Let $O$ be the circumcenter of $ABC$. Then triangles $OBQ$ and $OAP$ are congruent, so there is a rotation at $O$ that sends one of these triangles to the other. As a corollary, $\angle POQ = \angle AOB$, so $O$ is on $(APQ)$. $OP = OQ$ means $OR$ is the perpendicular bisector of $PQ$. So $\angle POR = \angle PCR$, which means $O$ is on $(CRP)$ as well. By Miquel's theorem, $(BQOR)$ is cyclic. Angle chasing shows that $PO_1RO_2$ is cyclic. Also note that since $\angle OAP = \angle OCP$, $(O_1)$ and $(O_2)$ have the same radius. Now note that $\angle O_1PO_2 = 180^\circ - \angle O_1RO_2 = 90^\circ + \angle OPR = 90^\circ + \angle OCB = \angle 180 - \angle BAC$. Since $B, O, P$ are collinear, we have $\angle BOR = \angle PCR = B$ due to cyclic quadrilateral $OPCR$. Now let's chase some lengths. We have $OR = 2 \cdot OO_2 \sin \angle OCR = 2 \cdot OO_2 \cos A$. Also, we have $O_1O_2 = 2 \cdot OO_2 \sin B = OR \cdot \frac{\sin B}{\cos A}$. By law of sines on $OBR$, we get $BR = OR \cdot \frac{\sin \angle BOR}{\sin \angle OBR}$, and we are done.

there is a post collection already named India EGMO TST 2022, if these problems are for EGMO 2023, shouldn't they be named with the year number 2023?

Nice problem! My solution is similar to the above one, though I am posting it since some parts are a bit different. The angle condition implies that the circumcenter $O$ lies on $BP$. It is easy to see that triangles $BQO$ and $APO$ are congruent, so $OP=OQ$ and $APOQ$ is cyclic. In addition, $\angle QPO= \angle QBO =\frac{\alpha} {2} $, so $BQ=QP=PA$. Redefine $R=(BQO) \cap (CPO)$; we have that $\angle PQR=90-\frac{\alpha}{2}=\angle QPR$, so $RQ=RP$ and $R$ lies on $BC$ by Miquel. Note that $\angle PQR= \angle QBR$ so $PQ$ is tangent to $(BQO)$ and similarly $PQ$ is tangent to $(CPO)$. Notice that $O_1, O, R$ are collinear, $PO_2 \perp PQ$, so $PO_2RO_1$ is a trapezoid. But notice that $\triangle CPO$ and $\triangle AQO$ are congruent, so their circumcircles are also congruent, i.e. $PO_1=PO_2=RO_2$, hence $PO_2RO_1$ is an isosceles trapezoid and thus $PR=O_1O_2$. It's only left to notice that $BR=RP$ by an easy angle chase, so we are done.

it s clear $P\in BO ; OQ\sim AOP$ thus $\angle BQO=\angle APO$ then $ O\in (APQ)$ we have then $RO$ is bisector of $PQ$ so $\angle O_1OP=\frac {\pi}{2} - \frac {A}{2}=\angle PCR$ hence $O \in (PCR)$. since $B $ is on the radical axis of $(O_1),(O_2)$ then $BR.BC=BO.BP=BQ.BA $ thus $QACR$ are cyclic then $\angle QRB=\angle A$ whence $ AQB \sim ABC$ and $ RQ\perp PB$ leads $ RQ\parallel O_1O_2(^1)$ and $BR=QR(^2)$ More $\angle QO_1R=\angle A$ ;$\angle ORO_2 =\frac {\pi}{2}-\angle RCO =\angle A$ then $O_1Q\parallel RO_2$ besides $(^1)$ we get $O_1O_2RQ$ is a parallelogram hence $O_1O_2= QR$ therefore reminding $(^2)$ we conclude $BR=O_1O_2$

