Let the line through $H$ parallel to $BC$ intersect $AB$ and $AC$ at $X$ and $Y$ respectively. Then $AH$ is perpendicular to $XY$. It is a well known fact that $OA$ is perpendicular to $DE$, and because $PE$ is perpendicular to $AB$ this implies $\angle QPH = \angle AED = \angle AHD = \angle AHQ$, and therefore $AH$ is tangent to the circumcircle of $PQH$. This implies that the circumcenter of $PQH$ lies on $XY$, and moreover by considering the power of $A$ to the circumcircles of $XHE$ and $PQH$ we find $$AP \cdot AQ = AH^2 = AE \cdot AX$$ Thus $E, Q, P, X$ are concyclic, implying $\angle XQP = \angle XEP = 90^{\circ}$, and $Q$ is the projection of $X$ onto $AO$. Analogously $P$ is the projection of $Y$ onto $AO$. Let $M$ be the midpoint of $XY$, then because $XQYP$ is a trapezoid with $PQ$ perpendicular to the bases, it is easy to verify that $MP = MQ$, and therefore $M$ is the circumcenter of $PQH$, because it lies on $XY$ and on the perpendicular bisector of $PQ$ and the circumcenter does too. Finally, notice that $AM$ goes through the midpoint of $BC$ because $XY$ and $BC$ are parallel, and therefore the circumcenter of $PQH$ lies on the $A$-median of $ABC$.

Really nice problem. Although too easy to bash.

