Problem 1435

1. **Consider an arbitrary point $Q$ on the plane and a circle $\Omega$ centered at $Q$ with radius $R$.

2. Calculate the average distance from a point on $\Omega$ to any of the $3n$ lines.**

- For a line in the plane, the distance from a point on the circle to the line can be approximated by considering the average distance over all points on the circle.
- The average distance from a point on the circle to a line is given by $R \cdot \frac{2}{\pi}$. This is derived from the integral of the absolute value of the sine function over one period:
$$\int_{0}^{2\pi} |\sin x| \, dx = 4$$
Dividing by the period $2\pi$, we get the average value:
$$\frac{4}{2\pi} = \frac{2}{\pi}$$
- Therefore, the average distance from a point on $\Omega$ to any of the $3n$ lines is approximately:
$$3n \cdot R \cdot \frac{2}{\pi} = \frac{6nR}{\pi}$$

3. **Calculate the average distance from a point on $\Omega$ to any of the $2n$ points.**

- The average distance from a point on the circle to any of the $2n$ points is approximately $R$, since the distance from a point on the circle to a fixed point inside the circle is roughly the radius $R$ for large $R$.
- Therefore, the average distance from a point on $\Omega$ to any of the $2n$ points is approximately:
$$2n \cdot R = 2nR$$

4. Compare the two sums of distances.

- The sum of the distances from a point on $\Omega$ to the $3n$ lines is:
$$\frac{6nR}{\pi}$$
- The sum of the distances from a point on $\Omega$ to the $2n$ points is:
$$2nR$$

5. **For sufficiently large $R$, show that the sum of the distances to the lines is less than the sum of the distances to the points.**

- We need to show:
$$\frac{6nR}{\pi} < 2nR$$
- Dividing both sides by $nR$ (which is positive for $n, R > 0$):
$$\frac{6}{\pi} < 2$$
- Since $\pi \approx 3.14159$, we have:
$$\frac{6}{\pi} \approx \frac{6}{3.14159} \approx 1.909 < 2$$
- This inequality holds true.

6. Conclusion:

- Therefore, for sufficiently large $R$, the sum of the distances from a point on $\Omega$ to the $3n$ lines is less than the sum of the distances from a point on $\Omega$ to the $2n$ points.

$\blacksquare$