Thankyou so much Atul for proposing!!! Saved me on Day 2 :plead: Also although I told to Atul before but circumcentre really gives away. Really great problem! Define $O$ as the circumcentre of $ABC$. Claim: $P=OB\cap AC$ Proof: Since $\angle PBC=90-A$, but since $M$ and $O$ are isogonals $\angle CBO=90-A\implies B-O-P.$ Define $Q'=CO\cap AB$. Since $O\in \text{ perpendicular bisector of } PQ'$. Moreover, we have $AP=AQ'$ and $AP=QB\implies Q'A=BQ\implies O\in \text{ perpendicular bisector of }QQ'$ [ as $O\in $ perpendicular bisector of $AB$] $\implies O$ is circumcentre of $\Delta Q'QP\implies O \in \text{ perpendicular bisector of }QP$ but $AO$ is the angles bisector of $\angle QAP$ $\implies O\in (AQP)$ and $O$ is midpoint of arc. Define $M$ as the midpoint of $QP$. Note that $M-O-R$. Claim: $O\in (RPC)$ Proof: We do angle chase, let $\angle QAP=A\implies \angle QOP=180-A\implies \angle MOP=90-A/2$. But $90-A/2=90-\angle OAC=\angle ACB$. And by Miquel theorem, we get $OQBR$ cyclic too. Claim: $QB=QP$ Proof: We will show $\angle QBP=\angle QPB$. And $\angle QBP=\angle BAO=\angle QPO=A/2$. Claim: $RB=RP$ Proof: We will angle chase, taking $\angle BAC=A$. We get $\angle PBC=90-A, \angle BRP=180-2A\implies RPB=90-A\implies RP=RB$. Claim: $R$ is the circumcenter of $\Delta BQP$. Proof: We got $BR=RP$ and $RQ=RP$. So $BR=RQ$. So it is enough to show $RQ=O_1O_2$, Claim: $QR$ is the perpendicular bisector of $BP$ Proof: As from earlier, we had $BQ=QP, RM=RP$. So $RQ\perp BP$. We also have $O_1O_2\perp OP\implies RQ||O_1O_2$. Claim: $\angle QO_1O_2=\angle QRO_2$ Proof: Note that $$\angle OO_1O_2=1/2\angle OO_1P=\angle OQP=A/2$$And $$\angle QO_1O=2\angle QPO=A.$$So $$\angle QO_1O_2=A+A/2.$$ Now, we compute $$\angle QRO_2=\angle QRP+\angle PRO_2=\angle QBR+\angle OCP+\angle PRO_2=A/2+A/2+\angle PRO_2=A+A/2.$$Since we earlier had $QR||O_1O_2$ and now we have $\angle QO_1O_2=\angle QRO_2\implies QRO_2O_1\text{ is a parallelogram}\implies BR=QR=O_2O_1$.

$P$ and $Q$ have the same power wrt $(ABC)$ thus $OR$ bisects $PQ \implies OP=OQ$. $$BQ \cdot BA=AP \cdot AC=AP^2 \frac{AC}{ AP}= PO \cdot PB \cdot \frac{AC}{ AP}=PO \cdot PB \cdot \frac{BO}{OP }=BO \cdot BP$$Thus $AQOP$ is cyclic. This also implies $BQOR$ and $PORC$ are cyclic by angle chasing. $\angle QBP=\angle QRO=\frac{1}{2} \angle QRP \implies BR=QR=RP$. It is also simple angle chasing that $\angle PO_1O=\angle ORO_2$ and $O_2P \perp QP$. Thus $PO_1RO_2$ is isosceles trapezoid and $PR=O_1O_2$.

