
3. Touching circles $k_{1}\left(S_{1}, r_{1}\right)$ and $k_{2}\left(S_{2}, r_{2}\right)$ lie in a right-angled triangle $A B C$ with the hypotenuse $A B$ and legs $A C=4$ and $B C=3$ in such way, that the sides $A B, A C$ are tangent to $k_{1}$ and the sides $A B, B C$ are tangent to $k_{2}$. Find radii $r_{1}$ a $r_{2}$, if $4 r_{1}=9 r_{2}$.

(Pavel Novotný)


Solution. The hypotenuse $A B$ has length $A B=5$ with respect to Pythagoras' theorem. Then for angles in the triangle there is $\cos \alpha=\frac{4}{5}, \cos \beta=\frac{3}{5}$,

$$
\begin{aligned}
& \cot \frac{\alpha}{2}=\sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}=3 \\
& \cot \frac{\beta}{2}=\sqrt{\frac{1+\cos \beta}{1-\cos \beta}}=2
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_5816354d2b4b35f9c465g-16.jpg?height=686&width=1239&top_left_y=1867&top_left_x=451)

Fig. 1

Since both circles $k_{1}, k_{2}$ whole lie in the triangle $A B C$, 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 $A B$ 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

$$
\left(r_{1}+r_{2}\right)^{2}=\left(r_{1}-r_{2}\right)^{2}+D E^{2}
$$

which follows $D E=2 \sqrt{r_{1} r_{2}}$.

An equality $A B=A D+D E+E B$ gives

$$
c=r_{1} \cot \frac{\alpha}{2}+2 \sqrt{r_{1} r_{2}}+r_{2} \cot \frac{\beta}{2}=3 r_{1}+2 \sqrt{r_{1} r_{2}}+2 r_{2},
$$

and since $r_{1}=\frac{9}{4} r_{2}$ we obtain

$$
\frac{27}{4} r_{2}+3 r_{2}+2 r_{2}=5
$$

which follows

$$
r_{2}=\frac{20}{47}, \quad r_{1}=\frac{45}{47}
$$

Remark. Both circles lie really in the triangle $A B C$ because an incircle of the triangle has diameter $\varrho=a b /(a+b+c)=1$, while both values $r_{1}, r_{2}$ are less than 1 .
