\(6\) points are taken on a plane such that the distance between any two of them is at most \(2\) units.Show that there are at least two points whose distance is less than \(\sqrt{2}\).

After you have solved this problem you can use a similar concept for this problem

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## Comments

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TopNewestI have yet to see an ordering of \(5\) points that satisfies the above conditions...

As for the proof, I've been trying to make a pigeonhole principle-based proof, but it just isn't working... :|

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Perhaps the given statement isn't the tightest possible one? Which then begs the question, how much tighter can you make it?

You can either reduce the number of points, or reduce the minimum distance.

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Sir, it would be immensely helpful if you go through the solution that I wrote to one of your problems Click here for the link

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You're going in the right direction.....

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I tried plotting the points in a pentagon, with an extra point in one of the 5 triangles. Then, the side of the pentagon would be 2. Then there are two possibilities. Either all the other 4 points are the farther vertices of the pentagon, or they are on the midpoint of each of the sides. That is as far as I have gotten to.

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