Nice problem,

bgn wrote: Let $O$ be the circumcenter of triangle $ABC$. Line $CO$ intersects the altitude from $A$ at point $K$. Let $P,M$ be the midpoints of $AK$, $AC$ respectively. If $PO$ intersects $BC$ at $Y$, and the circumcircle of triangle $BCM$ meets $AB$ at $X$, prove that $BXOY$ is cyclic. Proposed by Ali Daeinabi - Hamid Pardazi Let $A'$ be the antipode of $A$, $X$ be the point on $\overline{AB}$ with $\overline{AC}$ tangent to $\odot(CXB)$. It suffices to prove $\angle XA'K=\angle ABC$. Apply inversion $\odot(A, AC)$. Clearly, we obtain the equivalent Lemma. Point $X$ lies on side $\overline{AB}$ of triangle $ABC$ with $AC^2=AX \cdot AB$. Point $L$ is defined so that $\angle CLA+\angle CBA=90^{\circ}$ while $\overline{CL} \perp \overline{BC}$. Let $Y$ be the foot of the perpendicular from $A$ onto $\overline{CX}$. Then $\angle BYL=90^{\circ}$. (Proof) It suffices to show $\triangle BYL \sim \triangle BCD$, where $\overline{CD}$ is an altitude in $\triangle ABC$. By spiral similarity, we just need $\triangle BYD \sim \triangle BLC$. Drop perpendiculars $\overline{LE}, \overline{YZ}$ on $\overline{BC}, \overline{BA}$, respectively. Then we have $\triangle DYZ \sim \triangle CLE$, so it suffices to prove $$\frac{DZ}{DB}=\frac{CE}{CB} \iff \frac{DZ}{CE}=\frac{DB}{CB}=\cos B.$$Finally, we remark that $DZ=CY \cos C=AC \cos B \cos C$ while $CE=AC \cos C$, hence the lemma must hold. $\blacksquare$ Remark. The solution is written backwards, just as it was discovered. Our main idea was in a "local to global" style, removing most of the unfamiliar structure initially, to get a new configuration, that was easier to comprehend. Our motivation for doing inversion (and spiral similarity afterwards) was because we had a floating point $K$; the glaring relation $AC^2=AX \cdot AB$; the weird angle $XA'K$.

bgn wrote: Let $O$ be the circumcenter of triangle $ABC$. Line $CO$ intersects the altitude from $A$ at point $K$. Let $P,M$ be the midpoints of $AK$, $AC$ respectively. If $PO$ intersects $BC$ at $Y$, and the circumcircle of triangle $BCM$ meets $AB$ at $X$, prove that $BXOY$ is cyclic. Proposed by Ali Daeinabi - Hamid Pardazi

