Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$. @above it is

Too easy for P2! Sketch: construct antipode of A to get rid of L and find concyclics kills it. Construct midpoint of arc $BC$ not containing ${A}$ and $A-$ antipode to be $N$ and $A'$. Let $AN$ cut $SB$ at $Y$ and $DA$ at $X$. Let ${AE}$ cut $SA'$ at $T$. $\angle L=\angle EBC=\angle BPD$ so $LPDB$ concyclic. Note that $A'S//AN$ and by Reim $PYXB$ concyclic. Since $\angle TEA'=90$ and $SE=SA'$, $A'S=TS$ so by homothety at ${D}$, $AY=YX$. Combining $\angle APX=90$, $PY=XY$, $\angle YPD= \angle YXD=\angle PBY$, so $PY$ is tangent to $\omega$.

Let $A',S'$ be the antipodes to $A,S$ on $(ABC)$. Defining $X = AS' \cap BS$, we want to show $PX$ is tangent to $(BPD)$ or $XD\cdot XB = XP^2$. First, note by Reim on $(LPDB), (PBEA')$ that since $LD||EA'$ we must have $P,D,A'$ are collinear. Next, note that $\angle DBA = \angle S'AE = \angle XAD$, so $XA$ is tangent to $(BAD)$ and thus $XD\cdot XB = XA^2$, so to finish we only need to show $XA=XP$. The locus of such $X$ is the $A$-midline of $\triangle AA'P$, so if we let $Y=AX\cap A'D$ we just need to show $X$ is the midpoint of $AY$. But if we let $Q = AE \cap SA'$ then from angle-chasing $QAA'$ is isosceles, so $AS\perp A'S$ implies $S$ is the midpoint of $QA'$. Taking homothety centered at $D$ therefore shows $X$ is the midpoint of $AY$, as desired $\blacksquare$ Motivation: The motivation for this solution is to first eliminate $L$ using Reim to get a nicer characterization for $P$. After this, dealing with $XP$ and $(BPD)$ directly is still not nice, but we bypass this by noticing $XA$ is tangent to $(BAD)$.

$\text{No more cyclic quadrilaterals}$

Let $A'$ denote the $A$ antipode. We have $\angle BPD = \angle EBC = \angle BCA' =\angle BPA'$, so $P,D,A'$ are collinear. Let $AI$ hit $PA'$ at $H$ and $BD$ at $F$. Let $AE$ and $SA'$ intersect at $J$. Then Note that, $FH \equiv AI\parallel SA'$, so that $PFHB$ is cyclic by Reims. Now note that $JA'E$ is a right triangle with $SA'=SE$ (as $A'E \parallel BC$), so $SA'=SJ$. From homothety at $D$ we get $FA=FH$. Now as $\triangle APH$ is right, we have $FP = FH$. So we get $\angle FPA'= \angle FHP= \angle SA'P = \angle SBP = \angle DBP$. So $FP$ is tangent to $\omega$, as desired. $\square$.

Let $M$ be the midpoint of minor arc $BC$, and let $\overline{PM}$ intersect $\overline{AE}$ at $K$. Then $\overline{SM} \parallel \overline{DK}$, so $K$ lies on $\omega$ by Reim's. Now redefine $F$ to be the intersection of $\overline{AM}$ and $\overline{BS}$. We will show that $\overline{FP}$ is tangent to $\omega$. Since $\angle KBE = \angle LDE = 90^\circ$, we have \[ \measuredangle BFA = \measuredangle SFM = 90^\circ + \measuredangle BMA = 90^\circ + \measuredangle BEA = \measuredangle BKA, \]so $AFKB$ is cyclic. Moreover, \[ \measuredangle KBF = \measuredangle KAF = \measuredangle EAM = \measuredangle SMA = \measuredangle SBA = \measuredangle FBA,\]so $FA = FK$. Finally, $\measuredangle APK = \measuredangle APM = \measuredangle ASM = 90^\circ - \measuredangle SMA = 90^\circ - \measuredangle KAF$, so $F$ is the circumcenter of $\triangle PAK$ and hence $FP^2 = FA^2 = FD \cdot FB$. Hence $\overline{FP}$ is tangent to $\omega$ as desired.

Without antipodes: Let $\omega$ meet $AE$ at $T\neq D$. The circumcenter of $\triangle APT$ is $Q$. Connect $PT,BT,TQ,AP$. then we have $\angle BPD=\angle BLD=90^{\circ}-\angle BED=\angle APB-90^{\circ},$ so $AP \perp PD$. hence $\angle QPT=90^{\circ}-\angle PAT=\angle PDA=\angle PBT,$which means $PQ$ is tangent to $\omega$. by $\angle AQT=360^{\circ}-2\angle APT=180^{\circ}-2\angle DBT=180^{\circ}-\angle ABT$ we know that $A,Q,T,B$ is concyclic. since $AQ=QT\Rightarrow Q$ lies on $BS$. we just need to prove $Q$ lies on the bisector. $\Leftrightarrow \angle QAS=90^{\circ},$ which is obvious since $\angle AQB=\angle ATB,\angle TBE=90^{\circ}$.

