First solution (power of a point) Let $\gamma$ denote the nine-point circle of $ABC$. [asy][asy] pair A = dir(125); pair B = dir(210); pair C = dir(330); pair M = midpoint(A--B); pair N = midpoint(A--C); pair O = origin; pair H = A+B+C; draw(A--B--C--cycle, blue); pair E = foot(B, A, C); pair F = foot(C, A, B); draw(B--E, lightblue); draw(C--F, lightblue); pair R = extension(E, F, B, C); pair Q = -A+2*foot(O, A, R); filldraw(unitcircle, invisible, blue); draw(A--R--E, heavygreen); pair P = extension(M, N, A, A+dir(90)*A); draw(A--P--N, red); filldraw(circumcircle(A, M, N), invisible, red); filldraw(circumcircle(A, E, F), invisible, heavygreen); filldraw(circumcircle(M, N, E), invisible, heavycyan); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(145)); dot("$N$", N, dir(20)); dot("$O$", O, dir(315)); dot("$H$", H, dir(H)); dot("$E$", E, dir(40)); dot("$F$", F, dir(F)); dot("$R$", R, dir(R)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); /* TSQ Source: A = dir 125 B = dir 210 C = dir 330 M = midpoint A--B R145 N = midpoint A--C R20 O = origin R315 H = A+B+C A--B--C--cycle blue E = foot B A C R40 F = foot C A B B--E lightblue C--F lightblue R = extension E F B C Q = -A+2*foot O A R unitcircle 0.1 lightcyan / blue A--R--E heavygreen P = extension M N A A+dir(90)*A A--P--N red circumcircle A M N 0.1 lightred / red circumcircle A E F 0.1 lightgreen / heavygreen circumcircle M N E 0.1 lightcyan / heavycyan */ [/asy][/asy] Note that $PA^2 = PM \cdot PN$, so $P$ lies on the radical axis of $\Gamma$ and $\gamma$. $RA \cdot RQ = RE \cdot RF$, so $R$ lies on the radical axis of $\Gamma$ and $\gamma$. Thus $\overline{PR}$ is the radical axis of $\Gamma$ and $\gamma$, which is evidently perpendicular to $\overline{OH}$. Remark: In fact, by power of a point one may also observe that $R$ lies on $\overline{BC}$, since it is on the radical axis of $(AQFHE)$, $(BFEC)$, $(ABC)$. Ironically, this fact is not used in the solution. Second solution (barycentric coordinates) Again note first $R \in \overline{BC}$ (although this can be avoided too). We compute the points in much the same way as before. Since $\overline{AP} \cap \overline{BC} = (0 : b^2 : -c^2)$ we have \[ P = \left( b^2-c^2 : b^2 : -c^2 \right) \](since $x=y+z$ is the equation of line $\overline{MN}$). Now in Conway notation we have \[ R = \overline{EF} \cap \overline{BC} = (0 : S_C : -S_B) =\left( 0 : a^2+b^2-c^2 : -a^2+b^2-c^2 \right). \]Hence \[ \overrightarrow{PR} = \frac{1}{2(b^2-c^2)} \left( b^2-c^2 , c^2-a^2 , a^2-b^2 \right). \]On the other hand, we have $\overrightarrow{OH} = \vec A + \vec B + \vec C$. So it suffices to check that \[ \sum_{\text{cyc}} a^2\left( (a^2-b^2) + (c^2-a^2) \right) = 0 \]which is immediate. Third solution (complex numbers) Let $ABC$ be the unit circle. We first compute $P$ as the midpoint of $A$ and $\overline{AA} \cap \overline{BC}$: \begin{align*} p &= \frac{1}{2} \left( a + \frac{a^2(b+c)-bc\cdot2a}{a^2-bc} \right) \\ &= \frac{a(a^2-bc)+a^2(b+c)-2abc}{2(a^2-bc)}. \end{align*}Using the remark above, $R$ is the inverse of $D$ with respect to the circle with diameter $\overline{BC}$, which has radius $\left\lvert \frac{1}{2}(b-c) \right\rvert$. Thus \begin{align*} r - \frac{b+c}{2} &= \frac{\frac14(b-c)\left( \frac1b-\frac1c \right)} {\overline{\frac{1}{2}\left( a-\frac{bc}{a} \right)}} \\ r &= \frac{b+c}{2} + \frac{-\frac{1}{2} \frac{(b-c)^2}{bc}}{\frac1a-\frac{a}{bc}} \\ &= \frac{b+c}{2} + \frac{a(b-c)^2}{2(a^2-bc)} \\ &= \frac{a(b-c)^2+(b+c)(a^2-bc)}{2(a^2-bc)}. \end{align*}Expanding and subtracting gives \[ p-r = \frac{a^3-abc-ab^2-ac^2+b^2c+bc^2}{2(a^2-bc)} = \frac{(a+b+c)(a-b)(a-c)}{2(a^2-bc)} \]which is visibly self-conjugate once the factor of $a+b+c$ is deleted. (Actually, one can guess this factorization ahead of time by noting that if $A=B$, then $P=B=R$, so $a-b$ must be a factor; analogously $a-c$ must be as well.)

