5. Maxim and Mina are playing the game "Champions of the Line". First, Maxim draws one line in the plane, then Mina draws a line different from the previously drawn one, then Maxim draws a line different from the two previously drawn ones, and so on alternately. The game ends when a total of 18 lines have been drawn (clearly, Mina will draw the last one). Maxim wins if there are more than 100 different intersection points of the drawn lines, and Mina wins otherwise (i.e., if there are 100 or fewer such points). Who has the winning strategy?

## Second Grade - A Category
## 5. Mina Wins.

Let's denote the lines drawn by Maxim and Mina as $a_{1}, a_{2}, \ldots, a_{18}$, in the order they appear in the game. When Maxim draws the line $a_{1}$, Mina draws the line $a_{2}$ as any line parallel to $a_{1}$. Further, in each of her moves, Mina needs to draw a line parallel to $a_{1}$ and $a_{2}$, and this will be described in the following.

Namely, we claim that Mina can draw her lines in such a way that the number of different intersection points after drawing the line $a_{2 i+2}$ increases by no more than $\max \{3 i-2,2 i+1\}$ compared to the number after drawing the line $a_{2 i}$. Let's show this.

First, assume that Maxim's line $a_{2 i+1}$ intersects with some of the earlier Maxim's lines except $a_{1}$ to form a new intersection point. Then Mina should draw her line $a_{2 i+2}$ so that it passes through that intersection point (if there are multiple such intersection points, through any one of them). Let's calculate how much the number of different intersection points has increased during these two moves. Maxim's line $a_{2 i+1}$ can form at most $2 i$ new intersection points. Then, Mina's line $a_{2 i+2}$ does not intersect with line $a_{1}$, nor with any of the lines $a_{2}, a_{4}, \ldots, a_{2 i}$ (since it is parallel to all of them); moreover, since line $a_{2 i+2}$ passes through the intersection point of two Maxim's lines, no new intersection points are formed at the intersection of $a_{2 i+2}$ with those two lines; thus, Mina's line $a_{2 i+2}$ can form new intersection points with at most $i-2$ Maxim's lines, i.e., it can form at most $i-2$ new intersection points. Therefore, the total number of newly obtained intersection points during these two moves is no more than $2 i+(i-2)=3 i-2$.

Now, assume that Maxim's line $a_{2 i+1}$ does not form a new intersection point with any of the earlier Maxim's lines, except possibly $a_{1}$. Then, after that move by Maxim, the number of different intersection points has increased by no more than $i+1$ (new intersection points can be formed at the intersection with all of Mina's lines, as well as at the intersection with $a_{1}$). Then Mina can draw her line $a_{2 i+2}$ arbitrarily (of course, parallel to her previous lines), and it forms new intersection points with at most $i$ Maxim's lines (all except $a_{1}$), so the total number of newly obtained intersection points during these two moves is no more than $(i+1)+i=2 i+1$.

This proves the claimed statement. It remains to count how many different intersection points can be obtained in this way. For $i=1,2$ we have $\max \{3 i-2,2 i+1\}=2 i+1$, and for $i \geqslant 3$ we have $\max \{3 i-2,2 i+1\}=3 i-2$. Therefore, since after drawing the line $a_{2}$ we have no intersection points, and since new intersection points are obtained according to the given constraints, the total number of different intersection points at the end of the game is no more than:

$\sum_{i=1}^{8} \max \{3 i-2,2 i+1\}=\sum_{i=1}^{2}(2 i+1)+\sum_{i=3}^{8}(3 i-2)=(3+5)+(7+10+13+16+19+22)=95$,

so Mina wins.