We consider the inversion $\Phi$ wrt $D$ that preserves $\Omega$. Under $\Phi$ we have $A\leftrightarrow E$ and $B\leftrightarrow S$. It's natural to find$$\Phi(L)=\Phi(BE)\cap\Phi(DL)=\odot ADS\cap DL$$which indicates $\angle AS~\Phi(L)=\angle ADL=90^{\circ}$, i.e., $\Phi(\omega)=\operatorname{line}S\Phi(L)=SA'$ where $A'$ is the antipodal point of $A$. From that, $\Phi(P)=\Phi(\omega)\cap\Phi(\Omega)=A'$, so $PDA'$ collinear. Let the antipodal point of $S$ be $T$. $\angle APD=\angle SBT=90^{\circ}\Longrightarrow Q:=AP\cap BT\in\omega$ But also $AS=ET\Longrightarrow \overset{\LARGE \frown}{AP}+\overset{\LARGE \frown}{ET}=\overset{\LARGE \frown}{PS}\Longrightarrow R:=AE\cap PT\in\omega$ Now we apply Pascal's theorem to $(PPQBDR)$ to get the desired result.

Let $A'$ be the antipode of $A$. We have $\angle BPD = \angle EBC = \angle BCA' =\angle BPA'$, so $P,D,A'$ are collinear. Let $AE$ intersect $SA'$ at $I$ and let $AF$ intersect $PA'$ at $H$. Since $SE=SA'$ and $\triangle IEA'$ is a right triangle, we get that $SA' = SE = SA$. Note that $AH \equiv AF || SA'$ so by Thales we get that $\frac{AF}{IS} = \frac{DF}{DS} = \frac{HF}{SA'}$ and so $AF = HF$. Note that $\triangle APH$ is a right triangle, so $PF = AF = HF$. Now note that $\angle DAF = \angle FAA' = \angle AA'S = \angle ABS = \angle ABF$ so $\triangle ADF \sim \triangle ABF$ and so $FP^2 = AF^2 = FD.FB$. So $FP$ is tangent to $\omega$.

Let $N$ be the midpoint of the other arc $BC$ and $K= AD \cap PN$ Then $\angle BDK=\angle BDE =90-\angle DBC=90-\angle SBC=\angle BSN=\angle A/2=\angle BAN=\angle BPN=\angle BPK$ and so $K$ lies on $\omega$. Let now $H= AN \cap BS$. We need to show that $HP$ is tangent to $\omega$. It is obvious that $AE$, $SN$ are parallel and so $\angle DAH=\angle DAN=\angle ANS=\angle ABS=\angle ABD$, so $AH$ is tangent to $ABD$, i.e. $AH^2=HB*HD$, so it remains to prove $AH=HP$. Now we have from the cyclic quadrilaterals and the parallel lines: $\angle AKB=\angle DKB=$ $=180-\angle DLB$ $=180-\angle EBC$ $=180-\angle EAC$ $=180-\angle (90-C)$ $=90+\angle C$ $=90+\angle ANB$ $=\angle HBN+\angle HNB$ $=\angle AHB$ and hence $BKHA$ is cyclic, which gives that $\angle AKH=\angle ABH=\angle ABD=\angle DAH=\angle KAH$ and so $HA=HK$. Finally: $\angle AHK=\angle AHB+\angle BHK=\angle HBN+\angle HNB+\angle BAK=90+\angle C+\angle BAK=90+\angle C+90-\angle B=2\angle C+\angle A=2(\angle C+\angle A/2)=2(180-(\angle B+\angle A/2))=2(180-\angle ABN)=2(180-\angle APN)= 2(180-\angle APK)$ , from where we conclude that $H$ is the circumcenter of $APK$ and $AH=HP$, as needed.

This problem is proposed by Tiago Mourão, Nuno Arala, Portugal.

Mediocre solution because I'm washed up. Let $T$ be the midpoint of arc $BC$ of $\Omega$ not containing $T$, let $R$ be the second intersection of $\omega$ with $AE$, and let $X = AT \cap BS$. [asy][asy] import geometry; unitsize(5cm); defaultpen(fontsize(10pt)); point A = dir(120); point B = dir(210); point C = dir(330); point I = incenter(A, B, C); point O = circumcenter(A, B, C); point T = circumcenter(I, B, C); point S = 2 * O - T; line l = bisector(S, T); point E = 2 * (projection(l) * A) - A; point D = extension(A, E, B, S); point L = intersectionpoint(line(E, B), parallel(D, line(B,C))); point R = 2 * circumcenter(B, D, L) - L; point P = extension(L, S, T, R); point X = extension(A, T, B, S); draw(A--B--C--cycle, lightgray); draw(S--L, gray); draw(P--T, dashed); draw(L--E--A); draw(T--S); draw(A--T, lightgray); draw(B--S); draw(A--P, gray); draw(A--B--R--X--cycle, gray); draw(circumcircle(A, B, C), blue); draw(circumcircle(D, B, L), red); dot(A); dot(B); dot(C); dot(S); dot(T); dot(E); dot(D); dot(L); dot(P); dot(R); dot(X); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, W); label("$E$", E, (0,-1)); label("$L$", L, W); label("$X$", X, (1, 0)); label("$R$", R, (1, 0)); label("$S$", S, N); label("$T$", T, (0,-1)); label("$P$", P, N); [/asy][/asy] Claim 1. $L$, $P$ and $S$ are collinear. Proof. Notice that the tangent to $\Omega$ at $S$ is parallel to $BC$ which is parallel to $DL$. Since $S, D, B$ are collinear, the converse of Reim's Theorem gives the result. Claim 2. $T$, $R$ and $P$ are collinear. Proof. Since $ST$ is a diameter of $\Omega$ we have $\angle TPS = 90^{\circ}$. On the other hand \[\angle RPS = 180^{\circ} - \angle RPL = 180^{\circ} - \angle RDL = 180^{\circ} - 90^{\circ}\] And this $\angle RPS = \angle TPS$, giving the desired collinearity. To finish, angle chase to find $\angle ABX = \angle XBR = \frac{\angle B - \angle C}{2} = \angle RAX$, so $AXRB$ is cyclic with $XA = XR$ and \[\angle XRA = \angle XBA = \angle RBX = \angle RBD\] And therefore $XR$ is tangent to $\omega$. It now suffices to show that $XR = XP$. By angle chase we have $\measuredangle RXA = 2\measuredangle RPA$, and since $XA = XR$ this implies that $X$ is the circumcenter of $\triangle APR$, finishing the problem.