My solution: It is well known that $R$ is on the radical axis of the circumcircle and the npc of $ABC$. (Proof: radical axis on the these circles and the circle with $BC$ as a diameter.). Now doing the same trick with the circle with $AO$ as diameter and the npc and circumcircle of $ABC$ does the job as circle with diameter $AO$ is tangent to the circumcircle of $ABC$ by a simple homothety at $A$ or using collinearity of their centers.

see here https://artofproblemsolving.com/community/c312162h1469210

What a beautiful problem! I add my sol Let $X\equiv MN\cap EF$, $Y\equiv ME\cap NF$; then $AX$ is the polar of $Y$ with respect to the nine-point circle by Brokard's theorem. In addition, we know by Pappus that $Y\in OH$, so $AX$ is perpendicular to the Euler line. Then since $R\in BC$ we have $\overline{PX}$ bisects $\overline{AR}$ and from $AP\parallel RX$ we know quadrilateral $APRX$ is a parallelogram, as desired.

Draw the $9$-point circle, $(N)$. Radical Axis Theorem on $(ABC), (AMN), (N)$ implies $P$ lies on the radical axis of $(ABC)$ and $(N)$. Radical Axis Theorem on $(AEFH), (ABC), (N)$ implies $R$ lies on the radical axis of $(ABC)$ and $(N)$. Then the radical axis of $(ABC), (N)$ is just $PR$, so $PR$ is perpendicular to the line connecting their centers, which is just the Euler Line! Done!

Let $X$ be the midpoint of $BC$ and $Y=MN\cap AR$. $Q$ is the Miquel point of cyclic $BFEC$ so $R\in BC$ and $XQ\perp QR$. Let $L$ be the antipode of $A$; by the tangent $\angle XLO=\angle PAY$ and we can chase that $\angle XOL=\angle YPA=|B-C|$ so $\triangle XOL\sim \triangle YPA$. However, $HL=2XL$ and $RA=2YA$ so $\triangle HOL\sim \triangle RPA$ and thus $XQ\perp QR\implies PR\perp OH$.

We will prove a more general claim In a $\Delta ABC$ with circumcenter $O$ and circumcircle $\Gamma$ and Let $K$ be a point on the line $AO$ and $K^*$ denote the isogonal conjugate of $K$ with respect to $\Delta ABC$ and let $O_1$ be the midpoint of $KK^*$ and $M,N$ are projection of $P$ in $AB,AC$ and $E,F$ are projection of $K^*$ in $AB,AC$ let tangent at $A$ to $\Gamma$ meet $MN$ at $P$ and $(AEF)\cap (ABC)=A,Q$ and $EF\cap AQ=R$ prove that $PR\perp OO_1$ PROOF By six point circle theorem we see that $(EFMN)$ is cyclic with center $O_1$ and denote it's circumcircle by $\gamma$ Since $K \in AO$ it is easy to see that $MN\parallel BC$ so $(ABC),(AMNK)$ are tangent to each other,on the other hand By power of a point $(RQ)(RA)=(RE)(RF)$ $PA^2=(PM)(PN)$ so $PR$ is the radical axis of $\Gamma,\gamma$,so $PR\perp OO_1$ and we are done $\blacksquare$ When,$K\equiv O$,we get the problem

enhanced wrote: ... $(PM)(PN)=PA^2$ this result doesn't occur unless $NM\parallel BC$

