China Mathematical Competition (Extra Test)

We take $BK=CH$ on $BE$ and join $OB$, $OC$ and $OK$. From the property of the circumcenter of a triangle, we know that $\angle BOC=2\angle A=120^{\circ }$. From the property of the orthocenter of a triangle, we get $\angle BHC=180^{\circ }-\angle A=120^{\circ }$. So $\angle BOC=\angle BHC$. Then four points $B$, $C$, $H$ and $O$ are concyclic. Hence $\angle OBH=\angle OCH$. In addition, $OB=OC$ and $BK=CH$. Therefore, $△BOK\cong △COH$. It follows that $\angle BOK=\angle COH$, and $OK=OH$. $$\text{So,}\begin{array}{rl}\angle KOH& =\angle BOC=120^{\circ },\\ \angle OKH& =\angle OHK=30^{\circ }.\end{array}$$ In $△OKH$, by the sine rule, we get $KH=\sqrt{3}OH$. In view of $BM=CN$ and $BK=CH$, we get $KM=NH$, and $$MH+NH=MH+KM=KH=\sqrt{3}OH.$$ Therefore, $$\frac{MH+NH}{OH}=\sqrt{3}.$$