[asy][asy] size(9cm); import olympiad; import geometry; pair A = dir(119); pair A_prime = -A; pair E = reflect((0,0), dir(90)) * A_prime; pair S = dir(90); pair D = 0.6*E + 0.4*A; pair B = 2 * foot((0,0), D, S) - S; pair P = 2 * foot((0,0), A_prime, D) - A_prime; pair Q = reflect((0,0), (P+A_prime)/2) * S; pair M = -S; pair X = extension(E,Q,S,D); pair Y = extension(P,Q,A,M); pair L = extension(D+E-A_prime, D, B,E); fill(X--M--Q--cycle, paleblue); fill(A--P--S--cycle, paleblue); draw(unitcircle, linewidth(1)); draw(S--X, red+linewidth(1.2)); draw(P--A_prime, linewidth(0.8)); draw(circle(B,D,P), linewidth(0.8)); draw(P--Q, linewidth(0.8)); draw(X--M--Q--cycle, blue+linewidth(0.8)); draw(A--P--S--cycle, blue+linewidth(0.8)); draw(A--M, linewidth(0.8)); draw(A--E--A_prime, gray+linewidth(0.7)); draw(S--M, gray+linewidth(0.7)); draw(D--L, gray+linewidth(0.7)); draw(M--A_prime); dot("$A$", A, dir(118)); dot("$A'$", A_prime, dir(-62)); dot("$E$", E, dir(-84)); dot("$S$", S, dir(89)); dot("$D$", D, dir(16)); dot("$B$", B, 1.5*dir(-4)); dot("$P$", P, dir(133)); dot("$Q$", Q, dir(17)); dot("$M$", M, dir(-82)); dot("$X$", X, dir(-139)); dot("$Y$", Y, dir(143)); dot("$L$", L, dir(165)); [/asy][/asy] Let $A'$ be the antipode of $A$ in $\Omega$. Moreover, let the tangent to $\omega$ at $P$ meet $\Omega$ again at $Q$. Let $M$ be the other midpoint of arc $BC$. Let $X=QE\cap MA'$, and $Y=PQ\cap AM$. We finish the problem in three steps. Since $\angle BPD = \angle BLD = 90^\circ - \angle C = \angle BPA'$, we get that $P,D,A'$ are collinear. Pascal on $QPA'MAE$ gives $PQ\cap AM=Y$, $PA'\cap AE = D$, and $A'M\cap EQ = X$ are collinear. We have $\triangle QMX$ and $\triangle PAS$ are homothetic center $Y$. This is just angle chasing. To spell out details, we have Reim on $PPQ$ and $BDS$ gives $SQ\parallel PD$, implying that $A'Q=SP$, or $\angle PAS=\angle QMX$, $EM=AS$, so $\angle APS = MQX$, and clearly, $MX\parallel AS$. Therefore, $S,Y,X$ are collinear. Hence, $S, Y, D, X$ are collinear. $B$ clearly lies on this line, done.

Let $K$ be the mid-point of arc $CB$ which is not containing $A$,suppose $AK\cap BS=T$,$PK\cap AE=E$ Note that $\angle BDE=\dfrac{1}{2}\angle A=\angle BPK$,so $P,B,Z,D$are cyclic Consider that $\angle ABD=\angle B+\dfrac{1}{2}\angle A-90^{\circ}=90^{\circ}-\dfrac{1}{2}\angle A-\angle C=\angle DBZ=\angle EAK$ Which means that $A,B,Z,T$ are cyclic,and $TA=TC$ Notice $\angle ZAT=\angle TBZ=\angle ABT$,so $TA^{2}=TB\cdot TD$ Since $2\angle APK=2\angle B+\angle A=360^{\circ}-\angle ATZ$ Hence $TA=TZ=TP$ So $TP^{2}=TB\cdot TD$ Which means that $TP$ is tangent to $\omega$

Let \(K\) denote the midpoint of the other arc \(BC\), let \(A'\) be the antipode of \(A\), and let the tangent at \(P\) to \(\omega\) intersect \(\Omega\) at \(Y\). The goal is to show \(\overline{DS}\), \(\overline{AK}\), \(\overline{PY}\) concur. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=lightred; pen sec=orange; pen tri=heavyblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,SS,K,EE,D,Ap,P,L,Y,X; A=dir(120); B=dir(210); C=dir(330); SS=dir(90); K=-SS; EE=-B*C/A; D=extension(A,EE,B,SS); Ap=-A; P=2*foot(origin,Ap,D)-Ap; L=2*foot(circumcenter(B,D,P),D,D+B-C)-D; Y=Ap*P/SS; filldraw(circumcircle(B,D,L),sfil,sec); draw(B--SS/*--L*/,sec+linewidth(.4)); draw(Ap--P,sec+linewidth(.4)); draw(D--L--EE,sec+linewidth(.4)); draw(A--EE,pri); filldraw(unitcircle,fil,pri); filldraw(A--B--C--cycle,fil,pri+linewidth(.6)); filldraw(D--A--P--cycle,tfil,tri+linewidth(.6)); filldraw(SS--K--Y--cycle,tfil,tri+linewidth(.6)); dot("\(A\)",A,A); dot("\(B\)",B,dir(240)); dot("\(C\)",C,C); dot("\(E\)",EE,EE); dot("\(S\)",SS,SS); dot("\(K\)",K,K); dot("\(L\)",L,W); dot("\(A'\)",Ap,Ap); dot("\(P\)",P,dir(120)); dot("\(D\)",D,dir(10)); dot("\(Y\)",Y,Y); [/asy][/asy] Note that By converse Reim's theorem on \(\Omega\) and \(\omega\), since \(\overline{LD}\parallel\overline{EA'}\) we have \(P\), \(D\), \(A'\) collinear. By Reim's theorem on \(\Omega\) and \(\omega\), we have \(\overline{SY}\parallel\overline{PD}\). But also \(\overline{AD}\parallel\overline{SK}\) and \(\measuredangle APD=\measuredangle APA'=90^\circ=\measuredangle KYS\), so \(\triangle DAP\) and \(\triangle SKY\) are homothetic. This implies the desired concurrence at the center of homothety.

