grant sniped

Let $AC$ intersect $DP$ and $BM$ and $X$ and $Z$, and let $BD$ intersect $AM$ and $CP$ at $Y$ and $W$. Also, let $AD$ and $BC$ meet at $E$. By the second angle condition, $CWXD$ is cyclic. The arc midpoint also gives $BZYA$ is cyclic. Claim: $BWZC$ and $DYXA$ are cyclic. Proof. We prove one and symmetry finishes. Note that $\angle AMB=\angle CPD$ and \[\angle MAD+\angle ADP=\angle MBC+\angle BCP\]so $PD\parallel BM$ and $AM \parallel CP$. Thus \[\measuredangle XAY=\measuredangle CAM=\measuredangle ACP=\measuredangle XCW=\measuredangle XDW=\measuredangle XDY \]as desired. $\blacksquare$ Now let the circles above be $\omega_1$ and $\omega_2$. Radical axis on $\omega_1, \omega_2$, and taking turns between $(CWXD)$, $(BZYA)$, and $(ABCD)$ finishes since all of $P,M,E$ lie on the radical axis of $\omega_2$ and $\omega_2$.

[asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -23.88782587247942, xmax = 28.11307823832208, ymin = -9.242139859667088, ymax = 6.863695719095153; draw(circle((0.5809291498702329,-0.01981345647294291), 5.292133555959725), linewidth(1)); draw((0.5542578591817454,5.272252890084698)--(12.770695651820194,-1.2314621937659902), linewidth(0.6)); draw((-1.172584231738878,-5.012995647684546)--(12.770695651820194,-1.2314621937659902), linewidth(0.6)); draw((-2.6235032789327573,4.1918715250220755)--(5.864154960092957,-0.3267378971930568), linewidth(0.6)); draw((-3.8232675432202528,-2.954047438739143)--(5.864154960092957,-0.3267378971930568), linewidth(0.6)); draw((-2.6235032789327573,4.1918715250220755)--(4.616982119434293,-3.4428171083827763), linewidth(0.6)); draw((-3.8232675432202528,-2.954047438739143)--(4.986206114864625,2.9127984230305017), linewidth(0.6)); draw((2.672898183950598,1.372204559170503)--(2.52153455609322,-1.2332843473134645), linewidth(0.6)); draw((4.986206114864625,2.9127984230305017)--(4.616982119434293,-3.4428171083827763), linewidth(0.6)); draw((1.0656729475516833,0.30184064265604144)--(12.770695651820194,-1.2314621937659902), linewidth(0.6) + linetype("4 4")); dot((-1.172584231738878,-5.012995647684546),dotstyle); label("$A$", (-1.6429946695254456,-5.883748135844466), NE * labelscalefactor); dot((0.5542578591817454,5.272252890084698),dotstyle); label("$B$", (0.4153744515271139,5.672008333222611), NE * labelscalefactor); dot((4.986206114864625,2.9127984230305017),dotstyle); label("$C$", (5.146012256051417,3.2886335614775266), NE * labelscalefactor); dot((4.616982119434293,-3.4428171083827763),dotstyle); label("$D$", (4.568224432598067,-4.294831621347743), NE * labelscalefactor); dot((12.770695651820194,-1.2314621937659902),linewidth(4pt) + dotstyle); label("$Z$", (13.01837135060331,-1.4420042430468083), NE * labelscalefactor); dot((5.864154960092957,-0.3267378971930568),linewidth(4pt) + dotstyle); label("$M$", (6.084917469163111,-0.06975816234509281), NE * labelscalefactor); dot((-2.6235032789327573,4.1918715250220755),linewidth(4pt) + dotstyle); label("$X$", (-3.3041346619538268,4.4080974694184), NE * labelscalefactor); dot((-3.8232675432202528,-2.954047438739143),linewidth(4pt) + dotstyle); label("$Y$", (-4.604157264723864,-3.4642616251335467), NE * labelscalefactor); dot((1.0656729475516833,0.30184064265604144),linewidth(4pt) + dotstyle); label("$P$", (0.7764918411854577,-0.5392107689009429), NE * labelscalefactor); dot((2.672898183950598,1.372204559170503),linewidth(4pt) + dotstyle); label("$E$", (2.5098553115455076,1.7719405249124727), NE * labelscalefactor); dot((2.52153455609322,-1.2332843473134645),linewidth(4pt) + dotstyle); label("$F$", (2.257073138784667,-2.019792066500162), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $BC\cap AD=Z.$ Let $DP,CP$ meet $\omega$ again at $X,Y$ respectively. We claim $BC\parallel MX$ and $AD\parallel MY.$ This follows from angle chasing, since the first condition gives that arcs $BX$ and $AY$ sum to arc $CD,$ and the second gives that arcs $BX$ and $AY$ are equal. Thus arc $BX$ equals arc $CM,$ and arc $AY$ equals arc $DM.$ Now let $E=MX\cap CY$ and $F=MY\cap DX.$ Pascal's on $CYMMXD$ gives $EF\parallel CD,$ since the tangent to $\omega$ at $M$ is parallel to $CD.$ Finally, we have $EM\parallel CZ,FM\parallel DZ,EF\parallel CD$ so $\triangle EFM$ and $\triangle CDZ$ are homothetic. The center of homothety must be $CE\cap FD=P,$ so $P,M,Z$ collinear as desired.

