Puzzles with numbers and things

[snookersfun]

So how are they positioned?

randomly

[davis_greatest]

randomly

So, you mean, not in a ring? Are they in a 2 dimensional horizontal plane or could we have, say, one above another?

[snookersfun]

let's just say 2 dimensional, but I will give extra points for a solution for a 3-dimensional situation

[davis_greatest]

let's just say 2 dimensional, but I will give extra points for a solution for a 3-dimensional situation

OK... and should we imagine that the guests and Oliver are "points" (i.e. dots of zero size)? Because otherwise, Oliver could easily have the 11 others all touching various parts of him without any of them touching each other, so he would receive 11 balls.

[snookersfun]

OK... and should we imagine that the guests and Oliver are "points" (i.e. dots of zero size)? Because otherwise, Oliver could easily have the 11 others all touching various parts of him without any of them touching each other, so he would receive 11 balls.

are we playing 10 questions?
How can you fit 11 people/apes around Oliver without touching eachother (we are not stickfigures, aren't we). Also the guests are supposed to have some distance from eachother (all different- but not zero). And probably, before you ask again, let's take the distances as from centerpoint to centerpoint.
Any further questions, just shoot

[davis_greatest]

, before you ask again, let's take the distances as from centerpoint to centerpoint.
Any further questions, just shoot

OK, where's my centrepoint? Do you think I look like a snooker ball?

(No need to answer that)

[davis_greatest]

Is more than 5 possible (treating the people / apes as points in 2 dimensions)?

[snookersfun]

Is more than 5 possible (treating the people / apes as points in 2 dimensions)?

now that sounds promising

[snookersfun]

OK, where's my centrepoint? Do you think I look like a snooker ball?

(No need to answer that)

well, how would I know otherwise? Can I offer half points (for anybody having successfully answered on this thread) for putting his picture up in the gallery?

[April madness]

well, how would I know otherwise? Can I offer half points (for anybody having successfully answered on this thread) for putting his picture up in the gallery?

the only ones who could do that (Oliver, Charlie, Gordon) are more interested in bananas than (half)points!

(of course, I meant DGs photo)

[snookersfun]

the only ones who could do that (Oliver, Charlie, Gordon) are more interested in bananas than (half)points!

(of course, I meant DGs photo)

anybody willing to donate bananas then? Although I doubt that these three are your regular banana bunch (they seem to have quite a few other interests). Also, we know, that D_G started to manage to upload pictures, Avatars and stuff recently.

[davis_greatest]

now that sounds promising

I've changed my mind. I think that at least 6 is possible. What I'm not sure, without attacking this problem properly, is whether more is possible - e.g. perhaps all 11 balls from the others, perhaps by some kind of spiral formation.

[snookersfun]

I've changed my mind. I think that at least 6 is possible. What I'm not sure, without attacking this problem properly, is whether more is possible - e.g. perhaps all 11 balls from the others, perhaps by some kind of spiral formation.

looking forward to the set-up

[davis_greatest]

I've changed my mind back to 5 (ignore the comment above about 6) - but I'm still not certain that it can't be bettered

[davis_greatest]

Well, 5 is easy. First, put Oliver in the centre and space 6 of his friends in a circle, equally spaced around him. So, at this stage, the seven of them are all equal distances from each other. Now send one of the friends out of the circle, together with the other 5 guests whom we haven't mentioned, to another planet (so they pass balls among themselves). We are left with Oliver in the middle of a circle, surrounded by 5 friends. Spread the 5 friends out slightly, by moving them around the circle (but keeping them ON the circle), and they will then each be closer to Oliver than to anyone else.

If we add anyone else, either inside or outside the circle, either the new person / ape will be closer to one of the 5 friends in the circle than he / she is to Oliver, or one of the 5 friends in the circle will be closer to the new addition than to Oliver.

However, this still doesn't prove that there can't be another, better solution, with the five not placed in a circle...