Solved with SatisfiedMagma. Solution: Let $O$ be the circumcenter of $\triangle ABC$. $O$ lies on $BP$ by the angle condition in the problem. [asy][asy] import graph; size(9.6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.5, xmax = 6.2, ymin = -3.579614086556227, ymax = 3.8619801318974827; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen yqqqqq = rgb(0.5019607843137255,0.,0.); pen ffttww = rgb(1.,0.2,0.4); draw((2.1357002305054333,3.3944953668920594)--(-0.6446756313344209,-2.8749796156879914)--(4.857851242143845,-2.9004788475217542)--cycle, linewidth(1.2) + yqqqyq); /* draw figures */ draw((2.1357002305054333,3.3944953668920594)--(-0.6446756313344209,-2.8749796156879914), linewidth(1.2) + yqqqyq); draw((4.857851242143845,-2.9004788475217542)--(2.1357002305054333,3.3944953668920594), linewidth(1.2) + yqqqyq); /* special point */ draw((-0.6446756313344209,-2.8749796156879914)--(3.2909742324878244,0.7229242373077858), linewidth(1.2)); draw(circle((1.4807730894735145,1.5257090920358725), 1.9802251642555977), linewidth(1.2) + ffxfqq); draw(circle((3.928553351192876,-1.1518516342412364), 1.9802251642555975), linewidth(1.2) + ffxfqq); draw((3.2909742324878244,0.7229242373077858)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2) + yqqqqq); draw((0.5353088497578561,-0.21423029557796058)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2) + yqqqqq); draw((1.8281694616553013,-1.628455605384008)--(1.6686650446165432,-1.635775235961833), linewidth(1.2) + yqqqqq); draw((1.8281694616553013,-1.628455605384008)--(1.8497329166185283,-1.4702460814711957), linewidth(1.2) + yqqqqq); draw((3.2909742324878244,0.7229242373077858)--(0.5353088497578561,-0.21423029557796058), linewidth(1.2)); draw((1.4807730894735154,1.525709092035872)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2)); draw((1.4807730894735154,1.525709092035872)--(3.9285533511928756,-1.1518516342412366), linewidth(1.2) + yqqqqq); draw((2.773633701370962,0.1114837822298249)--(2.6141292843322037,0.10416415165199998), linewidth(1.2) + yqqqqq); draw((2.773633701370962,0.1114837822298249)--(2.7951971563341895,0.269693306142637), linewidth(1.2) + yqqqqq); draw((-0.6446756313344209,-2.8749796156879914)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2) + yqqqqq); draw((2.9830891114772173,-2.8917910218550693)--(4.857851242143845,-2.9004788475217542), linewidth(1.2) + yqqqyq); draw(circle((1.4693356998519826,-0.9423865810104978), 2.4681221738786703), linewidth(1.2) + linetype("4 4") + blue); draw((3.9285533511928756,-1.1518516342412366)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2) + ffttww); draw((3.407016139142441,-2.111637419080401)--(3.3480419220963475,-1.963255217417028), linewidth(1.2) + ffttww); draw((3.407016139142441,-2.111637419080401)--(3.563600540573743,-2.080387438679278), linewidth(1.2) + ffttww); draw((1.4807730894735154,1.525709092035872)--(0.5353088497578561,-0.21423029557796058), linewidth(1.2) + ffttww); draw((0.9592358774230816,0.5659233071967078)--(0.9002616603769881,0.7143055088600809), linewidth(1.2) + ffttww); draw((0.9592358774230816,0.5659233071967078)--(1.1158202788543843,0.5971732875978305), linewidth(1.2) + ffttww); draw((2.1183522081785666,-0.34906677951315035)--(4.857851242143845,-2.9004788475217542), linewidth(1.2)); draw((2.1183522081785666,-0.34906677951315035)--(2.1357002305054333,3.3944953668920594), linewidth(1.2)); /* dots and labels */ dot((2.1357002305054333,3.3944953668920594),dotstyle); label("$A$", (2.1929432629550827,3.5893942630896545), NE * labelscalefactor); dot((-0.6446756313344209,-2.8749796156879914),dotstyle); label("$B$", (-0.9690528152157342,-3.0071837620597877), NE * labelscalefactor); dot((4.857851242143845,-2.9004788475217542),dotstyle); label("$C$", (5.041465591996896,-3.020813055500179), NE * labelscalefactor); dot((2.1183522081785666,-0.34906677951315035),linewidth(1.25pt) + dotstyle + invisible,UnFill(0)); label("$O$", (1.7022886991009902,-0.28132507398150575), NE * labelscalefactor); dot((3.2909742324878244,0.7229242373077858),linewidth(4.pt) + dotstyle); label("$P$", (3.4604675529114877,0.8907941618921555), NE * labelscalefactor); dot((0.5353088497578561,-0.21423029557796058),linewidth(4.pt) + dotstyle); label("$Q$", (0.13491995345597343,-0.32221295430268), NE * labelscalefactor); dot((2.9830891114772173,-2.8917910218550693),linewidth(4.pt) + dotstyle); label("$R$", (2.901666521855438,-3.1979938702252673), NE * labelscalefactor); dot((1.4807730894735154,1.525709092035872),linewidth(4.pt) + dotstyle); label("$O_1$", (1.5387371778162928,1.7494396486368142), NE * labelscalefactor); dot((3.9285533511928756,-1.1518516342412366),linewidth(4.pt) + dotstyle); label("$O_2$", (4.114673638050277,-1.1127119738453817), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Claim: $O$ lies on both $\odot(APQ)$ as well as $\odot(CRP)$. Proof: $\triangle BQO \cong \triangle APO$. This means that $O$ lies on the perpendicular bisector of $QP$ since $\overline{OP} = \overline{OQ}$. $O$ lies on $\odot(APO)$ because it is is the intersection of angle bisector of $\angle QAP$ and and the perpendicular bisector of $\overline{QP}$. For $O \in \odot(PRC)$, \[\angle O_1OP = 90^\circ - \angle OAC = \angle PCR\]and the claim is proven. $\square$ Claim: $O_1PO_2R$ is cyclic with $O_2$ arc midpoint of $PR$ not containing $O_1$ and $R$ is the circumcenter of $\triangle QPR$. Proof: $O_1PO_2R$ is cyclic because \[\angle PO_2R = 2\angle PCR = \angle BAC = \angle OO_1P.\]Arc midpoint follows from $\overline{O_2P} = \overline{O_2R}$. Also since \[\angle BPR = \angle OPR = \angle OCR = \angle RBP \]we get $\overline{RB} = \overline{RP} = \overline{RQ}$ as claimed. $\square$ Claim: $O_1O_2QR$ is a parallelogram. Proof: Since \[\angle OBA = \angle OAQ = \angle OPQ\]we have $\overline{QP} = \overline{QB}$. This also implies $Q$ is the arc midpoint of $BP$ in $\odot(BPQ) \implies RQ \perp BP$. Also, since $OP$ is the radical axis of $\odot(PRC)$ and $\odot(APQ)$, we get $O_1O_2 \perp BP$. Finally \[\angle(O_1O_2,BP) = \angle(QR, BP) = 90^\circ\]we get $QR \parallel O_1O_2$. At, last we will show $\triangle O_2PC \cong \triangle O_1AQ$. As both of them are isosceles with $\overline{AQ} = \overline{PC}$, it is sufficient to prove $\angle CO_2P = \angle AO_1Q$. Let $\angle ABP = \theta \implies \angle BAC= 2\theta$. Now, \[\angle APQ = 180^\circ - (\angle QPB + \angle BPC) = 180^\circ - (\theta + 3\theta) = 180^\circ - 4\theta\]And, \[\angle PRC = 180^\circ - \angle PRB = 2(\angle RBP) = 180^\circ - 4\theta. \]This gives $\angle APQ = \angle PRC$. It also implies $\angle AO_1Q = \angle PO_2C$ and we are done proving the claim. $\square$ From the parallelogram, we get \[\overline{O_1O_2} = \overline{RQ} = \overline{BR}\]and the solution is complete. $\blacksquare$

