Under which condition does the following equality sequence hold:

$$
\frac{a+\frac{a b c}{a-b c+b}}{b+\frac{a b c}{a-a c+b}}=\frac{a-\frac{a b}{a+2 b}}{b-\frac{a b}{2 a+b}}=\frac{\frac{2 a b}{a-b}+a}{\frac{2 a b}{a-b}-b}=\frac{a}{b}
$$
Simplify the expressions separated by equality signs.

$$
\text { the first: } \frac{\left(a^{2}+a b\right)(a-a c+b)}{\left(a b+b^{2}\right)(a-b c+b)}=\frac{a(a+b)(a-a c+b)}{b(a+b)(a-b c+b)}=\frac{a(a-a c+b)}{b(a-b c+b)}
$$

the second: $\frac{\left(a^{2}+a b\right)(2 a+b)}{\left(a b+b^{2}\right)(a+2 b)}=\frac{a(2 a+b)}{b(a+2 b)}$,

the third: $\frac{\left(a b+a^{2}\right)}{\left(a b+b^{2}\right)}=\frac{a}{b}$.

In this process, we have extended or simplified with the following expressions:

$$
a-b c+b, \quad a-a c+b, \quad a+b, \quad a+2 b, \quad 2 a+b, \quad a-b
$$

The first set of conditions we need is that none of these can be 0. If these conditions are met, the last two expressions are identical, equal, if $b \neq 0$.

It is sufficient to find the condition that the first two expressions are each equal to the last. In this case, they are also equal to each other. The difference between the first and the last expression is

$$
\frac{a}{b}\left(\frac{a-a c+b}{a-b c+b}-1\right)=\frac{a}{b} \cdot \frac{(b-a) c}{a-b c+b}
$$

This is 0, and with it, the two expressions are equal if and only if at least one of the numbers $a, c$, and $b-a$ is 0. The possibility of $b-a$ was excluded in (1), so it remains that at least one of $a$ and $c$ is 0. Similarly, for the second and fourth expressions,

$$
\frac{a}{b}\left(\frac{2 a+b}{a+2 b}-1\right)=\frac{a}{b} \cdot \frac{a-b}{a+2 b}=0
$$

also only the condition $a=0$ can be maintained. Combining these two results, we see that the condition $c=0$ is not sufficient for the sequence of equalities to hold without $a=0$, but with $a=0$ it is not necessary, so only $a=0$ needs to be added to the above conditions.

With $a=0$, the requirements for the expressions in (1) simplify to:

$$
b(1-c) \neq 0, \quad b \neq 0 \quad(5 \text { expressions, and then independently })
$$

Therefore, the conditions for the sequence of equalities to hold are

$$
a=0, \quad b \neq 0, \quad c \neq 1
$$

Langer László (Budapest, Apáczai Csere J. gym. II. class)