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Do there exist $100$ lines in the $xy$-plane, no three concurrent such that they intersect exactly in $2017$ points?

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    I *knew* I just saw this! http://math.stackexchange.com/questions/2090430/combi-problem – 2017-01-10
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    Possible duplicate of : http://math.stackexchange.com/questions/2090430/combi-problem , however, this problem doesn't have an answer. – 2017-01-10

1 Answers 1

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Yes, it's possible. In fact, there's more than one way to do it.

Here's one way ...

Let $\;a = 2,\;\; b = 4,\;\; c = 77,\;\; w = 17$.

Note that $a + b + c + w = 100$.

Create sets $A,B,C,W$ of lines, with cardinalities $a,b,c,w$, respectively, such that

  • The lines of $A$ are parallel to each other.

  • The lines of $B$ are parallel to each other.

  • The lines of $C$ are parallel to each other.

  • All other pairs of lines intersect in exactly one point.

Then the number of intersection points is

$$(ab + bc + ca) + w(a + b + c) + {w \choose 2}$$ $$ = (470) + (1411) + (136)$$ $$ = 2017$$

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    [+1] for 1) the idea, and 2) having found the good values. But, your redaction should be a little clarified. It would suffice to consider the families of lines to be $x=k, \ k=1..2$, $y=\ell, \ \ell=1..4$, $x+y=m\sqrt{2}, m=1..17$. As there are a finite number N of points of intersection of these 3 families, there are at most $\binom{N}{2}$ lines through them. Thus among the complementary set of lines, which has an infinite number of elements, pick 17 of them for the fourth family. – 2017-01-10
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    Some other solutions $$(a, b,c,w)= (1, 4, 111,12), (1 ,4,138, 9), ( 2, 4,179 ,5), ( 2, 4,119,10).$$ – 2017-01-10