I like radical axis. [asy][asy] size(7.5cm); defaultpen(fontsize(10pt)); import olympiad; import geometry; pair A = dir(-20); pair B = dir(-130); pair C = dir(110); pair D = dir(70); pair M = dir(90); pair X = 2*foot(-M,C,B) - C; pair Y = 2*foot(-M,A,D) - D; pair P = extension(C,Y,D,X); pair T = 2*foot((0,0),M,P) - M; draw(circle(A,B,C), linewidth(0.7)); draw(A--B--C--D--cycle, linewidth(1)); draw(C--Y--A, black); draw(D--X--B, black); draw(arc(circumcenter(C,X,P),circumradius(C,X,P),70,-80,CW), gray+0.7); draw(arc(circumcenter(D,Y,P),circumradius(D,Y,P),130,-90,CCW), gray+0.7); dot("$A$", A, dir(5)); dot("$B$", B, dir(-163)); dot("$C$", C, dir(75)); dot("$D$", D, dir(111)); dot("$M$", M, dir(89)); dot("$X$", X, dir(-66)); dot("$Y$", Y, dir(-100)); dot("$P$", P, dir(7)); dot("$T$", T, 1.5*dir(-19)); [/asy][/asy] Let $X = DP\cap BC$ and $Y=CP\cap AD$, so the angle condition implies $\odot(CDXY)$ is cyclic. Consider circles $\odot(CPX)$ and $\odot(DPY)$. Clearly, $AD\cap BC$ lies on the radical axis, so it suffices to show that $M$ does. To that end, let $T$ be the intersection of those circles. Observe that \begin{align*}\angle CTD &= \angle CTP + \angle DTP = \angle CXP + \angle DYP \\ &= \angle(BC, AD) - \angle CPD = \angle(BC, AD) - \angle BDA \\ &= 180^\circ - \angle CMD, \end{align*}so $T\in\odot(ABCD)$. Moreover, $\angle CTP = \angle CXP = \angle DYP = \angle DTP$, so $M\in PT$. This implies the conclusion.

Claim 1: $PD \parallel BM$ and $AM \parallel PC$ Proof: We know $\angle DPC = \angle AMB$, also $\angle MAD = \angle MBC$ and $\angle PCB = \angle ADP$ $\implies \angle (AM,PD) = \angle (PC, BM)$ Let $F$ be a point such that $DPCF$ is a parallelogram. Noticed that $DF \parallel PC \parallel AM \implies \angle EDF = \angle EAM$ Analogously, $\angle ECF = \angle EBM$. Since $\angle EDF = \angle DAM = \angle MBC = \angle ECF$, by isogonality lemma, we have $\angle DEF = \angle PEC$. Also, since $\angle EDF = \angle EAM = \angle MCD = \angle MDC$ and analogously, $\angle ECF = \angle MCD$, we have $F$ and $M$ are isogonal conjugates and this implies $\angle DEF = \angle MEC$ Finally, $\angle PEC = \angle DEF = \angle MEC \implies E-M-P$. $\square$

Let $CP$ and $DP$ meet $AD$ and $BC$ at $T,Q$. Note that $\angle PTD = \angle DPC - \angle PDT = \angle PDB = \angle PCA = \angle DPC - \angle PCQ = \angle PQC$ so $TDCQ$ is cyclic. Note that $\angle DAC = \angle ATC + \angle ACT = \angle DTP + \angle CQP$ so if circles $PDT$ and $PCQ$ meet at $K$ we have that $K$ lies on $ABCD$. Using Radical axis we have that $TD,QC,KP$ are concurrent to we need to prove $M$ lies on $KP$ which is true since $\angle DKM = \frac{\angle DAC}{2} = \angle DTP = \angle DKP$.

Our aim is to show $M$ lies on the radical axis of the circumcircles of $\triangle PAD$ and $\triangle PBC$. This is sufficient as $P$ lies on both circles and $\overline{AD}\cap\overline{BC}$ lies on the radical axis by power of a point. To compute the power of $M$ with respect to the two circumcircles, let lines $\overline{MD}$ and $\overline{MC}$ intersect the circumcircles of $\triangle PAD$ and $\triangle PBC$ at points $D'$ and $C'$, respectively. We want to show $MC\cdot MC' = MD \cdot MD'$, which is equivalent to $CC' = DD'$ as $MC = MD$ by definition. We go about computing $CC'$ and $DD'$ by using the trigonometric variant of Ptolemy's theorem for the cyclic $PADD'$ and $PBCC'$: \begin{align*} DD'\sin\angle PDA &= DA \sin\angle PDD' - DP\sin\angle ADD'\\ CC'\sin\angle PCB &= CB \sin\angle PCC' - PC \sin\angle BCC'. \end{align*}Luckily, $\angle PDA = \angle PCB$, so now we just want to show \[DA \sin \angle PDM - DP \sin\angle ABM = CB \sin\angle PCM - PC\sin\angle BAM.\]At this point, it's only logical that $\triangle PDC \sim \triangle MAB$. Indeed, assume it's not and let $P'$ be such that $\triangle DP'C \sim \triangle AMB$ and $P'$ lies in the same halfplane as $A$ and $D$ with respect to line $\overline{CD}$. Then clearly $\angle DP'C = \angle AMB = \angle DPC$, and furthermore: \[\angle ADP' = \angle ADC - \angle P'DC = \angle ADC - \angle MAB\]\[\angle BCP' = \angle BCD - \angle P'CD = \angle BCD - \angle MBA\]As $\angle ADC - \angle MAB = \angle AMB - \angle MAC = \angle BCD - \angle MBA$, we get $\angle P'DA = \angle P'CB$. It's now obvious that $P \equiv P'$ as \[\angle P'DA = \frac{1}{2}(\angle DP'C+\angle DXC) = \frac{1}{2}(\angle DPC + \angle DXC) = \angle PDA.\]We are now ready to prove the line from above as $\angle PDM = \angle DAB$, $\angle PCM = \angle CBA$, $DP = AM\cdot \frac{CD}{AB}$, and $CP = BM \cdot \frac{CD}{AB}$ from $\triangle DPC \sim \triangle AMB$. Denote $\angle DAC = x$, $\angle ADB = y$, $\angle CAB = z$, and $\angle DCA = t = \pi-x-y-z$. Then \begin{align*} DA \sin \angle PDM - DP \sin\angle ABM &= DA \sin(x+z) - AM \cdot \frac{CD}{AB}\sin (x/2+y+z)\\ &= \frac{R}{\sin y}\left(\sin (x+y+z) \sin(x+z) \sin y - \sin^2(x/2+y+z) \sin x\right)\\ &=\frac{R}{4\sin y} (\sin(2x + 2z) - 2 \sin(x) - \sin(2y + 2z) + \sin(2y)) \end{align*}where $R$ is the radius of the circumcircle of $ABCD$. Analogously, \begin{align*} CB\sin\angle PCM - PC\sin\angle BAM &= CB\sin (y+z) - BM\cdot\frac{CD}{AB} \sin(x/2+z)\\ &=\frac{R}{\sin y}\left(\sin z \sin(y+z) \sin y- \sin^2(x/2+z)\sin x\right)\\ &=\frac{R}{4\sin y} (\sin(2x+2z) - 2\sin(x) - \sin(2y+2z)+\sin(2y)). \end{align*}With this, the solution is complete.

