[ Pythagorean Theorem (direct and inverse). ]  [ Two tangents drawn from one point ]

Find the legs of a right triangle if it is known that the radius of the circumscribed circle around the triangle is $R$, and the radius of the inscribed circle in it is $r$. For what ratio $\frac{R}{r}$ does the problem have a solution?
Let a circle with center $O$, inscribed in a right triangle $ABC$ with a right angle at $C$, touch the hypotenuse $AB=2R$ at point $K$, and the legs $AC$ and $BC$ at points $M$ and $L$ respectively. Denote $AC=x$.

The quadrilateral $OMCL$ is a square, so

$$
\begin{aligned}
& CM=OL=r, \quad AK=AM=AC-CM=x-r, \\
& BL=BK=AB-AK=2R-(x-r)=2R+r-x, \\
& BC=BL+CL=2R+r-x+r=2(R+r)-x
\end{aligned}
$$

By the Pythagorean theorem $AB^2=AC^2+BC^2$, or $4R^2=x^2+(2(R+r)-x)^2 \Leftrightarrow x^2-2x(R+r)+4Rr+2r^2 \Leftrightarrow$

$$
\Leftrightarrow x=r+R \pm \sqrt{R^2-2rR-r^2}
$$

Thus, either $AC=r+R+\sqrt{R^2-2rR-r^2}$ and $BC=r+R-\sqrt{R^2-2rR-r^2}$, or $AC=r+R-\sqrt{R^2-2rR-r^2}$ and $BC=r+R+\sqrt{R^2-2rR-r^2}$. The problem has a solution if $R^2-2rR-r^2 \geqslant 0$, or $\left(\frac{R}{r}\right)^2-2 \cdot \frac{R}{r}-1 \geqslant 0$, from which $\frac{R}{r} \geqslant 1+\sqrt{2}$.

## Answer

$$
r+R \pm \sqrt{R^2-2rR-r^2}, \quad \frac{R}{r} \geqslant 1+\sqrt{2}
$$