# Lines, lines, lines galore

Let $n$ be a positive integer. On each side of a square $n$ points are chosen. (No point is at a vertex of the square.)

(a) How many segments are there whose endpoints are two of the above points and such that no segment lies along a side of the square?

(b) What is the maximal possible number of intersection points of the segment from part (a)?

Give proof. Note by Sharky Kesa
5 years, 8 months ago

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@Julian Poon As I said, do it in a similar way for a). Of course, that assumes that you did a) in a "smart" way.

So, my way of doing a is: There are ${4n \choose 2 }$ segments, of which $4 \times { n \choose 2 }$ of them lie on a side of the square. Hence, there are $\frac{ 4n (4n-1) } { 2} - 4 \times \frac{ n(n-1) } { 2} = 6n^2$ pairs of such lines.

Use a similar approach to do b:

Hint: Given any 4 points, there is at most 1 way to get an intersection point.
Hint: How many sets of 4 points are there, which involve 3 or more points on the same line segment?
Hint: If in a set of 4 points, no 3 of them lie on the same line segment, then there is exactly 1 way to get an intersection point, and this doesn't involve a line segment on the side of the square.

${ 4n \choose 4 } - 5 \times { n \choose 3 } \times 3n - 4 \times { n \choose 4 } = \frac{1}{2} n^2 ( 17n^2 - 18n + 3 ).$

Staff - 5 years, 8 months ago

Oh... that's a great approach!

- 5 years, 8 months ago

I'm getting $6n^{2}$ for a) and as for b), I'm getting overly complicated formulas which are hopefully correct.

- 5 years, 8 months ago

Don't over complicate b. Do it in a similar way to a.

Staff - 5 years, 8 months ago

Note that b) is maximized when any two segments that don't share an endpoint intersect.

- 5 years, 8 months ago

UPDATE: I'm so done with this question. Through systematic counting, the solution for b) is

$\left(9n^2\sum _{k=1}^{n-1}k\right)-\left(\sum _{k=1}^{n-1}\sum _{j=1}^{3n}kj\right) + \left(n\sum _{k=1}^n\sum _{j=1}^{2n}\left(j-1+3\left(k-1\right)\right)\right)-\left(\sum _{k=1}^n\sum _{j=1}^{2n}\left(k-1\right)j\right) + \left(n\sum _{k=1}^n\sum _{j=1}^n\left(2j+\left(3k-5\right)\right)\right)-\left(\sum _{k=1}^n\sum _{j=1}^n\left(k-1\right)j\right)$

Goodbye and good luck.

I can elaborate on how I counted it if you want, but there are many other ways to do so.

- 5 years, 8 months ago

Elaborate please. Your formula might be able to be simplified via peer help.

- 5 years, 8 months ago

Upon simplification, the answer becomes $\frac{1}{2}n^2\left(17n^2-18n+3\right)$

To get the answer, I first split the question up into $3$ different groups, group A, B and C. For each group, the numbers represent the order in which I would draw the lines. I would draw the lines in Group A first, then group B and finally group C. When drawing these lines, I would count the lines it would intersect. So for example, the first point I am going to entertain would be Point 1 in group A. After drawing all the lines that are connected to that point, I would get 0 intersections.

Next, I would entertain the second point in group A. After drawing all the lines connecting to that point, I would get <Some function in terms of n> intersections.

And so on.

After doing this for Group A, B and C, I get that the number of intersections for

Group A: $\left(9n^2\sum _{k=1}^{n-1}k\right)-\left(\sum _{k=1}^{n-1}\sum _{j=1}^{3n}kj\right)$

Group B: $\left(n\sum _{k=1}^n\sum _{j=1}^{2n}\left(j-1+3\left(k-1\right)\right)\right)-\left(\sum _{k=1}^n\sum _{j=1}^{2n}\left(k-1\right)j\right)$

And Group C: $\left(n\sum _{k=1}^n\sum _{j=1}^n\left(2j+\left(3k-5\right)\right)\right)-\left(\sum _{k=1}^n\sum _{j=1}^n\left(k-1\right)j\right)$

I have "verified" the formula for accuracy by drawing out and manually counting the number of intersections for case $n=1,2,3$. The numbers matched.

- 5 years, 8 months ago