Claim. $DP \parallel BM$ and $CP \parallel AM$. Proof. Let $E = AC \cap BD$. We have \begin{align*} \angle ADB &= \angle CPD \\ &= 180^{\circ} - \angle PDB - \angle PCA - \angle BDC - \angle ACD \\ &= 180^{\circ} - 2 \angle PDB - \angle AED \\ &= \angle AEB - 2 \angle PDB. \end{align*}This means that \begin{align*} 2 \angle PDB &= \angle AEB - \angle ADB \\ &= \angle DAE \\ &= \angle DAC \\ &= \angle DBC. \end{align*}However, by the Incenter-Excenter Lemma, $BM$ is the angle bisector of $\angle DBC$, so $2 \angle DBM = \angle DBC$, which gives $\angle PDB = \angle DBM$, implying that $PD \parallel BM$. Analogously, we obtain $PC \parallel AM$, as desired. $\blacksquare$ Now, let $PD$ meet $\omega$ again at $E$, $PC$ meet $\omega$ again at $F$. It follows that $DEBM$ and $AMCF$ are isoceles trapezoids and $BE = DM = MC = AF$. Claim. $ADMF$ and $BCME$ are isoceles trapezoids. Proof. Since $\angle DAM$ and $\angle AMF$ are half the lengths of equal arcs, they must equal. Thus, $AD \parallel MF$, but $ADMF$ is cyclic, so $ADMF$ is an isoceles trapezoid. Similarly, $\angle MEC = \angle ECB$, so $BCME$ is also an isoceles trapezoid and $BC \parallel ME$. $\blacksquare$ Let $AF$ and $BE$ meet at $X$, $AD$ and $BC$ meet at $T$. Claim. $M$, $P$, $X$ are collinear. Proof. Since $FP \parallel AM$, $PE \parallel MB$, and $FE \parallel AB$, $\Delta FPE$ and $\Delta AMB$ are homothetic. Thus, $AF$, $BE$, $MP$ concur at the center of homothety $X$, giving the claim. $\blacksquare$ Claim. $M$, $T$, $X$ are collinear. Proof. $FME$ and $ATB$ are homothetic triangles, so $TM$, $AF$, $EB$ concur at the center of homothety $X$ similar to above. $\blacksquare$ The two claims above imply that $T,M,P$ are collinear, which is exactly the desired condition, so we are done. $\blacksquare$

This was P1 on the first Romanian IMO TST this year. I solved it with homothety and Pappus's theorem - will post a full write up later.

Let $AD\cap BC = E, CP\cap AB =X, DP\cap AB = Y$. We'll prove $\overline{EMP}$ is the radical axis $\rho$ of $\odot ADY$ and $\odot BCX$. The identity $ED.EA=EC.EB$ means $E\in \rho$. A simple angle chase gives that $\angle MDP = \angle DAY$ which means $MD$ is the tangent to $\odot ADY$ at $D$. Therefore, $\operatorname{Pow}_{\odot ADY}(M)=MD^2$; similarly $\operatorname{Pow}_{\odot CBX}(M) = MC^2$. Since $MC=MD$, $M\in\rho$. Notice that \[\angle YXC = \angle BXC = 180^{\circ}-\angle CBA-\angle PCB=\angle ADC - \angle ADP = \angle PDC = \angle YDC,\]meaning $XYCD$ is cyclic. Then $PX.PC=PY.PD\implies P\in\rho$, which finishes the solution.