Great problem! Here is my solution, which doesn’t seem to have been posted above.

Suppose that the internal angle bisector of $\angle{BAC}$ intersects $BS$ at $J;$ $AD$ intersects $(BLD)$ again at $Q$. Note that $D, L, B, Q$ lie on a circle, so $(BE, BQ) \equiv (DL, DQ) \equiv \dfrac{\pi}{2} \pmod \pi$. Hence, we have $$(JA, JB) \equiv (AJ, AS) + (SA, SB) \equiv \dfrac{\pi}{2} + (EA, EB) \equiv (BE, BQ) + (EA, EB) \equiv (QA, QB) \pmod \pi$$Then $A, J, Q, B$ lie on a circle. We also have $$(BQ, BD) \equiv (BE, BS) + (BQ, BE) \equiv (BC, BS) + (BE, BC) + \dfrac{\pi}{2} \equiv (CS, CB) + (AE, AC) + \dfrac{\pi}{2}$$$$\equiv (CS, CB) + (CB, CA) \equiv (CS, CA) \equiv (BS, BA) \pmod \pi$$Therefore, $BS$ is internal angle bisector of $\angle{ABQ},$ then $JA = JQ$. Since $$(LD, LP) \equiv (BD, BP) \equiv (CS, CP) \equiv (CS, CB) + (CB, CP) \equiv (BC, BS) + (SB, SP) \equiv (BC, PS) \equiv (LD, LS) \pmod \pi$$we have $L, P, S$ are collinear. Hence $$(PQ, PA) \equiv (PQ, PD) + (PD, PS) + (PS, PA) \equiv (LQ, LD) + (QD, QL) + (PS, PA) \equiv \dfrac{\pi}{2} + (PS, PA)$$$$\equiv \dfrac{1}{2}(2 \pi - \pi + (\overrightarrow{BQ}, \overrightarrow{BA})) \equiv \dfrac{1}{2}(2 \pi - (\overrightarrow{JQ}, \overrightarrow{JA})) \pmod \pi$$So $P \in (J, JA)$. But $(AD, AJ) \equiv (BQ, BJ) \equiv (BJ, BA) \pmod \pi$ then $\triangle ADJ \stackrel{+}{\sim} \triangle BAJ$. Hence $\dfrac{JA}{JD} = \dfrac{JB}{JA}$ or $JA^2 = JD \cdot JB$. Therefore, $JP^2 = JD \cdot JB$ or $JP$ tangents $(DLB)$ at $P$

1. Let $O$ be the circumcenter of $\Omega$, $SQ$ is the diameter of $(O)$. $AQ$ meet $BD$ at $F$, $AO$ meet $(O)$ at $G$, $DG$ cut $AQ$ at $R$. 2. We have: $(PB,PD) \equiv (LB,LD) \equiv (BE,BC) \equiv (AE,AC) \equiv (AB,AG) \equiv (PB,PG) \pmod \pi$ Then $P$ lie on $DG$. And $(DF,DA) \equiv (DB,DE) \equiv (SB,SQ) \equiv (AB,AQ) \equiv (AB,AF) \pmod \pi$ Give us $FA^2=FD \cdot FB$ 3. Let $(O)$ be the unit circle, $A(a^2),B(b^2),C(c^2)$ such that $s=bc$ Then $e=\frac{-b^2c^2}{a^2}$, $q=-bc$,$g=-a^2$ We have $$d=\frac{a^2e(b^2+s)-b^2s(a^2+e)}{a^2e-b^2s}$$so $$d=\frac{a^2b^2c+a^2bc^2+a^4b-b^3c^2}{a^2b+a^2c}$$We also have: $$f=\frac{a^2q(b^2+s)-b^2s(a^2+q)}{a^2q-b^2s}$$Then $$f=\frac{2a^2b^2+a^2bc-b^3c}{a^2+b^2}$$Now note that: $$\frac{f-o}{d-g}=\frac{a^2b(b+c)}{(a^2+bc)(a^2+b^2)} \in \mathbb{R}$$so that $OF \parallel DG$ 4. From 3 we have $F$ is the midpoint of $AR$, and note that $\triangle APR$ is a right triangle at $P$ so that $FA=FP$, and from 2 we have $FP^2=FB\cdot FD$, hence that $FP$ is a tangent of $(BLD)$, complete the solution.

