Draw circles with the radius 1. No two circle have a common point. so $1389\pi\le \pi + 4d+4$ where $d$ is the longest distance. Done.

Actually I got $694 \sqrt 3$ on the exam, which is more than $1000$ obviously. I will post the solution if needed.

ArefS wrote: Draw circles with the radius 1. No two circle have a common point. so $1389\pi\le \pi + 4d+4$ where $d$ is the longest distance. Done. why is this true?

Solution of amparvardi's (hey I'm psychic): Assume the line is the x-axis. By PHP, there is a side in which there are at least 695 points in it. Since the width of the area that can be occupied on one side of the line is 1 cm while the distance between two points in on the line is at least 2 cm, then the x-coordinate difference between two adjacent points is at least $\sqrt{3}$ cm by Pythagorean Theorem. Thus since there are 694 such gaps, the x-distance between the two farthest points on that side is at least $694\sqrt{3}$ cm.

sahilsharma94 wrote: ArefS wrote: Draw circles with the radius 1. No two circles have a common point. so $1389\pi\le \pi + 4d+4$ where $d$ is the longest distance. Done. why is this true? this was my solution during the exam. we only need to find a figure such that each circle lies in the figure. not hard to show that it is what I've shown below.

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You can also partition the given 2 by 1000 block into blocks of 1 by sqrt 3 and notice the only way 2 points are in the same square is if they are diagonally opposite then u can consider 4 such blocks making a bigger block and you can easily break this bound of 1000.