## Task B-4.5.

Let

$$
A=\left\{(x, y) \mid \log \left(1+x^{2}+y^{2}\right) \leqslant 1+\log (x+y)\right\}
$$

and

$$
B=\left\{(x, y) \mid \log \left(2+x^{2}+y^{2}\right) \leqslant 2+\log (x+y)\right\}
$$

be sets of points in the plane. What is the ratio of the area of set $B$ to the area of set $A$?
## Solution.

Apply the properties of logarithms to the given inequalities to determine the sets $A$ and $B$.

$$
\begin{aligned}
& \log \left(1+x^{2}+y^{2}\right) \leqslant 1+\log (x+y) \\
& \Leftrightarrow \log \left(1+x^{2}+y^{2}\right) \leqslant \log (10(x+y)) \\
& \Leftrightarrow 1+x^{2}+y^{2} \leqslant 10(x+y) \\
& \Leftrightarrow x^{2}+y^{2}-10 x-10 y+1 \leqslant 0 \\
& \Leftrightarrow(x-5)^{2}+(y-5)^{2} \leqslant 49 .
\end{aligned}
$$

Similarly,

$$
\begin{aligned}
& \log \left(2+x^{2}+y^{2}\right) \leqslant 2+\log (x+y) \\
\Leftrightarrow & 2+x^{2}+y^{2} \leqslant 100(x+y) \\
\Leftrightarrow & (x-50)^{2}+(y-50)^{2} \leqslant 4998 .
\end{aligned}
$$

We conclude that the sets $A$ and $B$ are circles with their respective circumferences. The center and radius of circle $A$ are $S_{1}(5,5), r_{1}=7$, and the center and radius of circle $B$ are $S_{2}(50,50), r_{2}=\sqrt{4998}$.

For the coordinates $(x, y)$ of points in both sets, the condition $x+y>0$ (due to the logarithm) must be satisfied, meaning all points in sets $A$ and $B$ must lie above the line $y=-x$. This is clearly satisfied for all $(x, y)$.

Therefore, the desired ratio is $\frac{4998 \pi}{49 \pi}=102$.