What a beautiful problem! Let $T$ be the center of $\odot(AH)$; since $H$ is the antipode of $A$ in $\odot(AH)$, notice that it is equivalent to showing that the midpoint of $PQ$ lies on the dilation of $\odot(ABC)$ at $H$ with ratio $\frac{1}{2}$, which is the nine-point circle. But this locus is just the projection of $T$ onto $PQ$, or the circle with diameter $\overline{TM}$, as desired.

Let $E$ and $F$ be the projections of $B$ and $C$ on the opposite sides of $\triangle{ABC}$.Because $BFEC$ is cyclic with diameter $BC$, so respectively with circumcenter $M$ we have the following: $$\angle{FEM} = \angle{FEC}-\angle{MEC}=\angle{A}+\angle{C}-\angle{C}=\angle{A}$$Respectively we have that $\angle{FEM}=\angle{MFE}=\angle{A}$ so $MF$ and $ME$ are tangent to $(AFE) \implies$ $(QP;FE)=-1$. Inverting with pole $A$ and radius $\sqrt{AF\cdot AB}$(that sends $F$ to $B$, $E$ to $C$, $(ABC)$ to $EF$ and $(AEF)$ to $BC$) and using the fact that this inversion will preserve $(QP;FE)=-1$, the problem becomes the following: Let $ABC$ be a triangle, let $F$ and $E$ be the projections of $B$ and $C$ on $ACB$ and $AB$ respectively. Let $Q$ and $P$ be on $BC$ such that $(QP;BC)=-1$, Let the perpendicular to $AQ$ at $Q$ meet $AP$ at $Y$ and let the perpendicular to $AP$ at $P$ meet $AQ$ at $X$. Prove that $EF$ passes through the Miquel point of $QXPY$.* Let $XY\cap PQ=S$. Let $T=PX \cap QY$ and let $M$ be the Miquel point of quadrilateral $QXPY$. Because $QXPY$ is cyclic with diameter $XY$, and using the well-know fact that $M$ is the image of $S$ under inversion wrt $(XPYQ)$ and that $T-M-A$(which is also well-known, because $PXQY$ is cyclic), we have that $M=AT \cap XY$ and $YM$ is perpendicular to $AT$, so $\triangle{MQP}$ is the orthic triangle of $\triangle{TAY}$, so $A$ is the $Q$-excenter of $\triangle{MQP}$.Let $\omega$ be the $Q$-excircle of $\triangle{MQP}$.Let $D$,$K$,$L$ be the tangency point of $\omega$ with $QP$,$QM$ and $PM$ respectively.Let $H$ be the orthocenter of $ABC$. Let $F'$ be the reflection of $F$ over $AB$. Note that $BDFAF'$ is cyclic with diameter $AB$. $$\angle{EDA}=\angle{EDH}=\angle{EBF}=\angle{EBF'}=\angle{F'DA} \implies$$$F'$, $E$ and $D$ are colinear, so after reflecting over $AB$ we have that $EF$ passes through $W$, where $W$ is a point of $\omega$ such that $W$ is different from $D$ and $BW$ is tangent to $\omega$. If we let $V$ be a point on $\omega$ such that $CV$ is tangent to $\omega$, repeating the same argument as above we have that $EF$ passes through $V$. Now inverting wrt $\omega$ we have that the image of $Q$ will be the midpoint of $KD$, the image of $B$ will be the midpoint of $WD$, the image of $P$ will be of $LD$ and the image of $C$ will be the midpoint of $DV$. Now let $Z`$ be the image of $Z$ under this very inversion. We have that $IQ`B`P`C`$ must be cyclic and that it also passes through $D`=D$ because $Q,B,P,D,C$ are colinear. But $-1=(QP,BC)=D(Q`P`,B`C`)=D(KL,WV) \implies WV$ passes through $M \implies EF$ passes through $M \blacksquare$ * I did not preserve the name of points after inversion. EDIT(1/7/2019): this solution is the most horrible thing I've ever done

