Let $H $ be the orthocenter of $ABC $ ,we are going to show that $MN $ passes through $H $. $\textbf {Claim}$:$DV\perp AO $ and $EU\perp AO $. Proof:Consider inversion at $\omega $.We need to show $DVOB $ is cyclic which is equivalent to show that inverse of $V $ is $BP\cap AO $.$\angle CUP= 2 \angle CBU=\angle COP $ so $U $ and $V $ are inverses of each other. Let $PM\cap AO=S $ and $PN\cap AO=T $.By easy angle chasing we get that $\triangle MVS \sim \triangle NUT $ $\implies $ $\frac{MV}{NU}=\frac {VS}{UT} $ and by Thales on $PS\parallel CH$ and $PT\parallel BH $ we can get that $\frac {UH\cdot VS}{VH\cdot UT}=\frac {VP\cdot UB}{UP\cdot VC} =1$ by sine law in $\triangle UVP ,\triangle UVB $($VB=UC $) so $\frac {VH}{UH}=\frac{MV}{NU} $ which means $M,N,H $ are collinear. so done

tenplusten wrote: Let $H $ be the orthocenter of $ABC $ ,we are going to show that $MN $ passes through $H $. $\textbf {Claim}$:$DV\perp AO $ and $EU\perp AO $. Proof:Consider inversion at $\omega $.We need to show $DVOB $ is cyclic which is equivalent to show that inverse of $V $ is $BP\cap AO $.$\angle CUP= 2 \angle CBU=\angle COP $ so $U $ and $V $ are inverses of each other. Let $PM\cap AO=S $ and $PN\cap AO=T $.By easy angle chasing we get that $\triangle MVS \sim \triangle NUT $ $\implies $ $\frac{MV}{NU}=\frac {VS}{UT} $ and by Thales on $PS\parallel CH$ and $PT\parallel BH $ we can get that $\frac {UH\cdot VS}{VH\cdot UT}=\frac {VP\cdot UB}{UP\cdot VC} =1$ by sine law in $\triangle UVP ,\triangle UVB $($VB=UC $) so $\frac {VH}{UH}=\frac{MV}{NU} $ which means $M,N,H $ are collinear. so done How to get $\frac {UH\cdot VS}{VH\cdot UT}=\frac {VP\cdot UB}{UP\cdot VC}$?

Let $T$ be the reflection of $O$ w.r.t $BC.$Then $OP=OB=BT=CT.$ One can find $\angle OUP=180-\angle OBP-\angle BOU=180-(90-\angle BCP)-(90-\angle OBC)=\angle OCP,$ Then we have $O,P,U,E,S$ and similarly $O,V,P,D,B$ are cyclic. Then we can find easily with Sine rule (apply:$\triangle TVC,VPM,PMD,DPO,TUB,UNP,PUO.$) $\frac{TV}{VM}=\frac{TU}{UN},$ Consequently we get $T,M,N$ are collinear.

DV' perp AO then V'PO+CPO=180-V'DO+90-CBP=180-AOD+CBP=180 by OD,OA perp PB,PC; similar for EU. PM,PN-AO=K,L so VM/VH=VM/VK.VK/VH =UN/UL.UL/UH=UN/UH (MVK~UNL and VK/VH=VP/VC=UP/UB(Mene)=UL/UH) so MN pass H fixed

Here's my solution: Let $H$ be the orthocenter of $\triangle ABC$. We claim that $H \in MN$. Animate $P$ projectively on $\omega$. Then $P \mapsto D,E,U,V$ are all projective maps (the first two follow using pole polar transformation), and so these $4$ points have degree $1$ (since all of them move on fixed lines). Note that when $P$ is at $C$, both $D,V$ coincide with $A$, and since the map $D \mapsto V$ is a projectivity from line $AB$ to $AO$ which fixes a point, it must in fact be a perspectivity. This gives that the degree of line $DV$ is $1$. Also, as $P$ has degree $2$, so the line through $P$ perpendicular to $AB$ also has degree two. Then, degree of $M$ is atmost $3$. Similarly, $\text{deg}(N) \leq 3$. So we just need to check our claim of $H \in MN$ for $(3+3+0)+1=7$ positions of point $P$. The result is obvious when $P=B,C$ and when $P$ coincides with the two points where line $AO$ meets $\omega$. For the fifth position, take $P$ as the antipode of $B$ in $\omega$. Then $$D=\infty_{AB},V=\infty_{AO},M=\infty_{BO},U=O$$And, $N$ lies on the perpendicular bisector of $CP$ such that $PO \parallel CN$, which gives that $N$ becomes the reflection of $O$ in line $CP$. Then, $M,N,C,H$ all lie on a line, as desired. Sixth position is the antipode of $C$, which follows similarly as above. For the seventh position, take $P$ as either $BH \cap \omega$ or $CH \cap \omega$ (it can be shown with some work that these work too, but I am too lazy to do so). Hence, done. $\blacksquare$ EDIT (30/05/2020): The previous solution had some flaws (no idea what I was thinking), which have now been rectified.

