Solution

Shorter Solution

Solution by @kimchiks926 Let $X$ is intersection of $AP$ and $\odot(ABC)$. Claim: Quadrilateral $AXMO$ is cyclic. Proof: By symmetry we have $\angle XOP = \angle POK $. On the other side we know that $\angle XAK = \angle XAM =\frac{1}{2} \angle XOK = \angle XOM$, hence the claim follows. By PoP we have that: $$ PX \cdot PA = PM \cdot PO = PB \cdot PC $$Therefore $OMBC$ is cyclic quadrilateral. This implies that $\angle OBC = \angle OCB = \angle PMB =\angle OMC $. Now to finish the problem note that: $$ \angle PMB + \angle BMC +\angle OMC=180^{\circ} \implies \angle OMB + \angle OMC = 180^{\circ} $$

Let $AP$ meet $\Gamma$ at $S$. $\angle OKP = \angle OSP = \angle 180 - \angle ASO = \angle 180 - \angle PAK \implies AOKP$ is cyclic. $\angle APO = \angle KPO \implies AO = OK \implies K$ lies on $\Gamma$. $\angle SOM = \angle KOM = \angle MAS \implies AOMS$ is cyclic. By Radical Axis Theorem we have $BOMC$ is cyclic. $\angle OMB = \angle OCB = \angle OBC = \angle 180 - \angle OMC \implies \angle OMB+\angle OMC=\angle 180$.

BarisKoyuncu wrote: Given an acute angle triangle $ABC$ with circumcircle $\Gamma$ and circumcenter $O$. A point $P$ is taken on the line $BC$ but not on $[BC]$. Let $K$ be the reflection of the second intersection of the line $AP$ and $\Gamma$ with respect to $OP$. If $M$ is the intersection of the lines $AK$ and $OP$, prove that $\angle OMB+\angle OMC=180^{\circ}$. Fun. Let $AP \cap \Gamma=D$ then since $O$ is circumcenter then $2 \angle DAM=\angle DOK$ but since $OM$ bisects $\angle DOK$ we have that $\angle DAM=\angle DOM$ so $OMDA$ is cyclic and by PoP $$PM \cdot PO=PD \cdot PA=PB \cdot PC \implies BMOC \; \text{cyclic}$$Since $BO=OC$ we have that those arcs are equal on $(BMOC)$ so $\angle OMC=\angle OCB$ and its clear that $\angle OMB+\angle OCB=180$ becuase BMOC is cyclic, thus we are done

I'm not as smart as #above so i found a different solution from others Let $\angle BAC=\alpha , \angle ABC=\beta , \angle ACB=\gamma , AP \cap \Gamma={K'}$ Claim1. $K \in \Gamma$

Proof

Let $\angle KAC=x.$ From $\Gamma$ we have: $\angle KAC=KBC=\angle KK'C=x$ We know that $\angle AOK=2 \angle ABK=2(\angle ABC+ \angle CBK)=2(\beta+x)=2\beta+2x \implies \angle AOK=2 \beta +2x$ Since $OA=OK$ we get that from triangle $\triangle AOK$ : $\angle OAK=\angle OKA=90-\beta-x$ From $\Gamma$: $\angle ABC+ \angle AK'C=180 \implies \angle ABC=180-\angle AK'C=\angle PK'C \implies \angle ABC= \angle PK'C \implies \angle PK'C=\beta$ Let $OP \perp KK' \implies \angle K'TP=90$ Claim2. Points $A,K,P$ and $O$ are $concyclic$

Proof

Claim3. The line $OK'$-is tangent to the circumcircle of the triangle $(\odot K'PM)$

Proof

Claim4. Points $O,A,K'$ and $M$ are $concyclic$

Proof

Claim5. Points $B,O,M$ and $C$ are $concyclic$

Proof

We know that $PK'=PK \implies \angle PKK'=\angle PK'K=\beta+x \implies \angle PKK'=\beta +x.$ $180-\beta-x=180-(\beta+x)=180-KK'P=\angle K'AB \implies \angle K'AB=180-\beta-x=\alpha+\gamma-x \implies \angle K'AB=\alpha+\gamma-x $ $\angle K'AB= \alpha+ \gamma-x, \angle K'AB=\angle BAK+\angle KAK' \implies \angle KAK'=\gamma-x$ $\angle MAK'=\angle MAC+\angle CAK'=x+\gamma-x=\gamma \implies \angle MAK'=\gamma$ From $\square OMAK$-is $cyclic$ we have: $\angle MAK=\angle MOK'=\gamma \implies \angle MOK'=\gamma$. From the triangle $\triangle AOP$ we have: $\angle OAP+\angle APO+\angle AOP=180 \implies \angle OAM+\angle MAC+ \angle CAK'+ \angle APO+\angle AOK'+\angle K'OP=180 \implies \angle AOK'=\beta-2 \gamma +x$ We know that: $\angle AOC=2\angle ABC=2\beta$ $\angle ABC=\angle AOK'+\angle K'OM+\angle COM \implies \angle \angle COM= \gamma =2x$.Since the quadrilateral $\square CBOM$-is $cyclic$ we have: $\angle COM=\angle CBM=\gamma-2x$ $\angle AOB=2 \angle BAC=2\alpha \implies \angle AOB=2\alpha$ since we know that $OB=OC$ we get from the triangle $\triangle AOB$ that: $\angle OBC=OCB=90-\alpha $ $\angle OBC=\angle OBM+\angle MBC \implies \angle OBM=\beta+2x-90$ From the triangle $\triangle MCB$ we have: $ \angle BMC+\angle MCB+ \angle MBC=180 \implies \angle BMC=2\alpha$ Claim 6.$\angle OMB+\angle OMC=180^{\circ}$

Proof

Took me $1$ hour to write this on AOPS

AP intersects Γ at X.K is the reflection of X to OP.We get OX=OK=radius then K is on Γ.<BAK=α,<MAO=β,<OAC=θ.From angle chasing we get <OCA=θ,<BCK=α,<CBK=β+θ,<OBC=<OCB=90-α-β-θ. From AXBKC is cyclic we get <PXK=90-β and <XPO=β. We get from similarity OA²=OM×OP=OB²=OC² then we get <OBM=<OPB=<OPC=<OCM so MBOC is cyclic.Then we get <OMB=180-<OCB=90+α+β+θ,<OMC=<OBC=90-α-β-θ.It is easy to check <OMB+<OMC=180.We are done.

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