In a triangle ABC,if $(II_1)^2+(I_2I_3)^2=\lambda R^2$,where I denotes incenter,$I_1,I_2,I_3$denotes centers of the circles escribed to the sides BC,CA and AB respectively and R be the radius of the circumcircle of triangle ABC.Find $\lambda$.
I know that $II_1=4R\sin\frac{A}{2}$ but i dont know what is $(I_2I_3)$,so could not solve further.Can someone help me in this question?
Let $I_1,I_2,I_3$ be the excenters opposite to $A,B,C.$
As $BI_2,CI_3$ intersect at $I.$
So $\angle I_2II_3=\angle BIC=\pi-\frac{B}{2}-\frac{C}{2}$,because they are vertically opposite angles.
Using the Cosine law in $\triangle II_2I_3$,
$(I_2I_3)^2=(II_3)^2+(II_2)^2-2(II_2)(II_3)\cos(\angle I_2II_3)$
$(I_2I_3)^2=(II_3)^2+(II_2)^2-2(II_2)(II_3)\cos(\pi-\frac{B}{2}-\frac{C}{2})$
$(I_2I_3)^2=(II_3)^2+(II_2)^2+2(II_2)(II_3)\cos(\frac{B}{2}+\frac{C}{2})$
As $II_3=4R\sin(\frac{C}{2}),II_2=4R\sin(\frac{B}{2})$,using this in the above equation.
$(I_2I_3)^2=(4R\sin(\frac{C}{2}))^2+(4R\sin(\frac{B}{2}))^2+2(4R\sin(\frac{C}{2}))(4R\sin(\frac{B}{2}))\cos(\frac{B}{2}+\frac{C}{2})$
$(I_2I_3)^2=16R^2[\sin^2\frac{C}{2}+\sin^2\frac{B}{2}+2\sin\frac{C}{2}\sin\frac{B}{2}(\cos\frac{B}{2}\cos\frac{C}{2}-\sin\frac{B}{2}\sin\frac{C}{2})]$
$(I_2I_3)^2=16R^2[\sin^2\frac{C}{2}+\sin^2\frac{B}{2}-2\sin^2\frac{B}{2}\sin^2\frac{C}{2}+2\sin\frac{C}{2}\sin\frac{B}{2}\cos\frac{B}{2}\cos\frac{C}{2}]$
$(I_2I_3)^2=16R^2[\sin^2\frac{C}{2}-\sin^2\frac{B}{2}\sin^2\frac{C}{2}+\sin^2\frac{B}{2}-\sin^2\frac{B}{2}\sin^2\frac{C}{2}+2\sin\frac{C}{2}\sin\frac{B}{2}\cos\frac{B}{2}\cos\frac{C}{2}]$
$(I_2I_3)^2=16R^2[\sin^2\frac{C}{2}\cos^2\frac{B}{2}+\sin^2\frac{B}{2}\cos^2\frac{C}{2}+2\sin\frac{C}{2}\sin\frac{B}{2}\cos\frac{B}{2}\cos\frac{C}{2}]$
$(I_2I_3)^2=16R^2[\sin\frac{B}{2}\cos\frac{C}{2}+\cos\frac{B}{2}\sin\frac{C}{2}]^2$
$(I_2I_3)^2=16R^2\sin^2(\frac{B}{2}+\frac{C}{2})$
$(I_2I_3)^2=16R^2\cos^2(\frac{A}{2})$
So $(II_1)^2+(I_2I_3)^2=16R^2\sin^2(\frac{A}{2})+16R^2\cos^2(\frac{A}{2})=16R^2$
I'll write $A^\prime$, $B^\prime$, $C^\prime$ (instead of $I_1$, $I_2$, $I_3$) for the excenters opposite respective vertices $A$, $B$, $C$. The relation you mention can then be written as $$|\overline{IX^\prime}| = 4 R \sin\frac{X}{2} \tag{$\star$}$$ for $X = A$, $B$, $C$. Now, observe that $\overline{BB^\prime}$ and $\overline{CC^\prime}$ cross at $I$, so that $$\angle B^\prime IC^\prime = \angle BIC = 180^\circ - \frac{1}{2}B - \frac{1}{2}C$$ By the Law of Cosines in $\triangle B^\prime I C^\prime$ and $(\star)$, we have $$\begin{align} |\overline{B^\prime C^\prime}|^2 &= |\overline{IB^\prime}|^2 + |\overline{IC^\prime}|^2 - 2 |\overline{IB^\prime}||\overline{IC^\prime}|\cos\angle B^\prime I C^\prime \\[4pt] &= 16 R^2 \left(\sin^2 \frac{B}{2} + \sin^2\frac{C}{2}- 2 \sin\frac{B}{2}\sin\frac{C}{2} \cos\left(180^\circ - \frac{1}{2}B - \frac{1}{2}C\right)\right) \\[4pt] &= 16 R^2 \left( \sin^2 \frac{B}{2} + \sin^2\frac{C}{2}+ 2 \sin\frac{B}{2}\sin\frac{C}{2} \cos\frac{B+C}{2} \right) \\[4pt] &= 16 R^2 \sin^2\frac{B+C}{2} \\[4pt] &= 16 R^2 \cos^2\frac{A}{2} \end{align}$$
Consequently,
$$|\overline{IA^\prime}|^2 + |\overline{B^\prime C^\prime}|^2 = 16 R^2 \sin^2 \frac{A}{2} + 16 R^2 \cos^2 \frac{A}{2} = 16 R^2$$
I suspect there's a more-direct route to this result.
