[ Tangent circles [ Opposite the larger side lies the larger angle ]

Two circles of radii $r$ and $p (r < p)$ touch each other externally and both touch internally a circle of radius $R$. It is known that the triangle with vertices at the centers of the circles is isosceles, and the angle between the lateral sides is greater than $\frac{\pi}{3}$. Find the length of the base of this triangle.
Against the larger angle of a triangle lies the larger side.

## Solution

Let the circle with center $O_{1}$ and radius $r$ and the circle with center $O_{2}$ and radius $p$ touch each other at point $C$, and the circle with center $O$ and radius $R$ touch at points $A$ and $B$ respectively.

Since the line of centers of two touching circles passes through their point of tangency, then $O_{1} O_{2}=r+p, O O_{1} = R-r, O O_{2}=R-p$.

The sides $O O_{1}=R-r$ and $O O_{2}=R-p$ cannot be the legs, because then $R-r=R-p \Rightarrow r=p$, which contradicts the condition $rR-p$. Therefore, $O O_{1}=R-r$ is the largest side of triangle $O_{1} O O_{2}$.

Thus, the base of the triangle is $R-r$.

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-47.jpg?height=778&width=806&top_left_y=423&top_left_x=635)

## Answer

$R-r$.

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