Straightforward areal coordinates with $\triangle{ABC}$ as reference triangle (Leaving the calculations to the reader, we get ) $P=(a^2(2S_{AC}+b^{2}S_{B}) : b^{2}S_{BC} : b^{2}S_{B}^{2})$ $X=(2c^{2}-b^{2} : b^{2} : 0)$ Let $(BXO) \cap BC = {Y}$, then $Y=(0 : b^{2}S_{B} : 2c^{2}S_{C}-b^{2}S_{B})$ We prove that $P,O,Y$ are collinear by evaluating the determinant: $\begin{vmatrix}a^2(2S_{AC}+b^{2}S_{B}) & b^{2}S_{BC} & b^{2}S_{B}^{2} \\a^{2}S_{A} & b^{2}S_{B} & c^{2}S_{C} \\0 & b^{2}S_{B} & 2c^{2}S_{C}-b^{2}S_{B} \end{vmatrix}$ Which in a bit of calculations gets to 0 $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.90882150443, xmax = 11.8370164443, ymin = -5.01012125464, ymax = 6.82185250768; /* image dimensions */ pen ffqqff = rgb(1.,0.,1.); /* draw figures */ draw(circle((2.72,1.58), 3.73363094052)); draw((4.74,4.72)--(6.09058139315,-0.0259828990882), red); draw((-0.724054214725,0.138302886859)--(6.09058139315,-0.0259828990882)); draw((2.72,1.58)--(6.09058139315,-0.0259828990882)); draw(circle((2.68886622651,0.288559476513), 3.41622642413)); draw((2.12354440067,2.5260600069)--(0.7,-1.56), ffqqff); draw((0.7,-1.56)--(4.6422216573,0.664117916228), ffqqff); draw((4.74,4.72)--(4.62643579367,0.00931441164022)); draw((0.7,-1.56)--(6.09058139315,-0.0259828990882)); draw((-0.724054214725,0.138302886859)--(0.7,-1.56)); draw((2.12354440067,2.5260600069)--(6.09058139315,-0.0259828990882), red); draw((2.12354440067,2.5260600069)--(4.6422216573,0.664117916228), ffqqff); draw((4.74,4.72)--(0.7,-1.56)); draw((2.12354440067,2.5260600069)--(4.74,4.72), red); draw((2.12354440067,2.5260600069)--(-0.724054214725,0.138302886859)); /* dots and labels */ dot((2.72,1.58),dotstyle); label("$O$", (2.80860273009,1.71367106411), NE * labelscalefactor); dot((4.74,4.72),dotstyle); label("$A$", (4.82811632407,4.87361586409), NE * labelscalefactor); dot((-0.724054214725,0.138302886859),dotstyle); label("$B$", (-0.636449871385,0.288132056598), NE * labelscalefactor); dot((6.09058139315,-0.0259828990882),dotstyle); label("$C$", (6.1823783812,0.121819172389), NE * labelscalefactor); dot((4.62643579367,0.00931441164022),dotstyle); dot((4.6422216573,0.664117916228),dotstyle); label("$K$", (4.73308039023,0.810829692685), NE * labelscalefactor); dot((5.41529069657,2.34700855046),dotstyle); label("$M$", (5.51712684436,2.49771751824), NE * labelscalefactor); dot((4.69111082865,2.69205895811),dotstyle); label("$P$", (4.78059835715,2.83034328666), NE * labelscalefactor); dot((0.128192450216,0.117757105511),dotstyle); label("$Y$", (0.21887353312,0.26437307314), NE * labelscalefactor); dot((3.43177220033,3.62303000345),dotstyle); label("$X$", (3.52137223385,3.75694364154), NE * labelscalefactor); dot((0.7,-1.56),dotstyle); label("$A'$", (0.789089136124,-1.42251475241), NE * labelscalefactor); dot((2.12354440067,2.5260600069),dotstyle); label("$Z$", (2.21462814363,2.66403040245), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $A'$ be the antipode of $A$, and $Z$ on $AB$ with $AZ\cdot AB = AC^2$; suffices to prove $\angle ZA'K=\angle B$ (since $X,O,P$ are midpoints of $AZ,AA',AK$ so $\angle B=\angle ZA'K=\angle XOP$.) We have $\angle KAC= \angle BAO=\angle BCA'$, and $\angle KCA=\angle OAC=\angle A'BC\implies \triangle AKC\sim \triangle CA'B$. From $\triangle AZC\sim \triangle ACB$, we get $\frac{ZA}{ZC}=\frac{AC}{CB}=\frac{AK}{CA'}$, or $\frac{ZA}{AK}=\frac{ZC}{CA'}$, combine $\angle ZAK=\angle ZCA' = 90^{\circ}-\angle B$, we have $\triangle ZAK\sim \triangle ZCA'\implies \triangle ZA'K\sim \triangle ZCA\sim \triangle CBA$ and we are done.

My solution: Since $XM$ is anti-parallel to $BC$, $AO \perp XM$ $\Rightarrow$ $\measuredangle XMO = \measuredangle OAC = \measuredangle ACO = \measuredangle AMP$ Note that $AK, AO$ are isogonal $\Rightarrow$ $P, O$ are isogonal conjugate wrt to $\triangle AXM$ $\Rightarrow$ $\measuredangle PXA = \measuredangle OXM$ $\Rightarrow$ $\triangle XPA \stackrel{+}{\sim} \triangle XOM$ $\Rightarrow$ $\triangle XPO \stackrel{+}{\sim} \triangle XAM$ $\Rightarrow$ $\measuredangle XOP = \measuredangle XMA = \measuredangle ABC$ $\Rightarrow$ $BXOY$ is cyclic, as desired

