$\angle PBA = \angle ACB$ so by the converse of the Alternate Segment Theorem, $PB$ is tangent to $(ABC)$. So $PB^2 = PA\cdot PC$. Thus $PD^2 = PA\cdot PC$ so $PD$ is tangent to $(ADC)$. By the Alternate Segment Theorem we can let $\angle PDA=\angle ACD=x$. Now $\triangle ABQ$ is similar to $\triangle ACB$ so $\angle AQD =\angle AQB=\angle ABC=\angle ARC$ so $DQCR$ is cyclic. Hence $\angle DRQ=\angle DCQ=x$. Let $\angle ACB=y$. $\angle QDR=\angle ADB=\angle PDA+\angle PDB=\angle PDA+\angle PBD=x+2y$. From $DQCR$ being cyclic, we get \[ \angle BQR = \angle DQR=\angle DCR=180-\angle QDR-\angle DCQ=180-(x+2y)-x=180-2x-2y. \] Also, note that $\angle QRB=\angle DRQ+\angle DRB=\angle DCQ+\angle ACB=x+y$. From the angle sum of $\triangle QBR$, $\angle QBR=x+y=\angle QRB$, so $QB=QR$, as desired.

Nice solution, Leminscate. This is my solution too, but I think maybe it is too easy for G4?

gobathegreat wrote: Nice solution, Leminscate. This is my solution too, but I think maybe it is too easy for G4? It's a bit on the easy side for G4, especially as this shortlist only has 6 geometry problems.

I came up with this solution: Let $\angle{PDA}=\theta$. Note that $\angle{PBA}=\angle{BCA} \Rightarrow PB^2=PD^2=PA \times PC \Rightarrow \triangle{PDA} \sim \triangle{PCD} \Rightarrow \angle{DCP}=\angle{ADP}=\theta$. Now we note that $\angle{DQC}=\angle{BQC}=180^{\circ}-B=180-\angle{ARC}=180-\angle{DRC} \Rightarrow$ points $D,Q,C,R$ are concyclic. So $\angle{DRQ}=\angle{DCQ}=\angle{DCP}=\theta$.Also note that $\angle{DRB}=\angle{ARB}=C$. Thus $\angle{QRB}=C+\theta$.Observe that $\angle{BPD}=180^{\circ}-4C$ and $\angle{BPA}=A-C \Rightarrow \angle{DPA}=A+3C-180^{\circ} \Rightarrow \angle{RBC}=\angle{RAC}=\angle{DAQ}=A+3C-180^{\circ}+\theta$. Finally $\angle{CBQ}=B-C \Rightarrow \angle{QBR}=\angle{QBC}+\angle{CBR}=B-C+A+3C-180^{\circ}+\theta=A+2C+B-180^{\circ}+\theta=C+\theta=\angle{QRB} \Rightarrow QB=QR %Error. "blackbox" is a bad command. $.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -7.029116190476198, xmax = 18.62749333333336, ymin = -7.128529523809536, ymax = 7.707683809523824; /* image dimensions */ /* draw figures */ draw(circle((6.189615238095245,-0.1287371428571429), 4.350468571428576)); draw((xmin, 3.756379750595548*xmin-10.48255296966338)--(xmax, 3.756379750595548*xmax-10.48255296966338)); /* line */ draw((xmin, 0.6947114015417101*xmin + 0.8685296177650087)--(xmax, 0.6947114015417101*xmax + 0.8685296177650087)); /* line */ draw((xmin, -0.6390328389919702*xmin + 5.813363381187939)--(xmax, -0.6390328389919702*xmax + 5.813363381187939)); /* line */ draw((xmin, 0.1356451804346471*xmin-2.301136778751466)--(xmax, 0.1356451804346471*xmax-2.301136778751466)); /* line */ draw(circle((-5.669572363431124,-3.070186944976369), 11.41778790815437)); draw((3.707482749049642,3.444160154548994)--(6.045875141860290,-1.481042954248480)); draw((xmin, 0.6805520907420449*xmin-3.532409088870188)--(xmax, 0.6805520907420449*xmax-3.532409088870188)); /* line */ draw((xmin, 0.2868738455051074*xmin-1.443734918709404)--(xmax, 0.2868738455051074*xmax-1.443734918709404)); /* line */ draw((5.305536171813876,0.07828464536528656)--(10.47467458279688,-0.8802976549737737)); draw((6.045875141860290,-1.481042954248480)--(9.367346986807440,2.842558487707804)); draw((9.367346986807440,2.842558487707804)--(3.707482749049642,3.444160154548994)); /* dots and labels */ dot((3.707482749049642,3.444160154548994),dotstyle); label("$B$", (3.819167619047623,3.608203809523816), NE * labelscalefactor); dot((10.47467458279688,-0.8802976549737737),dotstyle); label("$C$", (10.59585904761906,-0.7143771428571439), NE * labelscalefactor); dot((2.259601203119495,-1.994632765843977),dotstyle); label("$A$", (2.369011428571431,-1.829881904761908), NE * labelscalefactor); dot((-5.669572363431124,-3.070186944976369),dotstyle); label("$P$", (-5.551072380952387,-2.889611428571433), NE * labelscalefactor); dot((6.045875141860290,-1.481042954248480),dotstyle); label("$Q$", (6.161727619047626,-1.300017142857145), NE * labelscalefactor); dot((5.305536171813876,0.07828464536528656),dotstyle); label("$D$", (5.408761904761910,0.2338019047619054), NE * labelscalefactor); dot((9.367346986807440,2.842558487707804),dotstyle); label("$R$", (9.480354285714295,3.022563809523815), NE * labelscalefactor); dot((2.605217658713962,-0.6963661105763183),dotstyle); label("$E$", (2.703662857142860,-0.5191638095238102), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] I will denote $\angle BAC, ABC, ACB$ as $A, B, C$ respectively. Let $\angle DAB = \alpha$ and let $\angle DCA = \gamma$. First I will show that $QDRC$ is cyclic. Clearly, $\angle DQA = B$ and $\angle DRC = \angle ARC = B$, so we have $DQCR$ cyclic. Therefore, $\angle DCQ = \angle DRQ$, so $\angle BRQ = C + \gamma$. We also have $\angle QBR = A+B-C - \alpha = 180^\circ - 2C - \alpha$. Now we shift our attention: we know that $PB^2 = PA \cdot PC = PD^2$, so the circumcircle of $CDA$ is tangent to $PD$. Hence, $\angle DCA = \angle PDA = \gamma$. Now let $E = AB \cap PD$. Clearly, from exterior angles, $\angle PEA = \alpha + \gamma$, but we also have $\angle PEA = PBA + \angle BPD = C + 180^\circ - 4C$ which tells us that $\alpha + \gamma = 180^\circ - 3C$. Now we see that $\angle BRQ = C + \gamma = C + 180^\circ - 2C - \alpha = \angle QBR$, so $QB = QR$, as desired.

