Problem 11.3. Find all positive integers $n$ for which there exists $n$ points in the plane such that any of them lies on exactly $\frac{1}{3}$ of the lines determined by these $n$ points.

Aleksander Ivanov, Emil Kolev
11.3. We shall prove that $n=6$. If we take 6 points in general position (no three are collinear), then the lines are 15 and any point lies on 5 lines, i.e. $n=6$ is a solution of the problem.

Denote by $k$ the number of the lines defined by the given $n$ points. Assume that there is a line $l$ containing 4 of the given points. Any of the points belongs to $\frac{k}{3}-1$ lines different from $l$ which means that there are at least $4\left(\frac{k}{3}-1\right)+1$ lines. Then

$$
4\left(\frac{k}{3}-1\right)+1 \leq k
$$

i.e., $k \leq 9$. On the other hand, any point lying no on $l$ belongs to at least four lines (the lines through the point and the four points on $l$ ) and hence $k \geq 12$, a contradiction. So any line contains at most 3 points. Let $a$ of the lines contain 2 points. Then each of the other $k-a$ lines contains 3 points.

The number of the points (any of them counted $\frac{k}{3}$ times) is equal to $2 a+$ $3(k-a)$ and then $2 a+3(k-a)=\frac{n k}{3}$. On the other hand, since $n$ points define $\frac{n(n-1)}{2}$ lines (some of them may coincide) and any line containing three points is counted three times, then $a+3(k-a)=\frac{n(n-1)}{2}$. Thus

$$
a=\frac{n(n-1)(9-n)}{2(2 n-9)}, \quad k=\frac{3 n(n-1)}{2(2 n-9)}
$$

Since $k \leq \frac{n(n-1)}{2}$, then $2 n-9 \geq 3$, i.e., $n \geq 6$. Now $a \geq 0$ implies that $n \leq 9$. For $n=7$ the values of $a$ and $k$ are not integers and hence $n=8$ or $n=9$. For $n=8$ one has that $k=12, a=4$ and for $n=9$ we get that $k=12, a=0$.

Denote by $l$ the maximal number of points in general position among the given $n$ points. Then the remaining points belong to lines defined by these $l$ points.

Case 1. Let $l=3$ and let the respective points be $A_{1}, A_{2}, A_{3}$. Any of the other points lies on one of the lines $A_{1} A_{2}, A_{1} A_{3}$ and $A_{2} A_{3}$. Since any line contains at most 3 points, then we have at most 6 points, a contradiction.

Case 2. Let $l=4$ and let the respective points be $A_{1}, A_{2}, A_{3}, A_{4}$. Since the total number of the points is at least 8 , we may find a point belonging to exactly one of the lines defined by $A_{1}, A_{2}, A_{3}, A_{4}$. We may assume that the point is $A_{5}$ and $A_{5} \in A_{1} A_{2}$. Then the points $A_{3}, A_{4}, A_{1}, A_{5}$ as well as $A_{3}, A_{4}, A_{2}, A_{5}$ are in general position. Hence all the points must belong to the lines defined by $A_{1}, A_{2}, A_{3}, A_{4} ; A_{3}, A_{4}, A_{1}, A_{5}$ and $A_{3}, A_{4}, A_{2}, A_{5}$. The only common lines are $A_{3} A_{4}$ and $A_{1} A_{2} A_{5}$, i.e., all the points lie on two lines. This is a contradiction to the fact that any line contains at most 3 points.

Case 3. Let $l=5$ and let the respective points be $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$. Any of the points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ belongs to exactly 4 lines. This means that $A_{6} A_{i}, i=1,2, \ldots, 5$, is one of these lines. We may assume that $A_{6} \in A_{1} A_{2}$. Then $A_{6} A_{3}$ is one of the lines $A_{3} A_{4}, A_{3} A_{5}$ or $A_{4} A_{5}$. Let us have, for example, $A_{6} \in A_{3} A_{4}$. Then $A_{6} A_{5}$ is a new line, a contradiction.