I spent 2 hours on this problem, WITH GEOGEBRA. We first remark that $P$ is unique, since $\angle ADP = \angle PCB$ must vary until we get that the angle between the two lines is $\angle BDA.$ Now given this, we will actually classify $P.$ Let the line through $D$ parallel to $BM$ and the line through $C$ parallel to $AM$ intersect at point $P'.$ We claim that $P' = P.$ First of all, by the parallel lines, it is clear that $\angle CP'D = \angle BMA = \angle BDA.$ Second, we see that $$\angle ADP' = \angle(AD, BM) = \angle ADB - \angle MAD = \angle ADB - \angle MAC = \angle(BC, AM) = \angle P'CB,$$so $P'$ satisfies the problem conditions and thus is equal to $P.$ Now we will show that $$\frac{[BPC]}{[APD]} = \frac{[BMC]}{[AMD]}.$$This finishes, since that will readily imply that $PM$ passes through $AD \cap BC,$ at which point we are done. We first compute the first ratio. The area on top is equal to $\frac{BC \cdot CP \cdot \sin \angle BCP}{2}$ while the area on bottom is equal to $\frac{AD \cdot DB \cdot \sin \angle ADP}{2}.$ Dividing the two and noting that $\angle ADP = \angle BCP,$ we get $$\frac{[BPC]}{[APD]} = \frac{BC \cdot CP}{AD \cdot DP}.$$Now we compute the second ratio. The area on top is equal to $\frac{BC \cdot BM \cdot \sin \angle CBM}{2}$ while the area on bottom is equal to $\frac{AD \cdot AM \cdot \sin \angle DAM}{2}.$ Dividing the two and noting that $\angle DAM = \angle MBC,$ we get $$\frac{[BMC]}{[AMD]} = \frac{BC \cdot BM}{AD \cdot AM}.$$It thus suffices to show that $\frac{PC}{PD} = \frac{BM}{AM}.$ In fact, $\angle CPD = \angle AMD$, and if $DP$ hits $(ABCD)$ again at point $Z,$ then $MD = ZB = MC,$ so $CB \parallel ZM$ and so $\angle PDC = \angle BAM.$ Thus $\triangle CPD \sim \triangle BMA,$ giving the desired length ratio, and we are done.

Let $BC\cap AD=E, BC\cap DP=Y, AD\cap CP=X.$ Then clearly $D,X,Y,C$ is cyclic and $AB \parallel XY$ And Let $M'$ is $AM' \parallel XP$, $BM' \parallel YP$ Then $\angle M'AC = \angle ACX = \angle ACY - \angle PCY = \angle DPC-\angle PCY =\angle DYC =\angle DXC =\angle DAM'$ Similarly $\angle M'BD=\angle CBM'$ Thus $M'=M$ Thus $\Delta MAB$ and $\Delta PXY$ is homothetic, and their homotheric center is $E \implies E-M-P$

GrantStar wrote: Let $ABCD$ be a cyclic quadrilateral with $\angle BAD < \angle ADC$. Let $M$ be the midpoint of the arc $CD$ not containing $A$. Suppose there is a point $P$ inside $ABCD$ such that $\angle ADB = \angle CPD$ and $\angle ADP = \angle PCB$. Prove that lines $AD, PM$, and $BC$ are concurrent. Let $\Omega$ be the circumcircle of $ABCD$ and let $A'=CP\cap AD,B'=NC\cap DP,X=BC\cap AD$ and $Y=A'B'\cap CD$. We have $\angle A'CB'=\angle PCB=\angle ADP=\angle A'DB'$. So $A',B',C$ and $D$ lie on a circle $\Gamma$. Let $O$ be the center of $\Gamma$. By angle chasing we get $\angle BB'D=\angle B'DB$ and thus $\angle COD=2\angle CB'D=\angle CBD$. So $|BB'|=|BD|$ and analogously $|AA'|=|AC|$ and $O$ lies on $\Omega$. Now $O$ and $M$ are two points on $\Omega$ and the perpendicular bisector of $CD$. We have $BO\perp DP$ since B and $O$ lie on the perpendicular bisector of $B'D$. We have $BM\parallel DP$ since $\angle PDB=\frac{1}{2}\angle CBD=\angle MBD$. So $O\neq M$ and $OM$ is a diameter of $\Omega$. By Thales we have $\angle MDO=\angle OCM=90^\circ$. So $MC$ and $MD$ are tangent to $\Gamma$. So $CD$ is the polar of $M$ w.r.t. $\Gamma$. By pole-polar duality $M$ lies on the polar of $Y$ w.r.t. $\Gamma$. By Brocard $PX$ is the polar of $Y$ w.r.t. $\Gamma$. Thus $BC,AD$ and $PM$ concur at $X$.

Let $CP$ and $DP$ meet $(ABC)$ at $C'$ and $D'$. Then since $\angle ADP = \angle PCB$, $D'C'AB$ is an isosceles trapezoid with $D'C' \parallel AB$. Now let $M'$ be the point such that $AM' \parallel CP$ and $BM' \parallel PD$. Then since $\angle AM'B = \angle CPD = \angle ADB$, $M'$ lies on $(ABC)$, and we also have $M'C = C'A = D'B = M'D$, so $M' = M$. Since $\triangle ABM$ and $\triangle C'D'P$ are homothetic, we see that $\overline{D'B}$, $\overline{C'A}$ and $\overline{PM}$ are concurrent at a point $Z$. Now Pascal's theorem on the hexagon $BD'DAC'C$ shows that $Z$, $P$ and $\overline{AD} \cap \overline{BC}$ are collinear so we are done.

The angle conditions give $CP\parallel AM$ and $DP\parallel BM$ (this has already been proved in previous solutions, so I won't write it). Now let $A'=AM\cap CD$, $B'=BM\cap CD$ and $G=PM\cap CD$. Then by Thales/homothety we get that$$GA'\cdot GD=GB'\cdot GC,$$so $G$ is on the radical axis of $(MA'D)$ and $(MB'C)$. Finally, the inversion with centre $M$ and radius $MC=MD$ maps $AD$ to $(MA'D)$ and $BC$ to $(MB'C)$, which means that $AD\cap BC$ also lies on the radical axis of $(MA'D)$ and $(MB'C)$. Done.

