2. Find all possible values of the ratio

$$
\frac{r+\varrho}{a+b}
$$

where $r$ is the radius of the circumscribed circle and $\varrho$ is the radius of the inscribed circle of a right-angled triangle with legs of lengths $a$ and $b$.
2. For the lengths of the segments of the sides of a general triangle $ABC$ to the points of tangency of the inscribed circle (denoted according to the diagram), the following formulas apply:

$$
|A U|=|A V|=\frac{b+c-a}{2},|B V|=|B T|=\frac{a+c-b}{2},|C T|=|C U|=\frac{a+b-c}{2},
$$

which can be easily obtained by solving the system of equations

$$
|A V|+|B V|=c, \quad|A U|+|C U|=b, \quad|B T|+|C T|=a .
$$

The points $C, T, U$ together with the center $S$ of the inscribed circle are generally the vertices of a kite, which in the case of a right angle $ACB$ is a square with side $\varrho=|S U|=|S V|$. Comparing with the above formulas for the lengths of the segments $C T, C U$ leads to the relationship

$$
\varrho=\frac{a+b-c}{2}
$$

According to Thales' theorem, in such a right triangle, $r=\frac{1}{2} c$. Together, we get

$$
r+\varrho=\frac{c}{2}+\frac{a+b-c}{2}=\frac{a+b}{2} .
$$

The ratio $(r+\varrho) /(a+b)$ therefore has only one possible value in any right triangle, equal to the number $\frac{1}{2}$.

The given solution can be modified, especially by starting from the equalities $|C T|=|C U|=\varrho$, from which it follows that $|A V|=|A U|=b-\varrho$ and $|B V|=$ $=|B T|=a-\varrho$, thus

$$
2 r=c=|A B|=|A V|+|B V|=(b-\varrho)+(a-\varrho),
$$

from which the conclusion is straightforward.

Another solution. For the area $P$ of a general triangle $ABC$, the formula

$$
2 P=\varrho(a+b+c) ;
$$

is valid; to derive it, it is sufficient to sum the areas of the triangles $ABS, ACS$, and $BCS$ with the same height $\varrho$ to the sides of the original triangle. In the case of $\gamma=90^{\circ}$, however, $2 P=a b$ and, as we have already mentioned above, $r=\frac{1}{2} c$. Together with the Pythagorean theorem $c^{2}=a^{2}+b^{2}$, we get

$$
\begin{aligned}
r+\varrho & =\frac{c}{2}+\frac{a b}{a+b+c}=\frac{a c+b c+c^{2}+2 a b}{2(a+b+c)}=\frac{a c+b c+a^{2}+b^{2}+2 a b}{2(a+b+c)}= \\
& =\frac{(a+b) c+(a+b)^{2}}{2(a+b+c)}=\frac{(a+b)(a+b+c)}{2(a+b+c)}=\frac{a+b}{2},
\end{aligned}
$$

thus arriving at the same conclusion as in the original solution.

For a complete solution, award 6 points. The known formulas for the lengths of the segments of the sides to the points of tangency of the inscribed circle do not need to be proven, nor does the relationship between the area, the radius of the inscribed circle, and the perimeter of the triangle. For incomplete solutions, award 1 point for the formula $r=\frac{1}{2} c$; for expressing $\varrho$ in the form $\varrho=\frac{1}{2}(a+b-c)$, award 3 points, of which 1 point for identifying the square $C U S T$, while for the formula $2 P=\varrho(a+b+c)$, award only 1 point (the last two scores cannot be added).