I'm working on an analytical solution to the crossed ladders problem.
The solution is almost done and already useable (see my answer to Crossed Ladders Problem for details).
However I'm left with a problem in the end: depending on which ladder I choose using to calculate $w$, the expression of $w$ seems not to be the same.
Let $\left(j,k\right)\in\lbrace\left(1,2\right),\left(2,1\right)\rbrace$, $m\in\lbrace1,2,3,4\rbrace$, and $n\in\lbrace-,+\rbrace$.
Any quantity with a $j$ or $k$ index below seems to depend on the choice of the ladder while, in all numerical applications I tried, I always have $w_{jmn}=w_{kmn}$ with $w_{jmn}=n\sqrt{l_k^2-h_{jm}^2}$ and $w_{kmn}=n\sqrt{l_j^2-h_{km}^2}$.
This leads to:
\begin{equation} \begin{split} w_{j1+}&=\frac{\sqrt{R_9+R_7+c_F}}{6}+\text{i}\frac{c_{jL}\sqrt{R_9-R_7-c_F}}{6}\\ w_{j1-}&=\frac{-\sqrt{R_9+R_7+c_F}}{6}+\text{i}\frac{-c_{jL}\sqrt{R_9-R_7-c_F}}{6}\\ w_{j2+}&=\frac{\sqrt{R_9+R_7+c_F}}{6}+\text{i}\frac{c_{jM}\sqrt{R_9-R_7-c_F}}{6}\\ w_{j2-}&=\frac{-\sqrt{R_9+R_7+c_F}}{6}+\text{i}\frac{-c_{jM}\sqrt{R_9-R_7-c_F}}{6}\\ w_{j3+}&=\frac{c_{jN}\sqrt{\left|2c_F-2R_7+R_{j8}\right|}}{6}+\text{i}\frac{c_{jO}\sqrt{\left|2c_F-2R_7+R_{j8}\right|}}{6}\\ w_{j3-}&=\frac{-c_{jN}\sqrt{\left|2c_F-2R_7+R_{j8}\right|}}{6}+\text{i}\frac{-c_{jO}\sqrt{\left|2c_F-2R_7+R_{j8}\right|}}{6}\\ w_{j4+}&=\frac{c_{jP}\sqrt{\left|2c_F-2R_7-R_{j8}\right|}}{6}+\text{i}\frac{c_{jQ}\sqrt{\left|2c_F-2R_7-R_{j8}\right|}}{6}\\ w_{j4-}&=\frac{-c_{jP}\sqrt{\left|2c_F-2R_7-R_{j8}\right|}}{6}+\text{i}\frac{-c_{jQ}\sqrt{\left|2c_F-2R_7-R_{j8}\right|}}{6}\\ \end{split} \notag \end{equation}
With (again see the answer to Crossed Ladders Problem for all details):
\begin{equation} \begin{split} R_{j8}&=6\ \text{sgn}\left(3h+R_{j2}\right)\sqrt{2c_G-R_6+12\left|h\right|^3\sqrt{ \begin{split} &36h^2-6\sqrt[3]{27h^2c_{jA}^2+3c_{jA}^2\sqrt{c_H}}\\ &-6\sqrt[3]{27h^2c_{jA}^2-3c_{jA}^2\sqrt{c_H}}\\ \end{split} }}\\ c_{jL}&=1-\left(-1\right)^j\ \text{sgn}\left(3h-R_{j2}\right)-\left|\ \text{sgn}\left(3h-R_{j2}\right)\right|\\ c_{jM}&=1+\left(-1\right)^j\ \text{sgn}\left(3h-R_{j2}\right)-\left|\ \text{sgn}\left(3h-R_{j2}\right)\right|\\ c_{jN}&=\frac{1+\ \text{sgn}\left(2c_F-2R_7+R_{j8}\right)}{2}\\ c_{jO}&=\frac{1-\ \text{sgn}\left(2c_F-2R_7+R_{j8}\right)}{2}\\ c_{jP}&=\frac{1+\ \text{sgn}\left(2c_F-2R_7-R_{j8}\right)}{2}\\ c_{jQ}&=\frac{1-\ \text{sgn}\left(2c_F-2R_7-R_{j8}\right)}{2}\\ \end{split} \notag \end{equation}
Thus I wonder if it is possible to prove that:
\begin{equation} \begin{split} R_{k8}&=R_{j8}\\ c_{kL}&=c_{jL}\\ c_{kM}&=c_{jM}\\ \end{split} \notag \end{equation}
$R_{k8}=R_{j8}$ would imply $c_{kN}=c_{jN}$, $c_{kO}=c_{jO}$, $c_{kP}=c_{jP}$, and $c_{kQ}=c_{jQ}$.
Proving $\ \text{sgn}\left(3h+R_{k2}\right)=\ \text{sgn}\left(3h+R_{j2}\right)$ would be enough to prove all these relations.
For $c_{kL}=c_{jL}$ and $c_{kM}=c_{jM}$, proving $\ \text{sgn}\left(3h-R_{k2}\right)=-\ \text{sgn}\left(3h-R_{j2}\right)$ would be enough.
In my attempts, and when working on the general solution, I came across the following equations, which I write in case they are helpful:
\begin{equation} \begin{split} R_{j1}^3&=27c_{jA}^2R_{j1}+54c_{jC}\\ R_{j2}R_{j3}&=54\left(hc_{jA}+h^3\right)\\ R_{j3}&=2\sqrt{R_{j1}^2+c_{jB}R_{j1}+9c_{jA}^2-108h^2c_{jA}+81h^4}\\ R_6&=\frac{R_{j1}^2+6c_{jA}R_{j1}-18c_{jA}^2}{9}\\ R_7&=\frac{6hR_{j2}+R_{j3}}{2}\\ R_{j8}&=2R_{j4}\left(3h+R_{j2}\right)\\ R_9&=\sqrt{\left(c_F+R_7\right)^2+R_{j5}^2\left(3h-R_{j2}\right)^2}\\ R_{j1}&\geq6c_{jA}-9h^2\\ R_{k1}&=-R_{j1}\\ \text{sgn}\left(R_{j1}\right)&=\ \text{sgn}\left(c_{jA}\right)\\ hc_{jA}+h^3=0&\implies R_{j1}-c_{jB}=0\\ R_{j3}&\geq\left|R_{j1}+12c_{jA}-18h^2\right|\\ R_{j3}&\geq-6hR_{j2}\\ R_{j5}=0&\implies3h-R_{j2}=0 \end{split} \notag \end{equation}
Thanks in advance for your answers.
