## Task A-4.5.

In a circle, a finite number of real numbers are arranged. Each number is colored red, white, or blue. Each red number is twice as small as the sum of its two neighboring numbers, each white number is equal to the sum of its two neighboring numbers, and each blue number is twice as large as the sum of its two neighboring numbers. Let $b$ be the sum of all white numbers, and $p$ be the sum of all blue numbers, where both sums are different from 0.

Determine the ratio $\frac{b}{p}$.
## 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 $\frac{b}{p}=-\frac{3}{2}$.