mofumofu wrote: The line parallel to $AM$ through $D$ meets segment $AB$ at $D$,... Typo?

∠POM=∠ACD

We only need to prove that angle ACD=angle POM. Using trigonometric functions

I complex bashed this during the contest... the computations should have been easy but somehow I keep messing up arithmetics :/ Hope I can receive a 7 on it lol.

I will prove its generalization below. Generalization: Given a $\triangle ABC$ with its circumcenter $O$, orthocenter $H$, and circumcircle $c$. $D$ is a point lying on arc $BC$ of $c$ and $D'$ is its reflection point WRT $BC$. $A'A$ is a diameter of $c$. A line passing through $O$ which is perpendicular to $HD'$ inter sects $AC$, $AB$ at $E$, $F$ respectively. Suppose $BD\cap CA'=P$, $CD\cap BA'=Q$. Prove that $\angle OEB+\angle OPB=\angle OFC+\angle OQC$. Proof. Suppose the reflection point of $H$ WRT $BC$ is $H'$, then $\angle BFO=\angle AHD'-\angle ABC=180^\circ-\angle AH'D-\angle ABC=\angle ABD-\angle ABC=\angle PBC$. Now by $\angle FBO=90^\circ-\angle ACB=\angle PCB$ we get $\triangle OBF\stackrel{-}{\sim}\triangle PCB$, so$\dfrac{FB}{CB}=\dfrac{BO}{CP}=\dfrac{CO}{CP}$. Then by $\angle OCP=\angle ABC$ we have $\triangle FBC\stackrel{-}{\sim}\triangle OCP$, so$\angle OPC=\angle FCB$. Similarly we have $\angle OQB=\angle EBC$, hence $\angle OEB+\angle OPB=\angle OEB+\angle BPC-\angle OPC=\angle OEB+\angle BOE-\angle FCB=180^\circ-\angle EBO-\angle FCO-\angle BCO$. By $\angle OBC=\angle OCB$ it's obvious to see $\angle OFC+\angle OQC$ has the same expression. So $\angle OEB+\angle OPB=\angle OFC+\angle OQC$. $\quad\Box$

Very nice problem . It is glad to see that there are finally some nice problems appearing on the CMO [asy][asy] size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.760433772373926, xmax = 18.737430145278708, ymin = -26.36714817636629, ymax = 11.432112570887025; /* image dimensions */pen zzttqq = rgb(0.6,0.2,0); draw((-12.88157172452285,-8.344715836958592)--(-1.7044176212692368,-11.395787089657654)--(-4.908070810890858,-4.502571184832435)--cycle, linewidth(0.8) + zzttqq); draw((-12.88157172452285,-8.344715836958592)--(-3.5699087015309274,-0.8567019520006959)--(2.7485023397422332,-8.942805405744245)--cycle, linewidth(0.8) + zzttqq); draw((2.7485023397422332,-8.942805405744245)--(-2.855762980493301,-14.913373749478474)--(-4.908070810890858,-4.502571184832435)--cycle, linewidth(0.8) + blue); draw((2.7485023397422332,-8.942805405744245)--(-12.88157172452285,-8.344715836958592)--(-0.44421935937243345,7.659347429652019)--cycle, linewidth(0.8) + blue); 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dot((2.7485023397422332,-8.942805405744245),dotstyle); label("$C$", (2.9079928914184077,-8.54407902655381), NE * labelscalefactor); dot((-5.246506306683519,-13.347018808213976),linewidth(4pt) + dotstyle); label("$M$", (-5.106407330309352,-13.00981447348669), NE * labelscalefactor); dot((-4.908070810890858,-4.502571184832435),linewidth(4pt) + dotstyle); label("$O$", (-4.74755358903796,-4.19796149337806), NE * labelscalefactor); dot((-3.5699087015309274,-0.8567019520006959),linewidth(4pt) + dotstyle); label("$D$", (-4.428572485685611,-0.29044297731178914), NE * labelscalefactor); dot((-0.44421935937243345,7.659347429652019),linewidth(4pt) + dotstyle); label("$E$", (-0.2818181421050788,7.963193071930232), NE * labelscalefactor); dot((-10.371648875976552,-11.465911408767088),dotstyle); label("$K$", (-10.21010498394693,-11.056055215453556), NE * labelscalefactor); dot((-8.629126642062003,-11.130810979168137),linewidth(4pt) + dotstyle); label("$P$", (-8.455708915509012,-10.816819387939294), NE * labelscalefactor); dot((-7.74995749121634,-14.726095898302848),linewidth(4pt) + dotstyle); label("$Q$", (-7.578510881290054,-14.405356800653216), NE * labelscalefactor); dot((-1.7044176212692368,-11.395787089657654),linewidth(4pt) + dotstyle); label("$P'$", (-1.5577425555144733,-11.095927853372599), NE * labelscalefactor); dot((-2.855762980493301,-14.913373749478474),linewidth(4pt) + dotstyle); label("$Q'$", (-2.714049055166737,-14.604719990248434), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Reflect $P,Q$ about $OM$. Let the images be $P',Q'$. CLAIM. $\triangle BP'C\sim \triangle BOD$. Proof. $$\angle OBD=\angle OBA=90^{\circ}-\angle ACB=\angle PCB=\angle P'BC$$$$\angle ODB=\angle \frac{A}{2}=\angle PBC=\angle P'CB$$$\blacksquare$ Hence by spiral similarlity, $$\triangle BOP'\sim\triangle BDC$$by symmetry $$\triangle COQ'\sim\triangle CEB$$Therefore, \begin{align*} \angle ODC-\angle OEB&=\angle BDC-\angle BEC\\ &=\angle BOP'-\angle COQ'\\ &=\angle COP-\angle BOQ\\ &=\angle COQ-\angle BOP\\ &=\angle COQ+\angle OCM-\angle BOP-\angle OBM\\ &=\angle OPB-\angle OQC \end{align*}as desired.

