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Here is a problem, which neither I nor my friends (very experienced in solving things like this) can't solve. But it was used for a competition several years ago and one guy solved it there, as far as I know. Unfortunately the solution (and the guy) is lost.

You have a village. It is shaped as a square 1 by 1. The headman lives at the exact center of the square. Other houses are spread throughout the area of the village, and there is only one person per house. There are finite number of houses, they all have size 0 and everyone is aware of their placement. The headman needs to organise a meeting, which collects all villagers. To do this, he goes to some other house and tells about the meeting, then go to the another, etc. Each villager is of course is guaranteed to be found in his house; once informed, he can either participate in the gathering or go to the meeting place at the headman house. They all travel at the speed of 1 and it takes 0 time to inform someone, once you've reached his(-er) house. You need to find the minimum time $T_{\text{MinMax}}$ in which all villagers (including headman) can be collected at the centre of village, regardless of their number and initial placement. The answer should come together with proof, i.e. you need to:

a) provide a strategy (who goes where) and prove that it will work for any village in time $t \le T_{\text{MinMax}}$

b) prove that for all other strategies there must a village, which will be "collected" in time $t \ge T_{\text{MinMax}}$.

Can anyone here solve it?

You can be sure, that $T_{\text{MinMax}} < \infty$. For example, the following strategy works in $T_{\text{Max}} < 3\sqrt{5}+3\sqrt{2}/2$:

1) The village area is divided into 4 square subareas 0.5 by 0.5.

2) The headman goes to a closest house in one subarea and assings its host to be a subheadman and asks him to do the same strategy in his subarea.

3) Then the headman goes to another, neighboring subarea and does the same. And repeats this with the rest of 2 subareas.

4) The headman comes back to the centre of village.

5) If some subarea has no houses inside the headman just skips it.

Let's say that in the worst case scenario it will take time $X$. Then for subheadmen it will take time $X/2$. The first subheadman is assigned in at most $\sqrt{2}/2$ time, the next in $\sqrt{2}/2+\sqrt{5}/2$ and the last in $\sqrt{2}/2+3\sqrt{5}/2$. Then at time $\sqrt{2}/2+3\sqrt{5}/2+X/2$ the last subarea is collected it its center and in $\sqrt{2}/4$ time it can be at village center. Which means $X < \sqrt{2}/2+3\sqrt{5}/2+X/2+\sqrt{2}/4$. So $T_{Max} = X < 3\sqrt{5}+3\sqrt{2}/2$.

It is easy to prove that $T_{\text{MinMax}} \ge 2+\sqrt{2}$ (consider a village with 4 houses in the corners), and I am almost sure that $T_{\text{MinMax}} = 2+\sqrt{2}$, it was the result of that guy from the competition.