube, the octahedron is truncated.

The volume of the truncated octahedron is the volume of the octahedron minus the 6 pyramids that are truncated. Since the volume equation of an octahedron with sides $s$ is $\frac{\sqrt{2}}{3}s^3$ , and $s = \frac{3\sqrt{2}}{4}$ , the volume of the octahedron is $\frac{9}{16}$ . Since the volume equation of a pyramid with a base area $B$ and height $h$ is $\frac{1}{3}Bh$ , and each pyramid has a height of $h = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$ , with a base that is a square with diagonals that are $2h = 2 \cdot \frac{1}{4} = \frac{1}{2}$ , and so $B = \frac{1}{8}$ ; the volume of each pyramid that will be truncated is $\frac{1}{96}$ . Therefore, the volume of the truncated octahedron is $\frac{9}{16} - 6 \cdot \frac{1}{96}$ $=$ $\frac{1}{2}$ $=$ $\boxed{0.5}$ .

Divide the cube into 8 smaller cubes. Each small cube has one of the original cube vertices and the center of the original cube. Half of the points in every small cube is closer to the center due to symmetry, so the volume that contains all points that closer to the center is half the volume of the unit cube. V= 0.5

Thumbs up to Stewart Gorden: Like others I took 1/8 of the cube with a corner centered on the origin. Thus the farthest corner is at (1/2,1/2,1/2). I took a generic point at (x,y,z). The distance from this point to the center is $d_c^2=x^2+y^2+z^2$ . The distance to the nearest vertex is $d_v^2=(1/2-x)^2+(1/2-x)^2+(1/2-z)^2$ . Now $d_c^2\leq d_v^2$ so $d_c^2- d_v^2\leq 0$ . With a little bit of algebra we can show that $x+y+z\leq 3/4$ . This is the equation of a plane that slices through the cube separating the two regions. I guess one could triple integrate, but I am too rusty at the moment. Instead I calculated the volume of the pyramid that is defined by the three points (3/4,0,0), (0,3/4,0), and (0,0,3/4): $V_P=9/128$ , then subtracted the three little pyramids that stick out past the limits of the cube. These all have 1/3 the linear dimensions of $V_P$ , but there are three of them so subtract off $V_p=1/128$ . So the volume in one octant is $V=8/128$ . Since there are 8 octants the sought after volume is 64/128 = 1/2.

I included a 3d view of the plane and cube below. The full shape is some sort of polyhedron. My picture indicates four facets after trimming, but I think the other octants would match up with the trimmed faces. So 8(1+3) is counting some twice, therefore I think there are 8+8*3/2 = 20 faces. I now believe 20 is wrong. I think 4 of the faces formed by trimming match up to form a square. Thus there are 8 + 24/4 = 14 faces. This would be a truncated octohedron.

Without figuring out the shape of the 3D region we're talking about, we can easily answer the question: Cut the cube into 8 smaller cubes with each side half the length of the original side. Consider any one of these little cubes. One vertex (call it A) is the original cube's center, and the opposite vertex (call it B) is a vertex of the big cube. So the question for this little cube is, what is the volume of points closer to A than B? There's no difference between A and B, in this regard; so the answer must be half of the little cube's volume. The same is true for all 8 little cubes, so the total volume of points closer to the center than the nearest vertex must be half the total volume.

Intuition is insufficient to prove the answer, but it can help.

In 1D, a “unit cube” is a line segment (say from 0 to 1) of length 1. The set of points that are (closer to the segment’s midpoint than the endpoints)OR(equidistant to the midpoint and the closest endpoint), is the line segment from 0.25 to 0.75, of length 1/2.

In 2D, a “unit cube” is a unit square (side length 1). The set of points that are (closer to the center of the square than to the vertices)OR(equidistant to the center and the closest vertex) is a square inscribed in the unit square whose vertices are at the midpoints of the sides of the unit square. The diagonal of this new square is equal to the side length of the unit square, 1, so the side length of the new square is 1/sqrt(2) and the area of the new square is thus 1/2.

Though the above cases do not prove anything about the cases in higher dimensions, they do strongly suggest that the set of points that are (closer to the center of the unit cube)OR(equidistant to the center and the closest vertex) is a polyhedron with volume 1/2.

Divide the cube, through the planes of symmetry, into 8 smaller cubes.

Looking at each smaller cube, the vertex of the larger cube and the center of the larger cube will be opposing vertices in this smaller one.

The volume of the points nearer to each of them will be the same (1/2 of the smaller cube). Since the 8 cubes are the equal and they make up the larger one, the total volume of the points nearer to the center of the larger cube will also be 1/2.