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Hellenic Mathematical Olympiad

Geometry Difficulty 6.8 National Olympiad Prove it Greece

A triangle $AB\Gamma$ is given with $AB < A\Gamma$ . Let $I$ be the point of intersection of its bisectors. Bisector $A\Delta$ meets the circumcircle $C$ of the triangle $B\Gamma$ at the point $N$ with $N \neq I$ .
(i) Determine the angles of the triangle $B\Gamma N$ with respect to the angles of the triangle $AB\Gamma$ ,
(ii) the center of the circle $C$ .

Solution

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

Figure 1

(ii) Since $\widehat{\Gamma BN} = 90^\circ - \frac{\widehat{B}}{2}$ , $BN$ is the bisector of the external angle of $\widehat{B}$ .
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 $C$ .

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