^ what the actual overkill

ELMO 2017/2 wrote: Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$ Proposed by Michael Ren Let $\overline{BE}$ and $\overline{CF}$ be the altitudes in $\triangle ABC$. It is well-known that $\overline{ME}, \overline{MF}$ are tangent to $\odot(AH)$. Let $N, L$ be midpoints of $\overline{PQ}$ and $\overline{AH}$. Claim: $N$ lies on the nine-point circle of $\triangle ABC$. (Proof) Note that $\measuredangle LEM=\measuredangle LFM=\measuredangle LNM=90^{\circ}$ hence $N$ lies on $\odot(MEF)$ as desired. $\blacksquare$ Construct parallelogram $PHQT$; since $N$ lies on the nine-point circle, $T$ lies on the circumcircle of $\triangle ABC$. However, $T$ is the orthocenter of $\triangle APQ$, hence we conclude that the result holds. $\blacksquare$

Here's a complex bash solution that (I think) works together with other solutions above. I did not actually turned in this solution but I enjoyed it. So set up usual coordinates - let $H'$ be the orthocenter of $APQ$ and we will use lowercase for the complex numbers. Denote $N$ as the midpoint of $AH$. Clearly, $h'=(a-n)+(p-n)+(q-n)+n = a+p+q-2n=p+q+a-(a+h) = p+q-h$. So it suffices to prove $|p+q-h|=1$. Now note $|p-n|=|q-n|=\frac{1}{2} |AH| = \frac{1}{2} |b+c|$. Also, note that $p-\frac{b+c}{2}$ and $q-\frac{b+c}{2}$ are parallel vectors, since $P, Q, M$ are colinear. Set $p'=p-\frac{b+c}{2}$ and $q'=q-\frac{b+c}{2}$. We now have $|p'-a|=|q'-a| = \frac{1}{2} |b+c|$ and $p', q'$ are parallel. Now what we need to show is that $p+q-h=p'+q'-a$ has magnitude $1$. Now we will completely look at this problem with vectors. Take $O$ as the origin. Take a point $A$ with coordinate $a$, and draw a circle with radius $\frac{1}{2} |b+c|$. Draw a line from $O$ and let it hit the circle at two points $P', Q'$. We know that $p', q'$ are the coordinates for $P', Q'$. This is because $|p'-a|=|q'-a| = \frac{1}{2} |b+c|$ and $p', q'$ are parallel. Now $p'+q'-a$ is just the reflection of $A$ wrt $P'Q'$, so $|p'+q'-a| =|a| = 1$, as desired.

Let $Y$ and $X$ be the circumcenter of $\odot ABC$ and $\odot AH$.Let $H_1$ be the orthocenter of $\triangle APQ.$ Let $K$ be the midpoint of $PQ$ and $L$ be the reflection of $M$ across $K$ It's well known that $AYMX$ , $HPH_1Q$ and $YXHM$ are parallelogram. (Basically the reflection of orthocenter across the midpoint of a side is the antipode of the other vertex) Now, See that $MH_1LH$ is a parallelogram ! $=>LH_1 \parallel HM => LH_1 \parallel XY$.Also $LH_1 = XY => LH_1YX$ is a parallelogram! $=> H_1Y = LX = MX = AY$.Done ! Motivation

antipode of $A$ in $\odot(ABC)$ is $O$, line through $A$ and perpendicular $PQ$ cut $(ABC),(AH)$ at $X,Y$ so $XO//YH$ so perpendicular bisector of $XY$ cut $HO$ at midpoint of $OH$ is $M$ so $M $ in perpendicular bisector of $XY$ so $P,Q$ in perpendicular bisector of $XY$ so $X$ is orthocenter of $APQ$