$M$ and $N$ are the midpoints of $AB$ and $AC$ so $NM\parallel BC$

@PROF65 I don't see any flaw in enhanced's proof,

Since there was some controversy, here is a little bit more explanation: (Basically v_enhance's solution) We have $AMON$ cyclic and internally tangent to $\Gamma$ as they share a diameter. Hence, we have from power of a point, $PA^2 = PM \cdot PN$. Additionally, let $\gamma$ denote the nine-point circle of $\triangle{ABC}$. One key lemma which was left out as it is very well known is that the nine-point circle of a triangle is the triangle passing through the three midpoints, the altitudes, and the midpoints of $AH, BH, CH$. It is easy to see from a homothety centered at $H$ with dilation factor $2$ that the center $N$ of the nine point circle is the midpoint of $OH$, hence $O, N, H$ are collinear. Also, it is ovious that $AEHF$ is cyclic, so from power of a point, we also have $RA \cdot RQ = RE \cdot RF$. The key observation is that $Pow_{\gamma}P = Pow_{\Gamma}P$, and similarly, $Pow_{\gamma}R = Pow_{\Gamma}R$. Thus, both points lie on the radical axis of circles $\Gamma, \gamma$, and as two points uniquely determine a line, $PR$ is the radical axis of $\Gamma$ and $\gamma$. The definition of the radical axis is that it is perpendicular to the line connecting the centers of the circles, so we see that $\overline{OH} \perp \overline{PR}$ and from the previous collinearity with $H$, the desired result follows. $\blacksquare$

Nice and easy. Here's my solution: Let $AP \cap BC=T$, and let $X$ be the point such that $ATRX$ is a parallelogram. Also let $K$ be the midpoint of $RX$. From here, we get that $X$ lies on $EF$, and so $K$ is the point where $EF$ and $MN$ meet. But, as $P$ is the midpoint of $AT$, we get that $APRK$ is also a parallelogram, or equivalently that $AK \parallel PR$. Restating the problem in terms of $\triangle DEF$, where $D$ is the foot of $A$-altitude, we are left with the following problem:- New problem wrote: Let $O$ be the circumcenter, $\omega$ be the circumcircle, $I$ be the incenter, $I_A$ be the $A$-excenter of $\triangle ABC$. Let $BI_A \cap \omega=M$ and $CI_A \cap \omega=N$. Finally let $BC \cap MN=K$. Then show that $I_AK$ is perpendicular to line $OI$. Let $BN \cap CM=L$. Note that, by Brokard's Theorem, $I_AK$ is the polar of $L$ wrt $\omega$, and so $I_AK$ is perpendicular to $OL$. So it suffices to show that $O,I,L$ are collinear. Let $BI \cap \omega=U$ and $CI \cap \omega=V$. Then by Pascal's Theorem on $UMCVNB$, we get the desired result. $\blacksquare$

Let \(A'\) denote the antipode of \(A\) in \(\Gamma\) and \(AP \cap BC= X\). Then, \(\Delta AHA' \sim \Delta ARX \Rightarrow \angle AHO= \angle PRX \therefore PR\perp OH.\)

Dear RC. very nice and elegant proof... Sincerely Jean-Louis

$PR$ is polar of $H$ wrt. $(ABC)$ .