We present a synthetic solution and a complex numbers solution. First solution by ratios (Zhao Ting-Wei) Let $X$ and $Y$ denote the midpoints of $\overline{PH}$ and $\overline{QH}$. Recall that $\triangle ABE \sim \triangle ACF$, and moreover that $\overline{AO}$ and $\overline{AH}$ are isogonal. Thus $H$ and $P$ on $\overline{BE}$ correspond to $Q$ and $H$ on $\overline{CF}$, or in other words that \[ BH : HX : XP : PE = CQ : QY : YH : HF. \]In other words \[ \frac{BX}{XE} = \frac{CY}{YF}. \][asy][asy]size(200); pair A = dir(115); pair B = dir(220); pair C = dir(320); pair H = orthocenter(A, B, C); pair E = foot(B, A, C); pair F = foot(C, A, B); pair D = foot(A, B, C); draw(A--B--C--cycle, red); pair P = extension(A, origin, B, H); pair Q = extension(A, origin, C, H); pair S = circumcenter(H, P, Q); pair M = midpoint(B--C); pair X = midpoint(H--P); pair Y = midpoint(H--Q); draw(B--H--F, red); draw(P--E, red); draw(Q--C, red); draw(P--H--Q--cycle, heavycyan); draw(X--S--Y, heavycyan); draw(A--M, dashed+heavygreen); draw(D--A--P, dotted+red); pair O = origin; dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, dir(270)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$D$", D, dir(D)); dot("$P$", P, dir(340)); dot("$Q$", Q, dir(Q)); dot("$S$", S, dir(10)); dot("$M$", M, dir(M)); dot("$X$", X, dir(110)); dot("$Y$", Y, dir(Y)); dot("$O$", O, dir(50)); /* TSQ Source: A = dir 115 B = dir 220 C = dir 320 H = orthocenter A B C R270 E = foot B A C F = foot C A B D = foot A B C A--B--C--cycle 0.1 lightred / red P = extension A origin B H R340 Q = extension A origin C H S = circumcenter H P Q R10 M = midpoint B--C X = midpoint H--P R110 Y = midpoint H--Q B--H--F red P--E red Q--C red P--H--Q--cycle 0.1 lightcyan / heavycyan X--S--Y heavycyan A--M dashed heavygreen D--A--P dotted red O = origin R50 */ [/asy][/asy] But for any $(X, Y) \in \overline{BE} \times \overline{CF}$ with that directed ratio, the intersection $S$ lies on the $A$-median. Indeed this follows by noting the result holds for $(E,F)$ and $(B,C)$, and then applying linearity. Second solution by similarity (shortlist packet) We begin with the following claim: Claim: Line $AH$ is tangent to $(APQ)$, and $\triangle HPQ \sim \triangle ABC$. Proof. By angle chasing $\measuredangle PQH = \measuredangle PHA = \measuredangle ACB$. Similarly $\measuredangle QPH = \measuredangle QHA = \measuredangle ABC$. This implies both parts. $\blacksquare$ Let $T = \overline{AA} \cap \overline{BC}$, and $M = \overline{AS} \cap \overline{BC}$. [asy][asy] size(300); pair A = dir(135); pair B = dir(200); pair C = dir(340); pair H = orthocenter(A, B, C); pair E = foot(B, A, C); pair F = foot(C, A, B); pair D = foot(A, B, C); draw(A--B--C--cycle, red); pair P = extension(A, origin, B, H); pair Q = extension(A, origin, C, H); pair S = circumcenter(H, P, Q); pair M = midpoint(B--C); draw(B--H--F, red); draw(P--E, red); draw(Q--C, red); draw(P--H--Q--cycle, heavycyan); draw(A--M, dashed+heavygreen); draw(D--A--P, heavycyan); pair O = origin; pair T = extension(B, C, A, A+dir(90)*A); draw(unitcircle, red); draw(circumcircle(H, P, Q), heavycyan); draw(A--T--B, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, dir(270)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$D$", D, dir(D)); dot("$P$", P, dir(340)); dot("$Q$", Q, dir(Q)); dot("$S$", S, dir(10)); dot("$M$", M, dir(M)); dot("$O$", O, dir(50)); dot("$T$", T, dir(T)); /* TSQ Source: A = dir 135 B = dir 200 C = dir 340 H = orthocenter A B C R270 E = foot B A C F = foot C A B D = foot A B C A--B--C--cycle 0.1 lightred / red P = extension A origin B H R340 Q = extension A origin C H S = circumcenter H P Q R10 M = midpoint B--C B--H--F red P--E red Q--C red P--H--Q--cycle 0.1 lightcyan / heavycyan A--M dashed heavygreen D--A--P heavycyan O = origin R50 T = extension B C A A+dir(90)*A unitcircle red circumcircle H P Q heavycyan A--T--B red */ [/asy][/asy] Then under the similarity mapping $HPQ$ to $ABC$ (i.e.\ $H \mapsto A$, $P \mapsto B$, $Q \mapsto C$) we also have $S \mapsto O$, $A \mapsto T$. Therefore \[ \measuredangle MAO = \measuredangle SAP = -\measuredangle OTB = \measuredangle MTO. \]Consequently, $AOTM$ is cyclic. This implies $\overline{OM} \perp \overline{BC}$ as desired. Third solution by complex numbers (Ankan Bhattacharya) Inscribe $\triangle ABC$ in the unit circle, so $h = a + b + c$. The reflection $H_B$ of $H$ over $\overline{AC}$ equals $-ac/b$ and the antipode $A'$ of $A$ is $-a$, so by chord intersection $P = \overline{BH_B} \cap \overline{AA'}$ is \begin{align*} p & = \frac{b(-ac/b)(a - a) + a^2(b - ac/b)}{-ac + a^2}\\ & = \frac{a(b^2 - ac)}{b(a - c)}\\ & = \frac{a(b^2 - ac) - a(a + b + c)(a - c)}{b(a - c)} + h\\ & = \frac{(bc - a^2)(b + c)}{b(a - c)} + h. \end{align*}Of course $q = \tfrac{(bc - a^2)(b + c)}{c(a - b)} + h$. These forms allow us to compute the required circumcenter easily: The circumcenter of $bc$, $ac$, $ab$ is $0$ (unit circle). The circumcenter of $0$, $c(a - b)$, $b(a - c)$ is $-bc$. The circumcenter of $0$, $\tfrac{(bc - a^2)(b + c)}{b(a - c)}$, $\tfrac{(bc - a^2)(b + c)}{c(a - b)}$ is $-\tfrac{(bc - a^2)(b + c)}{(a - b)(a - c)}$. The circumcenter $O_1$ of $h$, $p$, $q$ is $a + b + c - \tfrac{(bc - a^2)(b + c)}{(a - b)(a - c)}$. Thus the vector $\overrightarrow{AO_1}$ equals \begin{align*} b + c - \frac{(bc - a^2)(b + c)}{(a - b)(a - c)} & = (b + c) \cdot \frac{(a - b)(a - c) - (bc - a^2)}{(a - b)(a - c)}\\ & = \frac{(b + c)(2a^2 - ab - ac)}{(a - b)(a - c)}\\ & = \frac{a(b + c)(2a - b - c)}{(a - b)(a - c)} \end{align*}and twice that for the $A$-median is $b+c-2a$. Thus we must show $\tfrac{a(b + c)}{(a - b)(a - c)}$ is real, which is clear since it is equal to its conjugate.

Let $S$ be the circumcenter of $\triangle HPQ$ and $M$ the midpoint of $BC$. It’s not hard to show (by angle chasing) that $SP, SQ$ are tangent to $(APB), (AQC)$ and $MB,MC$ are also tangent to $(APB),(AQC)$. Thus $A,S,M$ lie on the radical axis of these two circles and they are collinear. QED

Let $X_A$ be the $A$-HM point; observe that $H(AX_A;PQ) = -1$ and that $\overline{AH}$ is tangent to $\odot(HPQ)$ via angle chasing, so $X_A$ belongs to the polar of $A$ with respect to $\odot(HPQ)$ which finishes the problem since $\angle AX_AH = 90^\circ$.

Let $M$ be the midpoint of $BC$ and let $K$ be the circumcenter of $\triangle PQH$. Since $H,O$ are isogonal conjugates, a bit of angle chasing shows that $\angle HQP=\angle C$ and $\angle HPQ =\angle B$, so $\triangle HPQ \sim \triangle ABC$. Furthermore, since $\angle PHA=\angle PQH= \angle C$, the line $AH$ is tangent to $(PQH)$. This implies that $KH \perp AH$. Let $M'$ be the foot of the perpendicular from $K$ to $PQ$, i.e. the midpoint of $PQ$. Clearly $AHKM'$ is cyclic, so $\angle KAM'=\angle KHM'=\theta$. But since $\triangle HPQ \sim \triangle ABC$, we have that $\triangle HKM' \sim \triangle AOM$, and so $\angle OAM=\theta$. Hence, $\theta = \angle OAK=\angle OAM$, so $A, K, M$ are collinear as desired.

