3. In the set of real numbers, solve the equation

$$
\left\lfloor\frac{x}{1-x}\right\rfloor=\frac{\lfloor x\rfloor}{1-\lfloor x\rfloor}
$$

where $\lfloor a\rfloor$ denotes the greatest integer that does not exceed the number $a$.

The school-based written part of the first round, category B, takes place

on Tuesday, January 25, 2005

so that it starts in the morning and the participants have 4 hours of pure time to solve the problems. For each problem, a participant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies, school MF tables, and calculators without a graphic display. These details will be communicated to the students before the start of the competition.
3. The expression $\left\lfloor\frac{x}{1-x}\right\rfloor$ is an integer, so $\frac{\lfloor x\rfloor}{1-\lfloor x\rfloor}=\frac{1}{1-\lfloor x\rfloor}-1$ is also an integer, which means that $1-\lfloor x\rfloor \in\{-1,1\}$, or $\lfloor x\rfloor \in\{0,2\}$.

Let $\lfloor x\rfloor=0$. Then $0 \leqq x < 1$ is equivalent to the inequalities $0 \leqq x < 1$ and $0 \leqq x-1+x$, or $x \geqq 2$. In this case, all $x$ from the interval $\langle 2,3)$ satisfy the given equation.

Conclusion. All solutions to the given equation form the set $\left\langle 0, \frac{1}{2}\right) \cup\langle 2,3)$.

For a complete solution, award 6 points. Award 1 point for recognizing that $\frac{\lfloor x\rfloor}{1-\lfloor x\rfloor}$ is an integer, another point for $\lfloor x\rfloor \in\{0,2\}$. For deriving the system of inequalities and solving it in each of the two cases, award 2 points each. If the correct conclusion that all $x \in\left\langle 0, \frac{1}{2}\right) \cup\langle 2,3)$ satisfy the equation is missing, deduct one point.