nice problem

Revenge after nearly a year! (Took a hint to construct the antipode though.) Let $M$ be the midpoint of arc $BC$ not containing $A,$ and let $X = BS \cap AM.$ We wish to show that $PX$ is tangent to $\omega.$ Let $G$ be the second intersection of $\omega$ with $AE,$ and let $Z$ be the point such that $ABCZ$ is an isosceles trapezoid. Note that $E$ and $Z$ are diametrically opposite each other. Claim 1: $B,G,Z$ are collinear. Proof: Follows from $\measuredangle EBG = \measuredangle LBG = \measuredangle LDG = 90^\circ$ and $DL \parallel BC.$ Claim 2: $ABGX$ is cyclic. Proof: We have $\measuredangle GAX = \measuredangle EAM = \measuredangle ZBS = \measuredangle GBX$ since $EM = SZ.$ Claim 3: $XA = XG$ and $XG$ is tangent to $\omega.$ Proof: The first part follows from $\angle ABX$ and $\angle XBG$ subtending equal arcs in $(ABGX)$ ($\angle ABS = \angle SBZ$). As for the second part, this results from $$\measuredangle XGD = \measuredangle XGA = \measuredangle XBA = \measuredangle SBA = \measuredangle ZBS = \measuredangle GBD.$$ Claim 4: $\angle APD = 90^\circ.$ Proof: Construct $A',$ the antipode of $A$ with respect to $\Omega.$ It amounts to show that $P,D,A'$ are collinear. This follows since $$\measuredangle BPD = \measuredangle BLD = \measuredangle EBC = \measuredangle BPA'$$and $BA' = CE$. Claim 5: $P,G,M$ are collinear. Proof: We have $$\measuredangle BPG = \measuredangle BDG = \measuredangle SDE = \measuredangle DSE + \measuredangle SED = \measuredangle BSE + \measuredangle EAM = \measuredangle BPM$$since $EM = AS.$ Now let $\overline{PDA'}$ meet line $AM$ at point $Y.$ Claim 6: $APGY$ is cyclic. Proof: This follows from $$\measuredangle GAY = \measuredangle EAM = \measuredangle MAA' = \measuredangle MPA' = \measuredangle GPY.$$ Since $\measuredangle APY = 90^\circ$ by Claim 4, we see that $(APGY)$ has diameter $AY.$ Moreover, by Claim 3, $XA = XG,$ so $X$ is in fact the circumcenter of $APGY.$ This implies that $XG = XP.$ Since $XG$ is tangent to $\omega,$ $XP$ must be too, and we are done.

Let $T$ be the antipode of $D$ and let $F$ be the reflection of $E$ about the perpendicular bisector of $BC$. Let $TD$ meet $(ABC)$ again at $G$. Let $\mathcal{J}$ denote the negative inversion about $D$ that preserves $(ABC)$. Claim: $P$, $D$, and $F$ are collinear $$\angle DPB=\angle DLB=\angle CBE=\angle FPB$$Claim: $SE$ is tangent to $(EDG)$ $$\angle SED=\angle SEA=\angle TGE=\angle DGE$$ Construct the point $X'$ along ray $SD$ such that $SX'\cdot SD=SE^2=SF^2$. Under $\mathcal{J}$, line $BS$ is preserved, angle bisector $AT$ is mapped to $(EDG)$, and the tangent to $(BDL)$ at $P$ is mapped to the circle tangent to $SF$ at $F$ and passing through $D$. By PoP, $X'$ lies on all three objects. The inverse $X$ thus lies on all three lines, as desired.

Let $M$ be the antipode of $S$ in $\Omega$. Let $E'$ be the reflection of $E$ in $SM$. Define $J=AS\cap BE'$. Claim. $JD\parallel BC$. Proof. Applying Pascal's on $AEE'SBC$, we get that $\underbrace{EE'\cap BC}_{=\infty_{BC}},SE'\cap AC,\underbrace{AE\cap SB}_{=D}$ are collinear. The rest of details are left as an exercise. [My proof involves complex numbers.] Claim. $L,P,S$ are collinear. Proof. Apply Pascal's on $SPE'BEA$ to get $SP\cap BE,\underbrace{PE'\cap EA}_{=D},\underbrace{E'B\cap AS}_{=J}$ are collinear. However, since $JD\parallel BC$, we have $J, L, D$ collinear. So, $L=BE\cap JD$, which means $L\in SP$. $\blacksquare$ Note, \[(B,M;E,S)\overset{A}{=}(B,K;D,S)\overset{P}{=}(B,KP\cap(BDP);D,L).\]So, if $P'=KP\cap(BDP)$, we need to show $P=P'$. It suffices to show \[\frac{\frac{BE}{EM}}{\frac{BS}{SM}}=\frac{\frac{BD}{BL}}{\frac{DP}{LP}}\] Claim. $\frac{DP}{LP}=\frac{BD}{DL}\frac{BS}{SM}\frac{EM}{BE}$ Proof. Note that by Power of Point, $DP=\frac{DB\cdot DS}{DE'}$. Let $r_2$ be the radius of $(BDL)$. Then, $LP=2r_2\sin(\angle PDL)$, while $\angle PDL=\angle DE'E$ as $EE'\parallel LD$, and since $DE\perp EE'$, we have $\sin(\angle DE'E)=\frac{DE}{DE'}$. Hence, \[\frac{DP}{LP}=\frac{DB\cdot DS}{2r_2\sin(\angle PDL)DE'}=\frac{DB\cdot DS}{2r_2DE}\]It suffices to show \[\frac{DB\cdot DS}{2r_2DE}=\frac{BD}{DL}\frac{BS}{SM}\frac{EM}{BE}\]\[\Longleftrightarrow \frac{DS}{2r_2DE}=\frac{BS\cdot EM}{DL\cdot SM\cdot BE}\]By sine rule, $DL=2r_2\sin(\angle LPD)$. Note, $\angle LPD=\pi-\angle E'ES=\angle EBS=B+\frac A2=\frac{\pi}2+\frac{B-C}{2}$. So, $DL=2r_2\cos\left(\frac{B-C}2\right)$. Similarly, $BS=SM\cos\left(\frac A2\right)$. Finally, $\frac{EM}{BE}=\frac{\sin\left(\frac{B-C}{2}\right)}{\cos(B)}$. Finally, $\frac{DS}{DE}=\frac{DA}{DB}=\frac{\sin\left(\frac{B-C}2\right)}{\cos(B)}$ Plugging in the ratios, we're done.

