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5 Ants are walking across a squared window, with the side length of 1 meter. I need to prove that in any given moment there are at least two ants the the distance between them is less than 75 cm.
I simplified the problem to the title: "How to prove that the max distance between 5 points on a unit square perimeter is 0.75", But I need to prove it without using calculus, I need to use Pigeonhole principle.

killer333ii
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  • See this very similar problem https://math.stackexchange.com/questions/741941/pigeonhole-principle-question-given-any-5-points-inside-a-square-of-side-length – Robert Z Nov 05 '22 at 12:23

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Divide the unit square into $4$ small squares of side $0.5$.

According to the Pigeonhole principle, there are at least 2 points in the same small square.

The maximum distance between two points in the small square is the distance of the diagonal, that is equal to $\sqrt{2}\cdot \frac{1}{2} \approx 0.71$ (cm) $< 0.75$ (cm)

NN2
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    The question asker said it was their job to prove it. You have not given them any room to do that. – Paul Nov 05 '22 at 14:23