Problem 1708

1. Let $\triangle ABC$ be the first right-angled triangle with the right angle at $C$, and sides $a = BC$, $b = CA$, $c = AB$. Let $\triangle A'B'C'$ be the second right-angled triangle with the right angle at $C'$, and sides $a' = B'C'$, $b' = C'A'$, $c' = A'B'$.

2. According to the problem, the inradius $r$ of $\triangle ABC$ equals the circumradius of $\triangle A'B'C'$. The circumradius of a right-angled triangle is half of its hypotenuse. Therefore, the circumradius of $\triangle A'B'C'$ is $\frac{c'}{2}$. Given that this is equal to $r$, we have:
$$r = \frac{c'}{2} \implies c' = 2r$$

3. By the Pythagorean theorem applied to $\triangle A'B'C'$:
$$c'^2 = a'^2 + b'^2$$
Substituting $c' = 2r$:
$$(2r)^2 = a'^2 + b'^2 \implies 4r^2 = a'^2 + b'^2$$

4. The area $S'$ of $\triangle A'B'C'$ is:
$$S' = \frac{1}{2} a' b'$$

5. Using the inequality $2xy \leq x^2 + y^2$ for any real numbers $x$ and $y$:
$$2a'b' \leq a'^2 + b'^2$$
Substituting $a'^2 + b'^2 = 4r^2$:
$$2a'b' \leq 4r^2 \implies a'b' \leq 2r^2$$
Therefore:
$$S' = \frac{1}{2} a'b' \leq \frac{1}{2} \cdot 2r^2 = r^2$$

6. By a known lemma, the area $S$ of $\triangle ABC$ is given by:
$$S = r^2 \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$$
Since $C = 90^\circ$, we have $\frac{C}{2} = 45^\circ$ and $\cot 45^\circ = 1$. Thus:
$$S = r^2 \cot \frac{A}{2} \cot \frac{B}{2}$$

7. The angles $A$ and $B$ are the acute angles of the right-angled triangle $\triangle ABC$, so $0^\circ < A < 90^\circ$ and $0^\circ < B < 90^\circ$. Consequently, $0^\circ < \frac{A}{2} < 45^\circ$ and $0^\circ < \frac{B}{2} < 45^\circ$.

8. Using the inequality $\sqrt{\tan \frac{A}{2} \tan \frac{B}{2}} \leq \tan \frac{\frac{A}{2} + \frac{B}{2}}{2}$:
$$\sqrt{\tan \frac{A}{2} \tan \frac{B}{2}} \leq \tan \frac{A + B}{4}$$
Since $A + B = 90^\circ$:
$$\tan \frac{A + B}{4} = \tan \frac{90^\circ}{4} = \tan 22.5^\circ = \sqrt{2} - 1$$
Therefore:
$$\sqrt{\tan \frac{A}{2} \tan \frac{B}{2}} \leq \sqrt{2} - 1$$

9. Squaring both sides:
$$\tan \frac{A}{2} \tan \frac{B}{2} \leq (\sqrt{2} - 1)^2 = 3 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$$

10. Hence:
$$S = r^2 \cot \frac{A}{2} \cot \frac{B}{2} = \frac{r^2}{\tan \frac{A}{2} \tan \frac{B}{2}} \geq \frac{r^2}{(\sqrt{2} - 1)^2} = r^2 (3 + 2\sqrt{2})$$

11. Combining this with $S' \leq r^2$:
$$\frac{S}{S'} \geq \frac{r^2 (3 + 2\sqrt{2})}{r^2} = 3 + 2\sqrt{2}$$

Therefore, we have proved that:
$$\frac{S}{S'} \geq 3 + 2\sqrt{2}$$

$\blacksquare$

The final answer is $\boxed{ \frac{S}{S'} \geq 3 + 2\sqrt{2} }$