Let $CP$ and $DP$ meet the circumcircle of $\square ABCD$ again at $Q,R$, respectively. Denote, $X,Y$ the intersections of $AB,CD$ and $AQ,BR$, respectively. By several angle chasing, we obtain that $MR\parallel BC$, $MQ\parallel AD$ and $QR\parallel AB$. Hence, $\triangle MQR$ and $\triangle XAB$ are homothetic, implying $Y,M,X$ are collinear. Applying Pascal to $ADRBCQ$, we obtain that $Y,P,X$ are also collinear, so we are done. $\square$

The angle conditions imply that $AM \parallel PC$ and $BM \parallel PD$. Define $X = BC \cap DP$, $Y = AD \cap CP$ and $Z = PM \cap (ABCD)$. By construction $CPXZ$, $DPYZ$ and $CDXY$ are all cyclic (!) so that radical axis finishes.

Let $CP$ meet $AD$ at $E$ and $DP$ meet $BC$ at $F$. As $\angle PDA=\angle PCB$, $EFCD$ is cyclic. We claim that $MC$ is tangent to $(EFCD)$ $$\angle MCD=\frac{1}{4}\overarc{CD}=\frac{1}{2}\overarc{AB}-\frac{1}{2}(\angle C+\angle D+\frac{1}{2}\overarc{AB}-180^{\circ})=\angle DPC-\angle ADP=\angle DEC$$Similarly, $MD$ is tangent to $(EFCD)$. We are done by Pascal's Theorem on $CCEDDF$.

GrantStar wrote: Let $ABCD$ be a cyclic quadrilateral with $\angle BAD < \angle ADC$. Let $M$ be the midpoint of the arc $CD$ not containing $A$. Suppose there is a point $P$ inside $ABCD$ such that $\angle ADB = \angle CPD$ and $\angle ADP = \angle PCB$. Prove that lines $AD, PM$, and $BC$ are concurrent. I have 2 short solutions

Gorgeous Define $Q = AD \cap BC$, $L = PD \cap (ABCD)$, $K = PC \cap (ABCD)$ and $J = AK \cap BL$. Let $\angle PDA = \angle PCB = \beta$ and $\angle PDB = \angle PCA = \alpha$. Because $\angle CPD = \alpha + \beta$ and $\angle ADP + \angle BCP = \angle CPD + \angle CQD \implies \angle DQC = \beta - \alpha$. Additionally, $\angle DBC = \angle ADB - \angle DQC \implies 2 \alpha$. This shows $MK \parallel AD$, $ML \parallel BC$ and $KL \parallel AB$ due to the fact that $\widehat{AK} = \widehat{BL} = \widehat{MC} = \widehat{MD}$. $J$ is the homothety center of $KLM$ and $ABQ$. This shows that $M \in JQ$. By Pascal Theorem on $KADLBC$, $P \in JQ$ which is sufficient.

Let $X = AD \cap CP$ and $Y = BC \cap DP$. Then $\angle XCY = \angle PCB = \angle ADP = \angle XDY$ gives $(XDCY)$ cyclic. Let $Q = AC \cap DP$ and $R = BD \cap CP$. Then $\angle QDR = \angle ADB - \angle ADP = \angle ACB - \angle PCB = \angle QCR$ gives $(QRCD)$ cyclic. Since \[ \angle DYB = \angle QYC = 180^\circ - \angle YQC - \angle QCY = 180^\circ - \angle YQC - \angle ADB = 180^\circ - \angle YQC - \angle CPD = \angle QCR = \angle QDR = \angle YDB, \]we have $BD = BY$, and similarly $AC = AX$. Then $\angle DBM = \frac12 \angle DBC = \frac12 (180^\circ - \angle DBY) = \angle YDB$, so $BM \parallel DY$ and similarly $AM \parallel CX$. Let $Z$ be the second intersection of $PM$ with $(ABCD)$. Then $\angle XPZ = \angle AMZ = \angle ADZ = \angle XDZ$ gives $(XDPZ)$ cyclic, and similarly $(YCPZ)$ is cyclic. The desired claim follows from radical axis on $(XDPZ)$, $(YCPZ)$, and $(XDCY)$.

Solution using the isogonal conjugate of $P$ in triangle $CDE$ where $E=AD\cap BC$. Let $Q$ be the isogonal conjugate. Then $$\angle CPD=180^\circ-\angle PCD-\angle PDC=\angle EDC-\angle ECD-180^\circ+2\angle QDC=2\angle QDC-\angle CED.$$As $\angle ADB=\angle E+\angle CBD$, we find $\angle E+\angle DBM=\angle QDC$. Now, let the perpendicular bisector of $CD$ intersect $(ABCD)$ at $N\neq M$ and $(CDE)$ at $U$ and $V$ with $U$ in the minor arc $CD$. Then $$\angle QDM=\angle QDC+\angle MDC=\angle E+\angle DBM+\angle CBM=\angle CED+\angle CND.$$Thus we find $\angle QDM=180^\circ-\angle DMC+180^\circ-\angle DUC=2\angle UDM$. I.e. $DU$ bisects $\angle MDQ$. Appolonius circle gives that $EU$ bisects $\angle EMQ$, and therefore $EM$ and $EQ$ are isogonal. Hence $EM$ pass through $P$ by construction of $Q$.

