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?

Easy Math Editor

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Comments

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 - 8 years, 2 months ago

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.

Rijad Muminović - 8 years, 2 months ago

(4n^3)-(3n^3)

Zobair Caca - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$