Problem 1586

1. Counting the Total Number of Regions:
We start by determining the total number of regions created by $3n$ lines in the plane. The formula for the number of regions $R$ created by $L$ lines, where no two lines are parallel and no three lines are concurrent, is given by:
$$R = \binom{L}{2} + \binom{L}{1} + 1 = \frac{L(L-1)}{2} + L + 1$$
For $L = 3n$, this becomes:
$$R = \frac{3n(3n-1)}{2} + 3n + 1 = \frac{9n^2 - 3n}{2} + 3n + 1 = \frac{9n^2}{2} + \frac{3n}{2} + 1$$

2. Defining Good and Bad Regions:
A region is considered "good" if all of its borders are red lines. Conversely, a region is "bad" if at least one of its borders is a blue line. We need to count the number of bad regions to determine the number of good regions.

3. Counting Bad Regions:
Each blue line intersects with the other $3n - 1$ lines, creating $3n$ segments. Each segment touches exactly two regions, making them bad regions. Since there are $n$ blue lines, the total number of bad regions is:
$$n \cdot 3n \cdot 2 = 6n^2$$

4. Adjusting for Overcounting at Intersections:
When two blue lines intersect, they create four regions around the intersection point. Each intersection is counted twice in the previous step, so we need to subtract these overcounted regions. The number of intersections between $n$ blue lines is:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$
Each intersection affects four regions, so we subtract:
$$4 \cdot \frac{n(n-1)}{2} = 2n(n-1)$$
Therefore, the adjusted number of bad regions is:
$$6n^2 - 2n(n-1) = 6n^2 - 2n^2 + 2n = 4n^2 + 2n$$

5. Calculating the Number of Good Regions:
The number of good regions is the total number of regions minus the number of bad regions:
$$\text{Good regions} = \left( \frac{9n^2}{2} + \frac{3n}{2} + 1 \right) - (4n^2 + 2n)$$
Simplifying this expression:
$$\text{Good regions} = \frac{9n^2}{2} + \frac{3n}{2} + 1 - 4n^2 - 2n = \frac{9n^2}{2} - \frac{8n^2}{2} + \frac{3n}{2} - \frac{4n}{2} + 1 = \frac{n^2}{2} - \frac{n}{2} + 1$$

6. Verifying the Minimum Number of Good Regions:
For $n \geq 2$, we need to check if the number of good regions is at least 2:
$$\frac{n^2}{2} - \frac{n}{2} + 1 \geq 2$$
Simplifying:
$$\frac{n^2 - n + 2}{2} \geq 2 \implies n^2 - n + 2 \geq 4 \implies n^2 - n - 2 \geq 0$$
Factoring the quadratic:
$$(n-2)(n+1) \geq 0$$
Since $n > 1$, the inequality holds for $n \geq 2$.

Therefore, there are at least two regions of the plane such that all of their borders are red.