We can do this without adding any new points. 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Now $\angle OCP=90^{\circ} - \angle OCA = \angle DBC$, thus $\triangle BCD\sim \triangle CPO$. Similarly $\triangle CBE\sim \triangle BQO$. Finally we can chase: \begin{align*} \angle OPB-\angle OQC&=(\angle OBP +\angle OPB)-(\angle OCQ + \angle OQC)\\ &=(180^{\circ}-\angle BOP)-(180^{\circ}-\angle COQ)\\ &=\angle COP-\angle BOQ\\ &=\angle BDC-\angle BEC\\ &=(\angle ODC+\angle ADE)-(\angle OEB+\angle AED)\\ &=\angle ODC-\angle OEB \end{align*} and we are done.

mofumofu wrote: In acute triangle $ABC (AB>AC)$, $M$ is the midpoint of minor arc $BC$, $O$ is the circumcenter of $(ABC)$ and $AK$ is its diameter. The line parallel to $AM$ through $O$ meets segment $AB$ at $D$, and $CA$ extended at $E$. Lines $BM$ and $CK$ meet at $P$, lines $BK$ and $CM$ meet at $Q$. Prove that $\angle OPB+\angle OEB =\angle OQC+\angle ODC$.

We start by noting the following: \[\angle PCB = \angle KCB = \angle KAB = \angle OAB = \angle OBA = \angle OBD,\]and \[\angle PBC = \angle MBC = \angle MCB = \angle MAB = \angle ODB.\]So, we have $\triangle BPC \sim \triangle DOB - (1).$ Similarly, $\triangle CQB \sim \triangle EOC - (2)$. We further observe that, as \[\frac{BC}{DB} = \frac{PC}{OB} = \frac{PC}{OC},\]here we get $\triangle CPO \sim \triangle BCD - (3)$, as $\angle OCP=90^{\circ} - \angle OCA = \angle DBC$. Likewise, $\triangle CBE \sim \triangle BQO - (4)$. Now, we note that as $\angle OQC = \angle BQC - \angle BQO$, and \begin{align*} \angle BQC \overset{(2)}{=} \angle COE = \angle COA + \angle AOE = 2\angle B + \angle AOD &= 2\angle B + \left(180^{\circ} - \overline{\angle ODA + \angle OAB} \right) \\&= 2\angle B + \cancel{180^{\circ}} - (\cancel{180^{\circ}} - \angle BAM) - (90^{\circ} - \angle C) \\&= 2\angle B + \angle A/2 - 90^{\circ} + \angle C \\&= 90^{\circ} + \angle B - \angle A/2 \end{align*}also, $\angle BQO \overset{(4)}{=} \angle EBC = 180^{\circ}- \angle C - (\angle A/2 + \angle BEO).$ Whence, \[\angle OQC + \angle ODC = 90^{\circ} - \angle A + \angle BEO + \angle ODC\quad(*).\]Next, \[\angle OPB + \angle OPC = \angle BPC \overset{(1)}{=} \angle DOB = 180^{\circ} - \angle OBD - \angle BAM = 90^{\circ}+\angle C - \frac{A}{2}\]yields \begin{align*} \angle OPB = 90^{\circ}+\angle C - \frac{A}{2} - \angle OPC &\overset{(3)}{=} 90^{\circ}+\angle C - \frac{A}{2} - \angle DCB \\&= 90^{\circ}+\angle C - \cancel{\frac{A}{2}} - \left\{180^{\circ}-\angle B - \left(\cancel{\frac{\angle A}{2}} + \angle ODC \right) \right\} \\&= 90^{\circ} - \angle A+\angle ODC. \end{align*}Adding $\angle OEB$ to both sides of the above equation, and comparing it with $(*)$ gets us done. $\blacksquare$ Remarks: $\triangle ADE$ is isosceles with $AD=AE$; also $\triangle FKM$ is isosceles, with $FK=FM$, where $F$ is $OD \cap CM.$ Further, if we let $BM \cap OD$ as $G$, we get a couple of cyclic quadrilaterals, namely, $(BKGO), (COKF), (BECG)$, and $(BDCF)$.

