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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
3 years, 12 months 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. Abhishek De · 3 years, 12 months ago

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@Abhishek De 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. Rijad Muminović · 3 years, 12 months ago

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(4n^3)-(3n^3) Zobair Caca · 1 year, 3 months ago

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