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 11.48, ymin = -1, ymax = 6.3; /* image dimensions */ pen ffqqff = rgb(1.,0.,1.); /* draw figures */ draw(circle((2.72,1.58), 3.87200206612)); draw((4.9,4.78)--(6.4602440518,0.57851388777)); draw((-0.985434663386,0.456543745676)--(4.9,4.78)); draw((-0.985434663386,0.456543745676)--(6.4602440518,0.57851388777)); draw((2.72,1.58)--(6.4602440518,0.57851388777)); draw(circle((2.73946095501,0.392004183645), 3.72545470017)); draw((3.43961585289,2.63631684828)--(6.4602440518,0.57851388777)); draw(circle((4.5901220259,1.07925694388), 1.93600103306), green); draw((2.72,1.58)--(1.98782637495,0.505249724701), ffqqff); draw((4.93112768863,2.8798073402)--(0.86030098045,0.4867793558)); draw((3.43961585289,2.63631684828)--(2.73740469421,0.517528816723), ffqqff); draw((3.43961585289,2.63631684828)--(0.86030098045,0.4867793558), ffqqff); draw((3.43961585289,2.63631684828)--(2.72,1.58), ffqqff); draw((3.43961585289,2.63631684828)--(4.9,4.78)); draw((4.9,4.78)--(4.9692260547,0.554089025521)); /* dots and labels */ dot((2.72,1.58),dotstyle); label("$O$", (2.8,1.7), NE * labelscalefactor); dot((4.9,4.78),dotstyle); label("$A$", (4.98,4.9), NE * labelscalefactor); dot((-0.985434663386,0.456543745676),dotstyle); label("$B$", (-0.9,0.58), NE * labelscalefactor); dot((6.4602440518,0.57851388777),dotstyle); label("$C$", (6.54,0.7), NE * labelscalefactor); dot((4.9692260547,0.554089025521),dotstyle); dot((4.96225537727,0.979614680403),dotstyle); label("$K$", (5.04,1.1), NE * labelscalefactor); dot((5.6801220259,2.67925694388),dotstyle); label("$M$", (5.76,2.8), NE * labelscalefactor); dot((4.93112768863,2.8798073402),dotstyle); label("$P$", (5.02,3.), NE * labelscalefactor); dot((0.86030098045,0.4867793558),dotstyle); label("$Y$", (0.94,0.6), NE * labelscalefactor); dot((3.79163264322,3.96579033583),dotstyle); label("$X$", (3.88,4.08), NE * labelscalefactor); dot((3.43961585289,2.63631684828),dotstyle); label("$Z$", (3.52,2.76), NE * labelscalefactor); dot((1.98782637495,0.505249724701),dotstyle); label("$D$", (2.06,0.62), NE * labelscalefactor); dot((2.73740469421,0.517528816723),dotstyle); label("$N$", (2.82,0.64), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the foot of perpendicular from $C$ to $AO$ be $Z$, $N$ be the midpoint of $BC$, and $AO$ meet $BC$ at $D$. Then $Z,M,C,N,O$ all lie on the circle with diameter $OC$. Now $(Y,N;D,C)=O(P,\infty ;A,K)=-1$. Since $\angle DZC=90^{\circ}$, $DZ$ bisects $\angle NZY$. Thus $\angle YZO=\angle OZN=\angle OCB =\angle YBO$ $\implies B,Y,O,Z$ cyclic. Finally $AZ\cdot AO=AM\cdot AC=AX\cdot AB$ $\implies X\in (BYOZ)$ and we are done.

Let $Z$ be feet of$\perp$ from $C$ onto $AO$. Let $C'$ be reflection of $C$ in $Z$. $\angle AC'C=\angle ACC'=\angle ABC\implies ABC'C$ is cyclic quadrilateral. Now simple angle chasing yields that $\triangle BC'C\sim\triangle AKO$. $P$ is midpoint of $AK$ and $Z$ is midpoint of $CC'$. $\therefore \triangle ZBC\sim\triangle POA\implies\angle YOZ=\angle AOP=\angle CBZ=\angle YBZ$. Hence $BOYZ$ is cyclic. Also $\angle AMO=\angle AZC=90^\circ$. So $MOZC$ is cyclic. Also $BXMC$ is cyclic $\implies AX.AB= AM.AC=AO.AZ\implies BXOZ$ is cyclic. $\therefore BXOY$ is cyclic as desired. Q.E.D.

Let $T$ be the projection of $C$ onto $AO$. Trig Ceva on $OTC$ wrt $Y$ and generalized angle bisector on $AOK$ wrt $P$ yields \[ \frac{ \sin \angle OTY}{\sin \angle YTC} = \frac{\sin \angle OBC}{\cos \angle OBC} \implies \angle OTY = \angle OBC \implies BOYT \text{ is cyclic}. \] On the other hand $AX \cdot AB = AM \cdot AC= AO \cdot AT$, thus $BXOT$ is cyclic. We conclude $BXOYT$ is cyclic.