Let $T$ be the harmonic conjugate of $D$ WRT $\overline{BC},$ $X$ be the circumcenter of $\triangle HPQ,$ and $H'$ be the reflection of $H$ across $\overline{AX}.$ Lemma: $\overleftrightarrow{AH}$ is tangent to $\odot (HPQ).$ Proof: $$\angle AQH = \angle QAC + \angle ACQ = \angle OAC + \angle ACH = (90^{\circ}-\angle B) + (90^{\circ}-\angle A)=\angle C = \angle AHP,$$implying the conclusion. By the lemma, $\overline{AH},\overline{AH'}$ are tangent to $\odot(HPQ).$ Thus $HPH'Q$ is a harmonic quadrilateral, and $$(T,D;B,C)=-1=(H,H';P,Q)\stackrel{H}{=}(D,\overleftrightarrow{HH'}\cap\overleftrightarrow{BC};B,C),$$so $T,H,H'$ are collinear and $\overleftrightarrow{TH}\perp \overline{AX}.$ To finish the problem, we note the well-known fact that $\overleftrightarrow{TH}\perp \overline{AM}$ implies $A,X,M$ collinear, as desired.

[asy][asy] size(7cm); pair A = dir(130); pair B = dir(200); pair C = dir(-20); pair F = foot(C,A,B); pair E = foot(B,A,C); pair H = extension(C,F,B,E); pair O = (0,0); pair P = extension(A,O,B,E); pair Q = extension(A,O,C,F); pair M = (B+C)/2; pair O_1 = circumcenter(H,P,Q); pair H_1 = orthocenter(H,P,Q); draw(A--B--C--cycle, red); draw(arc(circumcenter(A,P,B), abs(circumcenter(A,P,B)-A), -110, 80), lightblue); draw(arc(circumcenter(A,Q,C), abs(circumcenter(A,Q,C)-A), 190, 280), lightblue); pair D = foot(A,B,C); draw(A--Q, orange); draw(B--E^^C--F^^A--D, orange); draw(circumcircle(A,B,C), blue); draw(O_1--P^^O_1--Q, orange); draw(A--M, mediumred+dotted); draw(circumcircle(H,P,Q), heavygreen); draw(P--H_1, orange); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, S); dot("$O$", O, NE); dot("$H$", H, S); dot("$P$", P, W); dot("$Q$", Q, S); dot("$E$", E, NE); dot("$F$", F, W); dot("$O'$", O_1, SW); dot("$H'$", H_1, S); dot("$O'$", O_1, SW); dot("$M$", M, S); [/asy][/asy] Let $E$ and $F$ be the foots of the altitudes from $B$ and $C$, respectively, in $\triangle ABC$. Additionally, let $H$ and $O$ be the orthocenter and circumcenter, respectively, of $\triangle ABC$. Since $H$ and $O$ are isogonal conjugates, $\angle QHP = 90^\circ - \angle EBA = \angle BAC$ and $\angle HPQ = 90^\circ - \angle OAC = 90^\circ - \angle BAD = \angle ABC$ so $\triangle ABC \sim \triangle HPQ$. Let $\omega_b = (APB)$ and $\omega_c = (AQC)$. Since $\angle QCA = \angle OCA = \angle OAC$ and $\angle PBC = \angle OBA = \angle BAO$, we see that $\omega_b$ and $\omega_c$ are both tangent to $BC$. Then if $M$ is the midpoint of $BC$, the radical axis of $\omega_b$ and $\omega_c$ is line $AM$. Let $O'$ and $H'$ be the orthocenter and circumcenter, respectively, of $\triangle HPQ$. Then it suffices to show that $O' \in AM$. Lemma: $O'P$ and $O'Q$ are tangent to $\omega_b$ and $\omega_c$, respectively. Proof: We have \[\angle O'PA = 180^\circ - \angle OPQ = 180^\circ - \angle HPH' = 90^\circ + \angle BAC \]but $\angle ABP = 90^\circ - \angle BAC$ showing that $O'P$ is tangent to $\omega_b$. Similarly, $O'Q$ is tangent to $\omega_c$. $\square$ Finally, since $O'$ is the circumcenter of $\triangle HPQ$, we have that $O'P=O'Q$ so by Lemma, $O'$ has equal power with respect to $\omega_b$ and $\omega_c$, showing that $O' \in AM$. $\blacksquare$

I have a clear impression that this is a problem from All-Russian in previous year.

