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There are $2$ out of each $6$ points in a rectangle of $3\times 4$ with a distance $\le \sqrt 5$.

I know you need to divide the rectangle into $5$ equal parts so that in one part you'll get at least 2 points by the pigeon hole principle.

And I thought making rectangles of $2\times1$ in the entire rectangle but then you'd divide the rectangle in six parts.

Can somebody give me a hint on how to proceed?

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  • $\begingroup$ try to add a triangle to each of the rectangles (it is possible if you add the triangle the right way). $\endgroup$ – Yanko Sep 27 '17 at 10:55
  • $\begingroup$ I only get even number of triangles. $\endgroup$ – Anonymous196 Sep 27 '17 at 10:57
  • $\begingroup$ As I did it, you need two rectangles of 2*1 , two rectangles with a triangle and one pentagon looking like an house. $\endgroup$ – Yanko Sep 27 '17 at 10:58
  • $\begingroup$ rectangles with a triangle= do you mean a triangle with an angle of 90° $\endgroup$ – Anonymous196 Sep 27 '17 at 11:02
  • $\begingroup$ added a picture. $\endgroup$ – Yanko Sep 27 '17 at 11:04
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Partition the rectangle $[0,4]\times[0,3]$ into $5$ regions by using the following 7 segments: $$(0,1)(1,2),\;(1,2)(1,3),\;(1,2)(2,1),\;(2,1)(2,0),\;\;(2,1)(3,2),\;(3,2)(3,3),\;(3,2)(4,1).$$ The diameter of each region is $\sqrt{2^2+1^2}=\sqrt{5}$.

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  • $\begingroup$ How can you come up with something like that? I've looked back into some books and found that picture too. There 3 pentagons and 2 half of those pentagons. $\endgroup$ – Anonymous196 Sep 27 '17 at 11:28
  • $\begingroup$ @AnonymousI I made several drawings keeping in mind that the diameter of the 5 regions should be at most the diagonal of a rectangle $2\times 1$. $\endgroup$ – Robert Z Sep 27 '17 at 11:32
  • $\begingroup$ Oh, yes that's something that's logical. I often lack of such insight. $\endgroup$ – Anonymous196 Sep 27 '17 at 11:33

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