$$
x \geq y^{2}+t y \geq x^{2}+t
$$

For which values of the real parameter $t$ does the system of inequalities have exactly one solution in the set of real number pairs?
I. Solution. Let's add the two inequalities and transform them as follows:

$$
\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leq \frac{1}{2}-2 t
$$

1. If $\frac{1}{2}-2 t < 0$, i.e., $t > \frac{1}{4}$, then the system of inequalities has no solution.
2. If $\frac{1}{2}-2 t=0, t=\frac{1}{4}$, then there is exactly one solution, which is $x=y=\frac{1}{2}$. We can verify by substitution that this is indeed a solution to both inequalities.
3. If $\frac{1}{2}-2 t>0, \quad t<\frac{1}{4}$. Then there can be exactly one solution if $t=\frac{1}{4}$. In this case, the inequality obtained by adding the original two inequalities,

$$
\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leq \frac{1}{2}-2 t
$$

has only $x=y=\frac{1}{2}$ as its solution; which is clearly a solution to the original two inequalities as well.