[asy][asy]size(7cm); pair B=(0,0),A=(7,11),C=(10,0),H,M,P,Q,X,Y,Z; H=orthocenter(A,B,C);M=(B+C)/2;X=(A+H)/2; Q=X+(A-X)*dir(45);Y=foot(X,Q,M);P=2*Y-Q; Z=orthocenter(A,P,Q); draw(circumcircle(A,P,Q)^^circumcircle(Y,X,M)^^circumcircle(A,B,C),green); D(MP("A",A,N)--MP("B",B,S)--MP("M",M,S)--MP("C",C,SE)--cycle); D(MP("H",H,S)--MP("X",X,NE)--A--MP("Q",Q,W)--MP("Y",Y,SSW)--MP("P",P,W)--M,magenta); D(X--Y,magenta);D(P--A--MP("Z",Z,S),magenta); dot(A^^B^^C^^H^^M^^X^^Q^^Y^^P^^Z); [/asy][/asy] Let $X$ be the midpoint of $AH$ (and hence the circumcenter of $\odot(AH)$), $Y$ be the foot of perpendicular from $X$ to $PQ$, and $Z$ be the orthocenter of $APQ$. Also let $k$ be the circle with diameter $XM$, which is therefore the nine-point circle of $ABC$. Let $f$ be the homothety centered at $H$ and ratio $2$; it sends $k\to\odot(ABC)$, and $X\to A$. Clearly $XY||AZ$ (both are $\perp$ to $PQ$), and $AZ=2\cdot XY$ by a well-known property of orthocenters. So $f$ sends $Y\to Z$. But $Y\in k$ since $\angle XYM=90^\circ$, so $Z=f(Y)\in f(k)=\odot(ABC)$, as desired. $\blacksquare$

Here's another way to complex bash ! Let $Y$ and $X$ be the circumcenter of $\odot ABC$ and $\odot AH$.Let $H_1$ be the orthocenter of $\triangle APQ$ We will work in the complex plane and use the general notations. Now, Set $m = 0$. It's well known that $AYMX$ and $HPH_1Q$ are parallelogram. (this is the main observation the rest is trivial !) (Basically the reflection of orthocenter across the midpoint of a side is the antipode of the other vertex) Now, $a = x+y => y = a-x$ Also $H$ is the reflection of $A$ across $X$ so $h = 2x - a$ Now, $h_1 = (p+q) - h => h_1 - y = (p+q) - (2x-a) -(a-x) = (p+q) -x$ It suffices to show that $|h_1-y| = |a-y| => |(p+q) -x| = |x|$ As $M-P-Q$ are collinear $=> \frac{p}{\overline{p}} = \frac{q}{\overline{q}}$ So , Eq. of line $PQ$ is $ \overline{z} = kz$ for some constant $k(\neq 0)$. Also see that $|k| = 1$. Eq. of circle $\odot AH$ is $|z -x| = |a-x| => |z|^2 - z\overline{x} - \overline{z}x + {|x|^2 - |a-x|^2} = 0$ Now, See that the quadratic equation $kz^2 - z(\overline{x} + kx)+ {|x|^2 - |a-x|^2} = 0$ has roots $p,q$. So, $p+q = \frac{\overline{x}+ kx}{k} => (p+q)-x = \frac{\overline{x}}{k} =>|p+q-x| = \frac{|\overline{x}|}{|k|} = |x|$ Hence, proved