Notice PD*PD=PB*PB=PA*PC,thus we have PDC is similar to PAD.Now,we have that triangles ABD and ABR are similar(<DBA=<QBA=<BRA=<BCA) so we have AB*AB=AD*AR=AQ*AC(ABQ and ABC are similar since <QBA=<BCA) and from this we obtain RCQD is a cyclic.The rest is just angle chasing.

Let us reformulate the problem as the following: let the circle with center $Q$ passing through $B$ be $\Gamma_{1}$ let the circle with center $P$ and passing through $B$ be $\Gamma_{2}$. Also, let $\omega$ be the circumcircle of $\Delta ABC$. We then need to prove that $A$, $R= \Gamma_{1} \cap \omega$ and $D=BQ \cap \Gamma_{2}$ are collinear. Simple angle chasing proves that the foot of the $B$ angle bisector (which we'll call $L$) belongs to $\Gamma_{2}$. Define $S$ and $T$ as the opposite points of $B$ in $\Gamma_{2}$ and $\Gamma_{1}$, respectively. Perform an inversion wrt $B$ with power $\sqrt{BA \cdot BC}$ followed by a simmetry wrt the bisector $BL$. This sends $AC$ into $\omega$ and vice versa (in particular it sends $A$ into $C$). Let $P'$, $L'$ and $Q'$ be the images of $P$, $L$ and $Q$ under this inversion: clearly they belong to $\omega$. Moreover, simple angle chasing shows that both $P'BC'A'$ and $A'Q'C'B$ are isosceles trapezoids with bases $A'C'$ and $A'B$, respectively. Now note that $\Gamma_{2}$ in sent into a line passing through $L'$ and $S'$; but $L'$ is the midpoint of arc $A'C'$ and $S'$ is the midpoint of $BP'$. This shows that $\Gamma_{2}$ gets sent into the axis of $A'C'$; a similar reasoning shows that $\Gamma_{1}$ gets sent into the axis of $BQ'$. Finally, note that $R'$ and $D'$ are now defined as the intersections of one of the diagonals of isosceles trapezoid $BC'Q'A'$ with the axis of the other diagonal; this construsction is obviously simmetrical wrt the axis of $BA'$, meaning that $BA'D'R'$ is also an isosceles trapezoid and is therefore cyclic. This concludes the proof.

