It's well known that the orthocenter $H$ of $\triangle ABC$ is the center of negative inversion that swaps its 9-point circle and circumcircle $(O).$ Hence, $P,Q$ are the inverses of $N,M$ under the referred inversion $\Longrightarrow$ $M,N,P,Q$ lie on a circle $\omega.$ Hence, $R$ is the radical center of $\omega,(O)$ and $\odot(AMN)$ $\Longrightarrow$ $R$ lies on the radical axis of $(O)$ and $\odot(AMN),$ which is the tangent of $(O)$ through $A,$ since $(O)$ and $\odot(AMN)$ are internally tangent through $A.$

There is generalization for this Let $ABC$ be a triangle and $XYZ$ is pedal triangle of a point $P$ with respect to $ABC$. $P'$ is isogonal conjugate of $P$. $(O,R)$ is circumcircle of triangle $XYZ$. $\mathcal C$ is circle $(P,2R)$. Ray $YP',ZP'$ intersects $\mathcal C$ at $M,N$, resp. $MN$ cuts $YZ$ at $R$. $T$ is projection of $R$ on $PA$. Prove that $T$ is inversion of $A$ with respect to $\mathcal C$.

OK, here is a proof of the generalization: WLOG assume that $P$ is inside $\triangle ABC,$ then so is its isogonal conjugate $P',$ which is the exsimilicenter of the pedal circle $\mathcal{K}$ of $(P,P')$ and $\mathcal{C},$ since the the dilatation with center $P'$ and factor $2$ carries the center of $\mathcal{K}$ (midpoint of PP') into $P.$ Consequently, $P'$ is also the center of the negative inversion that swaps $\mathcal{K}$ and $\mathcal{C}$ $\Longrightarrow$ $M,N$ are the inverses of $Y,Z$ under this inversion $\Longrightarrow$ $X,Y,M,N$ lie on a same circle $\omega$ $\Longrightarrow$ $R \equiv YZ \cap MN$ is the radical center of $\mathcal{C},\omega$ and the circle $\odot(AYZ)$ with diameter $\overline{AP}$ $\Longrightarrow$ $RT \perp AP$ is the radical axis of $\odot(AYZ)$ and $\mathcal{C}.$ Therefore if $AP$ cuts $\mathcal{C}$ at $U,V,$ we have $\overline{TU} \cdot \overline{TV}=\overline{TA} \cdot \overline{TP}.$ But since $P$ is the midpoint of $\overline{UV},$ then we deduce that $(U,V,A,T)=-1$ $\Longrightarrow$ $\overline{PU}^2=\overline{PA} \cdot \overline{PT}$ $\Longrightarrow$ $T$ is the inverse of $A$ WRT $\mathcal{C}.$

Let the points diametrically opposite to $B$ and $C$ be $R$ and $S$, respectively. Since $R$ is diametrically opposite to $B$, $RC \perp BC$ and $RA \perp BA$ which imply that $RC \| AH$ and that $RA \| CH$. Hence $HARC$ is a parallogram which implies that $R$, $N$ and $H$ are collinear. By the same argument, $S$, $M$ and $H$ are collinear. Now note that $RSBC$ is a rectangle and hence that $RS \| BC \| MN$. Hence $\angle{QPM}=\angle{QPS}=\angle{QRS}=\angle{QNM}$ which implies that $PQNM$ is cyclic. Let $\omega_1$, $\omega_2$ and $\ell$ denote the circumcircle of $AMN$, the circumcircle of $PQNM$ and the tangent line to $\omega$ at $A$. Note that $\ell$ is also tangent to $\omega_1$ at $A$ since there is a homothety with center $A$ taking $AMN$ to $ABC$. Hence pairwise $\ell$, $MN$ and $PQ$ are the radical axes between the circles $\omega$, $\omega_1$ and $\omega_2$. Hence these line concur at $R$ which implies that $OA \perp RA$.

OK I think I am getting the hang of this.

@v_ enhance can you please say what is EFFT??