Probably the worst solution: Let $H'$ be the orthocenter of $\triangle APQ$. As $P, Q$ lie on the circle with diameter $AH$, $\angle APH = \angle AQH = 90^{\circ}$. So, $AP\perp PH$ and $AQ\perp QH$. As $H'$ is the orthocenter of $\triangle APQ$, $H'P \perp AQ$ and $H'Q \perp AP$. $\Rightarrow PH' || HQ, QH'||HP$. So, $PH'QH$ is a parallelogram. Now we apply coordinate bash. Let $B=(-1,0), C = (1,0), A(a_1,a_2)$. So $M=(0,0)$. Coordinates of $H$: Slope of $AC = \frac{a_2}{a_1 - 1}$. As $AC\perp BH$, slope of $BH = \frac{1-a_1}{a_2}$. Equation of $BH = y - 0 = (x-(-1))\left(\frac{1-a_1}{a_2}\right)$ or $y = (x+1)\left(\frac{1-a_1}{a_2}\right)$. As $BC$ is parallel to the $x-axis$, $H$ has the same x-coordinate as that of $A$. So, $$H = \left(a_1,\frac{1-a_1^2}{a_2}\right)$$. Equation of circle with diameter $AH$: Let $O$ be the center of circle with diameter $AH$. As $O$ is the midpoint of $AH$, $O = \left(a_1, \frac{1-a_1^2+a_2^2}{2a_2}\right)$. Let $R = \frac{AH}{2}$. So, equation of circle with diameter $AH$ is: $$(x-a_1)^2 + \left(y-\frac{1-a_1^2+a_2^2}{2a_2}\right)^2 = R^2$$. Coordinates of $P, Q$: As $M$ is the origin, let the equation of line $PM$ be $y = mx$. So, let $$P = (p,mp), Q = (q,mq), \text{where} p\neq q$$Simplification of coordinates of $P, Q$ As $P, Q$ lie on the circle with diameter $AH$, $(p-a_1)^2 + \left(mp-\frac{1-a_1^2+a_2^2}{2a_2}\right)^2 = R^2$ $(q-a_1)^2 + \left(mq - \frac{1-a_1^2 + a_2^2}{2a_2}\right)^2 = R^2$. Subtracting these equations and using the identity $a^2 - b^2 = (a-b)(a+b)$, $(p-q)(p+q-2a_1) + (p-q)\left(p+q - 2\left(\frac{1-a_1^2 + a_2^2}{2a_2}\right)\right) = 0$ As $p \neq q$ we get after simplifying, $$p+q = \frac{m(1-a_1^2+a_2^2)+2a_1a_2}{a_2(m^2+1)}$$Coordinates of $H'$: As $PH'QH$ is a parallelogram, $H' = \left(p+q - a_1,mp+mq - \frac{1-a_1^2}{a_2}\right)$. Substituting the value for $p+q$ we get, $$H' = \left(\frac{m(1-a_1^2+a_2^2)+a_1a_2(1-m^2)}{a_2(m^2+1)}, \frac{m^2a_2^2+2ma_1a_2-1+a_1^2}{a_2(m^2+1)}\right)$$Equation of circumcircle of $ABC$: Let $O'$ be the center of circle $(ABC)$. As $M$ is on the y-axis and $BC||x-axis$, the x-coordinate of $O'$ is $0$. Let the y-coordinate of $O'$ be $y_1$. Let $B'$ be the midpoint of $AC$. So, $B' = \left(\frac{a_1+1}{2},\frac{a_2}{2}\right)$. As $O'B'\perp AC$, the product of the slopes of the two lines is $-1$. So, $\frac{\frac{a_2}{2}-y_1}{\frac{a_1+1}{2}} = -1$. Solving for $y_1$ we get $y_1 = \frac{a_1^2 + a_2^2 - 1}{2a_2}$. So, $O'=\left(0,\frac{a_1^2+a_2^2-1}{2a_2}\right)$. $(\text{Radius of} (ABC))^2 = OM^2 + MB^2 = \left(\frac{a_1^2+a_2^2-1}{2a_2}\right)^2 + 1 = \frac{(1+m^2)^2\cdot((a_1^2+a_2^2-1)^2+4a_2^2)}{(2a_2)^2(1+m^2)^2}$. So equation of $(ABC)$ is $$x^2 + \left(y-\frac{a_2^2+a_1^2-1}{2a_2}\right)^2 = \frac{(1+m^2)^2\cdot((a_1^2+a_2^2-1)^2+4a_2^2)}{(2a_2)^2(1+m^2)^2}$$ Putting the coordinates of $H'$ in the equation of the circle: We need to show: $$x^2 + \left(y-\frac{a_2^2+a_1^2-1}{2a_2}\right)^2 = \left(\frac{2m(1-a_1^2+a_2^2)+2a_1a_2(1-m^2)}{2a_2(m^2+1)}\right)^2 + \left(\frac{(m^2-1)(a_2^2+1-a_1^2+4ma_1a_2}{2a_2(m^2+1)}\right)^2 = \frac{(1+m^2)^2\cdot((a_1^2+a_2^2-1)^2+4a_2^2)}{(2a_2)^2(1+m^2)^2}$$We cancel the denominators. $$ LHS = (2m(1-a_1^2+a_2^2)+2a_1a_2(1-m^2))^2 + ((m^2-1)(a_2^2+1-a_1^2)+4ma_1a_2)^2 = 4m^2(1-a_1^2 + a_2^2)^2 + 4a_1^2a_2^2(1-m^2)^2 + 8ma_1a_2(1-a_1^2+a_2^2)(1-m^2) + (m^2-1)^2(a_2^+1-a_1^2)^2 + 16m^2a_1^2a_2^2+8ma_1a_2(a_2^2+1-a_1^2)(m^2-1) $$Simplifying and taking similar terms together we get, $$LHS = (m^2+1)^2((1-a_1^2+a_2^2)^2+4a_1^2a_2^2) = (m^2+1)^2((a_1^2+a_2^2-1)^2+4a_2^2) = (\text{Radius of (ABC)})^2$$So, $H'$ lies on $(ABC)$. Proved