$\measuredangle$ refers to directed angles. Claim 1: $\triangle HPQ\stackrel{-}{\sim}\triangle ABC$. Proof. \begin{align*} \measuredangle HPQ &= \measuredangle BPA = \measuredangle PBA + \measuredangle BAP = \measuredangle HBA + \measuredangle BAO\\ &= 90^{\circ} + \measuredangle CAB + 90^{\circ} + \measuredangle BCA = -\measuredangle ABC. \end{align*}Similarly, $\measuredangle PQH = -\measuredangle BCA$. Claim 2: $\overline{AH}$ is tangent to $\odot(HPQ)$. Proof. \[\measuredangle AHQ = \measuredangle AHB + \measuredangle PHQ = -\measuredangle ACB - \measuredangle BAC = -\measuredangle ACB = \measuredangle HPQ.\] Let $T\in \overline{BC}$ such that $AT$ is tangent to $\odot(ABC)$. Let $K$ be the circumcenter of $\triangle HPQ$ and $M = \overline{AK}\cap\overline{BC}$. Let $\tau$ be the spiral similarity that maps $\triangle HPQ$ to $\triangle ABC$. Claim 3: $O, A, T, M$ are concyclic. Under $\tau$, $A$ gets sent $T$ by the tangency. Therefore, \[\measuredangle OAM = \measuredangle PAK = - \measuredangle BTO = \measuredangle OTM.\] Main Proof. Now, $\measuredangle OMT = \measuredangle OAT = 90^{\circ}$. Therefore, $M$ is the midpoint of $\overline{BC}$.

A similar problem appeared on CGMO 2016. It may seemingly not similar but try to compare the solutions to these two problems.

EulerMacaroni wrote: observe that $H(AX_A;PQ) = -1$ Forgive me if this question is extremely dumb but why is this true?

Let $I$ be the center of $(HPQ)$ ,$M$ the midpoint of $CB$ ; remark $HPQ\sim ABC$ and $\angle C=\angle AHP=\angle HQP$ thus $AH$ is tangent to $(HPQ)$ then the perpendicular to $AI$ through $H$ is the $H$-symmedian of $HPQ$ say $\delta '$ we have $\angle (\delta ',HP)=\angle IAC$ but if we denotes the $A$-symmedian as $\delta$ we deduce $\angle (MA,AC)=\angle (AB,\delta)=\angle (\delta, 'HP) =\angle IAC$ hence $A,I,M$ are colinear . RH HAS

Vrangr wrote: EulerMacaroni wrote: observe that $H(AX_A;PQ) = -1$ Forgive me if this question is extremely dumb but why is this true? It's well-known; essentially, $\overline{HX_A} \cap \overline{BC}$ is the harmonic conjugate of $\overline{AH} \cap \overline{BC}$ with respect to $\overline{BC}$ (Brokard's theorem).

I was able to find another proof (not of the problem but of the lemma above). Let $E=\overline{BH}\cap\overline{CA}$ and $F=\overline{CH}\cap\overline{AB}$. Then note that $X_A\in\odot(BHC)$. $HX_A\cap BC$ is the radical centre of $\odot(HEF)$, $\odot(BHC)$ and $\odot(EFBC)$, thus, $\overline{HX_A}\cap \overline{BC}\in \overline{EF}$. Now, by ceva's and menelaus \[(\overline{HX_A}\cap \overline{BC}, D; B, C)=-1\]

Let $K$ be the circumcenter of $\triangle HPQ$, and $\odot (APB) \cap \odot (AQC) = H_A$. Now, $\angle KPB = 90^{\circ} - \angle PQH = \angle BAQ \Rightarrow KP$ is tangent to $\odot (ABP)$ Also, $\angle CBP = \angle CAH =\angle BAO \Rightarrow \odot (ABP)$ is tangent to $BC$. Similarly, $BC$ and $KQ$ are tangent to $\odot (ACQ) \Rightarrow H_A$ is the $A$-Humpty point of $\triangle ABC$. And, $Pow_{\odot (ABP)}(K) = KP^2 = KQ^2 = Pow_{\odot (ACQ)}(K) \Rightarrow K$ lies on $AH_A$. But, $AH_A$ bisects $BC \Rightarrow K$ lies on the $A$-median. $\blacksquare$

Let $K$ and $M$ be the $A$-$HM$ point of $\bigtriangleup ABC$ and $M$ the midpoint of $\overline{BC}$, respectively. We know $A,\ K,\ M$ are collinear. Observe that, $$\angle PQH=90^\circ-\angle BAO=90^\circ-\angle CAH=\angle C=\angle AHP$$so $AH$ is tangent to $(PHQ)$. Further, $\angle MKC=\angle C$, so $AKQC$ is cyclic. Similarly, $APKB$ is cyclic. Let $F=\overline{CH}\cap \overline{AB},\ E=\overline{BH}\cap \overline{AC},\ D= \overline{AH}\cap \overline{BC}$ and $\mathcal{T}$ be the inversion centered at $A$ with radius $r=\sqrt{AK\cdot AM}$, then $\mathcal{T}(C)=E,\ \mathcal{T}(B)=F,\ \mathcal{T}(H)=D$ and obviously $\mathcal{T}(K)=M$. Also, observe that $\mathcal{T}[(AKQC)]=ME$ and $\mathcal{T}[(APKB)]=MF$, so $Q'=\mathcal{T}(Q)=\overline{ME}\cap \overline{AO}$ and $P'=\mathcal{T}(P)=\overline{MF}\cap \overline{AO}$. Since $B,\ H,\ P,\ E$ all lie on a same line, we infer that $FDP'C$ is cyclic, and since $MF=MQ$, it must be an isosceles trapezoid, so $MD=MP'$. Analogously, $EQ'DB$ is an isosceles trapezoid and $MD=MQ'$, so $M$ is the circumcenter of $\bigtriangleup DQ'P'$. We can obtain that, $$\angle DP'Q'=\angle DP'A=\angle PHA=\angle PQH,\qquad \angle DQ'P'=\angle DHQ=\angle HPQ $$thus, $HP\parallel DQ'$ and $HQ\parallel DP'$, so $\bigtriangleup HPQ$ and $\bigtriangleup DQ'P'$ are homothetic with center $A$. Hence, $M$, the circumcenter of $\bigtriangleup HPQ$ and $A$ must be collinear, as desired.

