1. The ratio of the radius $R$ of the circumscribed circle and the radius $r$ of the inscribed circle in triangle $\triangle A B C$, which is isosceles, is $k$. Calculate the value of $\cos \alpha$ if $\alpha$ denotes the angle at the base of $\triangle A B C$.
Solution. Let $O$ be the center of the inscribed circle in $\triangle ABC$, and $A_{1}, B_{1}, C_{1}$ be the points of tangency of the inscribed circle with the sides of the triangle, and $\frac{R}{r}=k$.

From $\triangle AOC_{1}$ we have $\frac{r}{2}=\frac{c}{2} \tan \frac{\alpha}{2}$. On the other hand, from the sine theorem we have

$$
\begin{aligned}
& \frac{c}{\sin \gamma}=2 R \\
& \frac{c}{\sin \left(180^{\circ}-2 \alpha\right)}=2 R \\
& \frac{c}{2}=R \sin \alpha
\end{aligned}
$$

Then

$$
\begin{aligned}
& r=R \sin 2 \alpha \tan \frac{\alpha}{2} \\
& \frac{R}{r}=\frac{1}{2 \sin \alpha \cos \alpha \tan \frac{\alpha}{2}}=\frac{1}{4 \sin ^{2} \frac{\alpha}{2} \cos \alpha}=\frac{1}{2 \cos \alpha(1-\cos \alpha)} \\
& 2 \frac{R}{r} \cos \alpha(1-\cos \alpha)=1 \\
& 2 k \cos \alpha-2 k \cos ^{2} \alpha-1=0
\end{aligned}
$$

By solving the last quadratic equation, we get

$$
\cos \alpha=\frac{2 k \pm \sqrt{4 k^{2}-8 k}}{4 k}=\frac{2 k \pm 2 \sqrt{k^{2}-2 k}}{4 k}
$$

that is, $\cos \alpha=\frac{1}{2} \pm \frac{\sqrt{k^{2}-2 k}}{2 k}$.