Are these lines going to meet in exactly 2002 points?

Question

There is a plane P.100 lines are on P.Is it possible to arrange them in a way such that they intersect in exactly 2002 points given that no three of them are concurrent?

Problem is in starting the question as no specific conditions about lines being parallel are given.

Any help is highly appreciated.

Answer 1

As others have pointed out, you need to start out by partitioning the lines into groups of parallel ones. Assume that the sizes of the groups are $n_1,n_2,\ldots, n_k$. In other words, there is a group of $n_1$ lines parallel to each other but non-parallel to the rest, another group of $n_2$ lines parallel to each other but non-parallel to the rest et cetera.

The number of points of intersections is then $$ 2002=\sum_{1\le i<j\le k}n_in_j. $$ Multiplying this by $2$ gives $$ \begin{aligned} 4004&=\sum_{1\le i\le k, 1\le j\le k, i\neq j}n_in_j\\ &=(\sum_{1\le i\le k}n_i)(\sum_{1\le j\le k}n_j)-\sum_{1\le i\le k}n_i^2\\ &=100^2-\sum_{i=1}^kn_i^2. \end{aligned} $$ Therefore to solve the problem you need to find natural numbers $k$ and $n_1,n_2,\ldots,n_k$ such that $$ \begin{cases}\sum_{i=1}^kn_i&=100,\\ \sum_{i=1}^kn_i^2&=5996. \end{cases} $$

There are solutions to this system: $$ \begin{aligned} n_1&=75, n_2=19, n_3=n_4=2, n_5=n_6=1,\qquad \text{(Brian Tung)}\\ n_1&=76, n_2=14, n_3=4, n_4=2, n_5=n_6=n_7=n_8=1,\\ n_1&=77, n_2=7, n_3=2, n_4=\cdots=n_{17}=1.\qquad \text{(JL)} \end{aligned} $$

Answer 2

Hint: If $a$ lines are parallel in one direction, the other $100-a$ lines which are in different directions create $a(100-a)$ intersections.

Answer 3

100 lines on a plane meeting your condition (that no three of them are parallel) are intersecting at least at 2400 points.

Suppose there are some lines on a plane such that no two of them are parallel. Take some line and rotate it in a way that it will be parallel to some other line. After rotation it still intersects with all the other lines except the one it is parallel to. Then it is clear that in order to minimise the number of total intersections you need to make as many pairs of parallel lines as possible.

Suppose now there are no lines on a plane and let $a_n$ denote number of added intersection points by adding $n$th line to an existing system of lines on a plane. If there are no lines on a plane $a_1 = 0$ of course. Now according to first paragraph we want to minimise the number of intersections so we let second line be parallel to first one. So $a_2 = 0$. Now the third line will intersect both of existing lines in 2 points and $a_3 = 2$. We can make fourth one parallel to third one in order to minimise number of intersection points, so $a_3 = 2$.

$$\vdots$$

By continuing this process you get sequence $a = (a_1, a_2, \dots, a_{99}, a_{100})$, where $a_{2k} = {2k}-2$ and $a_{2k - 1} = a_{2k}$ for all $k=1,2,\dots,50$. The sum of all $a_n$'s is then $$2 \cdot (0+2+4+\dots+48) = 2 \cdot \frac{0+48}{2} \cdot 50 = 2400.$$ And this is the least number of intersections given your conditions.