$\left[\begin{array}{l}{\left[\begin{array}{l}\text { Tangent circles } \\ {\left[\text { Right triangle with an angle of } 30^{\circ}\right.}\end{array}\right]}\end{array}\right]$

In an equilateral triangle, a circle is inscribed. Three smaller circles are tangent to this circle and the sides of the triangle. Find the side of the triangle if the radius of the smaller circle is $r$.
Drop a perpendicular from the center of one of the smaller circles to the radius of the larger circle, drawn to the point of tangency with the side of the considered angle.

## Solution

Let $R$ be the radius of the circle inscribed in the given triangle. Drop a perpendicular from the center of the smaller circle to the radius of the larger circle, drawn to the point of tangency with the side of the considered angle. We obtain a right triangle with the hypotenuse $R+r$, the leg $R-r$, and an angle of $30^{\circ}$ opposite this leg. Therefore, $R+r=2(R-r)$. From this, we find that $R=3r$. Thus, the side of the triangle is $6r\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-16.jpg?height=398&width=546&top_left_y=1&top_left_x=756)

## Answer

$6r\sqrt{3}$.

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