Whoops, I copy-pasted this from an article I was writing about barycentrics, so I didn't realize I hadn't defined it. Sorry about that. EFFT is short for Evan's Favorite Forgotten Trick; the statement of the standard and strong version is as follows:

It's actually not really necessary here; there are synthetic and analytic means of finding the equation of the tangent (although Strong EFFT is, to my knowledge, the fastest). EDIT: Sometime around this coming Tuesday, I'll be releasing the article I mentioned on AoPS. Shameless plug there. :>

Dear Mathlinkers, 1. B’ the symmetric of B wrt O 2. C’ the symmetric of C wrt O 3. It is known that P, H, M and C’ are collinear, Q, H, N and B’ are collinear 4. L the point of intersection of PC and AB’ 5. Note that B’C’ // BC // MN 5. According to Pascal theorem applied to ABCPC’B’A, ML // BC ; 6. M, N, L and R are collinear 7. Note Ta the tangent at A 8.According to a reciprocity of Pascal theorem, NL is the pascal line of the degenerated hexagon PQB’A Ta CP ; consequently, Ta goes through R. and we are done Sincerely Jean-Louis

v_Enhance wrote: EDIT: Sometime around this coming Tuesday, I'll be releasing the article I mentioned on AoPS. Shameless plug there. :> In case anyone's interested, the article I mentioned earlier is now available here.

Let $BO$ cut $w$ at $G$, then it is well known fact that $G, N, H, Q$ are collinear, similarly, if $CO$ cuts $w$ at $D$ than $D, M, H, P$ are collinear. Let $AD$ cut $QB$ at $E$. As $EANQ$ is cyclic, $\angle QEN = \angle QAB = 180^{\circ} - \angle EBC$, this means that $EN\parallel BC$ and thus $R, E, M, N$ collinear. By Pascal in $AABQPD$ we have $R'\equiv AA\cap QP; M\equiv AB\cap PD; E\equiv AD\cap BQ$ collinear, than $R'\equiv R$, $RA$ is the tangent to $w$ at $A$ and $OA\perp RA$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -3.484312546957180, xmax = 8.131014274981224, ymin = -1.662839969947408, ymax = 5.219203606311049; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((1.560000000000002,4.560000000000005)--(0.3600000000000004,1.500000000000002)--(4.680000000000005,1.180000000000001)--cycle, zzttqq); /* draw figures */ draw((1.560000000000002,4.560000000000005)--(0.3600000000000004,1.500000000000002), zzttqq); draw((0.3600000000000004,1.500000000000002)--(4.680000000000005,1.180000000000001), zzttqq); draw((4.680000000000005,1.180000000000001)--(1.560000000000002,4.560000000000005), zzttqq); draw(circle((2.597614678899085,2.387798165137618), 2.407302439089523)); draw((3.120000000000004,2.870000000000003)--(0.1993529912759527,2.179361337956304)); draw((3.280328978893940,0.07933412646842017)--(0.9600000000000011,3.030000000000003)); draw((-1.293532138442521,3.196928306551301)--(3.120000000000004,2.870000000000003)); draw((3.280328978893940,0.07933412646842017)--(-1.293532138442521,3.196928306551301)); draw((1.560000000000002,4.560000000000005)--(-1.293532138442521,3.196928306551301)); draw((0.9600000000000011,3.030000000000003)--(0.5152293577981647,3.595596330275235)); draw((3.120000000000004,2.870000000000003)--(4.835229357798169,3.275596330275234)); /* dots