NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Let $B_1$ be the symmetrical of the point $B$ with respect to the line $IC$, and let $C_1$ be the symmetrical of the point $C$ with respect to the line $IB$. Denote by $B_2$ the midpoint of the segment $I{B}_1$, and by $C_2$ the midpoint of the segment $I{C}_1$. The circles ${\gamma }^{\prime }$ and ${\delta }^{\prime }$ have diameters $I{B}_1$ and $I{C}_1$, hence their centers are $B_2$ and $C_2$. The line passing through the meeting points of circles ${\gamma }^{\prime }$ and ${\delta }^{\prime }$ is perpendicular to the centers line. Since $I$ belongs to both circles ${\gamma }^{\prime }$ and ${\delta }^{\prime }$, the requirement comes to $OI\perp B_2{C}_2$. Since $B_2{C}_2$ is midline in triangle $I{B}_1{C}_1$, one has $B_1{C}_1\parallel B_2{C}_2$, and the requirement writes $OI\perp B_1{C}_1$. It is enough to show $O{B}_1^2-O{C}_1^2=I{B}_1^2-I{C}_1^2$. Let $D$ be the projection of $I$ onto the line $BC$. $D$ is the touching point of the incircle of triangle $ABC$ with the side $BC$, and $BD=p-b$, $DC=p-c$. From symmetry considerations we have $I{B}_1=IB$, $I{C}_1=IC$, and $$\begin{array}{rl}I{B}_1^2-I{C}_1^2& =I{B}^2-I{C}^2=\\ & =D{B}^2-D{C}^2=\\ & =(p-b{)}^2-(p-c{)}^2=\\ & =a(c-b).\end{array}$$ Denote by $M$ the midpoint of the side $AC$, and by $N$ the midpoint of the side $AB$; then we have $$\begin{array}{rl}O{B}_1^2-O{C}_1^2& =\\ & =(O{M}^2+M{B}_1^2)-(O{N}^2+N{C}_1^2)=\\ & =(M{B}_1^2+O{A}^2-M{A}^2)-\\ & (N{C}_1^2+O{B}^2-N{B}^2)=\\ & =(M{B}_1^2-M{A}^2)-(N{C}_1^2-N{B}^2),\end{array}$$ since $OA=OB$. As $M{B}_1^2-M{A}^2=M{B}_1^2-M{C}^2=(M{B}_1-MC)(M{B}_1+MC)=a(a-b)$, and analogously $N{C}_1^2-N{B}^2=a(a-c)$, it follows that $O{B}_1^2-O{C}_1^2=a(c-b)$, which is what was left to prove.