The main claim is that the second intersection of $\odot APQ$ and $\odot CRP$ is the circumcentre of $\triangle ABC$. We will prove this in two steps. Claim: The circumcentre of $\triangle ABC$ lies on $\odot AQP$ Proof: Let the angle bisector of $\angle BAC$ meet $\odot AQP$ again at $O$. We claim that $O$ is the circumcentre. Indeed, $BQ=AP$, $\angle APO = \angle BQP$, by cyclic quadrilateral, and $PO=QO$, as they subtend equal angles. This gives $\triangle APO \cong \triangle BQO$. Thus $AO=BO$, and by isosceles triangle and the fact that $AO$ is the angle bisector, we are done. Claim: $\odot POC$ meets BC again at a point $R'$ such that $PR'=QR'$ Proof: Firstly, in $\odot AQP$, by sin law, we have, $\frac{PO}{\sin PAO} = 2R_1$ and in $\odot CR'P$, we have $\frac{PO}{\sin PCO} = 2R_2$. Hence we conclude $R_1 = R_2$. We will call it $R$. Then by sin law, we get, $\frac{PQ}{\sin A} = 2R$, and, $\frac{PR'}{\sin C} = 2R$. Also, $\angle R'PQ = \angle OPR' + \angle OPQ = \angle OCR' + \angle QAO = \angle ABC$. Combining all three facts, we get $\triangle ABC \sim \triangle QRP$, and we are done. Hence, we conclude $R=R'$ As $\angle PBC = 90 - \angle A$, we know that $O$ lies on $PB$. Hence, $\odot AQP$, $\odot CRP$, and $PB$ are concurrent at $O$. Now, we get $O_1O_2 = 2R \sin OPO_2 = 2R \sin B$, and, by POP, $BR.BC = R_0BP$, where $R_0$ is the circumradius of $\triangle ABC$. Using sin law, we can get, $BR = \frac{BP}{2 \sin A}$. Again in $\triangle BPA$, $BP = AP \sin A \sec C$. By a short angle chase, $\angle AOP = 180 - \angle AOB = \angle A$. So, $AP = 2R \sin A$. Combining all these facts, $BR = \frac{AP}{2\cos C}$. Hence, to show $O_1O_2 = BR$, it is equivalent to $AP = 4R \sin B \cos C$ But, $4R \sin B \cos C = 2R \sin A = AP$, by sin law, and we are done.

Original write-ups are too tedious, so I'll leave a quick sketch for this one. [asy][asy] import graph; size(9.6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.5, xmax = 6.2, ymin = -3.579614086556227, ymax = 3.8619801318974827; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen yqqqqq = rgb(0.5019607843137255,0.,0.); pen ffttww = rgb(1.,0.2,0.4); draw((2.1357002305054333,3.3944953668920594)--(-0.6446756313344209,-2.8749796156879914)--(4.857851242143845,-2.9004788475217542)--cycle, linewidth(1.2) + yqqqyq); /* draw figures */ draw((2.1357002305054333,3.3944953668920594)--(-0.6446756313344209,-2.8749796156879914), linewidth(1.2) + yqqqyq); draw((4.857851242143845,-2.9004788475217542)--(2.1357002305054333,3.3944953668920594), linewidth(1.2) + yqqqyq); /* special point */ draw((-0.6446756313344209,-2.8749796156879914)--(3.2909742324878244,0.7229242373077858), linewidth(1.2)); draw(circle((1.4807730894735145,1.5257090920358725), 1.9802251642555977), linewidth(1.2) + ffxfqq); draw(circle((3.928553351192876,-1.1518516342412364), 1.9802251642555975), linewidth(1.2) + ffxfqq); draw((3.2909742324878244,0.7229242373077858)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2) + yqqqqq); draw((0.5353088497578561,-0.21423029557796058)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2) + yqqqqq); draw((1.8281694616553013,-1.628455605384008)--(1.6686650446165432,-1.635775235961833), linewidth(1.2) + yqqqqq); draw((1.8281694616553013,-1.628455605384008)--(1.8497329166185283,-1.4702460814711957), linewidth(1.2) + yqqqqq); draw((3.2909742324878244,0.7229242373077858)--(0.5353088497578561,-0.21423029557796058), linewidth(1.2)); draw((1.4807730894735154,1.525709092035872)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2)); draw((1.4807730894735154,1.525709092035872)--(3.9285533511928756,-1.1518516342412366), linewidth(1.2) + yqqqqq); draw((2.773633701370962,0.1114837822298249)--(2.6141292843322037,0.10416415165199998), linewidth(1.2) + yqqqqq); draw((2.773633701370962,0.1114837822298249)--(2.7951971563341895,0.269693306142637), linewidth(1.2) + yqqqqq); draw((-0.6446756313344209,-2.8749796156879914)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2) + yqqqqq); draw((2.9830891114772173,-2.8917910218550693)--(4.857851242143845,-2.9004788475217542), linewidth(1.2) + yqqqyq); draw(circle((1.4693356998519826,-0.9423865810104978), 