Too easy for G4 Solution : Let $\angle ABC= \angle B, \angle ACB = \angle C, \angle DAC= \alpha, \angle DCA =\beta$ Obviously $PB$ is tangent to $\odot ABC$, $\implies PB^2=PA \cdot PC \implies PD^2 = PA \cdot PC$ $\implies PD$ is tangent to $\odot ADC$ $ \implies \angle PDA = \beta $ Also since $\angle C = \angle ABQ = \angle ARB$ $\implies$ $AB$ is tangent to $\odot BDR$ and $\odot BQC$ $\implies AD \cdot AR = AB^2 = AQ \cdot AC$ $\implies RCQD$ is cyclic $\implies \angle QRD = \beta$ In $\triangle ABD$, We have $\angle A- \alpha +3 \angle C + \beta=180$ $\implies \beta \angle B-2 \angle C + \alpha$ We have $\angle QBR = \angle B-\angle C+\alpha$ $\angle QRB= \angle C + \beta = \angle B-\angle C+\alpha$ $\implies \angle QBR = \angle QRB$ $\implies QB= QR$ QED

thecmd999 wrote: From $PD^2=PB^2=PA\cdot PC$ we deduce that $\angle ADP=\angle ACD$. Keeping in mind that quadrilateral $CQDR$ is cyclic, angle chasing yields $\angle QBR=\angle QRB\implies QB=QR$ as desired. The best solution

easy enough for a G4! obviously $PB$ is a tangent.... which i realized after a long time now thus $PB^2=PD^2=PC.PD$ and thus triangles $PAD,PCD$ are similar. and hence $\angle DCP=\angle PAD$ after some trivial angle chase now $\angle BQA=B=\angle ARC$ implying $DRCQ$ is cyclic quad. and hence $\angle DRQ=\angle DCP=\angle PAD$ now $\angle BRQ=\angle BRA+\angle ARQ=\angle C+\angle PDA$ also, $\angle RBQ=\angle RBA-\angle QBA$ $=180-\angle RCQ-\angle C$ $=\angle RDQ-\angle C$ $=\angle BDA-\angle C$ $=2\angle C+\angle PDA-\angle C$ $=\angle PDA+\angle C=\angle BRQ$ and hence $QB=QR$ we are done

We get that $ PB $ is tangent to the circumcircle so $ PB^2 =PA.PC=PD^2 $ so, $ \angle PDA = \angle DCA $ and $ AB^2 =AQ.AC =AD.AR$ as $\angle ARB = \angle BCA $ This implies $ DRQC $ is a cyclic quadrilateral so $ \angle PDA = \angle DCA =\angle DRQ $ . let $ \angle PBA =\theta $ and $\angle PDA = \alpha $, we get $ \angle BDR =180 -2\theta + \alpha $ and $ \angle BRD =\theta $, this implies $\angle QBR=\angle QRB= \theta +\alpha \Longrightarrow QB=QR $. Which was to be proved..........

Let the internal angle bisector of $\angle ABC$ cut $AC, \odot (ABC)$ at $E, M$, respectively, and let the external angle bisector of $\angle ABC$ cut $AC, \odot (ABC)$ at $F, N$, respectively. Suppose that the line parallel to $AC$ through $B$ cuts $\odot (ABC)$ at $B'$, and let $P_{\infty}$ be a point at infinity on line $AC.$ Note that $Q$ lies on $\overline{AC}$, so using directed angles, we have $\measuredangle ABQ = \measuredangle BCA \implies \triangle AQB \sim \triangle ABC \implies \measuredangle CQD = \measuredangle AQB = \measuredangle CBA = \measuredangle CRA = \measuredangle CRD \implies C, D, Q, R$ are concyclic. Therefore $\triangle AQR \sim \triangle ADC \implies \tfrac{QR}{QA} = \tfrac{DC}{DA}.$ Also $\triangle AQB \sim \triangle ABC \implies \tfrac{QB}{QA} = \tfrac{BC}{BA}.$ Thus, to show that $QB = QR$, we need only prove that $\tfrac{BC}{BA} = \tfrac{DC}{DA}$, i.e. $D$ lies on the $B$-Apollonius circle, $\omega.$ Recall that $\omega$, which is the locus of points $X$ that satisfy $\tfrac{BC}{BA} = \tfrac{XC}{XA}$, is the circle with diameter $\overline{EF}.$ Also, $B$ lies $\omega$, so $\omega$ is just the circle centered at the midpoint of $\overline{EF}$ passing through $B.$ Keeping in mind that $PB = PD$, we need only prove that $P$ is the midpoint of $\overline{EF}$ (i.e. the center of $\omega$), and we will be able to conclude that $D$ lies on $\omega.$ To see this, note that $\measuredangle PBA = \measuredangle BCA \implies PB$ is tangent to $\odot (ABC)$ by the Alternate Segment Theorem. Then since $M, N$ are the midpoints of $\widehat{ABC}, \widehat{BC}$, respectively, $MN$ is a diameter of $\odot (ABC)$ and $MN \perp BB'.$ Therefore $\tfrac{MB}{MB'} = \tfrac{NB}{NB'} = 1 \implies BMB'N$ is harmonic $\implies -1 = B(M, N; B, B') = \left(E, F; P, P_{\infty}\right) \implies P$ is the midpoint of $\overline{EF}$, as desired. $\square$

