Three concentric circles have radii of 1, 2, and 3 units, respectively. A point is chosen on each of the three circles such that they form the vertices of an equilateral triangle. What can be the side length of this equilateral triangle?
Solution. Let the circles be $k_{1}, k_{2}$, and $k_{3}$, with their common center $O$, and the three selected points be $A_{1}, A_{2}$, and $A_{3}$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_299b6d8c09b5872bc093g-1.jpg?height=637&width=593&top_left_y=248&top_left_x=755)

Since $A_{1} A_{2} A_{3}$ is an equilateral triangle, there is a $60^{\circ}$ rotation (positive or negative direction) $\varphi$ around $A_{2}$ that maps $A_{3}$ to $A_{1}$. If the image of $O$ under this $\varphi$ rotation is the point $P$, then the triangle $A_{2} O P$ is equilateral, so $O P = O A_{2} = 2$, which means $P$ lies on the circle $k_{2}$. The image of the circle $k_{3}$ under $\varphi$ is a circle $k_{3}^{\prime}$ centered at $P$ with radius 3, which passes through the point $A_{1}$. Thus, $A_{1}$ is a common point of the circles $k_{1}$ and $k_{3}^{\prime}$. Since

$$
P O + O A_{1} = 2 + 1 = 3 = P A_{1}
$$

the point $O$ lies on the segment $P A_{1}$—otherwise, by the triangle inequality, $P O + O A_{1} > P A_{1}$ would hold.

Therefore, the angle $A_{2} P A_{1}$ is equal to the $60^{\circ}$ angle $A_{2} P O$. In the triangle $A_{2} P A_{1}$, $A_{2} P = 2$ and $P A_{1} = 3$, so by the cosine rule, the third side of the triangle is

$$
A_{2} A_{1} = \sqrt{A_{2} P^{2} + P A_{1}^{2} - 2 \cdot A_{2} P \cdot P A_{1} \cdot \cos 60^{\circ}} = \sqrt{7}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_299b6d8c09b5872bc093g-1.jpg?height=306&width=336&top_left_y=1511&top_left_x=884)

Thus, the side length of the equilateral triangle can only be $\sqrt{7}$. It is also clear from the above that such a triangle indeed exists.