Vrangr wrote: I was able to find another proof. Let $E=\overline{BH}\cap\overline{CA}$ and $F=\overline{CH}\cap\overline{AB}$. Then note that $X_A\in\odot(BHC)$. $HX_A\cap BC$ is the radical centre of $\odot(HEF)$, $\odot(BHC)$ and $\odot(EFBC)$, thus, $\overline{HX_A}\cap \overline{BC}\in \overline{EF}$. Now, by ceva's and menelaus \[(\overline{HX_A}\cap \overline{BC}, D; B, C)=-1\] Sorry, but I don't get how this proves the given problem? @below ok thanks

math_pi_rate wrote: Vrangr wrote: I was able to find another proof. Let $E=\overline{BH}\cap\overline{CA}$ and $F=\overline{CH}\cap\overline{AB}$. Then note that $X_A\in\odot(BHC)$. $HX_A\cap BC$ is the radical centre of $\odot(HEF)$, $\odot(BHC)$ and $\odot(EFBC)$, thus, $\overline{HX_A}\cap \overline{BC}\in \overline{EF}$. Now, by ceva's and menelaus \[(\overline{HX_A}\cap \overline{BC}, D; B, C)=-1\] Sorry, but I don't get how this proves the given problem? It doesn't, he was just giving a proof of the lemma that I used.

Wow!! Let $Z$ be the $A$-antipode. From $BHCZ$ parallelogram, we see $HQ\parallel BZ,HP\parallel CZ$ so $(BPZ),(CQZ)$ are tangent to $(PQH).$ Now invert at $A$ and denote the images of points with a $'.$ Let $D,E$ be points on $AC',AB'$ respectively such that $\angle AB'D=\angle AC'E=90^\circ.$ Letting $M$ be the midpoint of $DE,$ by Three Tangents lemma on $\triangle ADE$ we get $M$ is the intersection of the tangents at $B',C'$ to $(AB'C').$ Then notice $(AP'D)$ has diameter $AD,$ so $P'D\perp AP'\perp B'C'.$ Thus $\measuredangle MB'C'=\measuredangle B'AC'=\measuredangle B'AD=-\measuredangle C'B'P'=\measuredangle P'B'C',$ so $M$ lies on $B'P'.$ Similarly, $M$ also lies on $C'Q'.$ But we have that $(B'P'Z')$ has diameter $B'P',$ so the center of $(P'Q'H')$ must lie on $B'P'$ because of our tangency. It also has to lie on $C'Q',$ so it is $M,$ and since $AM$ is a symmedian of $\triangle AB'C'$ we finish by inverting back

Very easy problem, headsolved in 5min Note $\triangle PQH \sim \triangle BCA$ by angle chasing. From $\measuredangle PHA = \measuredangle BCA$ we have that $HA$ is tangent to $(PQH)$. Let $O'$ be circumcenter of $\triangle PQH$ and $H'$ be $H$-dumpty point of $\triangle PQH$ . Let $R = HH' \cap PQ$. hence from tangent property we know that $O,H',A$ are collinear. $$ (A,R;P,Q)\stackrel{H}{=}(A,H';E,F)=-1$$ which give us $H'$ is $A$-humpty point of $\triangle ABC$. Which give us $AO'$ pass through midpoint of $BC$.

Let $E= BH \cap AC$ and $F= CH \cap AB$ and $K$ the center of $PQH$. I will show that $AK$ bisects $BC$. Let $HK$ intercect $AB,AC$ at $M,N$ respectively. We have that: $\angle PQH = \angle QAC + \angle QCA = \angle OAC + \angle HCA = 90 - \angle ABC + 90 - \angle BAC = \angle BCA = 180 - \angle AHB = \angle AHP$, so $AH$ is tangent to $(PQH) => AH \perp HK => HK \parallel BC$. Using that $HF$ is the projection of $H$ in $AM$ and $AHM$ is a right angle triangle we get that $AF \times AM = AH^2 = AP \times AQ => MFPQ$ is cyclic. Analogously, $NEPQ$ is cyclic. Hence, we get that $ \angle NQP = \angle PEA = \angle HEA = 90$. Now let $ Y = NQ \cap (PQH)$. Then we have that $\angle PQY = 180 - \angle NQP = 90$ so $Y$ is the antidiametric point of $P$ in $(PQH)$, i.e. the symmetric of $P$ with respect to $K$. From the cyclic quadtilaterals we get that: $ \angle MPQ = \angle MFQ = \angle BFC = 90 = \angle NQP => MP \parallel NQ$, as they are both $\perp AO$. Also, $K$ is the midpoint of $PY$ and hence $MN$ bisects $PY$. So, we conclude that $MPNY$ is a parallelogram, so $K$ is the midpoint of $MN=> AK$ bisects $MN$ and so $AK$ bisects $BC$, because $MN \parallel BC$.