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/3.%20Perpendiculaires%20a%20la%20droite%20de%20Euler.pdf p. 24... Sincerely Jean-Louis

FISHMJ25 wrote: $PR$ is polar of $H$ wrt. $(ABC)$ . No.....Obviously, (Well-known) that $P,R$ lie on the orthic axis WRT $\Delta ABC$, But, clearly, orthic axis isn't exactly the polar of $H$ WRT $\Delta ABC$, So your argument fails here! Indeed, this helps us to provide another solution that isn't mentioned in the above posts that, Let $AP \cap BC=T$, Well-known that, midpoint of $AT$ is $P$ which lies on orthic axis of $\Delta ABC$, and also $R \in BC$ and $R$ lies on orthic axis WRT $\Delta ABC$, It is also well-known, that the orthic axis WRT $\Delta ABC$ is perpendicular to euler line, we're done!

AlastorMoody wrote: FISHMJ25 wrote: $PR$ is polar of $H$ wrt. $(ABC)$ . No.....Obviously, (Well-known) that $P,R$ lie on the orthic axis WRT $\Delta ABC$, But, clearly, orthic axis isn't exactly the polar of $H$ WRT $\Delta ABC$, So your argument fails here! Indeed, this helps us to provide another solution that isn't mentioned in the above posts that, Let $AP \cap BC=T$, Well-known that, midpoint of $AT$ is $P$ which lies on orthic axis of $\Delta ABC$, and also $R \in BC$ and $R$ lies on orthic axis WRT $\Delta ABC$, It is also well-known, that the orthic axis WRT $\Delta ABC$ is perpendicular to euler line, we're done! Sorry my mistake.

How much time is allowed?

Let $X$ be the intersection of $EF$ and $MN$. Now by Radical Center we know that $AX\perp OH$, it suffices to show that $AX\parallel PR$. But by Radical Center again we see that $R$ lies on $BC$. Thus $MN$ bisects $AR$. Finally, $RF$ and $PA$ are both perpendicular to $AO$, thus $RF\parallel PA$ so $AXRP$ is a parallelogram, done.

Remarks: This problem took me way too long (around three hours). Amusingly enough, I finally found this solution while watching a ballet performance . Let $\omega$ denote the Nine-Point Circle of $ABC$ and $N_9$ be the Nine-Point center of $ABC$. It's clear that $$Pow_{\Gamma}(P) = PA^2 = Pow_{(AMN)}(P) = PM \cdot PN = Pow_{\omega}(P)$$and $$Pow_{\Gamma}(R) = RA \cdot RQ = Pow_{(AEF)}(R) = RE \cdot RF = Pow_{\omega}(R)$$so $PR$ is the Radical Axis of $\Gamma$ and $\omega$. Because $H, N_9, O$ are collinear on the Euler Line, the desired result follows immediately from properties of Radical Axes. $\blacksquare$

We Claim that $PR$ is the radical axis of $(ABC), (EFM)$. For the proof, notice that $P$ is the radical center of $(AMN), (ABC), (EFM)$. Therefore the radical axis of $(ABC), (EFM)$ passes through $P$. Similarly $R$ is the radical center of $(BCF), (ABC), (EFM)$. So the radical axis of $(ABC), (EFM)$ passes through $R$. $\blacksquare$ It is very well known that the center of $(ABC)$ and $(EFM)$ are collinear with $H$. Hence done.

$K$ is midpoint $BC$.$D$ feet of the altitudes from $A$ $K,H,Q$ collinear. $AQEF$,$EFDK$,$AQDK$ cyclic. by radical axis $AQ,EF,DK$ concurrent. $R,B,C$ collinear. Carnot theorem. $$HP^2-OP^2=HR^2-OR^2$$$$( AP^2 +AH^2 - 2AP*AH*cos(\angle PAH))-(PA^2+OA^2)=RD^2+HD^2-RK^2-OK^2 $$This expression can be clearly proved because it will all be expressions related to $ABC$.

