Spring Mathematical Tournament

a) Since $\angle RAH=\angle QBH$ and $$\angle ARH=180^{\circ }-\angle CRQ=180^{\circ }-\angle CQR=\angle BQH,$$ then $△ARH\sim △QBH$. Hence $$\frac{AH}{BH}=\frac{AR}{BQ}=\frac{AP}{BP}$$ and $HP$ is the bisector of $\angle AHB$. Then $$\angle RHP=\angle RHA+\angle AHP=\angle QHB+\angle BHP=\angle QHP$$ which implies that $PH\perp RQ$.

b) If $AC=AB$, then $N\equiv P\equiv M$ and the points $I,O$ and $N$ lie on the bisector of $AB$. Let now $AC\ne AB$. Since $PH\perp RQ$ and $CI\perp RQ$, it follows that $HP\parallel CI$. On the other hand, $CH\parallel IP$ and hence $CHPI$ is a parallelogram. Then $CH=IP$. If $M$ is the midpoint of $AB$, then $CH=2OM=2R\cos \gamma$ and so $IP=2OM$. Since $AP=BN$, then $M$ is the midpoint of $PN$. Therefore $O$ is the midpoint of $IN$.