The main idea is that $Q$ is the circumcenter of $\triangle PCD.$ First let $H$ be the midpoint of arc $AB$ not containing $C$ on $(ABC).$ Then $CH$ bisects $\angle BCP,$ but $\triangle BCP$ is isosceles so $CH\perp BD.$ Similarly $DH\perp AC$ since $\triangle BCP\sim\triangle ADP$ from $ABCD$ cyclic. Thus $H$ is the orthocenter of $\triangle PCD,$ so the $(ABCDH)$ and $(PDC)$ are congruent. Next we show $SP$ passes through the circumcenter of $\triangle PCD.$ This is because there exists an negative inversion at $P$ swapping $A$ and $C$ and also swapping $B$ and $D.$ Then $S$ is sent to the antipode of $P$ on $(PCD),$ so $SP$ passes through this point and thus passes through the center of $(PCD).$ Together these imply that $Q$ is the center of $(PCD).$ To finish, we have \begin{align*}\measuredangle AEB&=\measuredangle(AE,QC)+\measuredangle CQD+\measuredangle(QD,BE)=90+90+\measuredangle CQD\\&=\measuredangle CQD=2\measuredangle CPD=2\measuredangle CPB=\measuredangle CPB+\measuredangle PBC=\measuredangle PCB\\&=\measuredangle ACB\end{align*}so $A,B,C,E$ are concyclic.

Let $E$ be on $\omega$ such that $CB=CE$. Claim: $DPCO$ is cyclic with center $Q$. Proof. The circumradius condition and spiral similarity at $C$ gives $CQ=QO=QP$. Then, $\measuredangle CPD=2\measuredangle CBP =2\measuredangle CBD =\measuredangle COD$ as desired. $\blacksquare$ Now, note that $P$ is the incenter of $ABE$ by fact $5$. Thus \[180^{\circ} -\angle(BE,DQ)=\angle EBP+\angle DPQ=\angle PBA+\angle DPQ=\angle PBS+\angle SPB=90 ^{\circ}\]and \[180^{\circ} - \angle(AE,CQ)=\angle EAC+\angle QCA=\angle APS+\angle SAP=90^{\circ}\]concluding.

It suffices to show that $\angle DQC = 180^\circ - \angle ACB$. In $\Delta ABC$, write $\angle A = \alpha, \angle B = \beta, \angle C = \gamma$. Note that $\Delta OBC \cong \Delta QPC$, so by spiral similarity we have $\Delta BCP \sim \Delta OCQ$. Let $R=AO \cap SP$. Since $\angle DAR = 90^\circ - \angle DBA$ and $\angle DPR = 180^\circ - \angle BPS = 180^\circ - (90^\circ - \angle DBA) = 90^\circ + \angle DBA$, we have that $(ADPR)$ is concyclic. Also, $\angle RDO = \angle ADO - \angle ADR = 90^\circ - \angle DBA - \angle CPQ = 90^\circ - (\beta - (90^\circ - \frac{\gamma}{2})) - (90^\circ - \alpha) = 90^\circ + \alpha - \beta - \frac{\gamma}{2} = \frac{3}{2}\alpha - \frac{\beta}{2}$ and $\angle RQO = \angle CQP - \angle CQO = 2\alpha - (90^\circ - \frac{\gamma}{2}) = \frac{3}{2}\alpha - \frac{\beta}{2} = \angle RDO$, so $(DOQR)$ is concyclic. Thus $\angle DQC = \angle DQO + \angle CQO = (180^\circ - \angle DRO) + \angle CBP = \angle CPB + \angle CBP = 180^\circ - \angle ACB$, hence proved.

