41. (6) A node is a point both of whose coordinates are integers. Inside the triangle $ABC$ with vertices at nodes, there are exactly $n>0$ nodes. What is the maximum number of nodes that can lie on the side $BC$?
Solution. Let there be $m$ nodes on side $B C$. Then they divide the side into $m+1$ equal segments (if the segments were not equal, there would be more than $m$ nodes on side $B C$). Denote $\bar{e}=\overline{B C} /(m+1)$, that is, $B C=(m+1) \cdot|\bar{e}|$. Consider among the nodes inside the triangle the nodes with the smallest distance to $B C$. Draw a line $l$ through them parallel to $B C$. Let line $l$ intersect $A B$ at point $B^{\prime}$, and $A C$ at point $C^{\prime}$. The node closest to $B^{\prime}$ among those located inside triangle $A B C$ and on line $l$ is denoted by $B_{1}$, and the node closest to $C^{\prime}$ among those located inside triangle $A B C$ and on line $l$ is denoted by $C_{1}$. Then $B^{\prime} B_{1} \leqslant|\bar{e}|, B_{1} C_{1} \leqslant(n-1) \cdot|\bar{e}|$ (since if we lay off from $B_{1}$ the vector $(n-1) \bar{e}$, we will get $n$ nodes on this vector, therefore, point $C_{1}$ must lie within or at the end of this vector) and $C_{1} C^{\prime} \leqslant \bar{e}$. Therefore, $B^{\prime} C^{\prime} \leqslant(n+1) \cdot|\bar{e}|$. On the other hand, line $l$ must lie between the midline (it can coincide with it) and side $B C$, since if it lies "above the midline", we can reflect $A$ relative to $B_{1}$ and obtain a node lying inside triangle $A B C$ and closer to $B C$. Therefore, $B^{\prime} C^{\prime} \geqslant B C / 2=(m+1)|\bar{e}| / 2$. Combining the obtained inequalities, we get $(m+1) \cdot|\bar{e}| / 2 \leqslant B^{\prime} C^{\prime} \leqslant(n+1) \cdot|\bar{e}|$. Thus, $m \leqslant 2 n+1$. An example for $2 n+1$ is as follows: $A(0 ; 2) ; B(0 ; 0) ; C(2 n+2 ; 0)$. On side $B C$ there are exactly $2 n+1$ nodes, and inside the triangle there are exactly $n$ nodes.