Key Concepts and Formulas

• Combinations: The number of ways to choose $k$ items from a set of $n$ distinct items, where order doesn't matter, is given by ${n \choose k} = \frac{n!}{k!(n-k)!}$. In this case, we use ${n \choose 2} = \frac{n(n-1)}{2}$ to find the total number of lines connecting pairs of stations.
• Circular Permutations: In a circular arrangement of $n$ objects, each object has two immediate neighbors. Therefore, there are $n$ connections between nearest neighbors.
• Problem Translation: Converting the word problem's conditions into a mathematical equation is crucial. In this case, the number of red lines is 99 times the number of blue lines.

Step-by-Step Solution

Step 1: Calculate the Total Number of Possible Lines

• What and Why: We need to find the total number of lines that can be drawn between any two stations. Since each pair of stations is connected by a line, we use combinations to choose 2 stations out of $n$. Order doesn't matter (station A to B is the same line as B to A).
• Calculation: Let $N_{total}$ be the total number of lines. $N_{total} = {n \choose 2} = \frac{n(n-1)}{2}$

Step 2: Determine the Number of Blue Lines

• What and Why: Blue lines connect nearest neighbor stations. In a circular arrangement, each station has two neighbors. Therefore, there are $n$ such connections.
• Calculation: Let $N_{blue}$ be the number of blue lines. $N_{blue} = n$

Step 3: Determine the Number of Red Lines

• What and Why: Red lines connect all non-nearest neighbor stations. To find the number of red lines, we subtract the number of blue lines from the total number of lines.
• Calculation: Let $N_{red}$ be the number of red lines. $N_{red} = N_{total} - N_{blue} = \frac{n(n-1)}{2} - n$ Simplifying the expression: $N_{red} = \frac{n(n-1) - 2n}{2} = \frac{n(n-1-2)}{2} = \frac{n(n-3)}{2}$

Step 4: Formulate the Equation from the Given Condition

• What and Why: The problem states that the number of red lines is 99 times the number of blue lines. We translate this into an equation using the expressions we derived for $N_{red}$ and $N_{blue}$.
• Equation: $N_{red} = 99 \times N_{blue}$ Substituting the expressions: $\frac{n(n-3)}{2} = 99n$

Step 5: Solve the Algebraic Equation for $n$

• What and Why: Now we solve the equation for $n$. Since $n > 2$, we know $n \neq 0$, so we can safely divide both sides by $n$.
• Solving: $\frac{n(n-3)}{2} = 99n$ Divide both sides by $n$ (since $n > 2$): $\frac{n-3}{2} = 99$ Multiply both sides by 2: $n-3 = 198$ Add 3 to both sides: $n = 201$

Common Mistakes & Tips

• Confusing Total Lines with Red Lines: Remember to subtract the blue lines from the total lines to get the number of red lines.
• Circular vs. Linear Arrangements: Understand that in a circular arrangement, each element has two neighbors, leading to $n$ connections.
• Algebraic Simplification: Be careful with algebraic manipulations to avoid errors.

Summary

We calculated the total number of lines, the number of blue lines, and the number of red lines in terms of $n$. Using the given condition that the number of red lines is 99 times the number of blue lines, we set up an equation and solved for $n$, finding that $n = 201$.

Final Answer

The final answer is $\boxed{201}$, which corresponds to option (A).