let $PM$ intersect $(ABCD)$ again at point $P'$, and let $AD$ intersect $PC$ at $A'$, and similarly define $B'$ since $ADP = PCB$, we have that $A'B'CD$ are concyclic, so we can just angle chase for $DB' \parallel BM$ and $CA' \parallel AM$ then, $PA'D=DAM=DP'P$ and $CB'D=CBM=CP'M$, so $A'DPP'$ and $B'CPP'$ are both cyclic thus, $PM$ is a radical axis of these two circles and since $A'B'CD$ is also cyclic, we have that $AD$, $PM$, and $BC$ are concurrent on $PM$

Let $T$ be the intersection of the line passing through $P$ and parallel to $BC$ with $BD$. Since $\angle{BCP}=\angle{CPT}$, then $\angle{TPD}=\angle{TDP}$. $\angle{BTP}=2\angle{TDP}$. It is also equal to $\angle{CBD}$. So $\angle{CBD}=2\angle{TPD}$. Since point $M$ is the midpoint, $\angle{MPD}=\angle{BDP}$. It can be seen that $MB||DP$ Since $\angle{BMA}=\angle{BDA}=\angle{CPD}$ it is $CP||MA$. Now we have to prove that $CR||MA$ for the point $R$, which is the intersection of the line passing through $D$ and parallel to $MB$ with the line passing through $BC\cap AD$ and $M$. Let $BC\cap AD=Z$. Let $AM\cap ZB=J$ and $BM\cap ZA=Q$. Since $\angle{CBM}=\angle{MAD}$, then $\angle{BQA}=\angle{BJA}$. $JQAB$ is cyclic. Thus $JQ||CD$. $\frac{|ZJ|}{|ZC|}=\frac{|ZQ|}{|ZD|}$. Also with $MQ||DR$, by the similarity in the triangle $ZDR$, $\frac{|ZQ|}{|ZD|}=\frac{|ZM|}{|ZR|}$. Since $\frac{|ZJ|}{|ZC|}=\frac{|ZQ|}{|ZD|}$ as we just found, $\frac{|ZJ|}{|ZC|}=\frac{|ZM |}{|ZR|}$ becomes. Thus, $JM||CR$ in the triangle $ZCR$. The proof ends.

Rather quick trig sol that I found. Let circle $(DCP)$ intersect $AD$ and $BC$ at $F$ and $G$ respectively and let $AD$ intersect $BC$ at $N$. Let the arcs $DC$, $AB$, $AD$ and $BC$ be equal to $4a$, $2b$, $2c$ and $2d$ respectively. We have that $FG$ is parallel to $AB$ and $P$ is the midpoint of arc $FG$. Quick angle chasing in $(DCGPF)$ gives $\angle PDC = d + a$ and $\angle PCD = a + c$. Also note that $\angle NDP = \angle NCP$. We apply trig ceva in $\triangle{NDC}$ with point $P$ $$\frac{\sin(\angle DNP)}{\sin(\angle PNC)}\cdot\frac{\sin(\angle NCP)}{\sin(\angle PCD)}\cdot\frac{\sin(\angle PDC)}{\sin(\angle PDN)}=1$$ So we get that: $$\frac{\sin(\angle DNP)}{\sin(\angle PNC)} = \frac{\sin(a+c)}{\sin(a+d)}$$ Another quick angle chase gives $\angle MDN = a+c$, $\angle MCN = a+d$ and $\angle DCM = \angle CDM$. So from trig ceva in $\triangle{DCN}$ with point $M$ we get: $$\frac{\sin(\angle DNM)}{\sin(\angle MNC)}\cdot\frac{\sin(\angle NCM)}{\sin(\angle MCD)}\cdot\frac{\sin(\angle CDM)}{\sin(\angle MDN)} = 1$$ So we get that: $$\frac{\sin(\angle DNM)}{\sin(\angle MNC)} = \frac{\sin(a+c)}{\sin(a+d)} = \frac{\sin(\angle DNP)}{\sin(\angle PNC)}$$ And since $\angle DNM + \angle MNC = \angle DNP + \angle PNC = \angle DNC$ we get that $\angle DNM = \angle DNP$ and $\angle MNC = \angle PNC$ so $N$, $M$ and $P$ are collinear. Done!

Let $\overrightarrow{\rm CP}$ and $\overrightarrow{\rm DP}$ meet $\overleftrightarrow{\rm AD}$ and $\overleftrightarrow{\rm BC}$ at $X$ and $Y$. Claim: $\Delta{AMB}$ and $\Delta{XPY}$ are homothetic. Proof. Angle chasing. We are done.

Let $CP\cap\omega,DP\cap\omega,AD\cap BC=X,Y,Z$. Angle chase to get $ZAMB,MXPY$ homothetic which finishes.

