https://artofproblemsolving.com/community/c6h2348815p19016791 just use this one

We have the following main Claim. Claim: $\angle ADQ=90^\circ-\angle A$. Proof: Let $\omega$ meet $AB$ again at $X$. Then, $BP \cdot BQ=BX \cdot BA=BD \cdot BC,$ hence $QPDC$ is cyclic. Therefore, $\angle ADQ=\angle QDC-90^\circ=\angle QPC-90^\circ=180^\circ-\angle a-90^\circ=90^\circ-\angle A,$ as desired $\blacksquare$ Back to the problem, if $PH$ meets $(ABC)$ again at $A'$, then $A'$ is the antipode of $A$ in $(ABC)$. Hence, we have $\angle HAR=\angle HPR=\angle A'PC=\angle A'AC=90^\circ-\angle B$ and, by Claim 1, $\angle HAS=\angle HQS=\angle HQS=\angle QHA-\angle ADQ=\angle QPA-\angle ADQ=\angle C-(90^\circ-\angle A)=90^\circ-\angle B,$ and so $\angle HAS=\angle HAR,$ which implies that $HS=HR$, as desired.

We have the following main Claim. Claim: $\angle ADQ=90^\circ-\angle A$. Proof: Let $\omega$ meet $AB$ again at $X$. Then, $BP \cdot BQ=BX \cdot BA=BD \cdot BC,$ hence $QPDC$ is cyclic. Therefore, $\angle ADQ=\angle QDC-90^\circ=\angle QPC-90^\circ=180^\circ-\angle a-90^\circ=90^\circ-\angle A,$ as desired $\blacksquare$ Back to the problem, if $PH$ meets $(ABC)$ again at $A'$, then $A'$ is the antipode of $A$ in $(ABC)$. Hence, we have $\angle HAR=\angle HPR=\angle A'PC=\angle A'AC=90^\circ-\angle B$ and, by Claim 1, $\angle HAS=\angle HQS=\angle HQS=\angle QHA-\angle ADQ=\angle QPA-\angle ADQ=\angle C-(90^\circ-\angle A)=90^\circ-\angle B,$ and so $\angle HAS=\angle HAR,$ which implies that $HS=HR$, as desired.

can @admins add shortlist on contest section?

Easier than G4 and maybe even than G3 (which came up on the contest). Here is a quick solution with only angle chasing and no new additional points. The key observation is that $D$, $Q$ and the foot $C_1$ of the $C$-altitude (which lies on $\omega$, since $\angle AC_1H = 90^{\circ}$) are collinear. If we show this, then the problem is reduced to $HR = HC_1$, which holds from $\omega$, as $\angle HAR = 90^{\circ} - \angle AHR = 90^{\circ} - \angle APR = 90^{\circ} - \angle APC = 90^{\circ} - \angle ABC = \angle BAH = \angle C_1AH$. Since $BC_1HD$ is cyclic, we have $\angle BC_1D = \angle BHD = 90^{\circ} - \angle HBD = \angle ACB$. On the other hand, $\omega$ and $k$ give $\angle AC_1Q = 180^{\circ} - \angle APQ = 180^{\circ} - \angle APB = \angle ACB$. Therefore $\angle AC_1Q = \angle BC_1D$ and the desired collinearity follows.

Let $PH\cap k=W$,$\omega \cap AB=S'$ Note that $\angle QAH=\angle QPH=\angle BAW=\angle DAC=\angle HS'D$ Which means that $A-S'-B$ So $S=S'$ Consider that $\angle RAH=\angle RPH=\angle CBW=\angle BAD$ So $HS=HR$

Let $S'$ be the altitude from $C$ to $AB$ and $M$ be the midpoint of $BC$. Pascal at $QS'S'AHP$ gives that $QF\cap AH\in BC\implies S'=S$. Pascal at $PRSSHH$ gives that $C,RS\cap HH,M$ are collinear hence $RS\parallel BC\iff HR=HS$ as desired.$\blacksquare$

Let $I$ be the $C$-altitude and $T$ the $B$-Altitude $Claim$: $S$ $coincides$ with $I$. $Proof$: $I$ clearly lies on $(APHR)$. So it suffices to prove that $Q, I, D$ are collinear which happens to be an easy angle chase: \[ \angle BIQ = \angle QPA = \angle BPA = 180^\circ - \angle C = \angle A + \angle B\overset{AIDC cyclic}{=}\angle B + \angle IDB = 180^\circ - \angle BID \].$\blacksquare$ $Claim$: $IR \parallel BC$ $Proof$: \[ \angle PRI = \angle PAI = \angle PAB = \angle PCB = \angle RCB \]$\blacksquare$ Moreover, \[ \angle AIR \overset{IR\parallel BC}{=} \angle B\overset{BITC cyclic}{=} \angle ATI = \angle ARI \]So $AI = AR$ and because $AH$ is the diameter, $HR =HS$.$\blacksquare$

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