## Procedimiento numérico

Procedimiento del punto medio

La posición del extremo de la cuerda es x0 se obtiene la ecuacón trascendente en θ0.

Configuración cóncava

$\frac{{x}_{0}}{C}=\frac{1}{1-{\alpha }^{2}}\left\{\frac{1}{\sqrt{{\alpha }^{2}-1}}\mathrm{ln}|\frac{\sqrt{1-{\alpha }^{2}}+\left(1+\alpha \right)\mathrm{tan}\left({\theta }_{0}/2\right)}{\sqrt{1-{\alpha }^{2}}-\left(1+\alpha \right)\mathrm{tan}\left({\theta }_{0}/2\right)}|+\frac{\alpha \text{\hspace{0.17em}}\mathrm{sin}{\theta }_{0}}{\mathrm{cos}{\theta }_{0}-\alpha }\right\}$

Configuración convexa

$\frac{{x}_{0}}{-\alpha \text{\hspace{0.17em}}d}=\frac{1}{{\alpha }^{2}-1}\left\{\begin{array}{l}\frac{2}{\sqrt{{\alpha }^{2}-1}}\mathrm{arctan}\left(\frac{\alpha +1}{\sqrt{{\alpha }^{2}-1}}\mathrm{tan}\left(\frac{{\theta }_{0}}{2}\right)\right)+\frac{\alpha \text{\hspace{0.17em}}\mathrm{sin}{\theta }_{0}}{\alpha -\mathrm{cos}{\theta }_{0}}\\ -\frac{2}{\sqrt{{\alpha }^{2}-1}}\mathrm{arctan}\left(\frac{\alpha +1}{\sqrt{{\alpha }^{2}-1}}\right)-1\end{array}\right\}$

• Cuando α=1

$\frac{{x}_{0}}{d}=\frac{1}{6{\mathrm{tan}}^{3}\left({\theta }_{0}/2\right)}-\frac{1}{2\mathrm{tan}\left({\theta }_{0}/2\right)}+\frac{1}{3}$

 `public abstract class Raiz { private final double CERO=1e-10; private final double ERROR=0.001; private final int MAXITER=200; public Raiz(double alfa, double x0, double lon) { this.alfa=alfa; this.x0=x0; this.lon=lon; } protected double puntoMedio(double a, double b) { double m, ym; int iter=0; do{ m=(a+b)/2; ym=f(m); if(Math.abs(ym)