We have 6 points in a 3*4 rectangle.prove that at least two points exist which their distance is less than $\sqrt2$.
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What have you tried? Where did you get stuck? Don't just ask us to do your homework for you. – Noah Schweber Dec 29 '16 at 18:26
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i want to try but i don't know from where ?! – Seyed Moein Ayyoubzadeh Dec 29 '16 at 18:33
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1This isn't true. See @shn's answer for a good counterexample. – cool.coolcoolcool Dec 29 '16 at 19:31
2 Answers
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You can chose 6 points whose distance is at least $\sqrt 2$ apart.
Doug M
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1@SeyedMoeinAyyoubzadeh It's not a proof, it's a counterexample: the statement you are trying to prove is false. Perhaps you meant "no greater than $\sqrt{2}$"? – Noah Schweber Dec 29 '16 at 18:38
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note that you shouldn't choose your point on the rectangle itself.!! – Seyed Moein Ayyoubzadeh Dec 29 '16 at 18:45
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1@SeyedMoeinAyyoubzadeh You should clarify that in your question. – Noah Schweber Dec 29 '16 at 19:26
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2Even so, my rectangle is only $2\times 3$ and can easily be embedded in a $3\times 4$ rectangle. – Doug M Dec 29 '16 at 19:27
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For another counterexample, take the points $(0;0)$, $(4;0)$, $(4;3)$, $(0;3)$, $(1.1;\ 2)$ and $(2.9;\ 2)$ in Cartesian plane.
rgm
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note that you shouldn't choose your point on the rectangle itself.!! – Seyed Moein Ayyoubzadeh Dec 29 '16 at 21:01
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Thus the points $(0.06;\ 0.07)$, $(0.07;\ 2.94)$, $(1.21;\ 2)$, $(3.93;\ 2.94)$, $(3.91;\ 0.07)$ and $(2.78;\ 1.98)$ suffice. This can be easily checked using Geogebra, for instance. – rgm Dec 29 '16 at 23:01