[gwnn]

1. Arrange 6 points in a plane so as any 3 points form an isosceles triangle.

2. Arrange 6 points in a plane so any point has at least 3 "neighbors".

neighbor=a point located exactly at a unit distance. (the same unit for all points)

for example, in a regular hexagon every point has 2 neighbors.

[jikl]

Interesting question, depends on definitions i am too lazy to look up, my easy answer without research is 5 points in xyz and one somewhere else.

Sean

[Blofeld]

Nice problems!

I think to make the second problem more interesting, you should specify that the points must be distinct (before I found a solution with them distinct, I wondered if this was a trick).

A similar problem I've seen (though with very little overlap in the solution): how many essentially different[1] ways are there to arrange four points in a plane such that the distance between pairs of points only takes two values?

[1] Up to translation, rotation and (uniform) rescaling (there's a word for that, but I can't remember what it is).

[Fluffy]

Blofeld, on Nov 8 2007, 09:32 AM, said:

I think to make the second problem more interesting, you should specify that the points must be distinct (before I found a solution with them distinct, I wondered if this was a trick).

I also though the same, put all the points on same position

[Fluffy]

I had to almost demostrate number 2 was not possible, squeeezing all posible conection options untill I came to the answer, hell it was so damn simple .

[gwnn]

Fluffy, on Nov 12 2007, 01:04 PM, said:

I had to almost demostrate number 2 was not possible, squeeezing all posible conection options untill I came to the answer, hell it was so damn simple .

That was my "method" too

But it took me about 3 weeks or so I think.