Way too easy for G4.. We have $\angle AQD = 180-\angle A - \angle ABQ = 180-\angle A - \angle C = \angle B = \angle ARC$, so $DQCR$ is cyclic. Let $\angle PDA = x$. Since $\triangle PBA \sim \triangle PCB$, we have $PD^2=PB^2=PC \cdot PA$, so $\triangle PDA \sim \triangle PCD$, so $\angle DCQ = \angle DRQ = x$. Now we have $$\angle QBR = \angle QBC + \angle CBR = (\angle B - \angle ABQ) + \angle CAR = \angle B - \angle C + \angle ADB - \angle AQD$$$$ = \angle B - \angle C + (x + \angle PDB) - \angle B = x-\angle C + \angle PBD = \angle C + x = \angle ARC + \angle QRA = \angle QRB$$which gives $QB = QR$ as desired.

We have $\angle BPA = \angle A - \angle C$ and $\angle PBQ = 2\angle C$. So $\angle PQD = \angle B = \angle DRC \implies CQDR$ is cyclic. Therefore, $\angle QRB = \angle DRB + \angle DRQ = \angle C + \angle DCQ = \angle C + \angle ADP$. The last equality follows from $PD$ being tangent to $\odot (ADC)$. Also, $\angle QBR = 360 - \angle BPD - \angle PDR - \angle DRB - \angle ABD = 360 - (180-4\angle C) - (180 - \angle ADP) - \angle C - 2\angle C = \angle C + \angle ADP.$ So $QB=QR$.

An easy problem for G4. (But nice...) Here is my solution. Since $\measuredangle PBA=\measuredangle ACB$ $\implies$ $PB$ is tangent to $\odot (ABC)$. So $PB^2=PA\cdot PC=PD^2$ $\implies$ $PD$ is tangent to $\odot (ADC) $. An easy angle-chasing gives $DQCR$ is cyclic $\implies$ again after an easy angle-chasing we get $QR=QB$.

Yeah, it was easy yet nice for G4. Solution: Let $\angle PBA$ = $\angle QBA$ = $\angle ACB$ = $x$. Join $BR$ and $DC$. Let the bisector of $\angle PDB$ meet $AB$ at $M$. Then, $\angle PBA$ = $\angle QBA$ = $\angle ACB$ = $\angle PDM$ = $\angle ARB$ = $\angle BDM$ = $x$ [as, $PB$ = $PD$]. Now, $PB$ is tangent to circle $ABC$. Also, $AB$ is tangent to circle $BQC$ and $BDR$. So, $AB^2$ = $AD\cdot AR$ = $AQ\cdot AC$. This implies that $DQCR$ is cyclic. So, $\angle DRQ$ = $\angle DCQ$.....($1$) Again, by power of a point theorem, $PB^2$ = $PA\cdot PC$, or, $PD^2$ = $PA\cdot PC$. Thus, $PD$ is tangent to circle $ADC$. So, $\angle PDA$ = $\angle ACD$...($2$) ($1$) and ($2$) gives, $\angle PDA$ = $\angle DRQ$. So, $\angle QRB$ = $x$ + $\angle DRQ$, = $x$ + $\angle PDA$ = $\angle ADM$....($3$) Now, $\angle MDB$ = $\angle BRD$ = $x$, or, $MD$ is tangent to circle $BDR$. Thus, by alternate segment theorem, $\angle ADM$ = $\angle QBR$....($4$) ($3$) and ($4$) gives $\angle QBR$ = $\angle QRB$. Hence, $QB$ = $QR$.

probably similar to the other solutions, but oh well. Recall that $\angle{ABP}=\angle{ACB}$ implies that $PB$ is tangent to $\omega$. By Power of a Point, we have $PB^2=PA*PC=PD^2$, so triangles $PDA$ and $PCD$ are similar. In particular, $\angle{PDA}=\angle{DCP}=x$. We claim that quadrilateral $DQCR$ is cyclic. This follows directly from the angle chase $$\angle{AQB}= 180- \angle{A}- \angle{C}-\angle{B}, \angle{DRC} = 180-\angle{B}.$$It suffices to show is that $\angle{RBQ}= \angle{QRB}=\angle{ARB}-\angle{ARQ}= \angle{C}-x$. Well, $$\angle{RBA}= \angle{RCQ}=\angle{QDA}=2\angle{C}-x \implies \angle{RBQ}= \angle{RBA}-\angle{C}=\angle{C}-x.$$

Kfp wrote: Now note that $\Gamma_{2}$ in sent into a line passing through $L'$ and $S'$. Anybody can explain to me this ????

