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# Doodling with points

$$6$$ 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

3 years, 8 months ago

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I 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... :|

- 3 years, 8 months ago

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.

Staff - 3 years, 8 months ago

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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.

- 3 years, 8 months ago