Let $T$ be diametrically opposite $S$ on $\Omega$, let $M$ be the intersection of $\omega$ with $AE$ other than $D$, and let $N$ be the intersection of $AP$ with $\omega$ other than $P$. Claim: $S$, $P$, and $L$ are collinear. Proof: This follows from \[\angle BPL = \angle BDL = \angle SBC = \angle SCB = 180^\circ - \angle BPS.\] Claim: $N$, $B$, and $T$ are collinear. Proof: First note that \[ \angle SPD = 180^\circ - \angle LPD = \angle LBD = 180^\circ - \angle SBE = \angle STE = \angle AST, \]so \[\angle APD = \angle APS + \angle SPD = \angle ATS + \angle AST = 180^\circ - \angle SAT = 90^\circ.\]This gives $\angle DBN = 180^\circ - \angle DPN = \angle APD = 90^\circ$, and since $\angle DBT = \angle SBT = 90^\circ$ the claim follows. Claim: $P$, $M$, and $T$ are collinear. Proof: We have $\angle DMP = \angle DBP = \angle SBP = \angle STP$; since $DM \parallel ST$, this gives $MP \parallel TP$, so $MP$ must be the same line as $TP$. Now Pascal's theorem on $PPNBDM$ gives $A$, $X$, and $T$ collinear, as desired.

what a P2!! let $O$ be the center of the circle $\Omega$ let $X,M$ be the intersection of $AO$ with $BC$ and $\overarc{CB}$ not containing $A$, let $F$ be the second intersection point $AE$ with $\omega$ we have that $MA'=ME=AP$ since $SM\parallel AE$ so we have $SAEM$ is a isosceles trapezium $$\boxed{\textbf{Claim:} M,K,P \text{are collinear }}$$proof:: $$\measuredangle KPB=\measuredangle KDB=\measuredangle BSM =\measuredangle BPM $$thus $M,K,P$ are collinear $\blacksquare$ $$\boxed{\textbf{Claim:} B,M,Y \text{are collinear }}$$ proof: let $D'$ be the antipode of $D$ we have that $DKD'L$ is a rectangle $\implies \measuredangle D'BK=90^{\circ}$ and since $O$ is the center of $\Omega$ we have that $$\measuredangle BSM=90^{\circ}\implies\measuredangle D'BK+\measuredangle BSM=180^{\circ} $$thus $B,M,Y$ are collinear $\blacksquare$ $$\boxed{\textbf{Claim:} D',P,A \text{are collinear }}$$Proof: $$\measuredangle KPL+\measuredangle MBS=180^{\circ}$$thus $D',P,A$ are collinear $\blacksquare$ $$\boxed{\textbf{Claim:} A,F,M \text{are collinear }}$$Proof: since $FP$ is a tangent we have that by Pascal Theorem that $A,F,M$ are collinear $\blacksquare$ now since $CM=MB\implies \measuredangle CAM=\measuredangle MAB$ hence we are done $\blacksquare$

Let $A', S'$ be antipode of $A$ and $S$, respectively. By angle chasing, we get $P, D, A'$ collinear. Since $A$ and $A'$ are antipodals, we have $PA \perp PA'$. Let $O$ be circumcenter of $ABC, BS \cap AS' = X, PA' \cap AS' = Y$ Suppose, we have $AX = XY$, it implies $XA = XP$. By angle chasing, we can obtain $\angle DBS = \angle DAS'$, which gives $AX^2 = XD \cdot XB = PX^2$, resulting in $PX$ tangent to $\omega$. Thus, it is suffice to show that $XO \parallel DA’$. We will proceed by cartesian. Let $A = (a,\sqrt{1-a^2}), B= (k,\sqrt{1 -k^2}), C= (-k,\sqrt{1-k^2}), S= (0,1), S' = (0,-1)$ and $ A' = (-a, -\sqrt{1-a^2})$ We have $$AE : x = a$$$$BS : y = \frac{\sqrt{1-k^2}-1}{-k} x +1$$$$AS' : y = \frac{\sqrt{1-a^2} +1}{a} x -1$$So, we have these points $$ D = (a, \frac{\sqrt{1-a^2}-a-k}{-k})$$$$ X = ( \frac{2ak}{k\sqrt{1-a^2} + a\sqrt{1-k^2} -a +k}, \frac{k\sqrt{1-a^2} - a\sqrt{1-k^2} +a +k}{k\sqrt{1-a^2} + a\sqrt{1-k^2} -a +k})$$$$\text{Slope} XO : \frac{k\sqrt{1-a^2} - a\sqrt{1-k^2} +a +k}{2ak}$$$$\text{Slope} DA’ : \frac{-k\sqrt{1-a^2} + a\sqrt{1-k^2} -a -k}{-2ak}$$Since Slope $XO = $Slope $DA’$, we get $XO \parallel DA’$, as desired.

Here we go again. Another day of resolving already solved problems thinking they are unsolved.