First we prove QDRC is cyclic. Since PBA=ACB PB is a tangent. Then PA*PC=PB^2=PD^2, consequently SAS yields sim triangles PDC and PAD. Triangles ABD and ARB are also similar by AA (ABD=PBA=ACB=ARB), and similar reasoning yields AQD and ARC are similar. Hence AB^2=AD*AR=AQ*AC, as claimed. Now, QRB=ARB+QRA=ACB+QCD=ACB+PDA and QBR=RBA-QBA=RDQ-ACB=ADB-ACB=2ACB+PDA-ACB, so they are equal, as desired. $\blacksquare$ i think this was not that bad for a g4 because its very easy to angle chase with so much information, and even with no ideas of which direction to go just easily find information and eventually chase in the right direction. agree with post#74

Did I not post my solution on this one or was it deleted? Also, what kind of retar d was I back then that I didn't use dangles for this problem lmao. Anyways, here is my soln written 1 year back in some forum. Let $\alpha = \angle DCB, \beta = \angle ACD$. So, $\angle PBA = \angle QBA = \alpha + \beta$. Now, $PA \cdot PC = Pow_{\odot ABC}(P) = PB^2 = PD^2 \implies PD$ is tangent to $\odot ADC \implies \angle PDA = \angle DCA = \beta$. And, $\angle BRD = \angle BRA = \angle BCA = \alpha + \beta = \angle ABD \implies AB$ is tangent to $\odot BDR \implies AB^2 = AD \cdot AR$. And, $\angle BCQ = \alpha + \beta = \angle ABQ \implies AB$ is tangent to $\odot BQC \implies AB^2 = AQ \cdot AC$. Thus, $AD \cdot AR = AB^2 = AQ \cdot AC \implies DQCR$ is cyclic. Also, $\angle PDB = \angle PBD = 2(\alpha + \beta)$. Thus, \begin{align*} \angle QBR &= \angle DBR\\ &= 180^{\circ} - (\angle DRB + \angle BDR)\\ &= 180^{\circ} - (\angle ARB + (180^{\circ} - \angle BDA))\\ &= 180^{\circ} - (\angle ACB + (180^{\circ} - (\angle BDP + \angle PDA)))\\ &= 180^{\circ} - (\alpha + \beta + (180^{\circ} - (2(\alpha + \beta) + \beta)))\\ &= 180^{\circ} - (\alpha + \beta + (180^{\circ} - 2 \alpha - 3\beta))\\ &= 180^{\circ} - (\alpha + \beta + 180^{\circ} - 2 \alpha - 3 \beta)\\ &= 180^{\circ} - (180^{\circ} - \alpha - 2 \beta)\\ &= 180^{\circ} - 180^{\circ} + \alpha + 2 \beta\\ &= \alpha + 2\beta. \end{align*} Finally, \begin{align*} \angle QRB &= \angle QRA + \angle ARB\\ &= \angle QRD + \angle ACB\\ &= \angle QCD + \alpha + \beta\\ &= \beta + \alpha + \beta\\ &= \alpha + 2 \beta\\ &= \angle QBR. \end{align*}and thus $QB = QR$ and we are done.

