there are 3 coplaner parallel lines. if any n points are taken on each of the lines, then what is the maximum number of triangles with vertices at those points?

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TopNewestLet the lines be \(a, b, c\) and the points \(a_1, ... , a_n, b_1, ... ,b_n, c_1, ... , c_n\). To ensure the maximal condition, we need to have no three \(a_i, b_i, c_i\) are collinear. Now we have a few options:

Choose \( a_i, b_i, c_i \implies (\dbinom{n}{1} \))^3 ways

Choose \(a_i, a_j , b_i \implies \dbinom{n}{2}*\dbinom{n}{1}\) ways. There are \(6\) such similar ways so total \(6*(\dbinom{n}{2}*\dbinom{n}{1})\) ways.

A generalization for arbitrary number of parallel lines would be interesting.

EDIT: Edited sad mistake; Thanks, Rijad, for pointing out. – Abhishek De · 3 years, 6 months ago

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And the final result will be \(n^{3} + 3 * n^{2} * (n-1)=n^{2} * (n+3 * (n-1))=n^{2} * (4 * n-3)\). Hope this helps and is right. – Rijad Muminović · 3 years, 6 months ago

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(4

n^3)-(3n^3) – Zobair Caca · 10 months, 3 weeks agoLog in to reply