3. In a right triangle $A B C$ with hypotenuse $A B$ and legs of lengths $|A C|=4$ and $|B C|=3$, there are mutually touching circles $k_{1}\left(S_{1}, r_{1}\right)$ and $k_{2}\left(S_{2}, r_{2}\right)$ such that $k_{1}$ touches sides $A B$ and $A C$ and $k_{2}$ touches sides $A B$ and $B C$. Determine the radii $r_{1}$ and $r_{2}$, given that $4 r_{1}=9 r_{2}$.
3. The hypotenuse $AB$ has a length of $|AB|=5$ according to the Pythagorean theorem. With the usual notation for the angles and sides, it follows that $\cos \alpha=\frac{4}{5}, \cos \beta=\frac{3}{5}$,

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

Since both circles $k_{1}, k_{2}$ lie entirely within the triangle $ABC$, they must have an external tangency - if they had an internal tangency (at the hypotenuse), the larger circle would be tangent to the leg touching the smaller circle. Let $D$ and $E$ be the points of tangency of the circles $k_{1}$ and $k_{2}$ with the side $AB$, and $F$ the perpendicular projection of the point $S_{2}$ onto the segment $S_{1}D$ (Fig. 1, according to the assumption $r_{1}>r_{2}$). According to the Pythagorean theorem for the triangle $F S_{2} S_{1}$, we have

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

from which $|D E|=2 \cdot \sqrt{r_{1} r_{2}}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_83739e88fb70e4de981eg-3.jpg?height=671&width=1236&top_left_y=1138&top_left_x=410)

Fig. 1

From the equality $|AB|=|AD|+|DE|+|EB|$, we have

$$
c=r_{1} \operatorname{cotg} \frac{\alpha}{2}+2 \cdot \sqrt{r_{1} r_{2}}+r_{2} \operatorname{cotg} \frac{\beta}{2}=3 r_{1}+2 \cdot \sqrt{r_{1} r_{2}}+2 r_{2}
$$

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

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

and thus

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

Note. Both circles with the determined radii indeed lie within the triangle $ABC$, as the circle inscribed in the triangle has a radius $\varrho=ab/(a+b+c)=1$, while both values $r_{1}, r_{2}$ are less than 1.

Another way to calculate the value of $\operatorname{cotg} \frac{1}{2} \alpha$ is provided by the right triangle $ATS$, where $S$ is the center of the circle inscribed in the triangle $ABC$ and $T$ is the point of tangency of this circle with the side $AB$. We have

$$
\operatorname{cotg} \frac{\alpha}{2}=\frac{|AT|}{|ST|}=\frac{b+c-a}{2 \varrho}=3 ;
$$

similarly,

$$
\operatorname{cotg} \frac{\beta}{2}=\frac{a+c-b}{2 \varrho}=2 .
$$

And another possibility: In the right triangle $XCB$ with legs of lengths $|XC|=c+b$, $|CB|=a$, the angle $BXC$ has a size of $\frac{1}{2} \alpha$ (by placing the point $X$ on the ray $CA$, an isosceles triangle $XBA$ is formed), and thus

$$
\operatorname{cotg} \frac{\alpha}{2}=\frac{c+b}{a} \quad \text{and similarly} \quad \operatorname{cotg} \frac{\beta}{2}=\frac{c+a}{b} .
$$

For a complete solution, award 6 points. Give one point for expressing $|DE|=2 \cdot \sqrt{r_{1} r_{2}}$, one point for the equalities $|AD|=r_{1} \operatorname{cotg} \frac{1}{2} \alpha,|BE|=r_{2} \operatorname{cotg} \frac{1}{2} \beta$, two points for calculating the values $\operatorname{cotg} \frac{1}{2} \alpha$ and $\operatorname{cotg} \frac{1}{2} \beta$, one point for forming the equality $c=r_{1} \operatorname{cotg} \frac{1}{2} \alpha+2 \cdot \sqrt{r_{1} r_{2}}+r_{2} \operatorname{cotg} \frac{1}{2} \beta$, one point for calculating the radii $r_{1}, r_{2}$. Verifying that the circles with these radii lie within the triangle $ABC$ can be missing in an otherwise complete solution, as can the initial observation about the external tangency of the circles being examined.