Note that $\angle DAC = \angle DBC = \angle CPB = \angle DPA$, so $DP = DA$. Now if $L$ is the midpoint of the arc $AB$ in $\omega$, not containing $C$, then $CL$ and $DL$ are angle bisectors of $\angle BCP$ and $\angle ADP$, so $LADP$ and $LBCP$ are kites. Hence $LB = LP = LA$ and $CP\perp DL$, $DP\perp CL$. Therefore $P$ is the orthocenter of $\triangle CDL$, and points $B$ and $A$ are the reflections of $P$ over $CL$ and $DL$, respectively. Working with complex numbers and $\omega$ as the unit circle, the above translates to $p = c+d+\ell$, $a = -\frac{d\ell}{c}$, $b = -\frac{c\ell}{d}$. Additionally, if $O$ is the circumcenter of $\omega$, then $PS\perp AB \perp OL$ and $OL = r = PQ$, so $PQOL$ is a parallelogram. Therefore, $q = p - \ell = c+d$. Regarding the condition of $E$ lying on $\omega$, this is equivalent to $\angle EAC = \angle EBC$, but as $EA\perp CQ$ and $EB \perp DQ$, we have: \[\angle EAC - \angle EBC = (90^{\circ} - \angle PCQ) - (\angle DBC - (90^{\circ} - \angle QDP)) = 180^{\circ} - \angle PCQ - \angle DBC - \angle QDP.\]However, note that $\frac{p-c}{q-c}\cdot\frac{d-b}{c-b}\cdot\frac{q-d}{p-d} = \frac{d+\ell}{d}\cdot\frac{d+\frac{c\ell}{d}}{c+\frac{c\ell}{d}} \cdot \frac{c}{c+\ell} = \frac{d^2+c\ell}{d(c+\ell)}$ is real, so the solution is complete.

Note that $\frac{BC}{\sin{A}} = \frac{PC}{\sin{PDC}}$ so $R_{ABC} = R_{DPC} = r$. Now since $PQ = r$ and $\angle QPC = \angle APS = 90 - \angle A = 90 - \angle PDC$ we have that $Q$ is center of $DPC$. Let $AE$ meet $CQ$ at $X$ and $BE$ meet $DQ$ at $Y$. $\angle AEB = \angle XQY = 180 - \angle DQC = 180 - (360 - 2(180 - \angle BPC)) = 180 - 2\angle BPC = \angle C$ so $E$ lies on $\omega$.

Step-by-step solution for 8th graders: https://dgrozev.wordpress.com/2024/07/17/g2-from-imo23-shortlist/

Claim. $\Delta QPC = \Delta OBC$ and $\Delta QDP = \Delta ODA$. Proof. We know that $PC=BC$ and $PQ = r = OB$. By angle chasing, we have \begin{align*} \angle QPC &= \angle APS \\ &= 90^{\circ} - \angle SAP \\ &= 90^{\circ} - \angle BAC \\ &= 90^{\circ} - \dfrac{\angle BOC}{2} \\ &= 90^{\circ} - \dfrac{180^{\circ} - 2 \angle OBC}{2} \\ &= \angle OBC \end{align*}and the first part of the claim follows by Side-Angle-Side congruence. Then, note that \begin{align*} \angle DAP &= \angle DAC \\ &= \angle DBC \\ &= \angle BPC \\ &= \angle APD \end{align*}so $\Delta DAP$ is isoceles and $DA = DP$. We also have \begin{align*} \angle DPQ &= \angle BPS \\ &= 90^{\circ} - \angle PBS \\ &= 90^{\circ} - \angle DBA \\ &= 90^{\circ} - \dfrac{\angle AOD}{2} \\ &= 90^{\circ} - \dfrac{180^{\circ} - 2 \angle OAD}{2} \\ &= \angle OAD \end{align*}and the second part of the claim follows by Side-Angle-Side congruence. $\blacksquare$ Now, we finish the problem by angle chasing to show that $AECB$ is cyclic, which implies the result. First note that $EYQX$ is cyclic since $\angle EYQ = \angle EXQ = 90^{\circ}$. Indeed, we have \begin{align*} \angle BEY &= \angle XQC \\ &= \angle DQC \\ &= \angle DQP + \angle PQC \\ &= \angle DOA + \angle BOC \\ &= 2 (\angle ABD + \angle BAC) \\ &= 2 \angle BPC \\ &= 180^{\circ} - \angle ACB \end{align*}but $\angle BEY = 180^{\circ} - \angle AEB$, so $\angle ACB = \angle AEB$, as desired. $\blacksquare$