Problem 2: Let $H_1$ be the orthocenter of triangle $APQ$ , $N$ be the midpoint of $PQ$ and $R$ be the midpoint of $AH$ $PQ$ is the simson's line of point $H$. So, $MH$ = $MH_1$. Since the reflection orthocenter in midpoint of side lies on the circumcircle of the triangle, $H_1NH$ is a straight line. $\angle RNP = \angle RNM = 90$. So, $M$ lies on the nine-point circle of triangle $ABC$. Now, consider homothety taking nine-point circle to the circumcircle of $\triangle ABC$, $N$ is taken to $H_1$ as desired.

Let $O$ the circumcenter of $\triangle ABC$, let $N$ the midpoint of $AH$, let $H'$ the orthocenter of $\triangle APQ$, let $M'$ the midpoint of $PQ$ and let $N'$ the midpoint of $AH'$, so from $H'M'=M'H$ and $H'N'=N'A$ we get $M'N'$ is the midline of $\triangle AH'H$. $$\Longrightarrow 2M'N'=HA=2OM$$So from $OM\parallel AH\parallel M'N'$ we get $OMN'M'$ is a paralellogram $\Longrightarrow$ $ON'\parallel MM'$, but $MM'\perp AH'$ $\Longrightarrow$ $ON'\perp AH'$ in $N'$, hence $OH'=OA$ $\Longrightarrow$ $H'\in \odot (ABC)$.

