<s>[INST] Equilateral triangle with vertices on 3 concentric circles Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?
My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ I will manipulate the formula afterwards,,,
 [/INST] Prompted by an additional point in a duplicate question asking whether the construction is always possible for arbitrary concentric circles, the following is about the general case of radii $\,A,B,C\,$ (where OP's question here is about $\,A=1,B=2,C=3\,$).
Let $\,a,b,c\,$ be points in the complex plane on the respective circles, and assume WLOG that $\,a=A \in \mathbb R\,$, since the problem is invariant to rotations. Let $\,\omega\,$ be a complex cube root of unity, then $\,\omega^2+\omega+1=0\,$, and the condition for the triangle $\,a,b,c,\,$ to be equilateral is:
$$
c-a + \omega(b-a)=0 \;\;\iff\;\; c = - \omega b + (1+\omega)a \tag{1} \;\;\iff\;\; c = - \omega b - \omega^2a
$$
Multiplying by the conjugate, and using that $\,\omega\overline\omega=|\omega|^2=1\,$, $\,\overline\omega=\omega^2\,$:
$$
\begin{align}
|c|^2 &= |b|^2+|a|^2 + a \left(\omega\overline\omega^2b+\overline\omega\omega^2\overline b\right) = |b|^2+|a|^2 + 2a \,\text{Re} \left(\omega^2 b\right)
\end{align}
$$
It follows that:
$$
\text{Re} \left(\omega^2 b\right) \,=\,\frac{|c|^2-|a|^2-|b|^2}{2a}=\frac{C^2-A^2-B^2}{2A} \tag{2}
$$
Since $\,\left|\omega^2 b\right|=|b|\,$, the condition for existence of a solution is:
$$
\left|\text{Re} \left(\omega^2 b\right)\right| \le |b| \quad\iff\quad \left(C^2-A^2-B^2\right)^2 \le 4 A^2 B^2 \tag{3}
$$
In that case:
$$
\text{Im} \left(\omega^2 b\right) \,=\,\pm \sqrt{|b|^2 - \left(\text{Re} \left(\omega^2 b\right)\right)^2}=\pm\sqrt{B^2-\left(\frac{C^2-A^2-B^2}{2A}\right)^2} \tag{4}
$$
When a solution exists according to $\,(3)\,$, equations $(2),(4)$ give $\,b\,$ (up to a reflection about the real axis), then $\,(1)\,$ gives $\,c\,$.
Quick verification for OP's case $\,a=A=1, B=2, C=3\,$, which satisfies $\,(3)\,$ so solutions exist.

*

*From $\,(2)\,$: $\;\text{Re}\left(\omega^2b\right)=\frac{9-1-4}{2}=2\,$.


*From $\,(4)\,$: $\;\text{Im}\left(\omega^2b\right)=\pm\sqrt{4-2^2}=0\,$.


*Then $\,\omega^2 b=\text{Re}(\omega^2b)+i\,\text{Im}(\omega^2b)=2\,$, so $\,b = 2\omega\,$.


*From $\,(1)\,$: $\;c=-\omega\cdot 2 \omega - \omega^2 \cdot 1 =-3\omega^2\,$.
Therefore the solution is $\,\big\{\,1, \,2 \omega, \,-3\omega^2 \,\big\}\,$ up to rotations and reflections, and it can be easily verified that the side length of the equilateral triangle is $\,|b-1|=\sqrt{7}\,$.

[ EDIT ] $\;$ Condition $\,(3)\,$ can be written as $\,\left(2AB\right)^2 - \left(A^2+B^2-C^2\right)^2 \ge 0\,$ and, after factoring the differences of squares twice, is equivalent to:
$$
(A+B+C)(A+B-C)(A-B+C)(-A+B+C) \ge 0 \tag{3'}
$$
This matches $\,(5)\,$ in Jean Marie's answer and, when $\,A \le B \le C\,$, is equivalent to $\,C \le A + B\,$ which matches the answer to another related question linked in  Jean Marie's comment.
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