Ngl drawing the diagram was one of the hardest parts of the problem. Turns out that if you draw $\triangle{ABC}$ with $\angle A > 90^\circ$ it's not too bad. Note that $\angle{ACB} = \angle{ARB}$, so $AB$ is not only tangent to $(QBC)$ but also $(DRB)$. Power of a point yields $AB^2 = AQ \cdot AC = AD \cdot AR$, which means that $QCRD$ is cyclic. Since $PB$ is tangent to $(ABC)$, power of a point yields $PB^2 = PA \cdot PC$ so $PD^2 = PA \cdot PC$. This means that $PD$ is tangent to $(ADC)$. Now, we angle chase. Let $\angle{ACD} = \theta$, $\angle{ACB} = \alpha$, and $\angle{QBC} = \beta$. It follows that $\angle{QRD} = \angle{QCD} = \theta$. So, $\angle{QRB} = \angle{QRD}+\angle{DRB} = \theta+\alpha$. Since $PD$ is tangent to $(ADC)$, we know that $\angle{PDA} = \theta$. We also know $\angle{PDB} = \angle{PBD} = 2\alpha$ since $PB = PD$. In addition, $\angle{AQB} = \angle{QCB}+\angle{QBC} = \alpha+\beta$. Therefore, we have $\angle{QAD} = \angle{ADB} - \angle{AQD} = \angle{PDA} + \angle{PDB} - \angle{AQB} = \theta+2\alpha-(\alpha+\beta) = \theta+\alpha-\beta$. This means that $\angle{QBR} = \angle{QBC} + \angle{CBR} = \beta + \angle{CAR} = \beta + \angle{QAD} = \beta+\theta+\alpha-\beta = \theta + \alpha = \angle{QRB}$ so $QB = QR$ as desired.

I actually don't think this is too easy even though it's ``just angle chasing"; it would fit fine as a hard 1/4 or easy 2/5. A lot of the difficulty of the problem is once you get to the angle chasing step, you need to actually frame the problem in terms of $\angle DCA$ instead of conventionally in terms of $A, B, C$; you get stuck otherwise, and I tried for a long time to get more angle information on the directionality of $\overline{AD}$ that ultimately isn't needed if you frame the problem correctly. That being said, the solution is very short when written out. First I show $DQCR$ is cyclic; this is as $\angle ABQ = \angle C = \angle ARB$ implies $AQ \cdot AC = AD \cdot AR = AB^2$. Now notice that $\overline{PD}$ is tangent to $(DAC)$, hence $$\angle QBR = \angle ABR - \angle ABD = \angle PBD + \angle ACD - \angle ABD = \angle DCA + 2\angle C = \angle QRB.$$

Well, it looks like I went around the bush but never mind. This is quite decent for a G4. We first make the following pieces of observation. Claim : Quadrilateral $QDRC$ is cyclic. Proof : Simply note that, \[\measuredangle DRC = \measuredangle ARC = \measuredangle ABC = \measuredangle BAC + \measuredangle ACB = \measuredangle BAQ + \measuredangle QBA = \measuredangle BQA = DQA\]which implies the required claim. Now, comes the barricade of similar triangles. Claim : The following pairs of triangles are all similar, \[\triangle ABQ \sim \triangle ACB, \triangle PAD \sim \triangle PDC, \triangle AQR \sim \triangle ADC \text{ and } \triangle PAB \sim \triangle PBC\] Proof : For the first, simply note that $\measuredangle ABQ = \measuredangle BCA$ and $\measuredangle QAB = \measuredangle CAB$. Thus, $\triangle ABQ \sim \triangle ACB$. Next, $\measuredangle APD = \measuredangle CPD$ and $PD^2 = PB^2=PA \cdot PA \implies \frac{PD}{PC}=\frac{PA}{PD}$ which implies that $\triangle PAD \sim \triangle PDC$. Then, $\measuredangle QAR = \measuredangle CAD$ and $\measuredangle ARQ = \measuredangle DCA$. Thus, $\triangle AQR \sim \triangle ADC$. Finally, $\measuredangle PBA = \measuredangle BCP$ and $\measuredangle APB = \measuredangle CPB$ as well. Now, note that since $\triangle ABQ \sim \triangle ACB$ we have, \[BQ = \frac{AB\cdot BC}{AC}\]Now, we do some length chase. \begin{align*} QR &= \frac{AR \cdot DC}{AC} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ (} \triangle AQR \sim \triangle ADC \text{ )}\\ &= \frac{AR \cdot AD\cdot PB}{AC \cdot AP} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ (} \triangle PAD \sim \triangle PDC \text{ )}\\ &= \frac{AQ \cdot AC \cdot PB}{AC \cdot AP} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ since $QDRC$ is cyclic - } AD \cdot AR = AQ \cdot AR\\ &= \frac{AQ\cdot PB}{AP}\\ &= \frac{AQ \cdot BC}{AB} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ (} \triangle PAB \sim \triangle PBC \text{ )}\\ &=\frac{AB \cdot BC}{AC} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ (} \triangle AQB \sim \triangle ABC \text{ )} \end{align*}But then, this means that in fact, \[BQ = \frac{AB \cdot BC}{AC} = QR\]which was indeed the required result.

