Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 93
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Well, yes, but I haven't solved that one yet.
Anyway, we should try to find the best solution for him in any case. Or prove that there's a solution in which he won't survive.
There has to be one answer.
And alsu has to prove that there can be no less.
Let it be TheXpert or MD... or Mislaid.
2 verybest: Justify and consider all options. So far it does not seem to be true.
Probably have to choose a point on a circle from which any flag is at least 100 metres away.
such a point may not exist. example: 4 flags inside a circle in the form of a square containing the centre of the circle.
The condition stated
Is it always possible for a Megamind to escape...?
In my solution, always yes.
With my solution, always yes.
In short, roughly speaking, the problem boils down to proving the fact that the centre of "mass" of the flags can always be approached closer than the points where they are located.
More precisely, there is always a point whose N distances are equal to the sum of the distances to the given N points. This point is defined by a simple procedure of averaging all the checkbox coordinates, and it is invariant with respect to the choice of origin. Consequently, 30 round trips are equivalent to 30 round trips to the geometric centre of the formation. At whatever point this centre is, we can always choose a point on the circle more than a radius away from it (100m), hence the total length of runs would be more than 100*30*2 = 6000m, which we are here to prove.
alsu:
Hence, 30 round-trips are equivalent to 30 round-trips to the geometric centre of the formation. Wherever this centre is located, we may always select a point on the circle more than a radius away from it (100m), hence the total length of the run would be more than 100*30*2 = 6000m, which we are required to prove.
No, that is not all. We still have to prove that (1) for the geometric centre in the circle centre it is also true, and to prove that the run to the points is at least not closer than to the geometric centre.
The only alternative is if the centre coincides with the centre of the circle. Then the runner would run in exactly 10 minutes. I guess in this case friendship wins! (More precisely, collaboration!))
In this case, there is a clarification that you can't put all the flags in the same point.
No, that's not all. We still have to prove that (1) is also true for the geometric centre at the centre of the circle, and prove that the escape to the points is at least not closer than the geometric centre.