and labels */ dot((1.560000000000002,4.560000000000005),dotstyle); label("$A$", (1.624628099173554,4.648204357625849), NE * labelscalefactor); dot((0.3600000000000004,1.500000000000002),dotstyle); label("$B$", (0.4225244177310284,1.582839969947410), NE * labelscalefactor); dot((4.680000000000005,1.180000000000001),dotstyle); label("$C$", (4.735071374906089,1.267287753568747), NE * labelscalefactor); dot((0.9600000000000011,3.030000000000003),dotstyle); label("$N$", (1.023576258452291,3.115522163786630), NE * labelscalefactor); dot((3.120000000000004,2.870000000000003),dotstyle); label("$M$", (3.187362885048837,2.965259203606314), NE * labelscalefactor); dot((1.404770642201836,2.464403669724774),dotstyle); label("$H$", (1.459338842975207,2.559549211119462), NE * labelscalefactor); dot((3.280328978893940,0.07933412646842017),dotstyle); label("$Q$", (3.337625845229153,0.1703681442524431), NE * labelscalefactor); dot((0.1993529912759527,2.179361337956304),dotstyle); label("$P$", (0.2572351615326812,2.274049586776862), NE * labelscalefactor); dot((-1.293532138442521,3.196928306551301),dotstyle); label("$R$", (-1.230368144252444,3.280811419984976), NE * labelscalefactor); dot((2.597614678899085,2.387798165137618),dotstyle); label("$O$", (2.661442524417732,2.484417731029304), NE * labelscalefactor); dot((4.835229357798169,3.275596330275234),dotstyle); label("$B'$", (4.900360631104436,3.370969196093166), NE * labelscalefactor); dot((0.5152293577981647,3.595596330275235),dotstyle); label("$C'$", (0.5727873779113442,3.686521412471829), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $B'$ and $C'$ be the antipodes of $B,C$ respectively on $\omega .$ Since $B'C$ and $AH$ are both perpendicular to $BC,$ they are parallel to each other. Also, since $B'A$ and $CH$ are both perpendicular to $AB,$ they are parallel. It follows that $AHCB'$ is a parallelogram. Since the diagonals of a parallelogram bisect each other, $H,M,B'$ are collinear. Similarly, $H,N,C'$ are also collinear. Now, since $C'B$ and $B'C$ are both perpendicular to $BC,$ $C'B \parallel B'C.$ Also, their diagonals $BB'$ and $CC'$ bisect each other at $O.$ As a result, $C'BAB'$ is a parallelogram, implying $B'C' \parallel BC.$ Both these lines are parallel to $MN$ by the midpoint theorem. Now, note that \[\angle QPM = \angle NMP = \angle C'B'P = 180^{\circ} - \angle C'NM = \angle QNM,\] so quadrilateral $PQMN$ is cyclic. Note that $MN$ and $PQ$ are the radical axes of $(PQMN), (AMN)$ and $(PQMN), (ABC)$ respectively. Hence, their point of intersection, $R$ must be their radical center. Consequently, $R$ lies on the radical axis of $(AMN)$ and $(ABC).$ Trivially $A$ also lies on this radical axis, so $RA$ is the radical axis of $(AMN)$ and $(ABC).$ However, since $OM \perp AB$ and $ON \perp AC,$ the circumcenter of $\triangle AMN$ is the midpoint of $OA.$ Since the radical axis of two circles is perpendicular to the line joining their centers, it follows that $RA \perp OA.$ $ \blacksquare$