WAT... this problem should easily be put before G1 (which was at the contest) and in general before any other G in the shortlist, as it uses only angles and straightforward PoP, while G1 requires radical axis, which is strictly more sophisticated... it's weird that the PSC has published an elementary solution in the shortlist and put G4, though ok, they might have realized it during leaders work before voting. Accidents happen Since $\angle ABQ = \angle ACB = \gamma$, then $\angle AQB = \angle QBC + \gamma = \angle ABC = \beta$ and now the circumcircle of $ABC$ yields $\angle DQC + \angle DRC = \angle DQC + \angle ARC = 180^{\circ} - \angle AQB + \angle ABC = 180^{\circ}$, i.e. the quadrilateral $CRDQ$ is cyclic. Next, from $\angle PBA = \angle ACB$ we have that $PB$ is tangent to the circumcircle of $ABC$, thus Power of a Point yields $PB^2 = PA \cdot PC$. Now $PB = PD$ gives $PD^2 = PA \cdot PC$, thus by Power of a Point $\angle ADP = \angle ACD = \varphi$, while the cyclic $CRDQ$ gives $\angle QRD = \angle QCD = \varphi$. Since $\angle PDB = \angle PBD = 2\gamma$ from $PB = PD$, we obtain $\angle BDR = 180^{\circ} - 2\gamma - \varphi$. On the other hand, $\angle QAD = \angle ADB - \angle AQB = 2\gamma + \varphi - \beta$, so $\angle CBR = \angle CAR = \angle QAD = 2\gamma + \varphi - \beta$. Hence $\angle QBR = \angle QBC + \angle CBR = \gamma + \varphi$ and $\angle QRB = \angle QRD + \angle ARB = \angle QRD + \angle ACB = \gamma + \varphi$, therefore $\angle QBR = \angle QRB$ and $QB = QR$, as desired.

Troll of the century. Note that we have $AQ \cdot AC = AB^2,$ and $\angle ARB = \angle ACB = \angle QBA = \angle DBA,$ so $AD \cdot AR = AB^2,$ so $AQ \cdot AC = AD \cdot AR,$ so $Q, C, R, D$ are concyclic. We also have $PA \cdot PC = PB^2 = PD^2,$ so we get $\angle PDA = \angle DCA = \angle DCQ = \angle QRD,$ and we have $\angle CBR = \angle CAR,$ and we have $\angle QBC = B - C,$ and we have $\angle ARB = C,$ so it suffices to show that $\angle QAR - \angle PDA = 2C - B$ (basically $QB = QR \iff \angle QBR = \angle QRB \iff \angle QBC + \angle KBR = \angle QRD + \angle ARB,$ and now we can use our previous calculations), which is equivalent to $\angle QPD = 2C - B,$ but we have $\angle QDP = 180 - 2C,$ and $\angle PQD = B$, which gives $\angle QPD = 2C - B,$ and we're done.

Easy for a G4 Skipping obvious angle chasing , we get $PA \cdot PC = PB^2 = PD^2$ $AQ \cdot AC = AB^2=AD \cdot AR$ From here observe $QCRD$ cyclic and we are done by trivial angle chasing.

Didn't expect an angle chasing problem to appear on G4... But still it is quite nice. Because $\angle PBA = \angle BCA$, $PD^2=PB^2=PA*PC$, so $\angle DCA = \angle PDA = \angle DAC - \angle DPA = \angle RAC - (\angle BDP - \angle BQP)=\angle RBC - 2\angle BCA+\angle ABC$. Also, because $\angle DRC = \angle ABC = \angle BQA$, so $RDQC$ is cyclic, so $\angle DRQ = \angle DCA = \angle RBC - 2\angle BCA + \angle ABC$, so $\angle BRQ=\angle BRD+\angle DRQ=\angle RBC-\angle BCA+\angle ABC = \angle RBC+\angle ABC - \angle ABQ = \angle QBR$, which we desired.

Claim: $D, Q, C, R$ are concyclic. Proof. Let $\measuredangle$ denote directed angles modulo $180^{\circ}$. $\measuredangle{CQD} = \measuredangle{AQD} = \measuredangle{AQB} = - (\measuredangle{QBA} + \measuredangle{BAQ}) = - (\measuredangle{ACB} + \measuredangle{BAC}) = \measuredangle{CBA} = \measuredangle{CRA} = \measuredangle{CRD}$, implying that $D, Q, C, R$ are concyclic as desired. Let $\measuredangle{QBA} = \measuredangle{ABP} = \measuredangle{ACB} = x$. $\text{Now notice that } \measuredangle{PBA} = \measuredangle{BCA} \implies PB \text{ is tangent to } (ABC) \implies PB^2 = PA \cdot PC \implies PD^2 = PA \cdot PC \implies PD \text{ is tangent to } (ADC) \implies \measuredangle{ADP} = \measuredangle{ACD} = y \text{ (say).}$ $\measuredangle{QRB} = \measuredangle{QRD} + \measuredangle{DRB} = \measuredangle{QCD} + \measuredangle{ARB} = \measuredangle{ACD} + \measuredangle{ACB} = x + y$. $\measuredangle{RBQ} = \measuredangle{RBD} = - (\measuredangle{BDR} + \measuredangle{DRB}) = - (\measuredangle{QDR} + \measuredangle{ARB}) = - (\measuredangle{QCR} + \measuredangle{ACB}) = - (\measuredangle{ACR} + \measuredangle{ACB}) = x + y$ Hence $\measuredangle{QRB} = \measuredangle{RBQ}$ as desired.

