107. Three spheres, two of which are identical, touch the plane $P$ and, in addition, touch each other pairwise. The vertex of a right circular cone lies on the plane $P$, and the axis of the cone is perpendicular to this plane. All three spheres are located outside the cone, and each touches its lateral surface. Find the cosine of the angle between the generatrix of the cone and the plane $P$, given that in the triangle with vertices at the points of contact of the spheres with the plane one of the angles is $150^{\circ}$.
107. Let $O_{1}, O_{2}, O_{3}$ be the points of tangency of the spheres with the plane $P$, where in point $O_{1}$ the sphere has radius $r$, and in $O_{2}$ and $O_{3}$ the radii are $\boldsymbol{R}, \boldsymbol{O}$ - the vertex of the cone (Fig. 15), $\varphi$ - the angle between the generatrix of the cone and the plane $P$. We can find that

$$
\begin{gathered}
\left|O_{1} O\right|=r \operatorname{ctg} \frac{\varphi}{2}, \quad\left|O O_{2}\right|=\left|O O_{3}\right|=R \operatorname{ctg} \frac{\varphi}{2} \\
\left|O_{1} O_{2}\right|=\left|O_{1} O_{3}\right|=2 \sqrt{R r},\left|O_{2} O_{3}\right|=2 R
\end{gathered}
$$

Since $\left|O_{1} O_{2}\right|=\left|O_{1} O_{3}\right|$, the angle $\widehat{O_{2} O_{1} O_{3}}$ can only be $150^{\circ}$, which means $R / r=4 \sin ^{2} 75^{\circ}=2+\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-069.jpg?height=417&width=461&top_left_y=488&top_left_x=198)

Fig. 14.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-069.jpg?height=497&width=280&top_left_y=417&top_left_x=761)

Fig. 15.

Further, if $L$ is the midpoint of $\mathrm{O}_{2} \mathrm{O}_{3}$, then

$$
\begin{gathered}
|O L|=\sqrt{\left|O O_{3}\right|^{2}-\left|O_{3} L\right|^{2}}=R \sqrt{\operatorname{ctg}^{2} \frac{\varphi}{2}-1} \\
\left|O_{1} L\right|=\sqrt{\left|O_{1} O_{3}\right|^{2}-\left|O_{3} L\right|^{2}}=\sqrt{4 R r-R^{2}}
\end{gathered}
$$

The point $O$ lies on the line $O_{1} L$, and it can be either on the segment $O_{1} L$ or on its extension beyond points $L$ or $O_{1}$ (as in the figure $O^{\prime}$ and $\left.O^{\prime \prime}\right)$. Accordingly, we get three relations

$$
\begin{gathered}
\left|O_{1} L\right|=\left|O O_{1}\right|+|O L|, \quad\left|O_{1} L\right|=\left|O_{1} O^{\prime}\right|-\left|O^{\prime} L\right| \\
\left|O_{1} L\right|=\left|O^{\prime \prime} L\right|-\left|O^{\prime \prime} O_{1}\right| .
\end{gathered}
$$

Replacing in each of these three relations $R=(2+\sqrt{3}) \mathrm{r}$, $\operatorname{ctg} \frac{\varphi}{2}=x$, in the first two we arrive at a contradiction ( $x=1$ or $x=-2 \sqrt{\overline{3}} / 3$ ), in the third case we find $x=2 \sqrt{3} / 3$.

Answer: $\cos \varphi=\frac{1}{7}$.