Verified answer

INSIDE (or "internal to") in this context means "belonging to the cube but not its surface". The same would be true if you asked about points INSIDE a circle. If you want to include the points on the surface (boundary), you phrase it another way (e.g. "of the cube", "not outside the cube", "not external to the cube", "belonging to the cube").

For every pair of lines that intersect where the vertices and midpoints are distinct, this defines 5 points (the two vertices, the two midpoints, and the intersection point) that must lie in a plane. From this it is easy to see that lines from opposite corners will not intersect internal to the cube.

Thus, we consider vertices that are adjacent to each other. Suppose a cube with vertices at (±1,±1,±1), let the unit vectors be u1=(1,0,0), u2=(0,1,0), u3=(0,0,1).

Let the two adjacent vertices be P = (-1,-1,-1) and R=(1,-1,-1) = P+2u1. Opposite corners are given by -P and -R so that the intersecting lines from these vertices are:

P + t(-P-un-P) and R + s(-R-un-R) for n in {2,3}. It's clear that these will intersect at s=t=½ (and since they are distinct lines, that intersection point is unique) => -½(un). That is to say, there are intersection points at (0,-½,0) and (0,0,-½).

By symmetry, there are intersection points at every ± permutation:

(±½, 0, 0), (0, ±½, 0), (0, 0, ±½) for a total of 6 internal intersection points.

Finally, consider two vertices that are opposite on a face, say

P=(-1,-1,-1) and R=(1,1,-1). There are two cases:

A) Suppose the lines go to -P-u3 and -R-u3. Then we have

P+t(-P-u3-P) = R + s(-R-u3-R) with equality by inspection at s=t=½, leading to the same points as above.

B) Suppose the lines go to -P-u1 and -R+u2 (-R+u1 fails the test of the 2nd paragraph): Then we have:

P + t(-P-u1-P) = R + s(-R+u2-R)

s(2R-u2)+t(-2P-u1) = (R-P)

s(2,1,-2) + t(1,2,2) = (2,2,0)

From which s=t=2/3 leading to an intersection point of

(-1,-1,-1) + (2/3)(1,2,2) = (-1/3, 1/3, 1/3)

By symmetry, all ± permutations of (1/3, 1/3, 1/3) will be intersection points leading to a total of 8: (±⅓, ±⅓, ±⅓)

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Regarding the intersection points on the surface of the cube,

besides the 8 from the vertices, and the 12 from the edge midpoints, the points internal to a face should be considered.

Clearly, lines from vertices not on the face strike the face at its edges, so we need only consider intersections of lines on the face. Lines from opposite corners of the face will intersect on a diagonal of the face, leading to 4 such intersections on a face or 24 total: all ± permutations of (1, ⅓, ⅓): (±1, ±⅓, ±⅓), (±⅓, ±1, ±⅓), (±⅓, ±⅓, ±1)

Finally, consider the intersection points internal to a face of lines from adjacent vertices on that face:

There are 4 due to both lines going to the midpoint closest to the adjacent vertex leading to 24 ± permutations of (0, ½, 1). There are 8 remaining points (one line to the midpoint nearest the adjacent vertex and the other line to the midpoint farther from the adjacent vertex) leading to 48 ± permutations of (1/5, 3/5, 1)

Total point count:

Internal from adjacent vertices: 6

Additional internal from opposite vertices on a face: 8

Vertices: 8

Edge midpoints: 12

Internal to a face from opposite vertices on a face: 24

Internal to a face from adjacent vertices: 24+48

Grand totals:

Internal: 14

Suface: 116

I did quite well in these maths competitions when I was in high school which is a long time ago now.

The set of points that make up the cube includes the points on the faces of the cube, so I would agree with you that the terminology and context indicates that a point on a face, the midpoints of the edges and the corners are included in the set of points 'inside' the cube.

In a different context (non-mathematical) inside might refer to only the points below the surface.

(edit)

Based on my assumption, the set of intersecting points includes all the corners, the midpoints on each edge (8 corners and 12 midpoints). On each of the 6 faces there are another 16 points (96 on the faces, excluding edges). Below the surface of the cube is a bit more trickey, there are 4 pairs of opposed corners each which I think have 3 intersecting rays each for 12 more points. There are 12 pairs of adjacent corners which I think have 2 more intersecting points each (24 more).

This is a pretty difficult question and as you don't have a lot of time in a maths quiz, and it's multiple choice, I would try to eliminate a few options then of the remaining ones try to establish which one I had the most confidence in and go with that.

(edit)

If I add up all the points - I get 152 of which 36 are below the surface.

(edit)

After reading Quadrill's response, I can see that the lines from opposite corners on the cube don't intersect. The proof is that if you have two line segments which originate from opposite corners to midpoints. The start and end points for these line segments will be on two non-intersecting edges. Any pair of edges that don't intersect, must be skew (as lines which are not intersecting and not parallel must be skew). 4 points on 2 skew lines can't be coplaner, so the 4 points cannot be on 2 intersecting lines.

Also taking another look at the lines from adjacent corners, they intersect, but more than 2 lines intersect at the same point. There are 12 pairs of adjacent corners. Each pair of adjacent corners has 2 intersecting lines each, but there are more than 2 lines intersecting at each point. As quadrill points out these points of intersection all occur half way between the centre of a face and the centre of the cube. As there are 6 faces, there can only be 6 unique points.

Then as Quadrill points out, pairs of corners which are opposite each other, but on the same face have intersecting lines also.

You'd have to be a real geometry wizard to get all that right under exam conditions.

I DID this maths comp u speak of and i guessed the ans to that question...i think i said 12...not sure

but those comps are worth nothing so don't worry about it