Here are the various official solutions provided by the organizers: First solution (Michael Ren): Let $R$ be the intersection of $(AH)$ and $(ABC)$, and let $D$, $E$, and $F$ respectively be the orthocenter of $APQ$, the foot of the altitude from $A$ to $PQ$, and the reflection of $D$ across $E$. Note that $F$ lies on $(AH)$ and $E$ lies on $(AM)$. Let $S$ and $H'$ be the intersection of $AH$ with $BC$ and $(ABC)$ respectively. Note that $R$ is the center of spiral similarity taking $DEF$ to $H'SH$, so $D$ lies on $(ABC)$, as desired. Second solution (Vincent Huang, Evan Chen): Let $DEF$ be the orthic triangle of $ABC$. Let $N$ and $S$ be the midpoints of $PQ$ and $AH$. Then $\overline{MS}$ is the diameter of the nine-point circle, so since $\overline{SN}$ is the perpendicular bisector of $\overline{PQ}$ the point $N$ lies on the nine-point circle too. Now the orthocenter of $\triangle APQ$ is the reflection of $H$ across $N$, hence lies on the circumcircle (homothety of ratio $2$ takes the nine-point circle to $(ABC)$). Third solution (Zack Chroman): Let $R$ be the midpoint of $PQ$, and $X$ the point such that $(M,X;P,Q)=-1$. Take $E$ and $F$ to be the feet of the $B,C$ altitudes. Recall that $ME,MF$ are tangents to the circle $(AH)$, so $EF$ is the polar of $M$. Then note that $MP \cdot MQ=MX \cdot MR=ME^2$. Then, since $X$ is on the polar of $M$, $R$ lies on the nine-point circle --- the inverse of that polar at $M$ with power $ME^2$. Then by dilation the orthocenter $2 \vec R - \vec H$ lies on the circumcircle of $ABC$. Fourth solution (Zack Chroman): We will prove the following more general claim which implies the problem: Claim: For a circle $\gamma$ with a given point $A$ and variable point $B$, consider a fixed point $X$ not on $\gamma$. Let $C$ be the second intersection of $XB$ and $\gamma$, then the locus of the orthocenter of $ABC$ is a circle Proof. Complex numbers is straightforward, but suppose we want a more synthetic solution. Let $D$ be the midpoint of $BC$. If $O$ is the center of the circle, $\angle OMX=90$, so $M$ lies on the circle $(OX)$. Then \[ H=4O-A-B-C=4O-A-2D. \]So $H$ lies on another circle. (Here we can use complex numbers, vectors, coordinates, whatever; alternatively we can use the same trick as above and say that $H$ is the reflection of a fixed point over $D$). $\blacksquare$ Fifth solution (Kevin Ren): Let $O$ be the midpoint of $AH$ and $N$ be the midpoint of $PQ$. Let $K$ be the orthocenter of $APQ$. Because $AP \perp KQ$ and $KP \perp HP$, we have $KQ \parallel PH$. Similarly, $KP \parallel QH$. Thus, $KPHQ$ is a parallelogram, which means $KH$ and $PQ$ share the same midpoint $N$. Since $N$ is the midpoint of chord $PQ$, we have $\angle ONM = 90^\circ$. Hence $N$ lies on the 9-point circle. Take a homothety from $H$ mapping $N$ to $K$. This homothety maps the 9-point circle to the circumcircle, so $K$ lies on the circumcircle.

Oops, one more I remembered just now: Sixth solution (Evan Chen, complex numbers): We use complex numbers with $(AHEF)$ the unit circle, centered at $N$. Let $a$, $e$, $f$ denote the coordinates of $A$, $E$, $F$, and hence $h = -a$. Since $M$ is the pole of $\overline{EF}$, we have $m = \frac{2ef}{e+f}$. Now, the circumcenter $O$ of $\triangle ABC$ is given by $o = \frac{2ef}{e+f} + a$, due to the fact that $ANMO$ is a parallelogram. The unit complex numbers $p$ and $q$ are now known to satisfy \[ p + q = \frac{2ef}{e+f} + \frac{2pq}{e+f} \]so \[ (a + p + q) - o = \frac{2pq}{e+f} \qquad \text{and}\qquad a - o = \frac{2ef}{e+f} \]which clearly have the same magnitude. Hence the orthocenter of $\triangle APQ$ and $A$ are equidistant from $O$.

v_Enhance wrote: The unit complex numbers $p$ and $q$ are now known to satisfy \[ p + q = \frac{2ef}{e+f} + \frac{2pq}{e+f} \] Where did this come from?