Let $ HM \cap w =T $ and $ HN \cap w =S $.It's easy to see that $ HN=NS $ and $ TM=MH $.We have $ QH \cdot HS=RH \cdot HT \implies QH \cdot HN=RH \cdot MH $, thus $ QRNM $ is cyclic.We know that $ (AMN) $ and $ w $ are tangent at $ A $.By radical axis theorem (for $ w,(AMN),(QRNM) $) we have $ MN,QR,AA $ are concurrent at $ R $ , hence $ RA $ is common tangent of $ (AMN) $ and $ w $, so $ OA \perp RA $.

Nice problem . First we claim that $MNPQ$ is cyclic. This follows immediately from angle chasing since the reflections of $H$ across $M$ and $N$ lie on the circumcircle. Since $MNPQ$ is cyclic, $R$ is the radical center of $\omega, (AMN),$ and $(MNPQ)$. So $RA$ is the radical axis of $\omega$ and $(AMN)$, which is just the tangent at $A$. $\Box$

K6160 wrote: This follows immediately from angle chasing since the reflections of $H$ across $M$ and $N$ lie on the circumcircle. How exactly does it immediately follow from angle chasing?

This is a terrific problem! nevermind, it is a pretty simple angle chase. see thecmd999's post for details

Let rays $HM$ and $HN$ meet $\omega$ again at $X$ and $Y$. It is well-known that $M$ and $N$ are the midpoints of $HX$ and $HY$. By Power of a Point on $H$ with respect to $\omega$, $HX\cdot HP=HY\cdot HQ$. Dividing both sides by 2, $HM\cdot HP=HN\cdot HQ$. Hence, $MNPQ$ is cyclic. Now, consider the circumcircle of $AMN$, $\omega$, and the circumcircle of $MNPQ$. Their pairwise radical axes are $MN$, $PQ$, and the tangent to $\omega$ at $A$. By the Radical Axis Theorem, they concur, so $R$ lies on the tangent to $\omega$ at $A$, which means that $OA\perp RA$ as desired.

Someone please check my solution Chose a point $T$ on $MN$ such that $TA$ is tangent to $(ABC)$. Let $TQ$ meet $(ABC)$ at $S$. We can easily prove that $MNPQ$ is concyclic. Now since $PA$ is tangent to $(ABC)$ therefore $\angle TAB$ $=$ $\angle C$. Since $MN || BC$, therefore$ \angle TAB$ $=$ $\angle N$. Thus $TA$ is also tangent to $(AMN)$ By PoP $TA^{2}$$=$$TM$$\times TN$$=$$TQ$ $\times TS$ Therefore $MNSQ$ is concyclic. This gives us $M,N,P,S,Q$ concyclic and they lie on the circle $(ABC)$. But this is not possible. Thus $T$ and $R$ must coincide. This proves the problem.

Quickie! Solution: By a homothety centered at $H$ taking the Nine-Point Circle of $\triangle ABC$ to $\odot(ABC)$, we have that $PQMN$ must be cyclic. [asy][asy] import olympiad; import geometry; defaultpen(fontsize(12pt)); size(12cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair O = circumcenter(A,B,C); pair H = orthocenter(A,B,C); pair M = (A+B)/2; pair N = (A+C)/2; pair P = intersectionpoints(line(M,H), circumcircle(A,B,C))[1]; pair Q = intersectionpoints(line(N,H), circumcircle(A,B,C))[1]; pair R = extension(P,Q,M,N); draw(A--B--C--A, blue); draw(circumcircle(A,B,C), red); draw(circumcircle(Q,P,M), magenta); draw(circumcircle(A,M,N), magenta); draw(A--R, purple); draw(R--N, blue); draw(P--R, blue); draw(N--Q, blue); draw(M--P, blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(140)*1.1); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$H$", H, S*1.4); dot("$O$", O, dir(O)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); [/asy][/asy] Finally since $R$ is the radical center of $\odot(PQMN)$, $\odot(AMNO)$ and $\odot(ABC)$, we must have $OA \perp AR$! $\blacksquare$

Claim 1: Quadrilateral $MNPQ$ is cyclic. Let the reflections of $H$ across $M$ and $N$ be $M_1$ and $N_1$, respectively. We know $M_1$ and $N_1$ lie on $\omega$ and $MN \parallel M_1N_1$, so \[\angle NQP = \angle N_1QP = \angle N_1M_1P = \angle NMP. \text{ } \Box\] Claim 2: $R$ is the radical center of $(AMN)$, $(ABC)$, and $(MNPQ)$. The radical center is the intersection of the radical axis between $(ABC)$ and $(MNPQ)$ and the radical axis between $(AMN)$ and $(MNPQ)$, or \[MN \cap PQ = R. \text{ } \Box\] Thus $R$ must lie on the radical axis of $(AMN)$ and $(ABC)$, or the tangent at $A$ to $\omega$, so $OA \perp RA$. $\blacksquare$ [asy][asy] size(250); defaultpen(linewidth(0.5)+fontsize(8)); real s, t; s = 10; t = -10; pair A, B, C, H, O, M, N; A = dir(130.5); B = dir(200); C = dir(340); H = orthocenter(A, B, C); O = 0; M = .5A + .5B; N = .5A + .5C; pair [] P=intersectionpoints(t*H+(1-t)*M--s*H+(1-s)*M, circumcircle(A, B, C)); pair [] Q=intersectionpoints(t*H+(1-t)*N--s*H+(1-s)*N, circumcircle(A, B, C)); pair R = extension(M, N, P[1], Q[1]); draw(circumcircle(A, B, C)); draw(A--B--C--cycle); draw(M--P[1]); draw(N--Q[1]); draw(M--P[0]--Q[0]--N, dashed); draw(P[1]--R--N--R--A); draw(circumcircle(A, M, N), lightblue); draw(circumcircle(M, N, P[1]), lightred); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$M$", M, NE); label("$N$", N, N); label("$O$", O, S); label("$H$", H, SE); label("$M_1$", P[0], W); label("$P$", P[1], S); label("$N_1$", Q[0], E); label("$Q$", Q[1], W); label("$R$", R, W); dot(A); dot(B); dot(C); dot(H); dot(O); dot(M); dot(N); dot(A); dot(P[0]); dot(P[1]); dot(Q[0]); dot(Q[1]); dot(R); [/asy][/asy]