Got recommended this by @MeineMeinung --- saying that this resembles the third problem of APMO 2021. I agree, albeit in a completely different manner. First Comment: A lot of available angles yet none of the four relates to each other (at first glance). Solution for this? Rotate existing same angles forming similar triangles. $\color{green} \rule{3.9cm}{2pt}$ $\color{green} \diamondsuit$ $\boxed{\textbf{Angle Spotting.}}$ $\color{green} \diamondsuit$ $\color{green} \rule{3.9cm}{2pt}$ We point out sets of angles which measure the same, and make the problem more-spiral-similarity-friendly. $\color{green} \rule{3.9cm}{0.2pt}$ Since $\overline{OD}, \overline{OE}$ are segments from the same line and $OD,OE \parallel AM$, then \[ \angle ODB = \angle MAB = \dfrac{\angle A}{2} = \angle CAM = \angle CEO. \]Also, $(ACMB)$ concylic and $\overline{AM}$ internal angle bisector implies \[ \angle MCB = \angle MBC = \angle MAC = \dfrac{\angle A}{2}. \]For simplicity, call $\frac{\angle A}{2} = \alpha$.

The next most important angles are the following: \[ \angle DBO = \angle PCB = 90 - \angle C, \angle ECO = \angle QBC = 90 - \angle B. \]From here, we can see that $\triangle DOB \sim \triangle BPC$, oppositely oriented. This also holds for its $E,C,Q$ counterpart. Since two similar triangles which are oppositely oriented and $\textsf{almost rotate-able}$ with each other are rare, let's transform this to a almost-too-good-to-be-true spiral! Namely, draw $P'$ so that $BCPP'$ is an isosceles trapezoid with bases $BC$ an $PP'$. The beauty of this transformation is not only the triangles behave relatively better with themselves, $\angle OPB$ can be substituted with $\angle OP'C$! Also do this with $Q$. Now, the problem is equivalent to

Let's proceed to the next part. $\blacksquare$

$\color{red} \rule{8cm}{2pt}$ $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Depersonalisation of} \ 90 - B \ \textbf{and} \ 90-C.}$ $\color{red} \clubsuit$ $\color{red} \rule{8cm}{2pt}$ Remove $A$, $E$ and $Q$ entirely from the picture, and consider the struture of the five points \[ B,O,C; \quad D,P' \]We Claim that given $BO = OC$, $\triangle DOB \sim \triangle CP'B$ and \[ \angle ODB = \angle P'CB = \alpha \](with $\angle A$'s definition to be half of $\angle BOC$ now), then \[ \angle OP'C - \angle ODC = 90-A. \]$\color{red} \rule{25cm}{0.2pt}$ $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{Proof.}}$ $\color{red} \spadesuit$ The bulk of the proof. Rotate the whole four-point configuration $D,O,C,B$ by $\angle DBC$ and resize it so that \[ \text{The image of} \ \triangle DOB = \triangle CP'B. \]Name the image of $D$ to be $D_R$ (this name intends to say that $D_R$ is $D$, rotated). Then add two fixed points of reference: let $F_1$ to be the point on $\overline{CP'}$ so that \[ (COB) \cap \overline{CP'} = \{C,F_1\} \]and $F_2$ also on $\overline{CP'}$ so that $BF_2 \parallel OF_1$. Simple angle chasing gives us \[ \angle CF_1O = \angle F_1F_2B = \angle F_1BF_2 = 90-A. \] Thus, disregarding $\angle ODC$ and replacing it with its image, we are left to prove \[ \angle D_RCP' = \angle P'OF_1. \]$\blacksquare$ $\blacksquare$ In fact, we Claim that $\triangle CD_R \sim OP'F_1$, directly implying the desired conclusion.

