62nd Czech and Slovak Mathematical Olympiad

The hypotenuse $AB$ has length $AB=5$ with respect to Pythagoras' theorem. Then for angles in the triangle there is $\cos \alpha =\frac{4}{5}$, $\cos \beta =\frac{3}{5}$, $$\cot \frac{\alpha }{2}=\sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }}=3,$$ $$\cot \frac{\beta }{2}=\sqrt{\frac{1+\cos \beta }{1-\cos \beta }}=2.$$

Since both circles $k_1,k_2$ whole lie in the triangle $ABC$, they are externally tangent—in the opposite case the leg tangent to the smaller circle intersects the greater circle. Let circles $k_1$ and $k_2$ touch the side $AB$ at points $D$ resp. $E$ and let point $F$ be orthogonal projection of the point $S_2$ to the element $S_{1}D$ (Fig. 1, under assumption is $r_1>r_2$). Using Pythagoras' theorem for a triangle $F{S}_2{S}_1$ we obtain $$(r_1+r_2{)}^2=(r_1-r_2{)}^2+D{E}^2,$$ which follows $DE=2\sqrt{r_1{r}_2}$. An equality $AB=AD+DE+EB$ gives $$c=r_1\cot \frac{\alpha }{2}+2\sqrt{r_1{r}_2}+r_2\cot \frac{\beta }{2}=3r_1+2\sqrt{r_1{r}_2}+2r_2,$$ and since $r_1=\frac{9}{4}{r}_2$ we obtain $$\frac{27}{4}{r}_2+3r_2+2r_2=5,$$ which follows $$r_2=\frac{20}{47},r_1=\frac{45}{47}.$$