Let $AD\cap BC=T$$\quad$ $\angle BDA=\angle BCA$ and $\angle PCB=\angle PDA$ then $\angle PCA=\angle PDB$ $\angle CPD+\angle PDC+\angle PCD=180^\circ=\angle BDA+ \angle BAD+\angle DBA$ $\angle DBA+ \angle DAB=\angle PCD+ \angle PDC$ $\implies$ $2\angle PDB=\angle CAD=2\angle CDM$ $\quad$ so, $\quad$ $\angle CDM=\angle PDB$ $\implies$ $\angle PDC=\angle BDM=\angle MCT$ $\quad$and analogously$\quad$ $\angle PCD=\angle MDT$ $$\frac{MT}{\sin \angle MCT}=\frac{MC}{\sin \angle MTC}\quad and \quad \frac{MT}{\sin \angle MDT}=\frac{MD}{\sin \angle MTD}$$ $$\frac{PT}{\sin \angle PCT}=\frac{PC}{\sin \angle PTC} \quad and \quad \frac{PT}{\sin \angle PDT}=\frac{PD}{\sin \angle PTD}$$ $$\frac{\sin\angle MDT}{\sin\angle MCT}=\frac{\sin\angle MTD}{\sin\angle MTC}\quad and \quad 1=\frac{\sin\angle PDT}{\sin\angle PCT}=\frac{\sin\angle PTD}{\sin\angle PTC}\cdot\frac{PC}{PD}$$ $$\frac{\sin\angle PTC}{\sin\angle PTD}=\frac{PC}{PD}=\frac{\sin\angle PDC}{\sin\angle PCD}=\frac{\sin\angle MCT}{\sin\angle MDT}=\frac{\sin\angle MTC}{\sin\angle MTD}$$$$\implies \frac{\sin\angle PTC}{\sin\angle PTD}=\frac{\sin\angle MTC}{\sin\angle MTD}$$ Let $\quad$ $\angle CTD=\alpha$, $\quad$ $\angle PTC=\beta$, $\quad \angle MTC=\theta$ $$\frac{\sin\angle \beta}{\sin\angle (\alpha-\beta)}=\frac{\sin\angle \theta}{\sin\angle (\alpha-\theta)}$$$$\implies \beta=\theta$$$$\angle PTC=\angle MTC $$$\implies$ $P,T,M$ are colleniar. Done!

Let $AD \cap BC = \{T\}$. Let the second intersections of $PC$ and $PD$ with $(ABCD)$ be $X$ and $Y$, respectively. We are going to show $\overline{TMP}$. By easy angle chasing, we conclude $\angle PDB = \angle PCA \implies arc(AX) = arc(YB)$. By angles, also we get $arc(AX) + arc(XY) + arc(YB) = arc(XY) + arc(DC) \implies arc(DM) = arc(MC) = arc(AX) = arc(YB)$ Hence, $XY || AB$, $XM || AD$ and $YM || BC$. So we conclude that $\triangle{ATB}$ and $\triangle{XMY}$ are homothetic. Thus, $AX$, $BY$ and $TM$ are concurrent. Let $Z$ be the intersection point of these lines. So we know that $\overline{TMZ}$. Now Pascal on $(ADYBCX)$ gives us $\overline{TPZ}$. $\overline{TMZ}$ and $\overline{TPZ}$ implies $\overline{TMP}$. $\blacksquare$

