2020 points, black and green

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2020 points, black and green

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Source: 2020 Austrian Federal Competition For Advanced Students, Part 2, p2

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#1 h Nov 22, 2020, 10:33 PM

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In the plane there are $2020$ points, some of which are black and the rest are green. For every black point, the following applies: There are exactly two green points that represent the distance $2020$ from that black point.
Find the smallest possible number of green dots.

(Walther Janous)

This post has been edited 1 time. Last edited by parmenides51, Nov 22, 2020, 10:33 PM

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#2 h Nov 23, 2020, 12:00 AM

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I was thinking of something to do with foci of ellipses.

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#6 h Jan 9, 2021, 7:09 PM

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I like this problem. It reminds me of Arthur Engel so much!
The answer is $\boxed {45}$ .
See that finding the minimal number of green points is the same as finding the maximal number of black points. Let us consider two green points in the plane $v_1$ and $v_2$ .There can be at most $2$ black points that use both of these points. Let us call an arbitrary third green point $v_3$ . Then $v_3$ and $v_2$ can have at most two black points that use them both and similarly goes for $v_3$ and $v_1$ . It would been a problem if for example one of the black points formed by $v_1$ and $v_2$ was the same as the one formed by $v_1$ and $v_3$ , but this can't happen since the third green point can be chosen anywhere in the plane. It goes in the same way for any green point we add. So for every $k$ green point we can have at most $2\tfrac {k(k-1)}{2}=k(k-1)$ black points. So if $m$ is the total number of points in the plane and $n$ is the minimal possible number of green points out of them we have that $n$ should be the smallest integer to satisfy $n+n(n-1)=n^2\geq m$ . So in the case of $m=2020$ , $n$ is $45$ .

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