This is do-able in three lines. Observe that $ABC$ and $HPQ$ are similar with $(HPQ)$ tangent to $\overline{AH}$ by angle chasing. Let the perpendicular to $AH$ at $H$ meet $AM$ and $AO$ at $O'$ and $X$ respectively. Let $AO \cap BC = Y$. Since $\angle O'HP = \angle BAO = 90 - C$, cevian $O'H$ in $HPQ$ corresponds to cevian $AO$ in $ABC$. We show $\frac{AO'}{AX} = \frac{DM}{DX} = \frac{AO}{AY}$ since $MO \parallel DA$ so we're done.

Proof by my goat scamsea. Let $\Delta DEF$ be the orthic triangle, and let $M$ be the midpoint of $\overline{BC}$. Claim: $\Delta HPQ \sim \Delta ABC$ and are oppositely oriented. Proof. Since $H$ and $O$ are isogonal conjugates, $$\measuredangle BAD = \measuredangle OAC=\measuredangle QAC = 90^{\circ} - \measuredangle CBA$$Then, $$\measuredangle PQH = \measuredangle QAC+\measuredangle QCA = 90^{\circ}-\measuredangle CBA + 90^{\circ} - \measuredangle BAC = \measuredangle ACB$$In a similar way, $\measuredangle HPQ = \measuredangle BAC$. Thus, the triangles are oppositely similar. Claim: $\overline{AD}$ is tangent to $(HPQ)$ Proof. $$\measuredangle PHA = \measuredangle EHA = 90 - \measuredangle HAE = \measuredangle ACB = \measuredangle PQH $$and the Tangent-Chord theorem finishes. Now, this means that $O_1H \parallel DM$, since they are both perpendicular to $\overline{AD}$. Claim: $A$, $O_1$, and $M$ are collinear. Proof. It's sufficient to prove that $\measuredangle AO_1H = \measuredangle AMD$. Let $X$ be the intersection of the tangent at $A$ to $(ABC)$ with $BC$. Note that by the first claim, $\measuredangle AO_1H = \measuredangle AOX$. We know that points $A$,$O$,$M$, and $X$ are concyclic, since $\measuredangle XAO = \measuredangle XMO = 90^{\circ}$, so $$\measuredangle AO_1H = \measuredangle AOX = \measuredangle AMX = \measuredangle AMD$$

Nice! Let $S$ be the $A$-antipode in $(ABC)$ and $T$ be the $H$-antipode in $(PQH)$. Let $M$ be the midpoint of $BC$ and $W$ be the circumcenter of triangle $PQH$. Since $W$ and $M$ are the midpoints of $HT$ and $HS$ respectively, it suffices to show that $AM \parallel TS$. From simple angle chasing we obtain $ABCS \stackrel{-}\sim HQPT$, so if $X = PT \cap BC$, then $\measuredangle QPT = \measuredangle SBC$ implies $B,P,X,S$ concyclic. Similarly, if $Y = QT \cap BC$, then $B,Q,Y,S$ are concyclic. But now $\angle BSX = \angle BPX = 90^\circ$ implies $AB \parallel TY \parallel XS$ and similarly $AC \parallel TX \parallel YS$, so $TXSY$ is a parallelogram, $TS$ is the $T$-median of $\Delta TXY$. By the homothety between $\Delta TYX$ and $\Delta ABC$, we obtain $TS \parallel AM$ as desired.

Angle chasing gives us that $(APB)$ and $(AQC)$ are tangent to $\overline{BC}$, so they both pass through the $A$-humpty point in $\triangle ABC$ (call this point $R$). More angle chasing with cyclic quads $APRB$ and $ARQC$ gives us that $\angle PRQ = 2 \angle A$. Even more angle chasing gives us that $\triangle HQP \sim \triangle ABC$ (inversely oriented), so $\overline{HR}$ is a symmedian in $\triangle HPQ$, so $R$ is the $H$-dumpty point of $\triangle HPQ$. So, since $\angle ARH = 90^{\circ}$, line $AR$ passes through the circumcenter of $\triangle HPQ$.

