3. In an equilateral triangle, a circle is drawn that touches the three sides of the triangle. The radius of this circle is equal to 10. A second, smaller, circle touches the first circle and two sides of the triangle. A third, even smaller, circle touches the second circle and two sides of the triangle (see the figure). What is the radius of the third circle?

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B3. $\frac{10}{9}$ In the figure, you see an enlarged view of the corner of the equilateral triangle. The large circle is not drawn, only the second and third circles. The two sides of the triangle that meet at vertex $A$ form an angle of 60 degrees. We will show that the radius of the second circle is three times as large as the radius of the third circle: $|B C|=3|D E|$.

We first consider triangle $A B C$. Angle $B A C$ is $\frac{60}{2}=30$ degrees, angle $A C B$ is 90 degrees, and angle $A B C$ is therefore $180-30-90=60$ degrees. This means that triangle $A B C$ and its reflection in line $A C$ together form an equilateral triangle. It follows that $|A B|=2|B C|$. The same applies to triangle $A D E$, and thus we find that $|A D|=2|D E|$. Now consider line segment $A S$. We have $|A S|=|A B|-|B S|=2|B C|-|B C|=|B C|$. On the other hand, we also have $|A S|=|S D|+|A D|=|D E|+2|D E|=3|D E|$. We see that indeed $|B C|=3|D E|$.

We can use exactly the same reasoning for the first and second circles. We then see that the radius of the first circle is three times as large as the radius of the second circle. The radius of the third circle is therefore equal to $\frac{1}{3} \cdot \frac{1}{3} \cdot 10=\frac{10}{9}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_732d64300714d2dca006g-3.jpg?height=708&width=640&top_left_y=2102&top_left_x=705)