math_pi_rate wrote: Here's my solution: Let $H$ be the orthocenter of $\triangle ABC$. We claim that $H \in MN$. Animate $P$ linearly on $\omega$. Then $P \mapsto D,E,U,V$ are all linear maps. This gives that the degree of line $DV$ and the line through $P$ perpendicular to $AB$ is also one. Then, degree of $M$ is atmost $2$. Similarly, $\text{deg}(N) \leq 2$. So we just need to check for $(2+2+0)+1=5$ positions of point $P$. The result is obvious when $P=B,C$ and when $P$ coincides with the two points where line $AO$ meets $\omega$. For the fifth position, take $P$ as the antipode of $B$ in $\omega$. Then $$D=\infty_{AB},V=\infty_{AO},M=\infty_{BO},U=O$$And, $N$ lies on the perpendicular bisector of $CP$ such that $PO \parallel CN$, which gives that $N$ becomes the reflection of $O$ in line $CP$. Then, $M,N,C,H$ all lie on a line, as desired. Hence, done. $M$ and $N$ move with degree $3$ so we need to check $7$ cases.But its not that bad since we can check $$P=B,P=C,P=AO \cap \omega =Q_1,P=AO \cap \omega =Q_2,P=BH \cap \omega,P=CH \cap \omega$$and antipods of $B$ and $C$.

FISHMJ25 wrote: math_pi_rate wrote: Here's my solution: Let $H$ be the orthocenter of $\triangle ABC$. We claim that $H \in MN$. Animate $P$ linearly on $\omega$. Then $P \mapsto D,E,U,V$ are all linear maps. This gives that the degree of line $DV$ and the line through $P$ perpendicular to $AB$ is also one. Then, degree of $M$ is atmost $2$. Similarly, $\text{deg}(N) \leq 2$. So we just need to check for $(2+2+0)+1=5$ positions of point $P$. The result is obvious when $P=B,C$ and when $P$ coincides with the two points where line $AO$ meets $\omega$. For the fifth position, take $P$ as the antipode of $B$ in $\omega$. Then $$D=\infty_{AB},V=\infty_{AO},M=\infty_{BO},U=O$$And, $N$ lies on the perpendicular bisector of $CP$ such that $PO \parallel CN$, which gives that $N$ becomes the reflection of $O$ in line $CP$. Then, $M,N,C,H$ all lie on a line, as desired. Hence, done. $M$ and $N$ move with degree $3$ so we need to check $7$ cases.But its not that bad since we can check $$P=B,P=C,P=AO \cap \omega =Q_1,P=AO \cap \omega =Q_2,P=BH \cap \omega,P=CH \cap \omega$$and antipods of $B$ and $C$. Sorry! Edited. Thanks a lot.

tenplusten wrote: Let $H $ be the orthocenter of $ABC $ ,we are going to show that $MN $ passes through $H $. $\textbf {Claim}$:$DV\perp AO $ and $EU\perp AO $. Proof:Consider inversion at $\omega $.We need to show $DVOB $ is cyclic which is equivalent to show that inverse of $V $ is $BP\cap AO $.$\angle CUP= 2 \angle CBU=\angle COP $ so $U $ and $V $ are inverses of each other. Let $PM\cap AO=S $ and $PN\cap AO=T $.By easy angle chasing we get that $\triangle MVS \sim \triangle NUT $ $\implies $ $\frac{MV}{NU}=\frac {VS}{UT} $ and by Thales on $PS\parallel CH$ and $PT\parallel BH $ we can get that $\frac {UH\cdot VS}{VH\cdot UT}=\frac {VP\cdot UB}{UP\cdot VC} =1$ by sine law in $\triangle UVP ,\triangle UVB $($VB=UC $) so $\frac {VH}{UH}=\frac{MV}{NU} $ which means $M,N,H $ are collinear. so done You don't need to add $S , T$ : It's easy to see that $CPUE , BDPV$ are cyclic so we have $UE||VD||BC$ . We have $\frac{VM}{UN}=\frac{VM}{PV}.\frac{UP}{UN}.\frac{PV}{PU}$ it's easy to see that is equal to $\frac{VH}{UH}$ (by use of sine law in $\triangle{UBV}$) and the else is the same .

math_pi_rate wrote: Here's my solution: Let $H$ be the orthocenter of $\triangle ABC$. We claim that $H \in MN$. Animate $P$ linearly on $\omega$. Then $P \mapsto D,E,U,V$ are all linear maps. This gives that the degree of line $DV$ is atmost $2$, and the line through $P$ perpendicular to $AB$ has degree one. Then, degree of $M$ is atmost $3$. Similarly, $\text{deg}(N) \leq 3$. So we just need to check for $(3+3+0)+1=7$ positions of point $P$. The result is obvious when $P=B,C$ and when $P$ coincides with the two points where line $AO$ meets $\omega$. For the fifth position, take $P$ as the antipode of $B$ in $\omega$. Then $$D=\infty_{AB},V=\infty_{AO},M=\infty_{BO},U=O$$And, $N$ lies on the perpendicular bisector of $CP$ such that $PO \parallel CN$, which gives that $N$ becomes the reflection of $O$ in line $CP$. Then, $M,N,C,H$ all lie on a line, as desired. Sixth position is the antipode of $C$. For the seventh position, take $P$ as either $BH \cap \omega$ or $CH \cap \omega$. Hence, done. $\blacksquare$ Why the line through $P$ perpendicular to $AB$ has degree one?

Anyone...

In solution, it's written degree two (not one). Reason for degree two: Line through $P$ perpendicular to $AB$ is basically the line joining $P$ and the point $\infty$ (the point at infinity in the direction perpendicular to $AB$). As $\infty$ is fixed, it has degree $0$. Hence line $P \infty$ has degree $$ \le \deg P + \deg \infty = 2 +0 = 2$$