........

Hii there is a lemma that I call it the Key Lemma. Lemma : Let S be a point out of the triangle ABC . Then we have : $Sin <BAS / Sin <SAC =( Sin <ABS / Sin <ACS) * BS / CS.$ Proof: $Sin <BAS / BS = Sin <ABS / AS.$ $Sin <SAC / CS = Sin <ACS / AS.$ SO : $Sin <BAS / Sin <SAC =( Sin <ABS / Sin <ACS) * BS / CS.$ Now let's back to the question. We have to prove that Sin<BAS = Sin <SAC. So we use the Lemma in ABSQ. We know : $Sin <BAS / Sin <SAQ = (Sin <ABS / Sin <AQS) * BS/SQ.$ $<ABS = <SRC = <SQC = 180 - <AQS => Sin <ABS = Sin <AQS.$ So we have to prove $BS=QS.$ $<PSB = <PRB = <QRC = <QSC.$ $ <CQS = <CRS = <PBS .$ And we know $BP=CQ.$ So $PBS = CQS => BS = QS.$ And we are done.

By construction, $S$ is the center of the spiral similarity that sends $BP$ to $QC$. However, since $BP = CQ$, this spiral similarity is actually a rotation centered in $S$. Thus we have: $$SB = SQ \text{ and } SP = SC$$Now, since $S$ is the center of the spiral similarity that maps $BP$ to $QC$, then we know that $ABSQ$ and $ACSP$ are cyclic quadrilaterals. Hence, $\angle BAS = \angle BQS = \angle RQS = \angle RCS$ and $\angle SAC = \angle SPC$. But $\angle SPC = \angle RCS$ since $SC = SP$. Then, it follows that $\angle BAS = \angle SAC$, as desired.

Let $\omega_1(O_1,r_1)$ and $\omega_2(O_2,r_2)$ be the circumcircles of $\triangle PRB$ and $\triangle QCR$, respectively. Let $\{T\}=\overleftrightarrow{SC} \cap \omega_1$. By extended law of sines in $\triangle QCR$ and $\triangle PRB$, $\frac{BP}{\sin \angle PRB}=2r_1=\frac{CQ}{sin}=2r_2 \implies r_1=r_2$. So $\frac{RS}{\sin \angle SBR}=2r_1=2r_2=\frac{RS}{\sin \angle RCS} \implies \angle SBR = \angle RCS$, since $\angle SBR + \angle RCS < 180^{\circ}$. Then $\angle SBR=\angle SPR=\angle RCS \implies SP=SC$. Note that $\angle QCS = \angle BRS = 180^{\circ}-\angle STB \implies BT \parallel AC$, thus $\angle PAC = 180^{\circ}- \angle TBP=\angle PST \implies ACSP$ is cyclic. But since $SC=SP$, $AS$ is the bisector of angle $BAC$. $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \square $

Dear Mathlinkers, a variation at 11-ième O.M. de St-Petersburg (1999), Round de sélection, problème 2. Sincerely Jean-Louis

That's famous situation

This problem was proposed by Burii.

Because $B, R$ and $Q$ are concyclic and $C, R, P$ are also concyclic, that means that we have that there exists a spiral similarity mapping $\triangle SBP$ to $\triangle SQC$. Because $BP=CQ$, then we know that we have that $\triangle SBP$ is congruent to $\triangle SQC$. In addition, by angle chasing, we get that $$\angle CSP = \angle CSR + \angle RSP = \angle AQR + \angle RBP = 180^{\circ} - \angle BAC.$$That shows us that $APSC$ is cyclic, and analogously, $AQSB$ is cyclic too. Finally, due to the fact that $\triangle SBP$ is congruent to $\triangle SQC$, we have that $PS=SC$. Since $APSC$ is syclic, then $\angle PAS = \angle SAC$ so $\angle SAB = \angle SAC$ meaning $S$ is on the angle bisector of $\angle ABC$ so we are done.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.099997253465038, xmax = 10.333668701128902, ymin = -8.057915244124446, ymax = 7.172037974209792; /* image dimensions */ /* draw figures */ draw((-2.590375969105767,4.069271180546538)--(-6.948194499531714,-0.8114855735304873), linewidth(0.8) + blue); draw((-6.948194499531714,-0.8114855735304873)--(0.4725479122793272,-3.824605814567835), linewidth(0.8) + blue); draw((0.4725479122793272,-3.824605814567835)--(-2.590375969105767,4.069271180546538), linewidth(0.8) + blue); draw((-6.948194499531714,-0.8114855735304873)--(-0.509674242709973,-1.2931877403271366), linewidth(0.8) + blue); draw((0.4725479122793272,-3.824605814567835)--(-5.139763875478851,1.2139567254087056), linewidth(0.8) + blue); draw(circle((-1.6598741152946308,-3.1957461366265285), 2.223215733621701), linewidth(0.8) + linetype("2 2") + red); draw(circle((-4.730730831640612,-0.9713075989400678), 2.2232157336216987), linewidth(0.8) + linetype("2 2") + red); draw((-2.590375969105767,4.069271180546538)--(-3.876433238801054,-3.0238338036672094), linewidth(0.8) + blue); draw((-3.876433238801054,-3.0238338036672094)--(-6.948194499531714,-0.8114855735304873), linewidth(0.8) + blue); draw((-3.876433238801054,-3.0238338036672094)--(-0.509674242709973,-1.2931877403271366), linewidth(0.8) + blue); draw((-3.876433238801054,-3.0238338036672094)--(-5.139763875478851,1.2139567254087056), linewidth(0.8) + blue); draw((-3.876433238801054,-3.0238338036672094)--(0.4725479122793272,-3.824605814567835), linewidth(0.8) + blue); /* dots and labels */ dot((-2.590375969105767,4.069271180546538),dotstyle); label("$A$", (-2.485788324375541,4.350661748511184), NE * labelscalefactor); dot((-6.948194499531714,-0.8114855735304873),dotstyle); label("$B$", (-6.841116672784102,-0.5251146415310723), NE * labelscalefactor); dot((0.4725479122793272,-3.824605814567835),dotstyle); label("$C$", (0.5821159210443259,-3.5382348825684216), NE * labelscalefactor); dot((-5.139763875478851,1.2139567254087056),dotstyle); label("$P$", (-5.03324452816168,1.4745015184300776), NE * labelscalefactor); dot((-0.509674242709973,-1.2931877403271366),linewidth(4pt) + dotstyle); label("$Q$", (-0.40399615784063136,-1.072954685356045), NE * labelscalefactor); dot((-2.5141717081341906,-1.1432199318993854),linewidth(4pt) + dotstyle); label("$R$", (-2.4036123178017945,-0.9359946743998018), NE * labelscalefactor); dot((-3.876433238801054,-3.0238338036672094),linewidth(4pt) + dotstyle); label("$S$", (-3.773212427364235,-2.7986508234047087), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\varphi$ be the spiral similarity centered at $S$, since we have that $BP=CQ$, we get that (under $\varphi$) $SB=SQ$ and that $SP=SC$. So now all that's left is to show that $CSPA$ is cyclic. But notice that: $$\angle SCQ=\angle SRB = \angle BPS$$meanins that $CSPA$ is cyclic.

