Five-pointed puzzle

[bonanova]

Toss out any three non collinear points and you've formed a triangle.

Add a fourth non coplanar point.

Each set of three points determines a triangle that intersects with the

other triangles to form six lines which are the edges of a tetrahedron.

Suppose now we add a fifth point such that no four of them are coplanar.

Each set of three points determines a triangle that can extend to an infinite plane.

What is the maximum number of lines that can result from the intersections

of these planes?

[Guest]

Toss out any three non collinear points and you've formed a triangle.

Add a fourth non coplanar point.

Each set of three points determines a triangle that intersects with the

other triangles to form six lines which are the edges of a tetrahedron.

Suppose now we add a fifth point such that no four of them are coplanar.

Each set of three points determines a triangle that can extend to an infinite plane.

What is the maximum number of lines that can result from the intersections

of these planes?

10

[Prof. Templeton]

10

that it would follow the triangular number sequence.

[Guest]

that it would follow the triangular number sequence.

that it would follow C(n,2). Tomayto tomahto.

[bonanova]

They give the same answers for 4 points but vastly different ones for 5 points.

So let me ask:

What is the minimum number of intersection lines for 5 points, no 4 of which are coplanar?

[if the max and min are different.]

[Guest]

What is the minimum number of intersection lines for 5 points, no 4 of which are coplanar?

[if the max and min are different.]

They give the same answer for 4 points but vastly different numbers for 5 points.

So let me ask:

I'm missing something. I don't see how there can be a max and min answer. There is more than one way of making the same intersection, but you still end up with the same line. O.o

[Guest]

I don't know about there being more than one approach to this.

10.

The blue was the original triangle. Green lines originate from the 4th point, making the tetrahedron. Red from the 5th point, with no 4 being coplanar.

The only way I could think to decrease the number of planar intersections was to place the 5th point on the same line as 2 others, but that always made for a 4th in an existing plane.

[bonanova]

I don't know about there being more than one approach to this.

10.

The blue was the original triangle. Green lines originate from the 4th point, making the tetrahedron. Red from the 5th point, with no 4 being coplanar.

The only way I could think to decrease the number of planar intersections was to place the 5th point on the same line as 2 others, but that always made for a 4th in an existing plane.

the intersection of plane ABC with plane CDE?

[Guest]

Huh... good clue there. Adding the labels helps a lot, actually. Like so many on BD, it's not as simple as it seems at first.

My head is starting to hurt from trying to visualize so many intersecting planes, but I think we're looking at something like 22 lines of intersection.

I'm done guessing on this one though. I'll just wait for the answer to be posted/confirmed by others.

[Guest]

the intersection of plane ABC with plane CDE?

Oooooh. Now I see.

that makes the max, uh... 90 45.

Edited March 20, 2009 by d3k3

[bonanova]

Oooooh. Now I see.

that makes the max, uh... 90.

Interesting but ... why?

Edit ... ok.

In the 4-point case, connecting the dots, 2C4 = 6 does it.

In the 5-point case, connecting the dots, 2C5 = 10 is wrong.

What is the right formula?

[Guest]

Interesting but ... why?

Edit ... ok.

In the 4-point case, connecting the dots, 2C4 = 6 does it.

In the 5-point case, connecting the dots, 2C5 = 10 is wrong.

What is the right formula?

C(n,3) * (C(n,3) - 1) / 2

[bonanova]

2C10 pairs of planes, or lines of intersection.

Four points: 3C4 = 4 planes => 2C4 intersections.

Coincidence that 4 points = 4 planes. The 6 face plane intersections coincide with the 6 polyhedron edges.

Five points: 3C5 = 10 planes => 2C10 intersections.

Face planes intersect at lines in addition to the 2C5 polyhedron edges.

N points: 2CN edges; 2C[3CN] intersections of faces.

That said, it seems max = min.

I wondered whether it mattered if the points formed a convex set or not.