Hey, people don't you see this beautiful problem.

I think nobody is interested in this interesting problem. I'll send the solution soon.

I don't think I'll get into all the details. Let $O_i,O$ be the centers of circles $i,\ i\in\overline {1,4}$ and the large circle, respectively. Let $\ell$ be the line parallel to $BC$ through $O_1$, and let $X_i,X$ be the projection of $O_i,O$ on $\ell$, respectively. Because circles $2,3$ are tangent to $BC$ and circle $4$, $O_2,O_3$ lie on the parabola having $O_4$ as focal point and $\ell$ as directrice (call the parabola $\mathcal P$). At the same time, the circles $2,3$ are tangent to the big circle and to circle $4$ , so the $O_2,O_3$ also lie on the ellipse having $O_4,O$ as foci and large axis $r_1+R$ (call the ellipse $\mathcal E$). It's easy to see that $d(O_4,\ell)+d(O,\ell)=r_1+R$, and in these conditions, it's also easy to see that if a point $P$ belongs to $\mathcal P\cap \mathcal E$, then so does its symmetric wrt the center of $\mathcal E$. $O_2,O_3\in\mathcal P\cap \mathcal E$, so the symmetric of $O_2$ wrt the midpoint of $O_4O$ also belongs to $\mathcal P\cap \mathcal E$. On the other hand, the vertex of $\mathcal P$, which is $M$, lies inside $\mathcal E$, so $\mathcal P\cap\mathcal E$ has at most two points, meaning that $O_2OO_3O_4$ is a parallelogram, so $OO_3=O_2O_4$, and since $OO_3=R-r_3,\ O_2O_4=r_2+r_1$, we get the conclusion. I hope that it's Ok, and that people can understand it .

This seems something like Sangaku problem in Edo era.

Yes, Omid Hatami ! I like this nice problem. Proof. Denote the projections $X,P,Y$ of the centers $O_{2},O_{4},O_{3}$ respectively to the line $BC\ .$ Prove easily that $OM=R-2r_{1}$ , $PX=2\sqrt{r_{1}r_{2}}$ , $PY=2\sqrt{r_{1}r_{3}}$ , $MX^{2}=4r_{1}(R-r_{1}-r_{2})$ , $MY^{2}=4r_{1}(R-r_{1}-r_{3})\ .$ Therefore, $(PX+PY)^{2}=(MX+MY)^{2}$ $\Longrightarrow$ $4r_{1}(r_{2}+r_{3})+8r_{1}\sqrt{r_{2}r_{3}}=4r_{1}(2R-2r_{1}-r_{2}-r_{3})+8r_{1}\sqrt{(R-r_{1}-r_{2})(R-r_{1}-r_{3})}$ $\Longrightarrow$ $r_{2}+r_{3}+2\sqrt{r_{2}r_{3}}=2R-2r_{1}-r_{2}-r_{3}+2\sqrt{(R-r_{1}-r_{2})(R-r_{1}-r_{3})}$ $\Longrightarrow$ $\sqrt{(R-r_{1}-r_{2})(R-r_{1}-r_{3})}-\sqrt{r_{2}r_{3}}=r_{1}+r_{2}+r_{3}-R\ .$ Denote $\boxed{R-r_{1}=t>0}\ .$ Thus, $\sqrt{(t-r_{2})(t-r_{3})}-\sqrt{r_{2}r_{3}}=r_{2}+r_{3}-t$ $\Longrightarrow$ $-t(r_{2}+r_{3}-t)=(r_{2}+r_{3}-t)\left[\sqrt{(t-r_{2})(t-r_{3})}+\sqrt{r_{2}r_{3}}\right]$ $\Longrightarrow$ $\left[t+\sqrt{(t-r_{2})(t-r_{3})}+\sqrt{r_{2}r_{3}}\right](r_{2}+r_{3}-t)=0$ $\Longrightarrow$ $t=r_{2}+r_{3}$ $\Longrightarrow$ $\boxed{\ r_{1}+r_{2}+r_{3}=R\ }\ .$

i guess inversion and go... but i'm too lazy to do it

gollywog wrote: i guess inversion and go... but i'm too lazy to do it i used inversion,with the intersection of circle$C(1)$,and arc$BC$, as the center, really i dont have enough time to post the solution,but im not lazy