Let's start with proving that APSC and AQSB are cyclic. ∠QSB = ∠RSQ + ∠RSB = ∠QCR + ∠RPA = 180 - ∠A ---> AQSB is cyclic. ∠PSC = ∠PSR + ∠RSC = ∠RBP + ∠RQA = 180 - ∠A ---> APSC is cyclic. ∠BAS = ∠BQS = ∠RCS and ∠CAS = ∠CPS = ∠RPS. It's easy to prove triangles SBP and SQC are congruent so SP = SC and ∠RCS = ∠RPS. We're Done.

It's clear that $S$ is the Miquel Point of $BPCQ$, so $SBP \overset{+}{\sim} SQC$. But we know $BP = CQ$, so the two triangles are actually congruent, yielding $SB = SQ$. Properties of Miquel Points imply $ABSQ$ is cyclic. Hence, $S$ is the midpoint of arc $BQ$, so $AS$ bisects $\angle BAQ \equiv \angle BAC$. $\blacksquare$

Note that $S$ is the miquel point of quad $APRQ$ (not $BPCQ$ because I don't like self intersecting quads ) so $SBP \sim SQC$ which along with the condition makes the two triangles congruent, so $SB = SQ$ and $SP = SC$ Now the fact that $ABSQ$ combined with sine rule on $\triangle SAB$ and $\triangle SAQ$ proves the desired result $\blacksquare$

Notice there is a spiral similarity at $S$ such that $\overline{BP}\mapsto\overline{CQ}.$ Since we also know $BP=CQ,$ we see that $\triangle SBP\cong\triangle SQC.$ Therefore, $$\delta(S,\overline{AB})=\delta(S,\overline{BP})=\delta(S,\overline{CQ})=\delta(S,\overline{AC}).$$$\square$

Let $D$, $E$ be the points in lines $AB$, $AC$ such that $SD \perp AB$ and $SE \perp AC$. We want to show that $SD=SE$. Let $X$ and $Y$ be the midpoints of $BQ$, $PC$. Note that $\angle PBS = \angle CRS = \angle CQS$, and $\angle BPS = \angle BRS= \angle SCQ$ together with $QC = BP$ implies $\triangle BPS \simeq SQC $. Therefore $SC = SP$ and $SB=SQ$ implies $ SX \perp BQ$ and $SY \perp PC$. So quadrilaterals $DBSX$ and $SYEC$ are cyclic. Hence: $$\angle DXS = \angle YXS = 180^{\circ}- \angle SBD + \angle SQC = 180^{\circ}$$Hence $D,X,Y$ are colinear. In a similar way we have $X,Y,E$ colinear. Therefore $\angle SDX = \angle SEX = \angle SBD$ and we are done since it follows that $SD = SE$.

Cute problem

After understanding that APSC is cyclic, we should notice that radii of (BPR) and (CQR) are equal. It is obvious by Law os sinuses to this triangles. Then angles SPR and SCR are equal, hence SP=SC. How we know APSC is cyclic then we can easily get that AS is angle bisector of angle PAQ. And we are done. (Sory I am too lazy for convert this to LaTeX)