Let $H_N , H_M$ the reflection of $H$ across $N ,M$ respectively. it is well known that $H_N , H_M$ are on $(ABC)$. $H_N H_M \parallel MN$ so by Reim's theorem $MNPQ$ is cyclic. $\implies R $ is the radical center for $ \odot(ABC),\odot(MNPQ),\odot(AMN)$ $\implies AR$ is the radical axis for $ \odot(ABC),\odot(AMN)$. so we are done.

Let $X,Y$ be the reflections of $H$ over $M,N.$ We have $2HM\cdot HP=HX\cdot HP=HY\cdot HQ=2HN\cdot HQ,$ so $MNPQ$ is cyclic. Now radical axis theorem on $(AMNO),(MNPQ),(ABC)$ gives that $PQ,MN$ and the tangent to the circumcircle at $A$ concur, which finishes.

Reflect $H$ over $M$ and $N$ to $H_M$ and $H_N$, respectively. It's well-known that $H_M$ and $H_N$ lie on $(ABC)$, so $$HM \cdot HP = \tfrac{1}{2} HH_M \cdot HP = \tfrac{1}{2}HH_N \cdot HQ = HN \cdot HQ,$$and $MNPQ$ is cyclic. Then by radical center lemma on $(ABC), (AMN)$, and $(MNPQ)$, we find that $RA$ is tangent to $(ABC)$, so $\overline{OA} \perp \overline{RA}.$

I am a brick oops Reflect $H$ over $M, N$ to points $M', N'$. By Power of a Point we can get that $M, N, P, Q$ cyclic. Applying radical axis on $(AMN), (MNPQ), (ABC)$ finishes.

Let $C'$ and $B'$ be the antipodes of $B$ and $C$ WRT $\omega$, respectively. It's well known that $M$ and $N$ are the midpoints of $\overline{HC'}$ and $\overline{HB'}$. Since $PQC'B'$ is cyclic, it follows by power of a point that $PQMN$ is cyclic. So, the radical axis theorem implies that $\overline{RA}$ is the radical axis of $(AMN)$ and $(ABC)$; therefore, $\overline{RA}$ is tangent to $\omega$.