Let $M$ be the midpoint of $BC$, $S$ be the circumcenter of $\triangle HPQ$, $T$ be the intersection of the tangent to $(ABC)$ at $A$ with $BC$ and $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, and $C$, respectively. Notice that $\triangle HPQ \sim \triangle ABC$ and also $AH$ is tangent to $(HPQ)$ by a quick angle chase. From the similarity, we have that $\angle SAH=\angle OTA$. Now notice that $ATMO$ is cyclic, hence $\angle OTA=\angle OMA=\angle MAH$ so $A$, $S$ and $M$ are collinear. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -6.4974336653974225, xmax = 28.256544027907935, ymin = -10.40214902498077, ymax = 7.443936938424266; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); draw((9.043957767876451,1.5088386351661338)--(7.,-5.)--(15.,-5.)--cycle, linewidth(0.8) + zzttqq); /* draw figures */ draw((9.043957767876451,1.5088386351661338)--(7.,-5.), linewidth(0.8) + zzttqq); draw((7.,-5.)--(15.,-5.), linewidth(0.8) + zzttqq); draw((15.,-5.)--(9.043957767876451,1.5088386351661338), linewidth(0.8) + zzttqq); draw(circle((11.,-2.680762783903952), 4.623728069915547), linewidth(0.8)); draw((7.718094898849119,-2.713277693501062)--(15.,-5.), linewidth(0.8)); draw((11.354090066782426,-1.0157032961128492)--(7.,-5.), linewidth(0.8)); draw((9.043957767876451,1.5088386351661338)--(11.581626409158005,-3.926534878779204), linewidth(0.8)); draw(circle((10.437916412506677,-3.129635797025964), 1.3939586446302261), linewidth(0.8)); draw((-4.897172507611289,-5.)--(7.,-5.), linewidth(0.8)); draw((9.043957767876451,1.5088386351661338)--(-4.897172507611289,-5.), linewidth(0.8)); draw((-4.897172507611289,-5.)--(11.,-2.680762783903952), linewidth(0.8)); draw((9.043957767876451,1.5088386351661338)--(11.,-5.), linewidth(0.8)); draw((9.043957767876451,1.5088386351661338)--(9.043957767876451,-5.), linewidth(0.8)); draw((11.,-2.680762783903952)--(11.,-5.), linewidth(0.8)); /* dots and labels */ dot((9.043957767876451,1.5088386351661338),dotstyle); label("$A$", (8.90118908016921,1.7943760105806144), NE * labelscalefactor); dot((7.,-5.),dotstyle); label("$B$", (6.616890076853366,-5.527618116119281), NE * labelscalefactor); dot((15.,-5.),dotstyle); label("$C$", (15.223802392918426,-5.28287179433544), NE * labelscalefactor); dot((11.,-2.680762783903952),linewidth(4.pt) + dotstyle); label("$O$", (11.185488083485057,-2.794617522866395), NE * labelscalefactor); dot((9.043957767876451,-3.129635797025963),linewidth(4.pt) + dotstyle); label("$H$", (8.33604723157005,-2.815013049681715), NE * labelscalefactor); dot((10.561313139898942,-1.7411495887371609),linewidth(4.pt) + dotstyle); label("$P$", (10.532831225394816,-1.4077216994246322), NE * labelscalefactor); dot((11.581626409158005,-3.926534878779204),linewidth(4.pt) + dotstyle); label("$Q$", (11.511816512530178,-4.4466551949073185), NE * labelscalefactor); dot((7.718094898849119,-2.713277693501062),linewidth(4.pt) + dotstyle); label("$F$", (7.228755881312968,-2.5498712010825546), NE * labelscalefactor); dot((11.354090066782426,-1.0157032961128492),linewidth(4.pt) + dotstyle); label("$E$", (11.430234405268898,-0.8570424754109911), NE * labelscalefactor); dot((11.,-5.),linewidth(4.pt) + dotstyle); label("$M$", (10.859159654439937,-5.548013642934601), NE * labelscalefactor); dot((10.437916412506677,-3.129635797025964),linewidth(4.pt) + dotstyle); label("$S$", (9.941360947750534,-3.284110166434076), NE * labelscalefactor); dot((-4.897172507611289,-5.),linewidth(4.pt) + dotstyle); label("$T$", (-5.09014231514034,-5.686827062488641), NE * labelscalefactor); dot((9.043957767876451,-5.),linewidth(4.pt) + dotstyle); label("$D$", (8.86039802653857,-5.527618116119281), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Let the orthic triangle of $\triangle DEF$ and let $M$ be the midpoint of $BC$, and $O'$ the circumcenter of $\triangle HPQ$. Then $\angle HPQ = \angle 180^{\circ} - \angle BOP - \angle PBO$. Since $\angle HBO = \angle B - 2\angle ABH = \angle B + 2\angle A - 180^\circ$ and $\angle BOP = 2\angle C$, so $\angle HPQ = 180^{\circ} - \angle A - \angle C = \angle B$. Then since $\angle BHC = \angle B$ we have that $AH$ is tangent to $(HPQ)$ which implies that $\angle HQP = \angle C$ and $\angle PHQ = \angle A$. Now consider the spiral similarity sending $\triangle HPQ \to \triangle ABC$. This spiral similarity sends $O' \to O$, and since $HH \cap PQ = A$ we have that $A$ goes to $AA \cap BC = Z$. Notice that $\angle ZAO = \angle ZMO = 90^\circ$ so $ZAOM$ is cyclic. By our spiral similarity, we have that $\angle PAO' = \angle BZO$ and $\angle BZO = \angle MAO'$ which proves our desired collinearity.

