[superprismatic]

It is easy to find a set of 4 points

in the box, {(x,y)|0≤x≤1 & 0≤y≤1},

such that, if every point is connected

to every other point by a straight

line segment, no two line segments

cross each other. Can you exhibit a

set of 5 or more such points?

Here's an example using a set of 3

such points: (0,0), (0.5,1), (1,0).

Here, the line segments form a

triangle, and so, no two of them

cross.

[Guest]

I don't think it's possible. Consider the following:

If three such points are created, they will always form a triangle.

You then have two possible places to put the fourth point: inside the triangle or outside the triangle. (not all points outside the triangle will work, but I'm assuming you choose one that does)

this leaves us with three cases for the remaining two points:

one inside and one outside

two inside

two outside

now consider each of those cases:

one inside and one outside: no matter what, the line between those two points will cross one of the sides of the original triangle.

two inside: the first one inside divides the triangle into three triangles, leaving you with three places to put the last point. However, whichever triangle you put it in, the line between the point that you just placed and the point that is not a vertex of that triangle will cross one of the sides of the triangle. (it essentially reduces to one inside and one outside)

two outside: because order of placement is irrelevant, the first one outside could just as easily be a vertex of the original triangle and one of the points that was a vertex of the original triangle is then a point inside the triangle dividing it into three triangles. Thus, this case reduces itself to either of the other two cases depending on your placement of the second point.

Therefore, there is no set of five such points. QED

[Guest]

The question asks

Can you exhibit a set of 5 or more such points?

No, "I" can't. And that is the correct answer(thank you Benny Hill)

Edited August 28, 2010 by Not Bob Dog

[Guest]

(0,0),(1,0),(0,1),(.5,.5),(1,1)

[Guest]

As far as I can conceive 4 is possible with the obvious triangle and then a point directly north of the triangle. As for five this method would fail since one of the points would be inside of the outer lines.

Edited August 28, 2010 by alphajonny

[plainglazed]

but i guess DA and DE "cross"

Edited August 28, 2010 by plainglazed

[Guest]

It is easy to find a set of 4 points

in the box, {(x,y)|0≤x≤1 & 0≤y≤1},

such that, if every point is connected

to every other point by a straight

line segment, no two line segments

cross each other. Can you exhibit a

set of 5 or more such points?

Here's an example using a set of 3

such points: (0,0), (0.5,1), (1,0).

Here, the line segments form a

triangle, and so, no two of them

cross.

[spoiler=]

(0,0),(1,0),(0,1),(.5,.5),(1,1)

(0,1),(1,0),(A,A),(B,B),(C,C),(D,D)...

Edited August 28, 2010 by The Allen

[Guest]

In my proof above, I assumed that if two line segments share one or more points, they are considered to be crossing. The original post is slightly ambiguous about this, but I think this is the case. If not, then I'm wrong...

[Guest]

In my proof above, I assumed that if two line segments share one or more points, they are considered to be crossing. The original post is slightly ambiguous about this, but I think this is the case. If not, then I'm wrong...

If it is true that no two line segments can share points, then I agree with magician, there is no possible answer.

[plainglazed]

yeah, me too. or otherwise might require some thinking outside the box... literally. and that would add a whole new dimension to the problem.

[Guest]

If 'crossing' is simply defined as two line segments intersecting where there exists at least one point of a line segment on both sides of the intersected line segment for both line segments, then one need only select 5 points on any one line segment within the box. The line segments formed may be colinear in part, but do not 'cross'.

(0, 1/5), (0, 1/4), (0, 1/3), (0, 1/2), (0,1)

[Guest]

If 'crossing' is simply defined as two line segments intersecting where there exists at least one point of a line segment on both sides of the intersected line segment for both line segments, then one need only select 5 points on any one line segment within the box. The line segments formed may be colinear in part, but do not 'cross'.

(0, 1/5), (0, 1/4), (0, 1/3), (0, 1/2), (0,1)

I think crossing means having an intersection. I think having an intersection means that both line segments have at least one point in common.

[Guest]

I think crossing means having an intersection. I think having an intersection means that both line segments have at least one point in common.

mmiguel, your definition contradicts the definition proposed in the problem, unless a line segment is defined to be both 'open' (the endpoints that bound the line are excluded as being part of the line and thus would not intersect with a line in which it was not in part colinear to) and 'closed' (the endpoints that bound the line are included with respect to the term connected and in regards to sharing a point). Without such a definition, each leg of a triangle would otherwise be, by the definition you have given, intersecting each other. I do not like such a definition, though, and would have perferred if superprismatic had stipulated that no two line segments were to be colinear and the box was of Euclidean space.

[superprismatic]

By "crossing" I mean that two line

segments have precisely one point in

common which is not an endpoint of

both segments. I also should have

added the stipulation that no three

points are collinear (there is no

straight line which contains all 3).

Magician correctly solved the problem

and gave a simple proof.

I must admit that Not Bob Dog also

got it right, but in a different way!

[Guest]

It is easy to find a set of 4 points

in the box, {(x,y)|0≤x≤1 & 0≤y≤1},

such that, if every point is connected

to every other point by a straight

line segment, no two line segments

cross each other. Can you exhibit a

set of 5 or more such points?

Here's an example using a set of 3

such points: (0,0), (0.5,1), (1,0).

Here, the line segments form a

triangle, and so, no two of them

cross.

i am working on more shapes right now

proof_one.bmp