Clearly, $\angle CPQ=90^\circ-\angle A$. By $SAS$ congruence, if $O$ is the circumcenter, $\triangle CPQ\cong \triangle CBO$. Thus $PQ=QC$. Let $F$ be the point on $(ABC)$ such $DC=CF$. Clearly $OC$ is a line through two centres of circles, hence it bisects $\angle FCD$. But, clearly, $\measuredangle DOC=2\measuredangle DBC=\measuredangle ACB$. By isosceles $\triangle DOC$, $\measuredangle DCO=\frac{\measuredangle ACB}{2}$. By symmetry, $\measuredangle DCF=\measuredangle ACB$. Now consider the rotation by $\angle C$ degrees. Clearly, $O\rightarrow Q, F\rightarrow D, B\rightarrow P$. Hence $OF=OC$ implies $QD=QC$. Thus $PQ=QC=QD$ imples $Q$ is the circumcenter of $\triangle PCD$. Angle chasing, as right angles obviously imply cyclic, \[\measuredangle(AE,EB)=\measuredangle(CQ,DQ)=2\measuredangle(CP,PD)-180^\circ=180^\circ-180^\circ-\angle C=\measuredangle(AC,CB) \]We are done by Bowtie.

GrantStar wrote: Let $E$ be on $\omega$ such that $CB=CE$. Claim: $DPCO$ is cyclic with center $Q$. Proof. The circumradius condition and spiral similarity at $C$ gives $CQ=QO=QP$. Then, $\measuredangle CPD=2\measuredangle CBP =2\measuredangle CBD =\measuredangle COD$ as desired. $\blacksquare$ Now, note that $P$ is the incenter of $ABE$ by fact $5$. Thus \[180^{\circ} -\angle(BE,DQ)=\angle EBP+\angle DPQ=\angle PBA+\angle DPQ=\angle PBS+\angle SPB=90 ^{\circ}\]and \[180^{\circ} - \angle(AE,CQ)=\angle EAC+\angle QCA=\angle APS+\angle SAP=90^{\circ}\]concluding. I think this is wrong, $DPCO$ doesn't seem to be concyclic. Where did you get $CQ=QO$?

Oops wait you’re right I forgot my solution that I submitted. Once you get CQ=QP you can angle chase to identify Q as the circumcenter of BCD. I will change this later

Beautiful! Claim: $Q$ is the circumcenter of $\triangle DPC.$ Proof: There are two steps to proving this claim. The first is showing that the circumradius of $\triangle DPC$ is $r.$ This is easy enough: if $r'$ is the circumradius of $\triangle DPC,$ the Law of Sines implies that $$r' = \frac{PC}{2 \sin \angle PDC} = \frac{BC}{2 \sin \angle BDC} = r.$$The second step is showing that $PQ$ passes through the circumcenter of $\triangle PCD,$ which follows since $PQ \perp AB,$ $ABCD$ is cyclic, and the orthocenter and circumcenter are isogonal conjugates in a triangle. These conditions pinpoint $Q$ down to one place, namely the circumcenter of $\triangle DPC,$ as claimed. Now let $O$ be the circumcenter of $\triangle ABC.$ Because $\triangle DPC$ has the same radius as $\omega,$ we see that $O$ and $Q$ are reflections across $CD.$ Furthermore, since $CO = OD$ and $CQ = QD,$ we see that $OCQD$ is a rhombus and thus a parallelogram. Hence $AE \perp DO$ and $BE \perp CO.$ Therefore, we need to show that there exists a point $E'$ on $\omega$ such that $AD = DE'$ and $BC = CE'.$ This is equivalent to showing that $f(AD) + f(BC) = f(CD),$ where $f$ measures the length of an arc. Indeed, $$\angle ABD + \angle BAC = \angle BAP + \angle ABP = \angle BPC = \angle PBC = \angle DBC,$$and we are done. why did \overarc stop working

Different solution: Let $M$ be the arc midpoint of arc $BCA.$ $\textbf{Claim: }$ $CMDP$ is a parallelogram. $\emph{Proof: }$ We have $\angle MCA=\angle MBA=\tfrac{180^\circ-\angle C}{2}=\angle CPD,$ so $\overline{CM}\parallel\overline{PD}.$ Moreover, since $CMDB$ is an isosceles trapezoid, $MD=CB=CP.$ $\blacksquare$ $\textbf{Claim:}$ $OMQP$ is a parallelogram. $\emph{Proof:}$ Both $\overline{OM}$ and $\overline{PQ}$ are perpendicular to $\overline{AB},$ so $\overline{OM}\parallel\overline{PQ}.$ Moreover, $PQ=r=OM.$ $\blacksquare$ These two claims together imply that $CODQ$ is a parallelogram. Hence, $\angle CQD=\angle COD=180^\circ-\angle C,$ which (by a phantom point argument) suffices.