First note that $PB$ is tangent to $(ABC)$, so $PD^2=PB^2=PA \cdot PC \implies PD$ is tangent to $(ADC)$. This implies that $\Delta PAD \stackrel{-}{\sim} \Delta PDC$ Also, $\Delta AQB \stackrel{-}{\sim} \Delta ABC$, so $\measuredangle CQD = \measuredangle AQB = \measuredangle CBA = \measuredangle CRD \implies (CQDR)$ concyclic. Now, we have, by the angle bisector theorem and similar triangles, $$\frac{QB}{QA}=\frac{PB}{PA}=\frac{PD}{PA}=\frac{DC}{AD}=\frac{QR}{QA},$$so $QB=QR$ as desired.

First notice by the angle condition that $PB$ is tangent to $(ABC)$ and $\Delta ABQ \sim \Delta ACB$ implying $\angle AQB = \angle AQD =\angle ARC= \angle ABC$ which gives us that $DQCR$ is cyclic. Also since $PB$ is tangent, we have that $PB^2=PA.PC=PD^2$ giving us that $PD$ is tangent to $(ADC)$. Now let $\angle PDA = \theta$ and express $\angle QRB$ and $\angle QBR$ in terms of angles of $\Delta ABC$ and $\theta$ using the cyclicity and tangency fact to get that both angles are equal implying $QB=QR$.

$PD$ tangent to $ADC$ $AB$ tangent to $BQD$ (after inscribed angle chase) 1 $AB$ tangent to $BQC$ 2 after length chase on 1 and 2 , we get that $DQCR$ cyclic after which we angle chase by alternate segment in the first statement and cyclic quad properties to get that $QBR = QRB$ QED

Easy for a G4 Claim 1: $\color{blue}{D,R,C,Q}$ are concylic. Proof: Let $\angle PBA = \angle QBA = \angle ACB=\theta, \angle{QBC}=\alpha$ and $\angle{CBR}=\beta$. Then we have, $\angle{BRA}=\angle{BCA}=\theta$. Thus due to alternate segment theorem, $AB$ is tangent to both $(BQC)$ and $(BDR)$. By Power of Point we have, $AB^{2}=AQ.AC$ and $AB^{2}=AD.AR \implies AQ.AC=AD.AR$ implying $D,R,C,Q$ are concylic as desired. $\blacksquare$ Claim 2: $\color{blue}{QB=QR}$ Proof: Note that since $PB=PD$, $\angle{PBD}=\angle{PDB}=2\theta \implies \angle{BPD}=180^{\circ}-4\theta$. As $\angle{PBC}=2\theta+\alpha$ and $\angle{PCB}=\theta$ by angle sum property, $\angle{BPC}=180^{\circ}-3\theta-\alpha$. As $\angle{BPD}=180^{\circ}-4\theta$ thus, $\angle{DPQ}=\angle{BPQ}-\angle{BPD}=\theta-\alpha$. Note that by alternate segment theorem, $PB$ is tangent to $(ABC)$ thus $PA.PC=PB^{2}=PD^{2}$ as $PB=PD$. Thus, $PD$ is tangent to $(DAC)$. Now, $\angle{DAC}=\angle{RAC}=\angle{RBC}=\beta$ and $\angle{DPA}=\theta-\alpha \implies \angle{PDA}=\angle{DAC}-\angle{DPA}=\alpha+\beta-\theta$. Hence, $\angle{DCA}=\alpha+\beta-\theta \implies \angle{DRQ}=\angle{DCQ}=\angle{DCA}=\alpha+\beta-\theta$ (since $D,R,C,Q$ are concylic). Now we have, $\angle{QBR}=\boxed{\alpha+\beta}$ and $\angle{QRB}=\angle{QRD}+\angle{DRB}=(\alpha+\beta-\theta)+(\theta)=\boxed{\alpha+\beta}$. Thus, $\angle{QBR}=\angle{QRB} \implies \boxed{QB=QR}$ as desired. $\blacksquare$