8,9 |

On a line, there are three points $A, B$, and $C$, such that $A B=B C=3$. Three circles of radius $R$ have centers at points $A, B$, and $C$.

Find the radius of the fourth circle that is tangent to all three given circles if a) $R=1$; b) $R=2$; c) $R=5$.
Let $x$ be the radius of the desired circle, $O$ be its center. The distances from the centers of the given circles to the point $O$ can be either $x+R$ (external tangency) or $|x-R|$ (internal tangency). All three segments cannot be equal. Therefore, there are two possible cases (up to symmetry).

1) $O A = O C = x + R, O B = |x - R|$ (left figure).

By the Pythagorean theorem, $(x + R)^2 - (x - R)^2 = 9$, from which $x = \frac{9}{4R}$.
2) $O A = O B = |x - R|, O C = x + R$ (right figure).

Let $D$ be the midpoint of $A B$. Then $(x + R)^2 - \frac{81}{4} = O D^2 = (x - R)^2 - \frac{9}{4}$, from which $x = \frac{9}{2R}$.

These conditions must be satisfied: $(x + R) + |x - R| \geq 6, (x + R) - |x - R| \leq 3$.

Solving these inequalities for $x = \frac{9}{2R}$, we get that $R \leq \frac{3}{2}$ or $R \geq 3$.
![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-33.jpg?height=466&width=1562&top_left_y=1&top_left_x=246)

## Answer

a) $\frac{9}{4}$ and $\frac{9}{2}$;
b) $\frac{9}{8};$
c) $\frac{9}{20}$ and $\frac{9}{10}$.