By Radical Axis on $(AEF),(BCEF),$ and $\Gamma,$ we see $R$ lies on $\overline{BC}.$ Then, the power of $R$ with respect to the Nine-Point circle of $\triangle ABC,$ $(N_9),$ is $$RF\cdot RE=RB\cdot RC=\textrm{pow}_{\Gamma}(R)$$as $\triangle RFB\sim\triangle RCE.$ Also, $\angle PAN=180-\angle CBA=\angle AMP$ so $\triangle AMP\sim\triangle NAP$ and $$\textrm{pow}_{(N_9)}=PM\cdot PN=PA^2=\textrm{pow}_{\Gamma}.$$Hence, $\overline{PR}$ is the radical axis of $(N_9)$ and $\Gamma.$ $\square$

$R$ is radical center of $\Gamma, (AFHE), (BCEF), (MFNE)$ and therefore $R=BC\cap EF.$ $P$ is radical center of $\Gamma, (AMN), (MFNE).$ Hence $PR$ is radical axis of $\Gamma, (MFNE).$ But $OH$ passes through centers of these circles, so we are done.

Another speedrun les goooo. By radax on $(AQEF), (BFEC), (ABC)$ we have that $R$ lies on $BC$ now let $D$ the projection from $A$ to $BC$, then its known that $DFMEN$ is cyclic becuase they lie on the nine point circle and since the nine point center lies on $OH$ is enough to show that $PR$ is the radical axis of the nine-point circle and $(ABC)$. Now by PoP $$RE \cdot RF=RB \cdot RC \implies R \; \text{lying inside the radax of} \; (ABC), (DFMEN)$$Now by angle chasing knowing that $D$ and $A$ are symetric w.r.t. $MN$ we have that $$\angle RAM=\angle ACB=\angle ANM=\angle DNM \implies PD \; \text{tangent to} \; (DFMEN) \implies PD^2=PN \cdot PM$$Wait, rememebr when i said $D$ and $A$ are symetric w.r.t. $MN$, hence $PD=PA$ meaning that $PA^2=PM \cdot PN$ hence $P$ lies on the radax of $(ABC), (DFMEN)$ Thus, we are done

Let $\omega$ denote the nine-point circle. Notice that since the center of $\omega$ is the centroid $G$, and $O$, $H$, $G$ are collinear, it suffices to prove that $PR$ is the radical axis of $\omega$ and $\Gamma$. Let $\gamma$ denote $(OMAN)$. Note that $\mathrm{Pow}(P, \omega) = PM \cdot PN = \mathrm{Pow}(P, \gamma)$. Now, since $A$ is the antipode of $O$ across $\gamma$, we have that $PA$ is tangent to $\gamma$ and therefore $\mathrm{Pow}(P, \gamma) = \mathrm{Pow}(P, \Gamma)$. It suffices to prove that $R$ lies on the radical axis. Note that $BFEC$ is cyclic. Therefore by radical axis theorem, $AQ, EF, BC$ all concur at point $R$. Because $R$ is the radical center, we are done.

Basically was told the fact that nuked the problem but I think typing this up will help me remember it: $P$ lies on the orthic axis of $\triangle ABC$, since $PA^2=PM\cdot PN$. More concretely, $\overline{PR}$ is the radical axis of $(ABC)$ and the 9-point circle, since $RE\cdot RF=RB\cdot RC$ (where $R$ lies on $\overline{BC}$ because of radical center) and $PA^2=PM\cdot PN$, so $\overline{PR}$ is perpendicular to $\overline{ON_9}$ which is just the Euler line. $\blacksquare$

