Three unit radius circles each pass through the point $P$, and their other intersection points are $A, B$, and $C$. What is the radius of the circumcircle of triangle $ABC$?
I. Solution. Let the centers of the circles be $X, Y$, and $Z$ as shown in Figure 1. Then the quadrilaterals $P X A Y, P X B Z$, and $P Y C Z$ are all rhombuses with unit side lengths, so the segments $X A, P Y$, and $Z C$ are parallel. Therefore, the quadrilateral $X A C Z$ is a parallelogram, which means $A C = Z X$. Similarly, from the parallelism of $Y A, P X$, and $Z B$, we get that $Y A B Z$ is also a parallelogram, so $A B = Z Y$. Furthermore, from the parallelism of $X B, P Z$, and $Y C$, we get that $X B C Y$ is also a parallelogram, which implies $B C = Y X$. Therefore, the triangle $A B C$ is congruent to the triangle $Z Y X$, so the radii of their circumscribed circles are also equal. A circle with a unit radius can be circumscribed around the triangle $Z Y X$, since all three vertices are a unit distance from $P$.

![](https://cdn.mathpix.com/cropped/2024_05_02_502172d7c0a99cf8acddg-1.jpg?height=534&width=1172&top_left_y=465&top_left_x=454)

Thus, the radius of the circle circumscribed around the triangle $A B C$ is also unit length.

II. Solution. Using the same notation as in Solution I. Scale down the triangle $A B C$ by a factor of $\frac{1}{2}$ from $P$. If two congruent circles intersect, then their common chord and the line connecting their centers are both symmetry axes of the figure formed by the two circles (Figure 2), so the images of $A, B, C$ under the scaling will be the midpoints of $X Y, X Z, Y Z$ respectively. The triangle formed by the midpoints of the sides is similar to the original triangle, and the similarity ratio is also $\frac{1}{2}$. Therefore, a circle with the same radius can be circumscribed around the triangle $A B C$ as around the triangle $X Y Z$. A unit circle can be circumscribed around the triangle $X Y Z$ with center $P$, so the radius of the circle circumscribed around the triangle $A B C$ is also unit length.

![](https://cdn.mathpix.com/cropped/2024_05_02_502172d7c0a99cf8acddg-1.jpg?height=304&width=447&top_left_y=1496&top_left_x=817)

Figure 2

Remark. As shown in Figure 1, the circles in the problem can be arranged in two different ways. Our solutions apply to both cases. However, some submitters only considered one case (usually the one shown on the left side of the figure), and their proofs were such that they would have worked for the other case with minor modifications. These solvers generally received 3 points.