1B. In the set of real numbers, solve the system of equations

$$
\left\{\begin{array}{l}
y x^{\log _{y} x}=x^{\frac{5}{2}} \\
\log _{4} y \cdot \log _{y}(3 x-y)=1
\end{array}\right.
$$
Solution. From the definition of the logarithmic function, it follows that $x>0$, $y>0$, $3 x-y>0$, and $y \neq 1$. If we take the logarithm of the first equation with base $y$ and use the fact that $\frac{1}{\log _{4} y}=\log _{y} 4$, we get

$$
\begin{aligned}
\begin{cases}y x^{\log _{y} x}=x^{\frac{5}{2}}, \\
\log _{4} y \cdot \log _{y}(3 x-y)=1, .\end{cases} & \Leftrightarrow\left\{\begin{array}{l}
1+\log _{y} x \cdot \log _{y} x=\frac{5}{2} \log _{y} x \\
\log _{y}(3 x-y)=\log _{y} 4
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
2\left(\log _{y} x\right)^{2}-5 \log _{y} x+2=0 \\
3 x-y=4
\end{array}\right.
\end{aligned}
$$

From the first equation of the last system, we get $\log _{y} x=2$ or $\log _{y} x=\frac{1}{2}$.

If $\log _{y} x=2$, then $x=y^{2}$ and substituting into $3 x-y=4$, we get the equation $3 y^{2}-y-4=0$, whose solutions are $y_{1}=\frac{4}{3}$ and $y_{2}=-1$. However, $y>0$, so in this case, the solution to the given system is only $x=\frac{16}{9}$, $y=\frac{4}{3}$.

If $\log _{y} x=\frac{1}{2}$, then $x=\sqrt{y}$, i.e., $y=x^{2}$, and substituting into $3 x-y=4$, we get the equation $x^{2}-3 x+4=0$, which has no real solutions.

Finally, the only real solution to the given system is $x=\frac{16}{9}$, $y=\frac{4}{3}$.

2AB. A regular triangular pyramid has a base edge $b$ and an angle between the lateral edges $\alpha$. Prove that the radius of the circumscribed sphere around this pyramid is $R=\frac{b}{4 \sin \frac{\alpha}{2} \sqrt{1-\frac{4}{3} \sin ^{2} \frac{\alpha}{2}}}$