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?

Let 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 \(a_i, a_j , b_i \implies \dbinom{n}{2}*\dbinom{n}{1}\) 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.

You should multiply the values in every case, not just add them up, then it would be:

Choose \(a_{i} , b_{i} , c_{i}\) (if none of these are colinear) \(\Rightarrow n^{3}\) ways

Choosing a pair of points from one line and the third point from another one would be \(\Rightarrow 6 * {n \choose 2} * {n \choose 1}=6 * \frac{n * (n-1) * n}{2} \). It's times 6 because there are \({3 \choose 2} = 3\) ways of picking the two lines on which a triangle will be constructed and \(2 * 3 = 6\) because you can either pick the first line to have two points on or the second one.

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.

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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.

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You should multiply the values in every case, not just add them up, then it would be:

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.

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(4

n^3)-(3n^3)Log in to reply