Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function for which we know that $f(1)=0$, and for any two real numbers $(x, y)$, $|f(x)-f(y)|=|x-y|$. Determine the function $f$!
By substituting 1 for $y$ in the condition, we get that for all $x \in \mathbf{R}$, $|f(x)|=|x-1|$, which means for all $x \in \mathbf{R}$,

$$
\text { either } f(x)=x-1, \quad \text { or } \quad f(x)=1-x
$$

We get one solution to the problem if for all $x \in \mathbf{R}$, the first case holds, meaning the function value at $x$ is $x-1$ for all $x \in \mathbf{R}$:

$$
f: x \mapsto x-1
$$

Indeed, in this case,

$$
|f(x)-| f(y)|=|(x-1)-(y-1)|=| x-y \mid
$$

Similarly, we get a solution if for all $x$, the second possibility in (1) holds, that is,

$$
f: x \mapsto 1-x
$$

Now, are there functions where both possibilities occur, meaning for some $x \in \mathbf{R}$, different from 1, $f(x)=x-1$, and for some $y \in \mathbf{R}$, also different from 1, $f(y)=1-y$?

We can show that such solutions do not exist. Indeed, applying the condition of the problem to this $(x, y)$ pair,

$$
|x-y|=|f(x)-f(y)|=|(x-1)-(1-y)|,
$$

which means

$$
|(x-1)-(y-1)|=|(x-1)+(y-1)|
$$

If the absolute value of the difference and the sum of two numbers, in this case $(x-1)$ and $(y-1)$, are equal, then one of the numbers must be 0, meaning either $x=1$ or $y=1$. This is impossible, as we have excluded both possibilities.

Therefore, the condition of the problem is satisfied by only two functions: one that takes the value $x-1$ at $x$, and the other that takes the value $1-x$ at $x$.

Remark. A large portion of the solutions contained the incorrect conclusion that if $|f(x)|=|x-1|$, then either $f(x)=x-1$ or $f(x)=-(x-1)$. This conclusion is only correct if these equalities involve specific values of the functions. However, the solvers were thinking about the functions themselves. Thus, the statement is not true: there are infinitely many functions $f$ for which $|f(x)|=|x-1|$. The method of solving equations by breaking down the absolute value, which is used in this context, leads to incorrect reasoning because here the function itself is the unknown, not specific elements of its domain. The notation $f(x)$ represents both the function and the value at a specific point. This is generally not confusing. However, it misled many, and a statement about specific values was formally identified with a statement about the function.