\(6\) 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

Post your solution in the comments :)

## 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... :| – Daniel Liu · 3 years, 3 months ago

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You can either reduce the number of points, or reduce the minimum distance. – Calvin Lin Staff · 3 years, 3 months ago

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Click here for the link – Eddie The Head · 3 years, 3 months ago

Sir, it would be immensely helpful if you go through the solution that I wrote to one of your problemsLog in to reply

– Eddie The Head · 3 years, 3 months ago

You're going in the right direction.....Log in to reply

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. – Nanayaranaraknas Vahdam · 3 years, 3 months ago

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