2.4681221738786703), linewidth(1.2) + linetype("4 4") + blue); draw((3.9285533511928756,-1.1518516342412366)--(2.9830891114772173,-2.8917910218550693), linewidth(1.2) + ffttww); draw((3.407016139142441,-2.111637419080401)--(3.3480419220963475,-1.963255217417028), linewidth(1.2) + ffttww); draw((3.407016139142441,-2.111637419080401)--(3.563600540573743,-2.080387438679278), linewidth(1.2) + ffttww); draw((1.4807730894735154,1.525709092035872)--(0.5353088497578561,-0.21423029557796058), linewidth(1.2) + ffttww); draw((0.9592358774230816,0.5659233071967078)--(0.9002616603769881,0.7143055088600809), linewidth(1.2) + ffttww); draw((0.9592358774230816,0.5659233071967078)--(1.1158202788543843,0.5971732875978305), linewidth(1.2) + ffttww); draw((2.1183522081785666,-0.34906677951315035)--(4.857851242143845,-2.9004788475217542), linewidth(1.2)); draw((2.1183522081785666,-0.34906677951315035)--(2.1357002305054333,3.3944953668920594), linewidth(1.2)); /* dots and labels */ dot((2.1357002305054333,3.3944953668920594),dotstyle); label("$A$", (2.1929432629550827,3.5893942630896545), NE * labelscalefactor); dot((-0.6446756313344209,-2.8749796156879914),dotstyle); label("$B$", (-0.9690528152157342,-3.0071837620597877), NE * labelscalefactor); dot((4.857851242143845,-2.9004788475217542),dotstyle); label("$C$", (5.041465591996896,-3.020813055500179), NE * labelscalefactor); dot((2.1183522081785666,-0.34906677951315035),linewidth(1.25pt) + dotstyle + invisible,UnFill(0)); label("$O$", (1.7022886991009902,-0.28132507398150575), NE * labelscalefactor); dot((3.2909742324878244,0.7229242373077858),linewidth(4.pt) + dotstyle); label("$P$", (3.4604675529114877,0.8907941618921555), NE * labelscalefactor); dot((0.5353088497578561,-0.21423029557796058),linewidth(4.pt) + dotstyle); label("$Q$", (0.13491995345597343,-0.32221295430268), NE * labelscalefactor); dot((2.9830891114772173,-2.8917910218550693),linewidth(4.pt) + dotstyle); label("$R$", (2.901666521855438,-3.1979938702252673), NE * labelscalefactor); dot((1.4807730894735154,1.525709092035872),linewidth(4.pt) + dotstyle); label("$O_1$", (1.5387371778162928,1.7494396486368142), NE * labelscalefactor); dot((3.9285533511928756,-1.1518516342412366),linewidth(4.pt) + dotstyle); label("$O_2$", (4.114673638050277,-1.1127119738453817), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] (diagram stolen from @2above) I denote $O$ as the circumcenter of $\odot(ABC)$. $\overline{B-O-P}$ (angle-chasing) $O\in\odot(APQ)$ (Fact 5) $O\in\odot(CPR)$ (angle-chasing) $O_1O_2=\dfrac{OP}{\tan\left(\dfrac{A}{2}\right)}$ (trig) $BO\cdot BP=BR\cdot BC$ (POP) $\dfrac{OP}{BR}=\dfrac{BC}{BO}-\dfrac{BO}{BR}=\dfrac{\sin(2A)}{\cos(A)}-\dfrac{\sin\left(\dfrac{3A}{2}\right)}{\cos\left(\dfrac{A}{2}\right)}$ (from previous claim, and some trig and angle chasing) $\dfrac{\sin(2A)}{\cos(A)}-\dfrac{\sin\left(\dfrac{3A}{2}\right)}{\cos\left(\dfrac{A}{2}\right)}=\tan\left(\dfrac{A}{2}\right)$ (some crazy trig manipulations).