[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 = (7.17699,14.19874); pair B = (0.24096,-13.95274); pair C = (38.92675,-14.03776); pair S = (19.65434,18.07435); pair X = (19.55821,-25.66198); pair O = (19.60628,-3.79381); pair D = (7.14014,-2.57086); pair K = (32.03556,-21.78637); pair P = (-0.94805,3.67199); pair L = (-9.82506,-2.53358); pair V = (19.58785,-12.17862); pair E = (7.09803,-21.73156); pair G = (10.60853,3.15108); pair M = (40.74894,1.79251); pair Q = (-9.83673,-7.84409); pair R = (7.12847,-7.88137); import graph; size(12cm); 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(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle(O, 21.86822), linewidth(0.6)); draw(B--S, linewidth(0.6)); draw(A--X, linewidth(0.6) + linetype("4 4") + blue); draw(A--K, linewidth(0.6) + linetype("4 4") + blue); draw(circle((-1.34829,-5.20748), 8.88849), linewidth(0.6)); draw(E--K, linewidth(0.6)); draw(A--E, linewidth(0.6)); draw(X--S, linewidth(0.6)); draw(L--D, linewidth(0.6)); draw(P--K, linewidth(0.6) + ffxfqq); draw((16.86098,-10.07389)--(15.59234,-10.35588), linewidth(0.6) + ffxfqq); draw((16.86098,-10.07389)--(16.81239,-8.77519), linewidth(0.6) + ffxfqq); draw((15.54375,-9.05718)--(14.27511,-9.33917), linewidth(0.6) + ffxfqq); draw((15.54375,-9.05718)--(15.49515,-7.75849), linewidth(0.6) + ffxfqq); draw(A--Q, linewidth(0.6) + linetype("4 4") + blue); draw(Q--R, linewidth(0.6)); draw(R--G, linewidth(0.6) + red); draw((8.15031,-2.42940)--(9.41983,-2.82985), linewidth(0.6) + red); draw((8.31717,-1.90043)--(9.58668,-2.30088), linewidth(0.6) + red); draw(S--M, linewidth(0.6) + ffxfqq); draw((31.51887,8.91672)--(30.25023,8.63474), linewidth(0.6) + ffxfqq); draw((31.51887,8.91672)--(31.47028,10.21542), linewidth(0.6) + ffxfqq); draw((30.20164,9.93343)--(28.93300,9.65144), linewidth(0.6) + ffxfqq); draw((30.20164,9.93343)--(30.15305,11.23213), linewidth(0.6) + ffxfqq); draw(P--G, linewidth(0.6) + red); draw((4.58316,4.08894)--(4.52321,2.75911), linewidth(0.6) + red); draw((5.13725,4.06396)--(5.07731,2.73414), linewidth(0.6) + red); draw(G--M, linewidth(0.6)); draw(Q--X, linewidth(0.6) + linetype("4 4") + blue); draw(P--X, linewidth(0.6) + linetype("4 4") + blue); draw(L--S, linewidth(0.6) + linetype("4 4") + blue); draw(L--E, linewidth(0.6) + linetype("4 4") + blue); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$S$", S, N); dot("$X$", X, dir(270)); dot("$O$", O, NE); dot("$D$", D, NW); dot("$K$", K, SE); dot("$P$", P, NW); dot("$L$", L, NW); dot("$V$", V, NE); dot("$E$", E, SW); dot("$G$", G, NW); dot("$M$", M, NE); dot("$Q$", Q, SW); dot("$R$", R, dir(0)); [/asy][/asy] Let $X$ be the midpoint of the arc $\widehat{BC}$ not containing $A$. Let $Q=AP\cap \odot(PLB)$. Let $R$ denote the second intersection of $AD$ and $\odot(PLB)$. Let $O$ denote the center of $\odot(ABC)$. Let $K=PD\cap \odot(ABC)$. Let $V=DK\cap SX$. Let $M=$ tangent at $P$ to $\odot(PLB)\cap \odot(ABC)$. We also redefine $G$ as $PP\cap RR$ and prove that it lies both on $BD$ and $AX$. Claim: $\overline{L-P-S}$ are collinear. Proof. Note that since $DL \parallel SS$, by the converse of Reim's, on $\left\{ \odot(PLB),\odot(ABC) \right\} $ with lines $\left\{ PL,DB \right\} $, we get that $\overline{L-P-S}$ are collinear. $\blacksquare$ Claim: $\overline{A-O-K}$ are collinear. Proof. Firstly note that since $AE \parallel SX$, we get that $AEXS$ is an isosceles trapezium. Now, \[ \measuredangle XKO=\measuredangle OXK=\measuredangle SXK=\measuredangle EXS=\measuredangle XSA = \measuredangle XKA. \]$\blacksquare$ Claim: $\overline{L-B-E}$ are collinear. Proof. $\measuredangle DBL=\measuredangle DPL=\measuredangle DPS =\measuredangle KPS = \measuredangle KXS = \measuredangle SXE = \measuredangle SBE = \measuredangle DBE$. $\blacksquare$ Claim: $V$ is the midpoint of $DK$. Proof. Note that $DE \parallel VX$ and also that $VX$ is actually the perpendicular bisector of $EK$. So by using Midpoint Theorem, we get that $V$ is indeed the midpoint of $DK$. $\blacksquare$ Claim: $SM \parallel PK$. Proof. By Reim's on circles $\left\{ \odot(LPBD),\odot(ABC) \right\} $ with lines $\left\{ PP,BD \right\} $, we get the desired result. $\blacksquare$ Claim: $\overline{P-R-X}$ are collinear. Proof. $\measuredangle BPR=\measuredangle BDR=\measuredangle SDA =\measuredangle DSO = \measuredangle BXS = \measuredangle BPX$. $\blacksquare$ Claim: $\overline{B-D-G}$ are collinear. Proof. Firstly, \[ (D,B;P,R)\overset{P}{=} (K,B;M,X) \overset{S}{=} (K,D;\infty_{SM},V) = -1. \]Now, since the quadrilateral $DPBR$ is harmonic, we must have that $PP$, $BD$ and $RR$ are concurrent. $\blacksquare$ Claim: $\overline{Q-B-X}$ are collinear. Proof. $\measuredangle DBQ=\measuredangle DPQ=\measuredangle DPA \measuredangle KPA = 90^{\circ} = \measuredangle XBS = \measuredangle XBD$. $\blacksquare$ Claim: $\overline{A-G-X}$ are collinear. Proof. By Pascal on $PRDBQP$, we get that $\overline{PR\cap BQ-RD\cap QP-DB\cap PP}$ are collinear, i.e., $\overline{X-A-G}$ are collinear. $\blacksquare$