Let $S=\overline{OM}\cap \overline{AK}$. Let $R$ be the reflection of $M$ over $O$. We have \begin{align*} \measuredangle MXA=\measuredangle MXB=\measuredangle ACB=90^\circ+\measuredangle MAK=\measuredangle MSA, \end{align*}thus $MSXA$ is cyclic quadrilateral, moreover $\overline{XS}\perp\overline{AB}$. Claim: $\triangle KOR\sim\triangle KSX$. Proof. Note that $\measuredangle KOR=\measuredangle COM=\measuredangle CBA=90^\circ+\measuredangle KAX=\measuredangle KSX$. On top of that, $\triangle ABC\sim\triangle KOS$ and $\triangle AXS\sim\triangle AMO$. Hence, \begin{align*} \frac{XS}{OR}=\frac{XS}{OM}=\frac{AX}{AM}=\frac{AC}{AB}=\frac{KS}{KO}, \end{align*}which means that $\triangle KOR\sim\triangle KSX$. $\square$ Now, $\measuredangle RXK=\measuredangle OSK=\measuredangle MXA$ and $\measuredangle XAM=\measuredangle SKO=\measuredangle XKR$, therefore $\triangle XKR\sim\triangle XAM$. By gliding principle, we have $\triangle XPO\sim\triangle XAM$. Hence, $\measuredangle XOY=\measuredangle XOP=\measuredangle XMA=\measuredangle XMC=\measuredangle XBY$, we are done. $\blacksquare$

Suprisingly easy from trig and length bashing. $OP \cap AC=L$, $T$ is the foot of $A$-altitude. $\textbf{Claim:} \triangle LOC \sim \triangle XOB.$ Proof: $(i)$ $XM \perp AO \implies \angle OCM=\angle OAM=\angle XMO$. $(ii)$ $\frac{MO}{MX}=\frac{MO}{ab/2c}=\frac{2cMO}{ab}$ and $\frac{CO}{CL}\overset{menelaus}=\frac{OC-OK}{b}$. Note that $CT=\frac{b^2+a^2-c^2}{2a}$ and $BT=\frac{a^2+c^2-b^2}{2a}$(these can be deduced from Pythagorean theorem). $CK=CO \frac{CT}{CN}=R\frac{b^2+a^2-c^2}{a^2}$ $\frac{MO}{MX}=\frac{CO}{CL} \iff 2cMO=a(OC-OK) \iff 2cMO=a(2OC-CK) \iff 2cMO=a(R-2R\frac{b^2+a^2-c^2}{a^2}) \iff \frac{MO}{R}=\frac{a^2+c^2-b^2}{2ac} \iff \cos(B)=\frac{a^2+c^2-b^2}{2ac} \iff \frac{BT}{c}=\frac{a^2+c^2-b^2}{2ac}$ which is true. $\square$ Now, $\angle BXO=\angle BXM - \angle MXO=\angle 180-C-\angle OLM=\angle OYB$ as desired.

). It follows that $$ \angle POX = \angle MOA = \angle B $$which completes the proof. $\blacksquare$

Let point $T$ be the foot of altitude from $C$ to the line $AO$ and suppose that $N$ is the midpoint of $BC$. So if $H$ is the foot of altitude from $A$ to $BC$ , then quadrilateral $AHTC$ is cyclic and we have $\angle BCT=\angle OAH$. Now while points $M , C , T , N , O$ lie on the circle with diameter $CO$ , then one can see that : $$NT=\sin(\angle BCT).OC=\sin(\angle OAH).AO=NH \implies \frac{NT}{BN}=\frac{NH}{CN}=\frac{KO}{OC}=\frac{KO}{AO}$$So since we have $\angle AOK=\angle COT=\angle CNT$ and $PA=PK$ , we can get $\angle YOT=\angle CBT$ and as the result , quadrilateral $BOYT$ is cyclic. Now since quadrilaterals $TOMC$ and $BXMC$ are cyclic too , we can get : $$AM.AC=AO.AT=AX.AB$$So points $B , X , O , Y ,T$ all lie on the same circle and we're done.