Let $O$ be the center of $(ABC)$. Redefine $E$ as the reflection of $P$ about $CD$ and $Q$ as the reflection of $O$ about $CD$. We show they satisfy the stated properties. Claim: $E\in (ABC)$ $\angle DEC=180^{\circ}-\angle BPC=180^{\circ}-\angle PBC$ Claim: $PQ=r$ Just reflect about $CD$. Claim: $PQ\perp AB$ $\angle BPQ-\angle ABP=180^{\circ}-\angle BDQ-\angle ABP=90^{\circ}-\angle ABP-\angle CAB+\angle DBC=90^{\circ}$ Claim: $AE\perp CQ$ Simply, $AX\perp OD\parallel CQ$. Claim: $BE\perp DQ$ By simple angle chase, $BX\perp OC\parallel DQ$.

Proposed by Mahdi Etesamifard, Iran

Define $E$ as the intersection between $\omega$ and the line perpendicular to $DQ$, and define $L$ as the intersection between $BE$ and $DQ$ . Define $K$ as the foot of the perpendicular from $CQ$ to $E$. It is sufficient to prove that $A$, $E$, and $K$ are collinear. $\angle CPQ = \angle APS = 90 - \angle SAP = 90 - \angle BAC = \angle OBC$, and since $PQ=r=OB$ and $BC=PC$, so $\Delta COB \sim \Delta CPQ$, in a spiral similarity centered at $C$. Thus $\Delta CPB \sim \Delta CQO$. Moreover, $\angle COD = 2 \angle CBD = 2 \angle CBP = 180 - \angle BCP = 180 - \angle QCO = 2\angle COQ$. Thus, $\angle QOC = \angle DOQ$, and since $DO=OC$, $DOQC$ is, thus, a parallelogram. We finish with an angle chase: $\angle BEK=\angle LEK = 180 - \angle LQK = 180 - \angle OCQ = 180- \angle BCP = 180- \angle BCA = 180 - \angle BEA$. Therefore, $K, E, A$ are collinear, which finishes up the problem.

Define $E$ as the point on $(ABC)$ with $\angle EAC = \angle BAC$. Because $PQ = BO, PC = BC, \angle OBC = 90 - \angle BAC =\angle APS = \angle QPC$, $QPC \cong OBC$. Additionally, $\angle PDC = \angle BAC = 2 \angle BOC = 2 \angle PQC$ gives $Q$ is the circumcenter of $PDC$. Lastly, $\angle ABE = \angle B -\angle A = 2 \angle ABP$. Therefore, $\angle BDQ + \angle DBE = 90 - \angle DCP + \angle ABD = 90$ which shows $EB \perp DQ$ and $\angle EAC + \angle QCA = \angle A + 90 - \angle A = 90$ which shows $AE \perp CQ$. $\square$

Let $R$ be the circumradius of $\triangle CPD$. By law of sines, we then get \[ 2R = \frac{CP}{\sin \angle PDC} = \frac{CB}{\sin \angle BDC} = 2r, \]so $R = r = PQ$. Moreover, $\angle QPD = \angle BPS = 90^\circ - \angle PBS = 90^\circ - \angle DCP$, so it follows that $Q$ is the circumcenter of $\triangle PDC$. Therefore, \[ \measuredangle AEB = \measuredangle CQD = 2\measuredangle CPD = \measuredangle CPB + \measuredangle PBC = \measuredangle ACB, \]so $E$ lies on $(ABC)$.

Let $O$ be center of $\omega$. We have $(PC, PQ) \equiv (PA, PS) \equiv \dfrac{\pi}{2} + (AC, AB) \equiv (BC, BO) \pmod \pi$. Combine with $PQ = OB = r$ and $CP = BC,$ we have $\triangle OBC \cong \triangle QPC$. Then $QP = QC = r$ and $(DP, DC) \equiv \dfrac{1}{2} (\overrightarrow{OB}, \overrightarrow{OC}) \equiv \dfrac{1}{2} (\overrightarrow{QP}, \overrightarrow{QC}) \pmod \pi$. So $D \in (Q, r)$. Hence $(EA, EB) \equiv (QC, QD) \equiv 2(PC, PD) \equiv 2(PC, PB) \equiv (CA, CB) \pmod \pi$ of $E \in \omega$

avisioner wrote: Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Lemma: $\angle ACB+\angle DQC=180$