Let $F$ be the minor arc midpoint. Let $X_{ook}$ be a point on the tangent from $S$ to $(ABC)$ on the same of $SF$ as $A$, then \[\angle PLD=\angle PBD=\angle PBS=\angle X_{ook}SP.\]Then we can show that this implies $L,P,S$ are collinear. Let $X=(BLD)\cap AD$. Then $\angle LDX=90^{\circ}$, so $\angle LPX=90^{\circ}$. Since $\angle SPF=90^{\circ}$, we get that $X$ lies on $PF$. Also, $\angle LBX=90^{\circ}$ too, so $\angle EBX=90^{\circ}$. But $\angle CBE=90^{\circ}-\angle C$, so $\angle XBC=\angle C$. So $BX,AC,SF$ concur, say at $Y$. Let $Q=AF\cap BS$. Note that $\angle YAX=90^{\circ}-\angle C=\frac12\angle BYC$, so $YA=YX$. Apply Pascal on $FACBSF$ to get that $Q$, $Y$, and $\infty_{BC}$ are collinear, which means $QY\parallel BC$. Since $AX\perp BC$, we then get that $QA=QX$ also. Now note that \[(QD)(QB)=\frac{QD}{QS}\cdot (QS)(QB)=\frac{AQ}{AF}\cdot (AQ)(AF)=QA^2=QX^2,\]so $QX$ is tangent to $(BDL)$. Since we want to prove that $QP$ is tangent to $(BDL)$, it suffices to show that $(PX;DB)=-1$. But in fact, \[(PX;DB)\stackrel{E}{=}(PE\cap (BDL),D;X,L),\]so we want to show that $PE\cap (BDL)$ is the reflection of $D$ over $LX$. This is equivalent to showing that $\angle XPE=\angle XPD$. Indeed, \[\angle XPE=\angle FPE=\angle FAE=\angle AFS=\angle QFY=\angle QBY=\angle DBX=\angle DPX\;\blacksquare\]

Not 'Too easy for P2', took me 2 attempts and a hint of L-P-S collinear (which was obvious).

Let $M$ be the midpoint of arc $BC$ and $MB\cap \omega=N,AB\cap \omega =R,AD\cap \omega=F$. Since $\measuredangle FPB=\measuredangle FDB=\frac{\measuredangle A}{2}=\measuredangle MPB$, we see that $P,F,M$ are collinear. Note that $\measuredangle FBC=\measuredangle C$ because $\measuredangle FBE=90$. \[\measuredangle FRN=\measuredangle FBM=\measuredangle FBC+\frac{\measuredangle A}{2}=\measuredangle C+\frac{\measuredangle A}{2}=\measuredangle NBR=\measuredangle NFR\]Thus, $NF=NR$. Pascal on $NPFRBN$ gives $NP\cap RB,M,AM_{\infty}$ are collienar because $RF\parallel AM$ by $\measuredangle FRM=\measuredangle FDB=\frac{\measuredangle A}{2}=\measuredangle MAB$. Hence $A,P,N$ are collinear. Pascal on $FPPNBD$ implies $M,PP\cap BD,A$ are collinear. So $PP\cap BD$ is on $AM$ or $KP$ is tangent to $\omega$ as desired.$\blacksquare$

Let \(M\) be the midpoint of the arc \(BC\) not containing \(A\) ,let \(A'\) be the antipode of \(A\) let \(F\) = \(BS \cap AA'\) let \(T\) = \(AM \cap BS\), let \(K\) = \(PM \cap AE\) label \(\angle ABC = 2b\) , \(\angle ACB = 2a\) \(\angle PMA = 2c\) i make my first claim: \(K\) belongs to the small circle. this is quite easy to prove: \(\angle PLD = \angle PBS = \angle PMA + \angle ABS = 2c + 90 - 2a - 2b\) now, \(\angle PAE = 4a + 2b - 90^\circ - 2c\) \(\angle MPA = \angle MPB = 180^\circ-2a-b\) and \(\angle PKA = 90^\circ + 2c - 2a - b = \angle PBS\) so its proven. next i claim \(L,P,S\) are collinear which just follows from \(\angle KBE = 90^\circ = \angle KBL = \angle KPL\) and we know \(\angle SPA = 90^\circ\) so its also done. now, its obvious that \(\angle DAT = \angle TAF\) and \(\angle MAA' = 90\) it follows from a known lemma that \((D, F, T, S) = -1\) project the line \(SD\) from \(A'\) to the line \(AM\) to get \((P_\infty, A, T, N) = -1\) meaning \(AT = TN\) . simple angle chasing gives us \(\angle PNA = \angle PBS\) now in the right triangle \(\triangle NPA\) , \(PT\) is the median on the hypotenuse so \(\angle TPN = \angle TNP\ = \angle PBS\) which means that \(TP\) is a tangent which finishes