Let $(ABC)$ be $\Omega$ and $r$ be radius of $\Omega$ Let $AD, BE$ and $CF$ be altitudes of $\Delta ABC$ We will first prove that $OP=OQ$ $\angle BHP=90=\angle CHQ$ ($PH\parallel AC$ and $QH\parallel AB$) $\angle PBH=\angle FBE=\angle FCE=\angle HCQ$ because $BFEC$ is cyclic $\implies \Delta BHP\sim \Delta CHQ$ $\frac{BP}{HP}=\frac{CQ}{HQ}$ $\frac{BP}{AQ}=\frac{CQ}{AP}$ $BP\cdot PA=CQ\cdot QA$ so powers of $P$ and $Q$ with respect to $\Omega$ are equal $OP^2-r^2=pow_\Omega(P)=pow_\Omega(Q)=OQ^2-r^2$ $\implies OP=OQ$ We will prove that $OP$ and $OQ$ are tangents to $(APQ)$ Case 1: $AB=AC$ $A, H, O$ are collinear $\angle BOD=\frac{\angle BOC}{2}=\angle BAC=\angle BPH \implies BPHO$ cyclic $\angle OPQ=180-\angle APQ-\angle BPO=180-\angle ABC-\angle BHO=180-2\cdot (90-\frac{\angle BAC}{2})=\angle BAC$ $\implies OP$ is tangent to $(APQ)$, but by symmetry same follows for $OQ$ Case 2: $AB\neq AC$ It is known that $AH$ and $AO$ are isogonal $AH$ is the $A$-median of $\Delta APQ$ so $AO$ is it's $A$-symmedian Thus $AO$ passes through intersection of tangents to $(APQ)$ at $P$ and $Q$, say at point $O'$ We see that $O'P=O'Q$ so $O'=O$ or rather $OP$ and $OQ$ are tangents to $(APQ)$ Let $X$ be midpoint of $PB$ $\angle OPQ=\angle BAC=\angle XPH$ $\angle PHB=90 \implies XP=XH$ Combining that with $OP=OQ$ we get $\Delta XPH\sim \Delta OPQ$ So $\frac{PX}{PH}=\frac{PO}{PQ}$ $\frac{PX\cdot PQ}{PH\cdot PO}=1$ $\frac{PX\cdot PQ}{QA\cdot QO}=1$ $\frac{PB\cdot PQ}{QA\cdot QO}=2$

Attachments:

Dear Mathlinkers, here Problem 4 Sincerely Jean-Louis