Let $E'\equiv AE\cap CQ , T\equiv DQ\cap BC$ $\angle E'BC=\angle E'AC=90-\angle QCA=90-(\angle QCB-\angle ACB)=90-(\angle TQC+\angle QTC-\angle ACB)=90-\angle QTC\Rightarrow BE'\perp DQ \Rightarrow E'\equiv E_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

Let $BE \cap DQ = G$ and $O$ be the center of $\omega$. Note that $\angle OAD = 90^\circ - \angle ABD = \angle DPQ$. Also, $\angle DAP = \angle DBC = \angle APD$ so $AD = DP$. Therefore, $\triangle DAO \cong \triangle DPQ$ and so $\triangle DAP \cong \triangle DOG$. This gives $DQ = PQ$ and $\angle DQP = \angle AOD = 2\angle ABD = 2\angle PCD$, implying $Q$ is the center of $(PCD)$ and so $DQ = CQ$. Thus, we have $DOCQ$ being a rhombus. Lastly, note that $\angle AEB = 180^\circ - 2\angle DQO = 180^\circ - 2\angle BPC = \angle ACB$. Hence, $E$ lies on $(ABC)$.

My first G2 ! Let us introduce $O$ the center of the circumcircle. Notice that the angles $\angle APS, \angle SPB, \angle BPC$ can be chased easily. The key point to find the other angles is to notice that $BOC$ is similar to $PQC$ (one angle and two sides), and $PQD$ is similar to $AOD$ (here, we would want to use again the one angle/two sides condition, but for that we need $AD = PD$. However, remark that $\angle DAP = \angle DBC = \angle PBC = \angle BPC = \angle APD$, as wished. ). The similarity of these triangles give access to many angles, enough to calculate $\angle DBE$ and $\angle DAE$, and see that they are equal. Hence $A,D,E,B$ are concyclic which means that $E \in \omega \huge{\blacksquare}$.

Pretty easy. Funnily, our national Olympiad also had a problem which involved circles with equal circumradii so I was prepared for the weird condition here. Let $O$ denote the circumcenter of $\triangle ABC$, and let $R$ and $T$ be the feet of the perpendiculars from $B$ to $\overline{DQ}$ and from $A$ to $\overline{CQ}$. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ real xmin = 2.5, xmax = 20.5, ymin = -9., ymax = 6.; pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O = (9.,-2.); pair A = (5.8,3.56); pair B = (4.,-6.); pair C = (14.00935,-6.00748); pair P = (7.49139,1.58879); pair S = (5.49946,1.96383); pair D = (8.79012,4.41167); pair Q = (13.79180,0.40034); pair T = (13.67545,3.82738); pair R = (10.95662,2.67414); pair E = (11.83131,3.76477); import graph; size(10cm); pen fueaev = rgb(0.95686,0.91764,0.89803); pen zzttqq = rgb(0.6,0.2,0.); pen fsfsff = rgb(0.94901,0.94901,1.); draw(A--B--C--cycle); filldraw(circle(Q, 6.41152),white+0.1*pri, pri); draw(circle(O, 6.41510), linewidth(0.6)); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw((xmin, -0.18828*xmin + 2.99930)--(xmax, -0.18828*xmax + 2.99930), linewidth(0.6) + linetype("2 2")); draw(B--D, linewidth(0.6)); draw(C--T, linewidth(0.6) + dotted); draw(T--A, linewidth(0.6) + dotted); draw(B--E, linewidth(0.6) + dotted); dot("$O$", O, dir((8.000, 20.000))); dot("$A$", A, dir((8.000, 20.000))); dot("$B$", B, dir(210)); dot("$C$", C, dir(320)); dot("$P$", P, dir(30)); dot("$S$", S, dir(40)); dot("$D$", D, dir(100)); dot("$Q$", Q, dir((8.820, 15.966))); dot("$T$", T, dir((8.455, 15.262))); dot("$R$", R, dir(0)); dot("$E$", E, dir((8.869, 15.523))); [/asy][/asy] Due to the right angles it is immediately clear that $ERQT$ is cyclic. Thus, it suffices to show the following key claim. Claim : $Q$ is the circumcenter of $\triangle DPC$. Proof : Let $\Gamma$ denote the circumcenter of $\triangle DPC$ and let $Q'$ be its circumcenter. First note that, \[2\measuredangle Q'PC = \measuredangle PQ'C = 2\measuredangle PDC = 2\measuredangle BDC = 2\measuredangle BAC\]so, $\measuredangle Q'PC = 90 + \measuredangle BAC$ from which it is clear that $Q'P \perp AB$. So, $Q'$ lies on the line $\overline{SP}$. Further, note that \[\measuredangle DPC = \measuredangle BPC = \measuredangle CBP = \measuredangle CBD\]so the segment $DC$ must subtend equal angles at the centers of $\omega$ and $\Gamma$. This then implies that the circles $\omega$ and $\Gamma$ are congruent, and in particular they have the same circumradius. Thus, $Q'P = r = QP$, which implies that $Q'\equiv Q$, implying the claim. Note that $DO=OC = r = QC=QD$ so it follows that $DOCQ$ is a rhombus. Now, we are left with a simple angle chase, \[\measuredangle BEA = \measuredangle REA = \measuredangle RQT = \measuredangle DQT = \measuredangle COD = 2\measuredangle CBD = \measuredangle BCP = \measuredangle BCA\]which implies that $E$ indeed lies on $\omega$ as desired.

