25 sats \ 3 replies \ @south_korea_ln OP 9 Oct
Visual representation of the problem.
The 5 points were chosen at random here.
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100 sats \ 2 replies \ @Undisciplined 9 Oct
I don't understand why it isn't just the square root of two. Can't the points be arbitrarily close to the corners?
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73 sats \ 1 reply \ @south_korea_ln OP 9 Oct
In that case, where would you put the 5th point?
Let me rephrase the problem, just in case:
The closest two points of five in a square cannot be further than X apart.
Find X.
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0 sats \ 0 replies \ @Undisciplined 9 Oct
That clarification helps. I'm still working on my coffee this morning.
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276 sats \ 5 replies \ @Undisciplined 9 Oct
I'll take half of the square root of 2. One point arbitrarily close to each corner and one in the center.
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269 sats \ 1 reply \ @SimpleStacker 9 Oct
I'll second this, with the added note that if the square is a closed set you don't have to say arbitrarily close to the corner :)
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20 sats \ 0 replies \ @Undisciplined 9 Oct
Indeed. I took "inside" as implying an open set.
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239 sats \ 2 replies \ @Scroogey 9 Oct
This assumes that the optimal placement is the four corners and the center.
The 'pigeonhole principle' can be used to prove your answer is correct no matter the placement:
Divide the square into four equal smaller squares (in the obvious way).
At least one of these squares must contain two points.
The maximum distance of two points in a square of length 1/2 is 1/sqrt(2).
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37 sats \ 0 replies \ @south_korea_ln OP 9 Oct
Was hoping for someone to bring up this pigeonhole principle.
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0 sats \ 0 replies \ @Undisciplined 9 Oct
I like it. I'm quite out of practice at doing proofs. Maybe these daily puzzles will get me back to peak form.
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