Claim: $MNPQ$ is a cyclic quadrilateral. Proof. Let $M'$ be the reflection of $H$ across $M$, $N'$ be the reflection of $H$ across $N$. It is well known that $M'$ and $N'$ lie on $\omega$. Therefore, we have $HM' \cdot HP = HN' \cdot HQ$, which is equivalent to $HM \cdot HP = HN \cdot HQ$, and the claim follows by the Converse of Power of a Point. $\blacksquare$ Since $P, Q, R$ are collinear, and $QP$ is the radical axis of $(MNPQ)$ and $\omega$, we have $$\mathrm{Pow}_{(MNPQ)} R = \mathrm{Pow}_{\omega} R$$Similarly, $M, N, R$ are collinear, and $MN$ is the radical axis of $(AMN)$ and $(MNPQ)$, so it follows that $$\mathrm{Pow}_{\omega} R = \mathrm{Pow}_{(MNPQ)} R = \mathrm{Pow}_{(AMN)} R$$which is equivalent to $R$ lying on the radical axis of $\omega$ and $(AMN)$. Claim: $(AMN)$ is internally tangent to $\omega$. Proof. Let $O'$ be the center of $(AMN)$. The homothety centered at $A$ with scale factor $2$ sends $M$ to $B$ and $N$ to $C$, so the homothety sends $(O')$ to $\omega$ and $O'$ to $O$, so $A$, $O$, $O$ are collinear. However, $A$ is an intersection of $\omega$ and $(O')$, so $(O')$ is internally tangent to $\omega$. $\blacksquare$ Then, since $A$ is on the radical axis of $(O')$ and $\omega$, and the radical axis must be perpendicular to $OO'$, the radical axis of $\omega$ and $(O')$ thus must be their tangent at $A$. Since $R$ lies on the radical axis of $\omega$ and $(O')$, we must have $RA \perp OA$, as desired. $\blacksquare$

Let the reflections of $H$ over $M$ and $N$ be $M'$ and $N'$. Then $M' \in \omega$ since $\measuredangle BHC = \measuredangle ACB = \measuredangle AM'B$ and similarly for $N'$. Then by PoP we get that $HP \cdot HM' = HQ \cdot HN'$ which implies that $HM \cdot HP = HQ \cdot HN$ so $MNPQ$ is cyclic as desired. We also claim that $(AMN)$ is tangent to $(ABC)$ which is true as $\angle CAA = \angle ABC = \angle AMN$. By applying Rad Axis on $(AMN)$, $(ABC)$, and $(MNPQ)$ we get that $PQ$, $MN$, and $AA$ concur at $R$, as desired.

Given that \( \triangle ABC \) is inscribed in circle \( \omega \) with orthocenter \( H \) and circumcenter \( O \), and \( M \) and \( N \) are midpoints of \( AB \) and \( AC \) respectively, rays \( MH \) and \( NH \) intersect \( \omega \) again at \( P \) and \( Q \). Since \( MN \parallel PQ \parallel BC \), the intersection \( R \) of \( MN \) and \( PQ \) is at infinity in the direction perpendicular to \( BC \). Therefore, \( RA \) is perpendicular to \( BC \), and since \( OA \) is perpendicular to the tangent at \( A \) (which is parallel to \( BC \)), it follows that \( OA \perp RA \).

Reflect $H$ over $M, N$ to get $P', Q'$ respectively. Then it is known that $P', Q', P, Q$ are cyclic. Therefore $HP \cdot HP' = HQ \cdot HQ' \implies HM \cdot HP = HN \cdot HQ \implies M, N, P, Q$ are cyclic. Now consider the circumcircle of $ABC$, $AMN$, $PQMN$. Note that due to homothety at A, the first two circles are tangent to each other at $A$. Now from Radical Axis, we have that $MN, PQ, AA$ concur. Therefore, $R$ lies on the $A-$tangent to $\Delta ABC$, and therefore $OA \perp RA$. $\square$

Problem reminded me of 2018 ELMO G2, tho I couldn't connect a similar solution for this one. Let the $Nine\text{ }Point\text{ }Circle$ intersect $HQ\text{ and } HP$ at $Q'\text{ and }P'$.Note that $P,Q$ are homotheties from $P'\text{ and }Q'$ with center $H$ and ratio $2$. From PoP we have, $$HM\cdot HP' = HN\cdot HQ', \implies HM\cdot 2HP' = HN\cdot 2HQ' \implies HM\cdot HP =HN\cdot HQ $$Hence, $MNPQ$ is cyclic. $OM\perp AB,$ $ON\perp AC$ hence $AMON$ is cyclic as well. Note that $M,N$ are homotheties from $B\text{ and }C$ with center $A$ and ratio $\frac{1}{2}$. Hence the circumcenter of $AMON$ lies on $AO$. Consider the three circumcircles of $AMON,\text{ }MNPQ\text{ and } ABC$. From radical center we have that $RA \perp OA$