Let $K$ be the foot of $A$ to $QC$. Let $AK$ intersects with the circumcircle of $\triangle{ABC}$ at $E$ at the second time. And let $O$ be the center of $(ABC)$. We are going to prove $BE \perp DQ$ to finish the problem. (So our $E$ would be same as the $E$ in the problem.) Claim1: $\triangle BOC \cong \triangle PQC$. Proof: Easily conclude that $\angle QPC = 90-\angle A = \angle OBC$. $PQ = OB$ and $PC = BC$. $\square$ Claim 2: $OC \perp EB$. Proof: By Claim 1, $\angle PQC = 2\angle A$. Also $(ASQK)$, hence $\angle SAK = 2\angle A \implies \angle PAE = \angle A$ That means $BC = CE$. $\square$ Now, we know that $DO = CQ$. If we prove $DO || QC$ $\implies DOQC$ is paralellogram $\implies DQ || OC \implies DQ \perp EB$ and we would be done. Finishing Claim: $DO || QC$. Proof: $DO || QC \iff DO \perp AE \iff AD = DE$. So we are going to prove the arcs $AD$ and $DE$ are equal. $\text{arc}(AE) = 360 - 4\angle A - 2\angle C = 2(\angle B - \angle A)$ $\text{arc}(AD) = 2\angle ABD = 2(\angle B - (90 - \frac{\angle C}{2})) = \angle B - \angle A$ $\text{arc}(AE) = 2.\text{arc}(AD)$ $\text{arc}(AD) = \text{arc}(DE)$ $\blacksquare$