Here's a complex solution (some calculations omitted, but I did do them): First note that by radical center on $(AEF)$, $(BCEF)$, and $(ABC)$, we find that $R$ is also the intersection of $\overline{AQ}$ and $\overline{BC}$. Since it is well-known that $Q$, the $A$-antipode, and the midpoint of $\overline{BC}$ are collinear, we set up the equation $$\frac{q+a}{q-\frac{b+c}{2}}=\frac{\frac{1}{q}+\frac{1}{a}}{\frac{1}{q}-\frac{\frac{1}{b}+\frac{1}{c}}{2}}.$$This can be neatly cross-multiplied, and since $q=-a$ is a factor we divide out $q+a$ and obtain a linear equation in $q$, which we can solve to yield $$q=\frac{2a+b+c}{2+\frac{a}{b}+\frac{a}{c}}.$$Now let $P'=\overline{AA} \cap \overline{BC}$. By complex intersection, $$p'=\frac{a^2(b+c)-2abc}{a^2-bc},$$and since $P$ is the midpoint of $\overline{AP}$ for homothety reasons we find that $$p=\frac{a^3+a^2b+a^2c-3abc}{2(a^2-bc)}.$$We then calculate $R$, which is surprisingly neat since a lot of $2+\tfrac{a}{b}+\tfrac{a}{c}$ terms die: $$r=\frac{a^2b+a^2c+ab^2+ac^2-2abc-b^2c-bc^2}{2(a^2-bc)},$$so $$p-r=\frac{a^3-ab^2-ac^2-abc+b^2c+bc^2}{2(a^2-bc)}.$$We would like to show that this is a pure imaginary multiple of $a+b+c$, so using the Euclidean algorithm we try to divide out $a+b+c$ from the numerator and find that it actually equals $(a+b+c)(a-b)(a-c)$, so we want to show that $$\frac{(a-b)(a-c)}{a^2-bc}=-\frac{(\frac{1}{a}-\frac{1}{b})(\frac{1}{a}-\frac{1}{c})}{\frac{1}{a^2}-\frac{1}{bc}},$$but this is obvious after multiplying the numerator and denominator of the RHS by $a^2bc$, so we're done. $\blacksquare$

It's easy to prove that $R\in BC,$ also note that $PR$ is the radical axis of $(O)$ and the Euler circle of $\triangle ABC$ $\Rightarrow PR\perp OH.$

Let $S$ be the mid-point of $\overline BC$, $D$ be the feet of altitudes from $A$ in $\triangle ABC$. Then, $M,N,D,E,F,S,$ are cyclic as they form the nine-point circle. Note that $R$ is the radical center of circles $(AQEF)$, $(MNDS)$, $(ABC)$. So, $R,B,C$ are collinear. Since $(BCEF)$ is cyclic, we have $RB\cdot RC=RE\cdot RF$. Since $(EFDS)$ is cyclic, we have $RE\cdot RF=RD\cdot RS$. Hence, we have $RD\cdot RS=RB\cdot RC$, which means $Pow_{(ABC)}(R)=Pow_{(MNDS)}(R)$ and hence $R$ lies on the radical axis of the circumcircle and the nine-point circle. Similarly, since $\overline{PA}$ is tangent to $(ABC)$ which is also tangent to $(AMN)$ because $(AMN)$ is just $(ABC)$ with homothety at $A$ with factor $\frac12$, we have $PA^2=PM\cdot PN$ which means $Pow_{(ABC)}(P)=Pow_{(MNDS)}(P)$ and hence $P$ lies on the radical axis of the circumcircle and the nine-point circle. So, $PR$ is the radical axis, which means $PR\perp ON_9$ and hence $PR\perp OH$. $\blacksquare$ Remark. If $U,S$ are the intersections of the $A-$median and the nine-point circle, then $\overline{PU},\overline{PD}$ are the tangents from $P$ to nine-point circle. So, $DU$ is the pole of $P$ in nine-point circle. Also, the mid-point of $\overline{AH}$, say $L$, lies on the pole of $R$.

Attachments:

Applying the radical axis theorem on $(ABC)$, $(AMN)$ and $(MNEF)$, it suffices to show that $R$ has equal power WRT $(ABC)$ and $(MNEF)$. This is true because $RQ \cdot RA = RF \cdot RE$.

Since $(AMN)$ is tangent to $(ABC)$ by homothety, $PA^2=PB\cdot PC,$ so $P$ lies on the radical axis of the circumcircle and the nine point circle. Since $(AEFQ)$ is cyclic, $RQ\cdot RA=RE\cdot RF,$ so $R$ lies on the radical axis of the circumcircle and the nine point circle. Thus $PR$ is perpendicular to the line through their centers which is $OH.$

It is well known that $R$ lies on the radical axis of the nine point circle and the circumcircle, also called the orthic axis. By radax on $(ABC),(AMN),(MNEF)$, we see that $P$ also lies on this radical axis and the line $OH$ is the line joining the centers of the $NPC$ and $(ABC)$ so it trivially follows that $PR \perp OH$.