Given a circle of radius $R$. Four circles of equal radii touch the given one externally, and each of these four circles touches two others. Find the radii of these four circles.

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The quadrilateral with vertices at the centers of four given equal circles is a square.

## Solution

Let $R$ be the radius of the given circle, $x$ be the radius of the other circles. Denote their centers as $O, O_{1}, O_{2}, O_{3}, O_{4}$, respectively.

The quadrilateral $O_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ is a rhombus with side length $2 x$. Since its vertices are at equal distances from point $O$, it is a square. Therefore, $2 x=(R+x) \sqrt{2}$. From this, we find that

$$
x=\frac{R}{\sqrt{2}-1}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-15.jpg?height=378&width=378&top_left_y=1666&top_left_x=840)

Answer

$\frac{R}{\sqrt{2}-1}$.