Solved with SilverBlaze_SY. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (24.81281,-5.45037); pair B = (-6.39088,-5.19529); pair C = (0.26530,21.43292); pair P = (18.77311,1.16403); pair Q = (18.92617,19.88811); pair O = (9.29559,5.02962); pair D = (27.95646,3.48480); pair F = (26.85608,19.23164); pair S = (18.71945,-5.40056); pair E = (26.32491,12.81529); pair X = (25.56886,3.68246); pair Y = (23.63508,11.33449); import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; pen ffxfqq = rgb(1.,0.49803,0.); draw((24.64484,9.50029)--(26.47904,10.51004)--(25.46928,12.34425)--Y--cycle, linewidth(0.6) + blue); draw((27.65551,3.50972)--(27.82825,5.59636)--(25.74160,5.76910)--X--cycle, linewidth(0.6) + blue); draw((15.86788,11.32006)--(17.00668,13.07706)--(15.24968,14.21586)--(14.11088,12.45886)--cycle, linewidth(0.6) + blue); draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle(O, 18.72470), linewidth(0.6)); draw(C--O, linewidth(0.6)); draw(O--D, linewidth(0.6) + ffxfqq); draw((20.10150,4.13506)--(19.29048,3.31085), linewidth(0.6) + ffxfqq); draw((20.10150,4.13506)--(19.43705,5.08142), linewidth(0.6) + ffxfqq); draw((18.62602,4.25721)--(17.81500,3.43300), linewidth(0.6) + ffxfqq); draw((18.62602,4.25721)--(17.96157,5.20357), linewidth(0.6) + ffxfqq); draw(D--Q, linewidth(0.6)); draw(A--F, linewidth(0.6)); draw(Q--S, linewidth(0.6)); draw(Q--F, linewidth(0.6) + ffxfqq); draw(B--E, linewidth(0.6)); draw(circle((27.14069,8.15005), 4.73602), linewidth(0.6) + linetype("4 4") + red); draw(C--Q, linewidth(0.6) + ffxfqq); draw((11.07121,20.53837)--(10.26018,19.71415), linewidth(0.6) + ffxfqq); draw((11.07121,20.53837)--(10.40676,21.48473), linewidth(0.6) + ffxfqq); draw((9.59573,20.66051)--(8.78471,19.83630), linewidth(0.6) + ffxfqq); draw((9.59573,20.66051)--(8.93128,21.60687), linewidth(0.6) + ffxfqq); draw(C--D, linewidth(0.6)); draw(O--Q, linewidth(0.6)); dot("$A$", A, SE); dot("$B$", B, SW); dot("$C$", C, NW); dot("$P$", P, W); dot("$Q$", Q, NW); dot("$O$", O, W); dot("$D$", D, SE); dot("$F$", F, NE); dot("$S$", S, dir(270)); dot("$E$", E, NE); dot("$X$", X, SW); dot("$Y$", Y, NW); [/asy][/asy] Claim: $\triangle QCP \cong \triangle OCB$. Proof. Note that $QP=r=OB$ and $PC=BC$ and $\angle QPC = \angle SPA=90^{\circ}-\angle A=\angle OBC$. $\blacksquare$ This gives us that $QC=CO=CD$. Claim: $Q$ is the reflection of $O$ over $CD$. Proof. Note that we already have $CQ=CO$. Now note that, \[ \angle OCQ=\angle OCA+\angle PCQ=90^{\circ}-\angle B+\angle BCO = 90^{\circ}-\angle B+90^{\circ}-\angle A=C .\]Now, \[ \angle OCD=\angle OCA+\angle ACD=90^{\circ}-\angle B + \angle ABD = 90^{\circ}-\angle B+(\angle B-\angle PBC) =90^{\circ}-\angle PBC=90^{\circ}- (90^{\circ}-\frac{\angle C}{2}) =\frac{\angle C}{2} .\]Now since $CO = CQ$, this gives us that $Q$ is indeed the reflection $O$ over $CD$. $\blacksquare$ Note that from this claim, we get $QD=OD$. It also gives us that $OQ\perp CD$. Combining all these information together, we get that $QCOD$ is a rhombus. This gives us that $CQ\parallel OD$ which further gives $OD\perp AE$. Let $X=AE\cap OD$ and $Y=BE\cap DQ$. Note that we get, $\angle DXE=\angle DYE=90^{\circ}\implies DXYE$ is cyclic. Now to finish, we have, \[ \measuredangle BEA=\measuredangle YEX=\measuredangle YDX =\measuredangle QDO=\measuredangle OCQ=\measuredangle C =\measuredangle BCA \]which gives us that $E$ indeed lies on $\odot(ABC)$ and we are done.

The problem is asking to prove that $QC=QD=QP$. Let $O$ be the circumcenter of $ABC$. Note that, $QPC$ is congruent to $OBC$, which implies $QP=QC=r$. Now let $AE$ intersects $CQ$ at $X$. Note that, $AXQS$ is cyclic. Now from this, and the cyclic quad $ABCD$ we can show, $\angle DPC =\frac{1}{2}\angle PQC$ proving our claim.

Note that the circumradius of $\triangle CPD$ is $r$ as: $$\tfrac{CP}{2 \sin(\angle BDC)} = r.$$Also notice that: $$\angle CPQ = \angle APS = 90^\circ - A =90^\circ - \angle PDC$$which implies $Q$ is the circumcenter of $\triangle PDC$. Notice that: $$\angle BPC = 90^\circ - \tfrac{C}{2} \implies \angle DQC = 2 \angle BPC = 180^\circ - C \implies \angle AEB = \angle ACB$$and we are done.

