Problem 1228

Solution:
Answer: $60$

First solution. From a total of $19$ points, there are $\binom{19}{2} = 171$ ways to choose two points. We consider lines that pass through more than $2$ points.
- There are $6 + 6 + 3 = 15$ lines that pass through exactly three points. These are: the six sides of the largest hexagon, three lines through the center (perpendicular to the sides of the largest hexagon), and the other six lines perpendicular to the sides of the largest hexagon.
- There are $6$ lines that pass through exactly four points. (They are parallel to the sides of the largest hexagon.)
- There are $3$ lines that pass through exactly five points. (They all pass through the center.)
For each $n = 3, 4, 5$, a line that passes through $n$ points will be counted $\binom{n}{2}$ times, and so the corresponding amount will have to be subtracted. Hence the answer is
$$171 - \binom{3}{2} \cdot 15 - \binom{4}{2} \cdot 6 - \binom{5}{2} \cdot 3 = 171 - 45 - 36 - 30 = 60$$

Second solution. We divide the points into $4$ groups as follows.
- Group $1$ consists of the center point.
- Group $2$ consists of the $6$ points surrounding the center.
- Group $3$ consists of the $6$ vertices of the largest hexagon.
- Group $4$ consists of the $6$ midpoints of the sides of the largest hexagon.
We wish to count the number of lines that pass through exactly $2$ points. Consider: all lines connecting points in group $1$ and $2$, $1$ and $3$, and $1$ and $4$ pass through more than $2$ points. So it is sufficient to restrict our attention to group $2$, $3$ and $4$.
- For lines connecting group $2$ and $2$, the only possibilities are those that the two endpoints are $120$ degrees apart with respect to the center, so $6$ possibilities.
- For lines connecting group $3$ and $3$, it is impossible.
- For lines connecting group $4$ and $4$, the two endpoints must be $60$ degrees apart with respect to the center, so $6$ possibilities.
- For lines connecting group $3$ and $2$. For each point in group $3$, the only possible points in group $2$ are those that are $120$ degrees apart from the point in group $3$. So $2 \cdot 6 = 12$ possibilities.
- For lines connecting group $4$ and $2$, the endpoints must be $150$ degrees apart with respect to the center, so $2 \cdot 6 = 12$ possibilities.
- For lines connecting group $4$ and $3$. For each point in group $4$, any point in group $3$ works except those that are on the side on the largest hexagon of which the point in group $4$ is the midpoint. Hence $4 \cdot 6 = 24$ possibilities.
Therefore, the number of lines passing through $2$ points is $6 + 6 + 12 + 12 + 24 = 60$, as desired.