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There are 50 lines in a plane, no two of them are parallel, no three of them meet at the same point, and no three points of intersections are collinear.

Let A A denote the number points of intersection, and B B denote the number of new lines formed by joining the points of intersection.

Find A + B A+B .

Clarification: A line can be formed by joining two points(created by the intersection of lines). A line so formed qualifies to be 'new' line if it does not lie on the pre-existing lines.


The answer is 692125.

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2 solutions

Sravanth C.
Dec 30, 2016

Since no two lines are parallel every line intersects every other line, which means the number of points created is A = ( 50 2 ) = 1225 A=\binom{50}{2}=1225 .

The total number of lines which can be created by A 'A' points are ( ( 50 2 ) 2 ) \displaystyle\binom{\binom{50}{2}}{2} .

But this doesn't give us the number of new lines created because some of the lines created by joining the new points lie on the existing lines. We need to subtract the unwanted ones. Observe that there will be 50 sets of 49 co-linear points. Thus number of new lines created are

B = ( ( 50 2 ) 2 ) 50 × ( 49 2 ) = 690900 \displaystyle B=\binom{\binom{50}{2}}{2} - 50\times\binom{49}{2}=690900

Thus A + B = 692125 A+B=692125 .

Mat Baluch
Jan 3, 2017

@Sravanth Chebrolu : How can we be sure that none of the points created are aligned - except in the obvious case were these points lie on one of the existing line ?

In the example below, it seems that i can construct 6 lines, no two of them being parallel, and no three of them meeting at the same point (to be checked). Yet, three of the points created are aligned (circled in red).

Do you agree ?

Yes, but how does it affect the question? I see that we get the same no. of points and lines.

Sravanth C. - 4 years, 5 months ago

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Name the three intersection points on the above figure A, B and C. With your method of couting, they would create 3 new lines: (AB), (AC), (BC). But in fact it is only 1 new line since they are aligned.

mat baluch - 4 years, 5 months ago

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Which is why I have subtracted the co-linear lines, there will be n sets of (n-1) co-linear points.

Sravanth C. - 4 years, 5 months ago

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@Sravanth C. I don't think so, because you only substract lines created on already existing lines. You do not account for the case where two newly created lines would be co-linear, without being co-linear to a pre-existing line.

mat baluch - 4 years, 5 months ago

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@Mat Baluch @Sravanth Chebrolu I agree with Mat. See my report.

Calvin Lin Staff - 4 years, 5 months ago

@Mat Baluch Alright, I get it now. So you are saying that there will 1 new line instead of 3. I guess a little modification is required.

Sravanth C. - 4 years, 5 months ago

Thanks. I have edited the problem accordingly.

Calvin Lin Staff - 4 years, 4 months ago

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