We uploaded our solution https://calimath.org/pdf/IndiaEGMOTST2023-6.pdf on youtube https://youtu.be/yyTjCQV9f2w.

Amazing angle chasing Problem!! Let $O$ be center of $\triangle ABC$ and for sake of niceness, let $\measuredangle BAE = 2\theta$. Now $$\measuredangle PBA = 90 - \theta - (90 - 2\theta) = \theta = 90 - (90 -\theta) = 90 - \measuredangle ACB$$Which implies $O$ lie on $BP$. Claim 1: $\triangle RPQ \sim \triangle ABC$ Subclaim 1: $O$ lie on $O_1R$. Power of $P,Q$ with respect to $(ABC)$ is $PA\cdot PC = QB\cdot QA$ which is equal. By definition of power of point $Pow(\omega,P)=OP^2-R^2$ we get $OP=OQ$ which implies $O$ lie on perpendicular bisector of $PQ$. Subclaim 2: $A,P,O,Q$ is cyclic. Let $L = OC \cap AB$ and $M$ be midpoint of $AB$. Now note $\measuredangle LOA = 2\theta$ and $\measuredangle MOA = 90 - \theta$ , so $\measuredangle MOL = 90 - 3\theta$. Now as $Q$ is just reflection of $L$ above $M$, as $AL=AP$. We get $\measuredangle QOM = 90 - 3\theta \Rightarrow \measuredangle QOA = 180 - 4\theta$. Note $\measuredangle AOP = \measuredangle OPC - \measuredangle OAP = 2\theta$. Hence $\measuredangle QOP = 180 - 2\theta = \measuredangle BAC$ implies $A,P,O,Q$ lie on same circle. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.59815809799398cm); 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 = 3.327155351047827, xmax = 33.92531344904181, ymin = -6.685865644955695, ymax = 9.439220019362754; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen ffdxqq = rgb(1.,0.8431372549019608,0.); /* draw figures */ draw((15.646782839349592,8.29029839036684)--(20.91162819737685,-4.395574575200807), linewidth(0.4)); draw((15.646782839349592,8.29029839036684)--(10.89466058873522,-4.5964119601328965), linewidth(0.4)); draw((10.89466058873522,-4.5964119601328965)--(17.78010985695094,3.1499545247979945), linewidth(0.4)); draw((13.721212362199086,3.0685747709241666)--(20.91162819737685,-4.395574575200807), linewidth(0.4)); draw((12.820231065885725,0.6253116593097757)--(15.794630983736669,0.9162227230108644), linewidth(0.4)); draw((12.820231065885725,0.6253116593097757)--(18.52278445716068,-4.443470220723561), linewidth(0.4) + red); draw((18.52278445716068,-4.443470220723561)--(17.78010985695094,3.1499545247979945), linewidth(0.4) + red); draw((17.78010985695094,3.1499545247979945)--(12.820231065885725,0.6253116593097757), linewidth(0.4)); draw((13.936093724706325,4.567479563921099)--(18.52278445716068,-4.443470220723561), linewidth(0.4) + qqwuqq); draw((17.78010985695094,3.1499545247979945)--(19.63864711598128,-0.5013023161122397), linewidth(0.4) + qqwuqq); draw((13.936093724706325,4.567479563921099)--(17.78010985695094,3.1499545247979945), linewidth(0.4)); draw((19.63864711598128,-0.5013023161122397)--(18.52278445716068,-4.443470220723561), linewidth(0.4)); draw(circle((14.04868308339892,-1.0480285222991659), 