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# Maximum number of triangles

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?

Note by Siddhartha Chakraborty
4 years, 1 month ago

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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 $$a_i, b_i, c_i \implies (\dbinom{n}{1}$$)^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. · 4 years, 1 month ago

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

1. Choose $$a_{i} , b_{i} , c_{i}$$ (if none of these are colinear) $$\Rightarrow n^{3}$$ ways
2. 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. · 4 years, 1 month ago