<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] Following the notations of the figure by @Dr. Mathva, knowing that $AD=3, BD=2, CD=1,$ and setting $x=AB=AC=BC$, let us express that the Cayley-Menger determinant of the four points $A,B,C,D$ is zero (see here):
$$\begin{vmatrix}0&x^2&x^2&3^2&1\\
    x^2&0&x^2&2^2&1\\
    x^2&x^2&0&1^2&1\\
    3^2&2^2&1^2&0&1\\
     1&1 &1 &1& 0\end{vmatrix}=0\tag{1}$$
Expanding and factoring, one obtains :
$$-2x^2(x^2 - 7)^2=0\tag{2}$$
therefore with unique "geometrically acceptable" answer $x=\sqrt{7}$.
Generalization, with radii $a, b, c$.
$$\begin{vmatrix}0&x^2&x^2&c^2&1\\
    x^2&0&x^2&b^2&1\\
    x^2&x^2&0&a^2&1\\
    c^2&b^2&a^2&0&1\\
     1&1 &1 &1& 0\end{vmatrix}=0\tag{2}$$
giving, by setting $X=x^2$, the equation :
$$2X(X^2 - sX+ \underbrace{(s^2-3r)}_q)=0 \ \text{where} \ r:=a^2b^2+a^2c^2+b^2c^2, \ s:=a^2+b^2+c^2\tag{3}$$
which, setting apart solution $X=0$, is a quadratic equation. Therefore, conditionned by the following constraint on the sign of its discriminant:
$$\Delta=3(4r-s^2)\ge 0\tag{4}$$
warranting real solutions.
$\Delta$ possesses a remarkable factorization:
$$\Delta=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\tag{5}$$
As a consequence, when $\Delta > 0$, $\sqrt{\Delta}$ is 4 times the area of the triangle with sides $a,b,c$, due to Heron formula.
The signs of the roots are governed by the sign of $q$ which is in fact always positive, meaning that there are two possible sidelengths for the equilateral triangle if $\Delta >0$,
$$x=\sqrt{\frac12 (s \pm \sqrt{\Delta})}\tag{6}$$
Graphical representation: We can, without loss of generality, work in a  plane with coordinates $(a/c,b/c)$ or said otherwise in a plane with coordinates $(a,b)$ with $c=1$. Therefore, equation $\Delta=0$ becomes the equation of a curve... which is plainly the union of four straight lines, 3 of them appearing in the first quadrant. Here is the result of a simulation : in green, resp. red: values of $(a/c, b/c)$ giving a (double) solution, resp. not yielding a solution.

Different remarks:

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*3 lines are materialized in this simulation : $a+b-1=0, a-b+1=0, and -a+b+1=0$


*the particular case $a=3, b=2, c=1$ considered in the first part corresponds to a limit case between the green and the red zone.


*Here is a case where we have two solutions $a=1.8, \ b=1.5, \ c=1$ with two different values of sidelength $x$.

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