Three unit-radius circles pass through a common point. Their second intersection points are $A, B$, and $C$. What is the radius of the circle $ABC$?
Let the centers of the three circles be $A_{1}, B_{1}$, and $C_{1}$, and their common point be $P$ as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_8fb063dafa131d6690c1g-1.jpg?height=523&width=612&top_left_y=218&top_left_x=751)

We will show that the midpoints of the segments $A A_{1}, B B_{1}, C C_{1}$ coincide (at point $O$). Draw vectors from the common point $P$ to the centers of the circles: $\overrightarrow{P A_{1}}=\mathbf{a}, \overrightarrow{P B_{1}}=\mathbf{b}, \overrightarrow{P C_{1}}=\mathbf{c}$. (Only the vectors $\overrightarrow{P A_{1}}$ and $\overrightarrow{P C_{1}}$ are shown in the figure.) These vectors are of unit length. The segments $A_{1} B, C_{1} B, C_{1} A, B_{1} A, B_{1} C, A_{1} C$ are all of unit length, so

$$
\mathbf{a}+\mathbf{c}=\overrightarrow{P B}, \quad \mathbf{a}+\mathbf{b}=\overrightarrow{P C}, \quad \mathbf{b}+\mathbf{c}=\overrightarrow{P A}
$$

The vector pointing to the midpoint of the segment $A A_{1}$ is:

$$
\frac{\overrightarrow{P A_{1}}+\overrightarrow{P A}}{2}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{2}
$$

We get the same vector for the midpoints of $B B_{1}$ and $C C_{1}$, so the three midpoints indeed coincide. The reflection of the triangle $A_{1} B_{1} C_{1}$ about $O$ is the triangle $A B C$. Since a unit circle can be circumscribed around the triangle $A_{1} B_{1} C_{1}$ with center $P$, the same can be done for the triangle $A B C$.

Based on the work of Stomfai Gergely (Budapest, ELTE Apáczai Cs. J. Gyakorló Gimn. és Koll., 10th grade)

Note. It can be seen from the solution that the point $O$ is the center of the Feuerbach circle of the triangle $A_{1} B_{1} C_{1}$, the point $P$ is the center of the circumcircle, and the reflection of the point $P$ about $O$ is the orthocenter of the triangle $A_{1} B_{1} C_{1}$, which, as we learned, is the center of the circumcircle of the triangle $A B C$. This can also be stated as follows: the pedal triangle $D E F$ of the triangle $A_{1} B_{1} C_{1}$ was dilated by a factor of two from the center $P$ of the circumcircle, resulting in the triangle $A B C$.