Three concentric circles ω1, ω2, ω3 with radius r1, r2, r3 such that r1 + r3 ≥ 2r2,construct a line that intersects ω1, ω2, ω3 at A, B, C respectively such that AB = BC.
I got the problem partially solved. Here may be two cases.
Case 1: when $r_1+r_3=2r_2$, this is easy to prove that all lines in the region $[\omega_3]-[\omega_1]$ touching the circumference of the three circles (not tangent to $\omega_1$) can be considered as our solution because in that case we have to have $r_3-r_2=r_2-r_1$ which is from $r_1+r_3=2r_2$ and we'll be done.
Case 2: when $r_1+r_3>2r_2$, The answer, I know, will be the lines again in the $[\omega_3]-[\omega_1]$ region and touching the circumferences of the three circles but will be tangent to $\omega_1$. I can not prove this yet. But I have tried many approaches. In case 2, it is known that the value of $BC$ is $\sqrt{r_2^2-r_1^2}$. And what is left is to prove that $AB$ equals that.
Refer to the following drawing. Since $AB=BC$, we have
$$ OA^{2}+OC^{2}=AB^{2}+BC^{2}+2OB^{2} $$
Substituting $OA=r_{1}$, $OB=r_{2}$, and $OC=r_{3}$ yields the following:
$$ AB=BC=\sqrt{\frac{r_{1}^{2}+r_{3}^{2}-2r_{2}^{2}}{2}} $$
The expression inside the square root is non-negative, since by using QM-AM inequality and $r_{1}+r_{3}\geq 2r_{2}$, we have
$$ \frac{r_{1}^{2}+r_{3}^{2}-2r_{2}^{2}}{2}\geq\left(\frac{r_{1}+r_{3}}{2}\right)^{2}-r_{2}^{2}\geq0 $$
Therefore, a method to construct the line is: choose an arbitrary point $B$ on $\omega_{2}$. Then draw a circle with radius $\sqrt{\frac{r_{1}^{2}+r_{3}^{2}-2r_{2}^{2}}{2}}$. $A$ and $C$ are the intersection between this new circle with $\omega_{1}$ and $\omega_{3}$, respectively.
Notice that $\frac{r_{1}^{2}+r_{3}^{2}-2r_{2}^{2}}{2}=0$ if and only if $r_{1}=r_{2}=r_{3}$, so all three circles coincide. In this case, any line intersecting the three circles will do.
Additional Explanation
The cosine rule states that
$$ \begin{aligned} OA^{2}&=OB^{2}+AB^{2}-2OB\cdot AB\cos{\left(\angle OBA\right)}\\ OC^{2}&=OB^{2}+BC^{2}-2OB\cdot BC\cos{\left(\angle OBC\right)} \end{aligned} $$
Because $AB=BC$ and $\angle OBA + \angle OBC = \pi$, we can sum the two equations and obtain
$$ OA^{2}+OC^{2}=AB^{2}+BC^{2}+2OB^{2} $$