5.616636665309231), linewidth(0.4) + ffxfqq); draw((13.936093724706325,4.567479563921099)--(19.63864711598128,-0.5013023161122397), linewidth(0.4) + red); draw((10.89466058873522,-4.5964119601328965)--(18.52278445716068,-4.443470220723561), linewidth(0.4) + red); draw((18.52278445716068,-4.443470220723561)--(20.91162819737685,-4.395574575200807), linewidth(0.4)); draw((13.270721714042406,1.8469432151169713)--(15.794630983736669,0.9162227230108644), linewidth(0.4)); draw(circle((13.936093724706328,4.5674795639211), 4.097052264921426), linewidth(0.4) + blue); draw(circle((19.63864711598128,-0.5013023161122397), 4.097052264921426), linewidth(0.4) + ffdxqq); /* dots and labels */ dot((10.89466058873522,-4.5964119601328965),dotstyle); label("$B$", (10.420038243481878,-4.369455788945584), NE * labelscalefactor); dot((20.91162819737685,-4.395574575200807),dotstyle); label("$C$", (20.97855758715586,-4.207845798991389), NE * labelscalefactor); dot((15.646782839349592,8.29029839036684),dotstyle); label("$A$", (15.717254580869334,8.469560079637592), NE * labelscalefactor); dot((15.794630983736669,0.9162227230108644),linewidth(4.pt) + dotstyle); label("$O$", (16.166171219630982,0.6045405685334926), NE * labelscalefactor); dot((17.78010985695094,3.1499545247979945),linewidth(4.pt) + dotstyle); label("$P$", (17.961837774677576,3.1903004078005934), NE * labelscalefactor); dot((12.820231065885725,0.6253116593097757),linewidth(4.pt) + dotstyle); label("$Q$", (12.395271454033134,0.8918472173409483), NE * labelscalefactor); dot((13.721212362199086,3.0685747709241666),linewidth(4.pt) + dotstyle); label("$L$", (13.293104731556431,3.1903004078005934), NE * labelscalefactor); dot((18.52278445716068,-4.443470220723561),linewidth(4.pt) + dotstyle); label("$R$", (18.698061062246683,-5.357072394221213), NE * labelscalefactor); dot((13.936093724706325,4.567479563921099),linewidth(4.pt) + dotstyle); label("$O_1$", (13.329018062657363,5.219403615003249), NE * labelscalefactor); dot((19.63864711598128,-0.5013023161122397),linewidth(4.pt) + dotstyle); label("$O_2$", (19.703634333072774,-0.36511937119167026), NE * labelscalefactor); dot((13.270721714042406,1.8469432151169713),linewidth(4.pt) + dotstyle); label("$M$", (12.664621437290123,2.076987143671703), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Subclaim 3: $O,P,C,R$ cyclic. We get this from $\measuredangle OAP = \theta \Rightarrow \measuredangle OO_1P = 2\theta \Rightarrow \measuredangle POR = 90 + \theta$ But as $\measuredangle PCR = 90 - \theta$, we get $O,P,C,R$ lie on circle. $$\measuredangle QPR = \measuredangle QPO + \measuredangle OPR = \theta + \measuredangle OCR = \theta + 90 - 2\theta = 90 - \theta$$ Claim 2: $O_1PO_2R$ is isosceles trapezium. From $\measuredangle RO_1P = 2\theta$ and $\measuredangle PO_2R = 180 - \theta = 2\theta$ we get the are cyclic. Now we show $PO_2 \parallel O_1R \perp PQ$. But as $\measuredangle QPR = 90 - \theta \Rightarrow \measuredangle QPO_2 = 90^{\circ}$. Claim 3: $\triangle BRQ$ is isosceles triangle. Cuz $\measuredangle RQP + \measuredangle PQA = 90 - \theta + 2\theta = \measuredangle CBA$. Hence we have $$BR=RQ=PR=O_1O_2$$$\blacksquare$