Example 26. The ratio of the radius of the circle inscribed in an isosceles triangle to the radius of the circle circumscribed around this triangle is $k$. Find the cosine of the angle at the base of the triangle.
Solution (Fig. 30). Let $\widehat{B C A}=x$ and introduce an auxiliary parameter: $|M K|=r$. Then

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-047.jpg?height=497&width=462&top_left_y=1833&top_left_x=363)

Fig. 30

$$
|O B|=r k, \cos x=|C K|:|B C| .
$$

From triangle $M C K$ we have: $|C K|=$ $=r \operatorname{ctg} \frac{x}{2}$. By the Law of Sines, we write:

$|B C|=2 r k \sin x$.

Thus, $\cos x=\frac{r \operatorname{ctg} \frac{x}{2}}{2 r k \sin x}$,

46
from which

$$
\cos x=\frac{\operatorname{ctg} \frac{x}{2}}{k \sin x}
$$

i.e., $(\cos x)_{1}=\frac{1+\sqrt{1-2 k}}{2},(\cos x)_{2}=\frac{1-\sqrt{1-2 k}}{2}$.

Investigation. By the problem's context, $0^{\circ}0$. Moreover, since the radius of the inscribed circle in a triangle reaches its maximum when the triangle is equilateral, and in this case, the ratio of the radius of the inscribed circle to the radius of the circumscribed circle is $1: 2$, the values of the parameter $k$ satisfy the system of inequalities $0-2 k \geqslant-1$, or $1>1-2 k \geqslant 0$, i.e., $1>\sqrt{1-2 k} \geqslant 0$, and therefore,

$$
2>1+\sqrt{1-2 k} \geqslant 1
$$

from which

$$
1>\frac{1+\sqrt{1-2 k}}{2} \geqslant \frac{1}{2}
$$

Thus, when $\frac{1}{2} \leqslant(\cos x)_{1}<1 \quad$ it will certainly be $0<(\cos x)_{1}<1$, i.e., the first of the found values of $\cos x$ satisfies the problem's condition.

Similarly (verify this independently), the value $\cos x=\frac{1-\sqrt{1-2 k}}{2}$ also satisfies the problem's condition.