Saudi Arabia Mathematical Competitions

Recall some preliminary facts. The nine-point circle of a triangle $ABC$ passes through the midpoints of the sides, the midpoints of the segments joining its vertices to the orthocenter $H$ and the pedal point (i.e., the feet of its altitudes to the sides). Its center is the midpoint $O_9$ of the segment joining the circumcenter $O$ and the orthocenter $H$ of the triangle. Its radius $\frac{1}{2}R$ is equal to half the circumradius $R$ of the triangle $ABC$ and it touches internally the incircle with radius $r$ (as well as all three excircles). The square of the length of the segment $OI$ is $$|OI{|}^2=R^2-2Rr=R(R-2r).$$ Also, it is clear that $O_{9}I=\frac{1}{2}R-r$. Let $M$ be the symmetric point of $O$ with respect to $I$. It follows that $$HM=2O_{9}I=2\left(\frac{1}{2}R-r\right)=R-2r.$$ Since $$IM=OI=\sqrt{R(R-2r)}>R-2r=HM,$$ we get that $\widehat{IHM}>\widehat{MIH}$. Hence $\widehat{MIH}<90^{\circ }$, so that $\widehat{OIH}>90^{\circ }$.