24th Hellenic Mathematical Olympiad

(i) $$\widehat{\Gamma BN}=\widehat{\Gamma IN}=\frac{\hat{A}}{2}+\frac{\hat{\Gamma }}{2}=\frac{180^{\circ }-\hat{B}}{2}=90^{\circ }-\frac{\hat{B}}{2}$$ $$\widehat{B\Gamma N}=\widehat{B\Gamma IN}=\frac{\hat{A}}{2}+\frac{\hat{B}}{2}=\frac{180^{\circ }-\hat{\Gamma }}{2}=90^{\circ }-\frac{\hat{\Gamma }}{2}$$ $$\widehat{BN\Gamma }=180^{\circ }-\left(90^{\circ }-\frac{\hat{B}}{2}+90^{\circ }-\frac{\hat{\Gamma }}{2}\right)$$ $$=\frac{\hat{B}+\hat{\Gamma }}{2}=90^{\circ }-\frac{\hat{A}}{2}$$ Figure 1

(ii) Since $\widehat{\Gamma BN}=90^{\circ }-\frac{\hat{B}}{2}$, $BN$ is the bisector of the external angle of $\hat{B}$. Therefore $\widehat{IBN}=90^{\circ }$ and $IN$ is a diameter of the circle $C$. Moreover, if $A\Delta$ intersects the circumcircle of the triangle $AB\Gamma$ at $M$, then $$\widehat{BIN}=\frac{\hat{A}}{2}+\frac{\hat{B}}{2}=\widehat{\Gamma BM}+\widehat{\Gamma BI}=\widehat{IBM}.$$ Thus the triangle $IBM$ is isosceles with $MB=MI$. Therefore $M$ lies on the perpendicular bisector of $BI$. Since $M$ belongs to the diameter of the circle $C$, it is the center of $C$.