Distinct midpoints

Let us count these points case by case, but what kind of segments are formed:

Out of these 28 such line segments, 12 of them are 'edge' segments; they are an edge of the cube. There are 12 such intersection points for these, since each cube has 12 edges.

Another 12 are face diagonal line segments; they connect to form a diagonal across one of the faces of the cube. (There are 12 because for each face, there are 2 such segments) This results in 6 more midpoints; the center of each face of the cube.

The final 4 segments are space diagonals segments; there are four space diagonals in a cube. All four of these segments meet at one specific point; the center of the cube. This adds on one more point.

Adding up our total, we have $12+6+1=\boxed{19}$ distinct points.

There are 28 segments. Which ones have overlapping midpoints? Well, there is the one point in the center of the cube, gotten by connected opposite corners. There are 4 such segments; subtract three. Then subtract another segment per face, for the two crossing the diagonals of each face have intersecting midpoints. Thus we have 28-3-6 = 19 distinct points.

Divide the line segments into 3 categories: Those connecting vertices that are both part of the same edge of the cube (category 1), those line segments going diagonally across a face of the cube (category 2), and those that go through the interior of the cube (category 3).

Category 1: A cube has 12 edges. So there are 12 distinct midpoints in category 1.

Category 2: A cube has 6 faces. There will be 2 line segments going across each face, however, as the intersect in the middle of each face, there are only a total of 6 distinct midpoints in category 2.

Category 3: This connects vertices on opposite corners of the cube. All of these will intersect in the center of the cube, so there is 1 distinct midpoint in category 3.

Since there is no overlap of midpoints between categories, the total number of midpoints is $12 + 6 + 1 = 19$

6 on the faces + 12 on the edges +1 at center

There are $12$ distinct edges and all edges are a pair of distinct vertices so there are $12$ midpoints on segments of length 1. (if the volume of the cube is 1, which we will assume it is.) Then, there are 2 diagonals on each face that cross at the center of the square so there is $1$ midpoint per a face, for a total of $6$ midpoints on the diagonals with length $\sqrt2$ . There are $4$ diagonals that go through the inside of the cube, but all of them cross at the exact center of the cube which happens to be all of the line's midpoints, so we have $1$ midpoint with a lenght of $\sqrt3$ in the center of the cube. Adding these up, we have $12+6+1=\boxed{19}$

In each face , mid points of the two diagonals are the same , so consider one out of two ==> there are six faces , so six points out of twelve points ==> Subtract 6 from 28 = 22 Four body diagonals have the same midpoint ==> consider only one out of four ==> Subtract 3 from 22 = 19

12 edges so 12 distinct mid points...Then 6 diffrent faces with 12 intersecting diagonals resulting in 6 more mid points. Than one at the center formed by intersecting diagonalof the cube. Total=12+6+1

So each EDGE will have 1 midpoint. There are 12 edges giving 12 midpoints for that case. There are are 2 diagonals on a FACE, but the midpoint of those two diagonals is 1 point. There are 6 such faces that have this so there are 6 midpoints for that case. There are 4 SPACE DIAGONALS, but where they all intersect is their midpoint, giving 1 point for that case. 12 + 6 + 1 = 19.