## Task B-1.7.

The length of the base of an isosceles triangle is $6 \mathrm{~cm}$, and the cosine of the angle at the base is $\frac{5}{13}$. Determine the radius of the circle that touches both legs of the triangle and the circle circumscribed around it.
## First Solution.

Let $P$ be the midpoint of the base $\overline{B C}$, $N$ the midpoint of the leg $\overline{A C}$, and $S$ the intersection of the angle bisectors of the isosceles triangle $A B C$, which is also the center of the circumscribed circle of the triangle with radius $R$.

![](https://cdn.mathpix.com/cropped/2024_05_30_c317f9bfd22736a3ecebg-10.jpg?height=548&width=534&top_left_y=660&top_left_x=178)

According to the conditions of the problem and the labels on the diagram, we have:

$$
\begin{aligned}
& \cos \beta=\frac{a}{2}, \text { that is, } \\
& \frac{5}{13}=\frac{3}{b}
\end{aligned}
$$

$$
\text { from which it follows that } b=\frac{39}{5} \text{ cm. }
$$

By applying the Pythagorean theorem to triangle $A P C$, we get the equation $|A P|^{2}+\left(\frac{a}{2}\right)^{2}=b^{2}$, from which it follows that $|A P|=\sqrt{\left(\frac{39}{5}\right)^{2}-3^{2}}=\frac{36}{5} \text{ cm}$.

It is true that $\triangle A N S \sim \triangle A P C$ by the $\mathrm{AA}$ similarity criterion because $\varangle A N S=\varangle A P C=90^{\circ}$ and $\varangle N A S=\varangle P A C=\frac{\alpha}{2}$.

From the similarity of the triangles, it follows that $\frac{\overline{2}}{R}=\frac{|A P|}{b}$, that is, $2 R=\frac{169}{20} \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_c317f9bfd22736a3ecebg-10.jpg?height=572&width=494&top_left_y=1730&top_left_x=198)

Let $r$ be the radius of the smaller circle and $x$ the distance from the center of this circle to vertex $A$.

Since $\sin \frac{\alpha}{2}=\frac{\frac{a}{2}}{b}=\frac{5}{13}$, it follows that $\frac{r}{x}=\frac{5}{13}$, that is, $x=\frac{13}{5} r . \quad 2$ points Note that $|A D|=2 R=x+r$, from which it follows:

$\frac{169}{20}=\frac{13}{5} r+r$, that is, $r=\frac{169}{72} \text{ cm}$.