## Solution.

Let $x_{1}, x_{2}, \ldots, x_{n}$ be the given numbers. For simplicity, denote $x_{n+1}=x_{1}$ and $x_{-1}=x_{n}$.

Let $c$ be the sum of all red numbers, and $S=c+b+p$ be the sum of all numbers.

If $x_{i}$ is a red number, then $x_{i-1}+x_{i+1}=2 x_{i}$.

If $x_{i}$ is a white number, then $x_{i-1}+x_{i+1}=x_{i}$.

If $x_{i}$ is a red number, then $x_{i-1}+x_{i+1}=\frac{1}{2} x_{i}$.

Thus, we have written the conditions from the problem as equations that connect three consecutive numbers. 2 points

Consider the sum of all these equations.

If we sum all these equations (for $i=1,2, \ldots, n$) on the right side of the equation, we will get

$$2 c+b+\frac{1}{2} p$$

while on the left side, we will get

$$\sum_{i=1}^{n} x_{i-1}+\sum_{i=1}^{n} x_{i+1}=2 S=2 c+2 b+2 p$$

Therefore, $2 c+2 b+2 p=2 c+b+\frac{1}{2} p$, 1 point

which simplifies to $\frac{b}{p}=-\frac{3}{2}$.