This is (thankfully) a lot easier than G1 ... the conditions of the problem basically solve themselves. Claim: [Main Claim] $Q$ is the circumcenter of triangle $PCD$. Proof: Note that by the Law of Sines, $2R_{PCD} = \frac{CD}{\sin \angle CPD} = \frac{CD}{\sin \angle BCD} = 2R_{ABC}$. Hence it suffices to show that the circumcenter $Q'$ of triangle $CPD$ lies on $\overline{PS}$. But this is clear because \[\angle CPQ' = 90^\circ - \angle CPD = 90^\circ - \angle A = \angle CPQ. \ \blacksquare\]So \[\angle BEA = 180^\circ - \angle CQD = 180^\circ - 2\angle CPB = \angle BCA\]which finishes.

Easier then G1 Also the condition seemed weird but it’s not too difficult to cope with avisioner wrote: Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$. First, by angle chasing it’s easy to see that $DP=DB$ so the condition of $C,D$ are in fact symmetrical wrt $A,B$ Also notice that $\measuredangle AEB=\measuredangle CQD$ so we only need to prove that $\measuredangle ACB=\measuredangle CQD$ (We can remove $E$ from our figure now) Now let $\measuredangle CBA=\beta$ and $\measuredangle DAB=\delta$ Applying law of sin on $\triangle CAP$ we know$CA=CP=2r\sin\beta$ Since $\measuredangle CPQ=\measuredangle BPS=90^{\circ}-\beta$, applying law of cos on $\triangle CPQ$ we know $CQ=r$ (Notice that $\cos(90^{\circ}-\theta=\sin\theta$) Then $\measuredangle PAC=\measuredangle CPA=\beta+\delta\implies \measuredangle BCA=180^{\circ}-2\beta-2\delta$ So $\measuredangle ACB=2\beta+2\delta=\measuredangle CQD$, Done

hi guys idk if i did directed angle for this right pls check if my directed angles are right........

deduck wrote: hi guys idk if i did directed angle for this right pls check if my directed angles are right........

You can't halve a directed angle.

im dumb $ $

Ok its not that deep there isnt even config issues

Nice question

A different solution with Power of Point and some laws of cossines and sines: Let $O$ be the circumcenter of $\omega$ and $\Gamma$ be the circle with center $Q$ and radius $r$. $\boxed{\text{Lemma:}}$ $Q$ is the center of $(CPD)$ Prove: We´ll take two power of points: one from $A$ in $\Gamma$ and other in $B$ in $\Gamma$ too$$C \in \Gamma$$$$\iff \text{Pot}_\Gamma A = AQ^2 - r^2=AP\cdot AC=AP^2+AP\cdot PC$$$$\iff AQ^2 -QP^2 -PA^2=AP\cdot PC$$But, notice that $\angle APS = 90 -\angle A \Rightarrow \angle APQ=90 +\angle A$, so$$AQ^2 -QP^2 -PA^2=-2PA\cdot QP\cdot \text{cos}(90+\angle A)=2rAP\cdot \text{sin} \angle A (\text{Law of cossines at the triangle APQ})$$$$=AP\cdot BC=AP\cdot PC \text{(Law of sines at the triangle ABC)}$$$$\Rightarrow C \in \Gamma$$Equivalently$$D \in \Gamma$$$$\iff \text{Pot}_\Gamma B=BQ^2 -QP^2=BP\cdot BD=BP^2 +BP\cdot PD$$$$\iff BQ^2 -QP^2 -PB^2=BP\cdot PD$$But, note that $\angle BPS=90 \angle SBP \Rightarrow \angle BPQ =90+ \angle SBP$, therefore$$BQ^2 -QP^2 -PB^2=-2QP\cdot PB\cdot \text{cos}(90+\angle SBP)=2rPB\cdot \text{sin}\angle SBP \text{(Law of cossines at the triangle BPQ)}$$$$PB\cdot AD=PB\cdot PD \text{(Law of sines at the triangle ACD)}$$$$\Rightarrow D \in \Gamma \blacksquare$$Finally, we note that $CQDO$ is paralelogram $\Rightarrow AC//DQ \Rightarrow BE \bot AC$ and then $\angle QPD=\angle QDP \Rightarrow \angle EBP=\angle PBA \Rightarrow \angle SPA=\angle SPC= 90 - \angle EBC \Rightarrow \angle A= \angle EBC \Rightarrow ABC=90 \Rightarrow E \in \omega$ $\blacksquare$