Well... My friend has generallise this problem! I want to post it here! Given the triangle $ABC$ and its circumcircle $(O)$. A circle $(M)$ passes through $B$ and $C$ intersects $AB$ and $AC$ at $E$ and $F$ respectively. A line passes through $M$ intersects the circle $(AEF)$ at two points $P$ and $Q$. Prove that the orthocenter of the triangle $APQ$ lies on the circle $(O)$.

whatshisbucket wrote: Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$ Proposed by Michael Ren Let $H'$ be the orthocenter of $\Delta APQ $ and let $A'$ be the antipode of $A $ in $\odot (ABC) $. Observe that it would now be enough to prove that $A'H' || MP $. For this, simply notice that $H'PHQ $ and $BHCA'$ are parallelograms (some easy angle chase) and then a homothety centered at $H $ with ratio $\frac {1}{2} $ will do the job.

Let $K$ be the orthocenter of $\triangle APQ,L$ be the midpoint of $PQ, BH\cap AC=E , CH\cap AB=F$ We have $K$ is the orthocenter of $\triangle APQ, AH$ is the diameter of $(APQ)\Rightarrow \overline{K,L,H}$ It's easy to prove that $ME,MF$ are the tangents of $(AH)$ Now note that $EPFQ$ is a harmonic quad $\Rightarrow \angle MFL=\angle MFP+\angle PFL=\angle FQP+\angle QFE$ Similarly $\angle MEL=\angle EQP+\angle QEF\Rightarrow \angle MFL+\angle MEL=180^{\circ}\Rightarrow EMFL$ is cyclic Consider the homothety $\mathcal{H}_H^{2},$ we get $K\in (ABC).$

The orthocenter of $\triangle APQ$ is the reflection of the antipode of $A$ in $\triangle APQ$, which is $H$, over the midpoint of $PQ$, which lies on the nine point circle of $\triangle ABC$.

Same as others. Let D be the midpoint of PQ, and E the midpoint of AH; note that $EDM=90\Rightarrow D\in(EM)$, and also that the orthocenter of APQ is the reflection of H over D (since H is antipode of A, and reflection of orthocenter over midpoint is antipode), which indeed lies on the circumcircle due to homothety at H taking ninepointcirc to circumcircle.

Define the intersections of the altitude from $A$ to $PQ$ with $(AH)$ and $PQ$ as $J$ and $K$, respectively. Also suppose $L$ is the orthocenter of $\triangle APQ$ and $A'$ as the antipode of $A$ on $(ABC)$. Then $HJ$ and $MK$ are both perpendicular to $AJ$. $K$ is the midpoint of $JL$ and $M$ is the midpoint of $HA'$. Hence we also have $A'L \perp AJ \implies \angle ALA' = 90$, so $L$ lies on $(ABC)$. $\blacksquare$

Let $N$ be the midpoint of $\overline{AH}$. It's well known that $N$ and $M$ are antipodes on the nine-point circle of $\triangle ABC$, so if $R$ be the foot from $N$ to line $PQ$, then $R$ lies on the nine-point circle. A homothety of factor $2$ sends $R$ to the orthocenter of $\triangle AQP$ (since it's the midpoint of $\overline{PQ}$), as well as sending the nine-point circle to $(ABC)$, hence the orthocenter of $\triangle AQP$ lies on $(ABC)$.

Let $O$ be the midpoint of $AH$ and $K$ the midpoint of $PQ$, as $H$ is the $A-$ antipode of $(AH)$ then the orthocenter of $APQ$ is the reflection of $H$ over $K$, so it suffices to prove that this point is on $(ABC)$, but as the homothety with factor $\frac{1}{2}$ with center $H$ sends $(ABC)$ to the nine-point circe, it suffices to prove that $K$ is on the nine-point circle, but this is true as $\angle OKM=90^\circ$