Observe (repeated in Notes 3) that $\angle BF_2F_1 = \angle BD_RP'$, implying $(BP'D_RF_2)$ cyclic. In turn, this will cause \[ \angle CF_2D_R = \angle P'F_2D_R = \angle P'BD_R = 90-A = \angle P'F_1O. \]Now we prove that \[ \dfrac{OF_1}{P'F_1} = \dfrac{CF_2}{D_RF_2}, \ \text{or} \ \dfrac{OF_1}{CF_2} = \dfrac{PF_1}{D_RF_2}. \]We can count the LHS manually: repeated sine rule on the circle $(COBF_1)$ yields \[ \text{LHS} = \dfrac{CF_1+F_1B}{F_1O} = \dfrac{\sin{(\alpha)}+\sin{(3\alpha)}}{\sin{(90-\alpha)}} = \dfrac{2 \sin{(2\alpha)} \cos{(\alpha)}}{\cos{(\alpha)}} = 2 \sin{A} = \dfrac{BF_2}{BF_1}. \]Finally, we end this by proving \[ \triangle P'F_1B \sim \triangle D_RF_2B \]which directly establishes this Section's Claim. Indeed, we know that $\angle BP'F_1 = \angle BD_RF_2$ by cyclicity and $\angle BF_1P = \angle BF_2D_R = \angle A$.

We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$

@Geoking and me have a unique solution. Let $F = BC \cap KM$. Note that $BPC \sim DOB$ so $$\frac{BC}{BP}=\frac{DB}{DO}$$Also note that $FBCM \sim ODEA$ so $$\frac{BF}{BC}=\frac{DO}{DE}$$So $$\frac{BF}{BP}=\frac{BF \cdot BC}{BC \cdot BP}=\frac{DO \cdot DB}{DE \cdot DO}=\frac{DB}{DE}$$and $\angle FBP= 180^\circ - \angle \frac{A}{2}= \angle BDE$. So, $FBP \sim BDE$. Similarly $FCQ \sim CED$. Note that by brocard $OFQP$ is an orthocentric system.So, $$\angle OPB + \angle OEB= \angle OPB+ \angle BPF=\angle OPF=180^\circ- \angle OQF=180^\circ-\angle CQF +\angle OQC=180^\circ-\angle EDC +\angle OQC=\angle OQC +\angle ODC$$

I'm going to admit that I did indeed peek at a few major claims of some of the solutions above, because bashing length/angle chasing isn't very intuitive when you have an entire ESSAY on it. Here is a very long and detailed elementary solution. Is there an alternative short solution that you could provide me with? Thanks, because I really hate these types of problems in geometry (or FEs!) where you can derive and see all this sim triangle and Brocard's I even saw myself but I couldn't piece it together at the end unless I spent another few hours but i Have to go soon, so i did need to see how to finish

Solved with kingu who apparently hates Chinese sim triangle problems. Claim : $\triangle PCO \sim \triangle BCD$ and $\triangle BQO \sim \triangle BEC$. Proof : Initially notice that $\angle OCP = \angle DBC$, now observe that as $\triangle DOB \sim BCP$, then $\frac{BD}{BC} =\frac{BO}{CP}= \frac{CO}{CP}$ implying the desired similarity. $\triangle BQO \sim \triangle BEC$ follows analogously. Now notice that \[\angle OPB - \angle ODC = \angle BPC - \angle OPC - \angle ODC = \angle BOD - \angle DCB - \angle ODC = \angle OBC\]and analogously, $\angle OQC - \angle OEB = \angle OCB$, as desired.

As far as I know every China MO Geo can be bashed out

This problem was really cool where i ended up angle chasing for 3 pages, in the end it was all simplified to $\triangle BOD\sim \triangle CPB$ and $\triangle BEC\sim \triangle QOB$, main motivation was just construction of reflection of different points as the figure was sorta symmetric (which is always a good thing) and simplifying the angle condition of the problem to match the similarity conditionsThe problem didn't use anything out of the blue, just plain old angle chasing(which felt like bash).