Set $(ABC)$ as the unit circle and WLOG $q=-1,p=1$ . $\longrightarrow$$c=\frac{1}{b}$ ,$r=1/2(a+c+1-ac) and s=\frac{a-1}{2}$ . Now it is easy to find that : $S=\frac{b-a}{r-a}.\frac{r-s}{b-s} \in R$.

Key claim: $SR=SB$

Now, since $AS$ bisects $\angle BAR$ and $SB=SR$, hence $S$ is the midpoint fo arc $ARB$ of $\odot(ARB)$, and hence lies on this circle, as desired. $\blacksquare$

This problem was proposed by Burii.

we just need to prove $$ARsin(\frac{A}{2})+c.sin(\frac{A}{2})=ASsin(A)$$which is obvious using $$bsin(\frac{A}{2})+csin(\frac{A}{2})=AQsin(A) , AR=\frac{|b-c|}{2}$$

have anyone thought about the problem in this way? please help me to prove synthetically that $RS$ is the $R$-symmedian in $\triangle BRC$

My solution: Let $BR$ cuts $(ABC)$ again at $X,$ $A'$ is the reflection of $A$ wrt $BX$ $\Rightarrow \angle PXR = 180^{\circ} - \angle PAB = \angle PAC = \angle PYR \Rightarrow P, X, Y, R$ are concyclic (lazy to use direct angle here) $\Rightarrow BXY = \angle RPY = \angle APR =\frac{1}{2} \angle A = \angle BXQ,$ so $X, Y, Q$ are collinear Since $RS \parallel YQ,$ by Reim theorem$, A, B, R, S$ are concyclic

Let $A',B'$ be the midpoints of $BC,AC$ ,$D$ foot of the $A$-bisector . Remark that $RA'$ is the simson line of $P$ thus $A'R\parallel AQ$ ; we have $ABQ\sim ADC$ then $ABS \sim ADB'$ hence to show that $ARSB$ cyclic it suffices to show $\angle BRA=\angle DB'A$ or $BR \parallel DB'$ but $\frac{CD}{CB'}=2\cdot\frac{CD}{CA}=2\cdot\frac{CA'}{CR}=\frac{CB}{CR}$ then the result follows . RH HAS

Here is my solution for this problem Solution Let $M$ be midpoint of $BC$ Since: $\widehat{AQP}$ = $\widehat{RCP}$ and $\widehat{PAQ}$ = $\widehat{PRC}$ = $90^o$, we have: $\triangle$ $PAQ$ $\sim$ $\triangle$ $PRC$ Hence: $\dfrac{AQ}{RC}$ = $\dfrac{PQ}{PC}$ = $\dfrac{CQ}{CM}$ = $\dfrac{BQ}{BM}$ or $\dfrac{SQ}{RC}$ = $\dfrac{BQ}{BC}$ But: $\widehat{BQS}$ = $\widehat{BRC}$, so $\triangle$ $BSQ$ $\sim$ $\triangle$ $BRC$ Then: $\widehat{SBQ}$ = $\widehat{RBC}$ or $\widehat{RBS}$ = $\widehat{CBQ}$ = $\widehat{CAQ}$ = $\widehat{RAS}$ So: $A$, $B$, $S$, $R$ lie on a circle

Consider $Hom\left(A , \frac{1}{2}\right)$ Let $B$ go to $X$, $R$ go to $Y$. Also $S$ goes to $Q$. Then the problem is equivalent to proving that $A$, $X$, $Q$, $Y$ lie on a circle. By Archimedes theorem, $CY = AB = BX$. Also $\angle QBX =B+\frac{A}{2} = \angle QCY$ and $QB = QC$. $\therefore \triangle QBX \cong \triangle QCY$ $\therefore \angle AXQ = \angle BXQ = \angle CYQ \implies AXYQ$ is cyclic as desired.

Claim1 : $CPR$ and $QPA$ are similar. Proof : $\angle PCR = \angle PQA$ and $\angle PRC = \angle 90 = \angle PAQ$. Claim2 : $POC$ and $BQC$ are similar. Proof : $PO = OC$, $BQ = QC$ and $\angle CPO = \angle CBQ$. Claim3 : $BSQ$ and $BRC$ are similar. Proof : $\angle RCB = \angle SQB$ and $\frac{RC}{SQ} = \frac{2RC}{AQ} = \frac{2PC}{PQ} = \frac{PC}{PO} = \frac{BC}{BQ}$. Now we have $\angle ARB = \angle ASB$.

Another nice synthetic solution at https://stanfulger.blogspot.com/2022/02/polish-mo-2018-2nd-round.html Best regards, sunken rock

Consider an inversion in $A$ with radius $\sqrt{AB \cdot AC}$ composed with the symmetry along the $AQ$. Let $D = BC \cap AQ, P'= BC \cap AP$. Then $P'$ is an image of $P$ in this inversion. Moreover, the image of $S$ is some point $S'$ so that $D$ is a midpoint of $AS'$. Denote $R'$ the image of $R$ and $X = AP \cap S'C, R^{''} = AB \cap S'C$. The statement is now equivalent to $R'=R^{''}$. Notice: $-1 = (B, C; D, P') = (R^{''}, C, S', X) = P'(R^{''}, D; S', A)$. However, since $D$ is a midpoint of $AS'$ and $R'P' \parallel AS'$, we also have: $P'(R', D; S', A)= -1$, so $R'=R^{''}$ follows.