Claim 1: (a) $AH$ is tangent to $(HPQ)$ (b) $\Delta{HPQ} \sim \Delta{ABC}$. Proof. Angle chasing. Now consider $(APB)$ and $(AQC)$. It is easy to see that they are both tangent to $BC$. Claim 2: $\overleftrightarrow{\rm AR}$ is the radical axis of $(APB)$ and $(AQC)$. Proof. Some angle chasing shows that $RP$ and $RQ$ are tangent to $(APB)$ and $(AQC)$ respectively. Hence, since $RP^2 = RQ^2$, $R$ has equal powers with respect to $(APB)$ and $(AQC)$. It is not hard to see that Claim 2 finishes the problem: Let $AR$ intersect $BC$ at $M$. Since $M$ is on the radical axis of $(APB)$ and $(AQC)$, it must have equal powers with respect to both circles, implying that $MB^2 = MC^2$, which in turn implies that $M$ is the midpoint of $BC$ as desired.

solid.

I don't think this solution is on the thread and in my opinion it's splendid.

Let all angles in this problem be directed. Claim 1: $\triangle HPQ \sim \triangle ABC$ Proof 1: Extend $BH$ and $CH$ to meet sides $AC$ and $AB$ at $E$ and $F$ respectively. Then notice that $$\angle HPQ= \angle EPA= 90-\angle PAE= 90- \angle BAH= \angle CBA. $$$$ \angle QHP= \angle FHB= 90- \angle HBF= \angle BAC. $$Hence by AA similarity our claim holds. Claim 2: AH is tangent to $(HPQ)$ Proof 2: This claim is equivalent to showing $\triangle APH \sim \triangle AHQ.$ This statement holds as $$\angle PHA= \angle BHJ= 90- \angle JBH= \angle ACB= \angle PQH$$and $\angle HAE= \angle HAE$ which gives us AA similarity. Claim 3: $DB$ and $DC$ are tangent to $(APB)$ and $(AQC)$ respectively. Proof 3: It is well-known and derivable by angle chasing that $\angle BAO= 90-\angle ACB.$ Furthermore, since $BH \perp AC,$ we get $\angle PBC= 90-\angle BCA.$ Hence $\angle DBP = \angle BAP$ so $DB$ is tangent to $(APB).$ By similar steps we get $\angle QCD= \angle QAC= 90- \angle CBA.$ Corollary of Claim 3: Since $A$ and $D$ have equal power to $(APB)$ and $(AQC)$ we conclude that $AD$ is the radical axis of the two circles. Claim 4: Let $O_1$ be the circumcenter of $HPQ.$ I claim that $O_1P, O_1Q$ are tangent to $(APB), (AQC)$ respectively. Proof 4: By circumcenter properties we know that $\angle HPO_1= 90- \angle PQH =90 - \angle ACB.$ Since $O$ and $H$ are isogonal conjugates we know that $\angle BAO= 90- \angle ACB.$ Hence $O_1P$ is tangent to $(APB).$ By similar steps we conclude that $O_1Q$ is tangent to $(AQC).$ This implies $O_1$ has equal power to $(APB),(AQC)$ which means $O_1$ lies on $AD$ (radical axis) which completes our proof.

I am way too lazy and dumb to finish synthetically. Let $D$ be foot of $A$ perpendicular to $BC$ define $E$ and $F$ similarly, let center of $(PQH)$ be $O_1$. $\angle BAD=\alpha=\angle OAC$; $\angle DAO=\beta$. From angle chasing you will find $HO_1$ is parallel to $BC$. Thus if we prove $\frac{HO_1}{AH}=\frac{DM}{AD}$ we are done. From LoS to $\triangle OPH$ and $\triangle AHP$ we will got $\frac{HO_1}{AH}=\frac{\sin{\beta}}{2\cos{(\alpha+\beta)} \cdot \cos{\alpha}}$. Let $AO \cap BC$ at $R$. From thales $\frac{MR}{OR}=\frac{DM}{AO}$. LoS to $\triangle OMR$ gives $DM=AO \cdot \sin{\beta}$. LoS to $\triangle ADC$ and $\triangle ABC$ give us $AD=AO \cdot 2\cos{(\alpha+\beta)} \cdot \cos{\alpha}$. By placing the terms we are done.

Let $O', M, N$ be the centers of $(\triangle HPQ), BC, PQ$ respectively. Define $D = AH \cap BC.$ Notice that $\angle PHQ = 180^\circ - \angle BHC = \angle A,$ and $\angle HPQ = \angle HBA + \angle BAO = 90^\circ - \angle A + \angle A - \angle OAC = 90^\circ - \angle OAC = \angle B.$ Therefore $$\triangle HPQ \sim \triangle ABC.$$ Also, observe that by our similar figures $\angle AHO = \angle AHP + \angle O'HP = \angle C + \angle OAB = \angle C + \frac12 (180^\circ - \angle AOB) = 90^\circ,$ so $AH$ is tangent to $O'.$ It follows that $O'H \parallel BC.$ and $AHO'N$ is cyclic. Thus $$\angle HAO' = \angle NHO' = \angle AMO = \angle DAM.$$Hence $A, O', M$ are collinear and we are done. QED