2016 Eighth Romanian Master of Mathematics

a) Let ${\ell }_1$ be the tangent to $\Omega$ parallel to $A_2{B}_3$, lying on the same side of $A_2{B}_3$ as ${\omega }_1$. The tangents ${\ell }_2$ and ${\ell }_3$ are defined similarly. The lines ${\ell }_1$ and ${\ell }_2$, ${\ell }_2$ and ${\ell }_3$, ${\ell }_3$ and ${\ell }_1$ meet at $C_3$, $C_1$, $C_2$, respectively (see Fig. ??). Finally, the line $C_2{C}_3$ meets the rays $X{A}_1$ and $X{B}_1$ emanating from $X$ at $S_1$ and $T_1$, respectively; the points $S_2$, $T_2$, and $S_3$, $T_3$ are defined similarly. Each of the triangles $△{\Delta }_1=△X{S}_1{T}_1$, $△{\Delta }_2=△T_{2}X{S}_2$, and $△{\Delta }_3=△S_3{T}_{3}X$ is similar to $△\Delta =△C_1{C}_2{C}_3$, since their corresponding sides are parallel. Let $k_i$ be the ratio of similitude of $△{\Delta }_i$ and $△\Delta$ (e.g., $k_1=X{S}_1/C_1{C}_2$ and the like). Since $S_{1}X=C_2{T}_3$ and $X{T}_2=S_3{C}_1$, it follows that $k_1+k_2+k_3=1$, so, if ${\varrho }_i$ is the inradius of $△{\Delta }_i$, then ${\varrho }_1+{\varrho }_2+{\varrho }_3=R$. Finally, notice that ${\omega }_i$ is interior to $△{\Delta }_i$, so $r_i\le {\varrho }_i$, and the conclusion follows by the preceding.

b) By part a), the equality $R=r_1+r_2+r_3$ holds if and only if $r_i={\varrho }_i$ for all $i$, which implies in turn that ${\omega }_i$ is the incircle of $△{\Delta }_i$. Let $K_i,L_i,M_i$ be the points where ${\omega }_i$ touches the sides $X{S}_i,X{T}_i,S_i{T}_i$, respectively. We claim that the six points $K_i$ and $L_i$ ($i=1,2,3$) are equidistant from $X$. Clearly, $X{K}_i=X{L}_i$, and we are to prove that $X{K}_2=X{L}_1$ and $X{K}_3=X{L}_2$. By similarity, $\angle T_1{M}_1{L}_1=\angle C_3{M}_1{M}_2$ and $\angle S_2{M}_2{K}_2=\angle C_3{M}_2{M}_1$, so the points $M_1,M_2,L_1,K_2$ are collinear. Consequently, $\angle X{K}_2{L}_1=\angle C_3{M}_1{M}_2=\angle C_3{M}_2{M}_1=\angle X{L}_1{K}_2$, so $X{K}_2=X{L}_1$. Similarly, $X{K}_3=X{L}_2$.