Solved with SilverBlaze_SY. We give two solutions, one with MMP (mine) and the other one by chasing angles (SilverBlaze's). [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 = (7.77952,50.31114); pair B = (-7.41116,-7.83103); pair C = (41.85666,-8.11885); pair D = (51.58016,23.65726); pair M = (50.74270,6.53777); pair P = (18.65367,15.52170); pair F = (104.39862,-8.48421); pair X = (37.38301,20.14939); pair Y = (32.01336,1.91008); import graph; size(15cm); 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((17.36719,16.75068), 34.90312), linewidth(0.6)); draw(A--D, linewidth(0.6)); draw(D--M, linewidth(0.6)); draw(M--C, linewidth(0.6)); draw(B--M, linewidth(0.6) + ffxfqq); draw((24.59535,0.07721)--(23.56486,-2.04245), linewidth(0.6) + ffxfqq); draw((24.59535,0.07721)--(22.69624,1.47304), linewidth(0.6) + ffxfqq); draw((21.66576,-0.64662)--(20.63528,-2.76630), linewidth(0.6) + ffxfqq); draw((21.66576,-0.64662)--(19.76666,0.74919), linewidth(0.6) + ffxfqq); draw(D--F, linewidth(0.6) + linetype("4 4") + red); draw(F--C, linewidth(0.6) + linetype("4 4") + red); draw(P--F, linewidth(0.6) + linetype("4 4") + red); draw(P--D, linewidth(0.6) + ffxfqq); draw((38.04650,20.31333)--(37.01602,18.19365), linewidth(0.6) + ffxfqq); draw((38.04650,20.31333)--(36.14740,21.70915), linewidth(0.6) + ffxfqq); draw((35.11691,19.58948)--(34.08643,17.46980), linewidth(0.6) + ffxfqq); draw((35.11691,19.58948)--(33.21781,20.98531), linewidth(0.6) + ffxfqq); draw(P--C, linewidth(0.6) + blue); draw((32.36897,1.54776)--(30.01987,1.35631), linewidth(0.6) + blue); draw((32.36897,1.54776)--(32.60426,3.89287), linewidth(0.6) + blue); draw((30.25517,3.70142)--(27.90607,3.50997), linewidth(0.6) + blue); draw((30.25517,3.70142)--(30.49046,6.04653), linewidth(0.6) + blue); draw(A--M, linewidth(0.6) + blue); draw((31.37491,26.27079)--(29.02581,26.07934), linewidth(0.6) + blue); draw((31.37491,26.27079)--(31.61021,28.61591), linewidth(0.6) + blue); draw((29.26111,28.42446)--(26.91201,28.23301), linewidth(0.6) + blue); draw((29.26111,28.42446)--(29.49641,30.76957), linewidth(0.6) + blue); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, NE); dot("$M$", M, NE); dot("$P$", P, NW); dot("$F$", F, NE); dot("$X$", X, N); dot("$Y$", Y, dir(270)); [/asy][/asy] The crux of the problem is to note that $MA\parallel CP$ and $MB\parallel DP$. Claim: $MA\parallel CP$ and $MB \parallel DP$. Proof. Let $X=MA\cap DP$ and $Y=MB\cap CP$. Firstly, note that, \[ \measuredangle YPX=\measuredangle CPD=\measuredangle ADB =\measuredangle AMB = \measuredangle XMY .\]Now note that, \begin{align*} \measuredangle PXM=\measuredangle DXA &= \measuredangle DAX+\measuredangle XDA\\ &= \measuredangle DAM+\measuredangle PDA\\ &=\measuredangle MBC+\measuredangle BCP\\ &=\measuredangle YBC +\measuredangle BCY\\ &=\measuredangle BYC=\measuredangle MYP .\end{align*}Now since the opposite angles of the quadrilateral $PYMX$ are equal, we must have that $PYMX$ is a parallelogram. This gives us our desired result. $\blacksquare$ Now we redefine $P$ as the intersection of line through $D$ parallel to $MB$ with the line through $C$ parallel to $MA$. This our problem statement can now be generalized as follows. Quote: Let $ABC$ be a triangle and let $M$ denote the midpoint of arc $\widehat{BC}$ not containing $A$. Let $D$ be any arbitrary point on $\odot(ABC)$. Let the line through $D$ parallel to $MB$ be $\ell_1$. Similarly, let the line through $C$ parallel to $MA$ be $\ell_2$. Define $P=\ell_1\cap \ell_2$. Prove that $AD$, $PM$ and $BC$ are concurrent. Fix $A$, $C$ and $D$. We animate $B$ projectively on $\odot(ACD)$. Firstly note that $M$ has degree $0$ (as it is also fixed). So, the line $MA$ is fixed and has degree $0$. Thus $\ell_2$ also has degree $0$. Furthermore note that the line at infinity is also fixed as $\infty_{AC}$ and $\infty_{AD}$ are fixed. Now note that as $MA$ is fixed, $CP$ is also fixed. Then we have, \[ B\mapsto MB\mapsto MB\cap \ell_{\infty}\mapsto D\infty_{MB}\mapsto D\infty_{MB}\cap CP \text{ via }\odot(ACD)\mapsto \mathcal{P}(M) \mapsto \ell_{\infty}\mapsto \mathcal{P}(D) \mapsto CP \]is projective. This means that $P$ has degree $1$. So the line $PM$ has degree $1+0=1$ Now note that, \[ B\mapsto CB\text{ via }\odot(ACD)\mapsto \mathcal{P}(C) \]is projective. This means that the line $CB$ has degree $1$. Thus finally, the condition that $AD$, $CB$ and $PM$ has degree $0+1+1=2$. Firstly note that when $B=C$, we can reduce the degree by $1$ using Zack's Lemma. So now we need to check for just two cases. $B=D$ -- Then note that $P=CP\cap DM$. So, $AD\cap PM\cap BC=D$. $B=A$ -- Then note that $DP\parallel AM\parallel CP$ which implies that $P\equiv \infty_{AM}$. Then clearly $AD\cap PM\cap BC=A$. And we are done.

oronsh wrote: 2023 g3 is good warmup geo Let $CP$, and $DP$ intersect $(ABCD)$ again at $C'$, and $D'$ respectively. Let $AD \cap BC = E$ . Let $BD' \cap AC' =X$. Let $C' M \cap DD'$, and $D'M \cap CC'$ be $F$ and $G$ respectively. $\angle ADP = \angle PCB$ implies $D'C' \parallel AB$ and arcs $BD'$, $AC'$, $DM$, and $CM$ have equal length. Therefore $ADMC'$ is an isosceles trapezoid along with $BCMD'$. Observe by pascals on $DD'BCC'A$, $X$, $P$, and $E$ must be collinear. By pascals on $CC'MMDD'$ $FG \parallel CD$. Therefore $(FGC'D')$ is cyclic. Observe that $\angle D'FC' = \angle C'FD = 180^\circ - \angle D'DA= \angle D'C'X$ therefore $XC'$ is tangent to $(GFC'D')$. Now Pascals on $C'FD'D'GC'$ gives $X$, $P$, $M$ collinear. This combined with $X$, $P$, $E$ collinear gives us the desired result.

Moral of the story: if you don't know what to do, apply Pascal and see what happens. We let $\overline{CP}$ intersect $\overline{AD}$ at $E$ and $\overline{DP}$ intersect $\overline{BC}$ at $F$. The key claim is the following: Claim: $BF=BD$ and $AC=AE$. Proof: We instead imagine picking points $E'$ and $F'$ on $\overline{AD}$ and $\overline{BC}$ satisfying the length conditions, then show that $P = \overline{CE'} \cap \overline{DF'}$ satisfies both angle conditions; obviously it satisfies the latter one as $FEDC$ is cyclic (see below). To see the former condition, $\angle CE'D = \frac 12 \angle CAD = \frac 12 \angle CBD = \angle CF'D$, so triangles $ACE'$ and $BDF'$ are similar. So \[\measuredangle CPD = \measuredangle(\overline{E'C}, \overline{F'D}) = \measuredangle(\overline{AE'}, \overline{DB}) = \measuredangle ADB\]as needed. $\blacksquare$ Now, note that the tangents at $C$ and $D$ to $(FEDC)$ meet at $M$ as $\angle MDC = \frac 12 \angle DAC = \angle DEC$. Thus by Pascal on $CEDDFC$ we have the desired concurrence. Edit: after writing this solution I realized that triangles $PEF$ and $MAB$ are homothetic, which just like solves the problem. Geometry is too hard.