Claim. Points $P,Q=AO\cap MX$ are isogonal conjugates wrt $AXOM.$ Proof. Since $XM,BC$ are antiparallel wrt angle $BAC$ we obtain $AO\perp XM.$ Therefore $Q$ has isogonal conjugate wrt $AXOM$ and it's suffice to check $\angle XAP=\angle MAQ$ and $\angle AMQ=90^\circ -\angle CAO=\angle COM=\angle PMO$ $\Box$ Now the conclusion follows from $\measuredangle XOY=\measuredangle XOP=\measuredangle QOM=\measuredangle XBY$.

Let $AO$ meet $BC$ at $Z$, and let $N$ be the midpoint of $BC$. We have that $(A,K;P,P_{\infty})=-1$, projecting through $O$ gives \begin{align*} (Z,Y;C,N)=-1 \implies \frac{ZY}{CY} = \frac{ZN}{CN} &\implies \frac{CY}{ZY} =\frac{BN}{ZN} \\ & \implies \frac{CY}{ZY} + 1 =\frac{BN}{ZN} + 1 \\ & \implies \frac{ZC}{ZY} = \frac{ZB}{ZN} \\ & \implies ZB \cdot ZY = ZC \cdot ZN \end{align*}Hence, $Z$ lies on the radical axis of $(BOY)$ and $(MONC)$. This means $AO$ is the radical axis of $(BOY)$ and $(MONC)$. Let these circles meet again at $T$. Then \[AO \cdot AT = AM \cdot AC = AB \cdot AX \implies T \in (BOX)\]Hence, $B,O,X,Y,T$ are concyclic as desired.

Let $PO\cap AC=Q$ $\textbf{Claim:} \ XOM\sim XPA$ $\angle OMX=90-\angle B=\angle PAX$ $\frac{1}{AK}.\frac{OM}{OA}=\frac{cos B}{AK}=\frac{sin A}{AC}=\frac{\frac{BC}{2.OA}}{AC}=\frac{BC}{AC}.\frac{1}{2.OA}\implies AK=\frac{2.OM.AC}{BC}\implies \frac{AP}{MO}=\frac{AC}{BC}=\frac{AX}{MX}\implies \frac{AX}{AP}=\frac{MX}{MO}$ and the result follows. $\textbf{Claim:} \ POX\sim ABC$ $\angle PXO=\angle PXM+\angle MXO=\angle PXM+\angle AXP=\angle AXM=\angle C$ $\frac{XO}{XP}=\frac{XM}{XA}=\frac{CB}{CA}$ Thus $POX\sim ABC$. $\textbf{Claim:} \ A,P,X,Q$ are cyclic. \[\angle XPQ=180-\angle OPX=180-\angle A=XAQ\] $\textbf{Claim:} \ X,O,M,Q$ are cyclic. \[\angle OMX=\angle PAX=\angle PQX=\angle OQX\] $\textbf{Claim:} \ B,X,O,Y$ are cyclic. \[180-\angle OXB=\angle AXO=\angle AXM+\angle MXO=\angle C+\angle MQO=\angle BYO\]As desired.$\blacksquare$

Let $N$ be the midpoint of $BC$ , $A'$ be the antipode of $A$, $H$ be the orthocenter of $ABC$. $H_A,H_B,H_C$ be reflections of $H$ across $BC,CA,AB$. $A'-N-H$ meet $(O)$ again at $Q$. Let $AA' \cap BC =D ; CQ \cap AH_A=Z$ and $A'Y$ meet $(O)$ again at $R$. Now we have $(R,Q;A,C)\stackrel{A'}{=}(Y,N;D,C)\stackrel{O}{=}(P,\infty_{AH_A};A,K)-1=(H_B,H_C;A,A')\stackrel{H}{=}(B,C;H_A,Q) \stackrel{Z}{=}(BZ \cap (O),Q;A,C)$. Thus $BR,AH_A,CQ$ concur at $Z$.Thus we project the involution pairs $((B,R);(C,Q);(A,H_A)) \stackrel{A'}{\leftrightarrow}((B,Y);(C,N);(D,\infty_{BC}))$.By radical axis theorem on $(OXB),(ONCM),(BXMC)$ , $A-O-D$ is radax of $(ONC),(OXB) \implies$ $((B,BD \cap (OXB));(C,N);(D,\infty_{BC}))$ are involution pairs. Thus $BD \cap (OBX)=Y.$