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Five-pointed puzzle


bonanova
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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?

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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?

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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

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

10.

post-12393-1237575393.jpg

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.

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I don't know about there being more than one approach to this.
10.
post-12393-1237575393.jpg

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?

post-1048-1237577278.gif

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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.

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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

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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.

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