The number $a$ is the $m$-th part of the sum of the numbers $b$ and $c$, and the number $b$ is the $n$-th part of the sum of the numbers $c$ and $a$. What fraction of the sum of the numbers $a$ and $b$ is the number $c$?
I. Solution: According to the conditions of the problem, $a, b$, and $c$ are unknowns, but there are certain (assumed known) values of $m$ and $n$ such that the following relationships hold:

$$
a=\frac{b+c}{m}, \quad b=\frac{c+a}{n}
$$

The goal is to determine the ratio of $a+b$ to $c$. To achieve this, we try to express $a$ and $b$ in terms of $c$. The equations can be written as:

$$
m a - b = c, \quad -a + n b = c
$$

From these, we can eliminate $b$ or $a$ by adding $n$ times the first equation to the second, or by adding the first equation to $m$ times the second:

$$
(m n - 1) a = (n + 1) c, \quad (m n - 1) b = (m + 1) c
$$

From the sum of these two, if $m n \neq 1$, the desired ratio is

$$
\frac{a+b}{c} = \frac{m+n+2}{m n - 1}
$$

If $m n = 1$, but one of $m$ and $n$ is not -1, then from the equations (2) it follows that $c = 0$, and in this case the question makes no sense.

Finally, if $m = n = -1$, then both equations (1) reduce to the equation

$$
a + b + c = 0
$$

From this, if $c \neq 0$,

$$
\frac{a+b}{c} = -1
$$

follows.

Remark. The solution to the problem was made possible by the fact that only two linear equations hold for the unknowns $a, b, c$, and in these equations, there are no terms that do not contain the unknowns. Such equations are typically called homogeneous linear equations. A system of such equations always has the solution where all values are zero. If there is a solution different from this (and in our case there is, because there are fewer equations than unknowns), then the system can determine at most the ratios of one unknown to the others (not necessarily uniquely). In our case, for example, we essentially determined the ratios $\frac{a}{c}$ and $\frac{b}{c}$.

II. Solution: Let the desired ratio be denoted by $x$, then we seek a relationship (independent of $a$, $b$, and $c$) between $m$, $n$, and $x$. If this is successful, we can then express $x$. The relationships for $m$, $n$, and $x$ are:

$$
m = \frac{b+c}{a}, \quad n = \frac{c+a}{b}, \quad x = \frac{a+b}{c}
$$

The symmetry of these equations in $a$, $b$, and $c$ can be made clearer by adding 1 to each equation:

$$
m + 1 = \frac{a+b+c}{a}, \quad n + 1 = \frac{a+b+c}{b}, \quad x + 1 = \frac{a+b+c}{c}
$$

From this, if none of $m$, $n$, and $x$ is -1, then

$$
\frac{1}{m+1} + \frac{1}{n+1} + \frac{1}{x+1} = \frac{a+b+c}{a+b+c} = 1
$$

From which the value of $x$ is

$$
x = \frac{m+n+2}{m n - 1}
$$

if $m n \neq 1$.

If $m n = 1$, then from the product of the first two equations (1),

$$
1 = \frac{a b + c(a+b) + c^2}{a b} = 1 + \frac{c(a+b+c)}{a b}
$$

From this, since we assumed that none of $a$, $b$, and $c$ is 0, it follows that

$$
a + b + c = 0
$$

and thus

$$
m = n = x = -1
$$

We also arrive at equation (2) if we assume that only one of $m$, $n$, and $x$ is -1.