$\left[\begin{array}{l}\text { [Tangent Circles] } \\ \text { Law of Cosines }\end{array}\right]$

In an isosceles triangle with an angle of $120^{\circ}$, the radius of the inscribed circle is $R$. Inside the triangle, there are two equal circles that are tangent to each other, each touching one of the lateral sides of the triangle and the inscribed circle of the triangle. Find the radii of these circles (find all solutions).
Let a circle of radius $R$ with center $O$ touch the lateral sides $AB$ and $AC$ of an isosceles triangle $ABC$ at points $P$ and $Q$ respectively, and the base $BC$ at point $M$. If equal circles of radius $x$ with centers $O_1$ and $O_2$ touch this circle internally (Fig.1) and the first touches the lateral side $AB$, while the second touches the lateral side $AC$, then their points of tangency with these sides coincide with $P$ and $Q$ respectively.

Point $M$ is the midpoint of the base $BC$ of the isosceles triangle, so $AO_1$ and $AO_2$ are the angle bisectors of $\angle BAM$ and $\angle CAM$, respectively, meaning,

$$
\begin{gathered}
\angle O_1AO_2=60^{\circ}, \angle O_1OO_2=\angle QOP=180^{\circ}-\angle QAP=180^{\circ}-120^{\circ}=60^{\circ}, \\
O_1O_2=2x, OO_1=OO_2=R-x .
\end{gathered}
$$

Triangle $OO_1O_2$ is equilateral, so $2x=R-x$, from which $x=\frac{1}{3}R$. Now let equal circles of radius $x$ with centers $O_1$ and $O_2$ touch the incircle of triangle $ABC$ externally and the first touches the lateral side $AB$ at point $E$, while the second touches the lateral side $AC$ at point $F$ (Fig.2). In triangle $AOO_1$, it is known that

$$
AO=\frac{OP}{\sin 60^{\circ}}=\frac{2R}{\sqrt{3}}, AO_1=\frac{O_1E'}{\sin \angle O_1AE}=\frac{O_1E'}{\sin 30^{\circ}}=2x, OO_1=R+x, \angle O_1AO=30^{\circ}.
$$

By the cosine rule,

$$
OO_1^2=AO_1^2+AO^2-2 \cdot AO_1 \cdot AO \cos 30^{\circ}
$$

or

$$
(R+x)^2=4x^2+\frac{4R^2}{3}-2 \cdot 2x \cdot \frac{2R}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \Leftrightarrow 9x^2-18Rx+R^2=0 \Leftrightarrow x=\left(1 \pm \frac{2}{3} \sqrt{2}\right)R
$$

The only root that satisfies the problem's condition is the smaller one, i.e., $x=\left(1 - \frac{2}{3} \sqrt{2}\right)R$.

## Answer

$x_1=\frac{1}{3}R, x_2=\frac{3-2\sqrt{2}}{3}R$