Hee hee hee haw, complex. We complex bash with respect to $(AEF)$. Set $a = 1$ and $h = -1$ and note $m = \frac{2ef}{e+f}$. Arbitrarily choose $p$ and $q$ and note that the collinearity condition is the same as, \begin{align*} \begin{vmatrix} p & 1/p & 1\\ q & 1/q & 1\\ \frac{2ef}{e+f} & \frac{2}{e + f} & 1 \end{vmatrix} &= 0\\ \iff p/q + \frac{2ef}{p(e + f)} + \frac{2q}{e+f} - q/p - \frac{2ef}{q(e + f)} - \frac{2p}{e + f} &= 0\\ \iff \frac{(p - q)(p + q)}{pq} + \frac{2ef}{e+f}\left(\frac{q - p}{pq} \right) + \frac{2}{e + f}(q - p) &= 0\\ \iff \frac{p + q}{pq} - \frac{2ef}{pq(e+f)} - \frac{2}{e+f} &= 0\\ \iff (p + q)(e + f) - 2ef - 2pq &= 0\\ \iff (p + q)(e + f) &= 2(ef + pq) \end{align*}Now note that the orthocenter of $\triangle APQ$ has coordinates $1 + p + q$, $B$ has coordinates $\frac{2ef + e - f}{e + f}$ and $C$ has coordinates $\frac{2ef - e + f}{e + f}$. Thus we need to show, \begin{align*} \frac{1 - \frac{2ef + e - f}{e + f}}{1 - \frac{2ef - e + f}{e + f}} \div \frac{1 + p + q - \frac{2ef + e - f}{e + f}}{1 + p + q - \frac{2ef - e + f}{e + f}} \in i\mathbb{R} \end{align*}Upon expansion we have, \begin{align*} \frac{1 - \frac{2ef + e - f}{e + f}}{1 - \frac{2ef - e + f}{e + f}} \div \frac{1 + p + q - \frac{2ef + e - f}{e + f}}{1 + p + q - \frac{2ef - e + f}{e + f}} &= \frac{2f - 2ef}{2e - 2ef} \div \frac{(1 + p + q)(e + f) - 2ef - e + f}{(1 + p + q)(e + f) - 2ef + e - f}\\ &= \frac{2f(1 - e)}{2e(1 - f)} \div \frac{(e + f) + 2(ef + pq) - 2ef - e + f}{(e + f) + 2(ef + pq) - 2ef + e - f}\\ &= \frac{f(1 - e)}{e(1 - f)} \div \frac{pq + f}{pq + e} \end{align*}Now taking the conjugate of this expression we find, \begin{align*} \overline{\left( \frac{f(1 - e)}{e(1 - f)} \cdot \frac{pq + e}{pq + f} \right)} &= \frac{1/f(1 - 1/e)}{1/e(1 - 1/f)} \cdot \frac{1/pq + 1/e}{1/pq + 1/f}\\ &= \frac{(e - 1)}{(f - 1)} \cdot \frac{ef + pqf}{ef + pqe}\\ &= \frac{f(1 - e)}{e(1 - f)} \cdot \frac{pq + e}{pq + f} \end{align*}and we're done.

Let $APQ$ be the unit circle. Since $M \in \overline {PQ}$, we have $m+pq \overline{m} = p+q$. Letting $H'$ be the reflection of $H$ over M, we want the orthocenter of $\Delta APQ$ to lie on the circle with diameter $\overline {AD}$. This is equivlent to $LA \perp LD$ where $L$ is the orthocenter of $\Delta APQ$. Note $H' = 2m+a$, so it suffices to show \[\frac{p+q}{p+q-2m} = \overline{\left(\frac{p+q}{p+q-2m}\right)} = \frac{p+q}{p+q-2pq \overline{m}}\]and we're done since $pq\overline{m} = p+q-m$.

Let $J$ be the orthocenter of $\triangle APQ$ and $K$ be the mid-point of $\overline{PQ}$. Here are a few well-known facts: (some of the outline proofs here) Fact 1: The $A-$antipode, mid-point $\overline{BC}$ and orthocentre are collinear. Hence, $K,J,H$ are collinear. Fact 2: $\overline{ME}$ and $\overline{MF}$ are tangents to the circle $(AEF)$, and $\angle MFE=\angle BAC$. Fact 3: If $Q_A$ is the $A-$Dumpty Point and $R$ is the reflection of $A$ in $Q_A$, then $R\in ABC$, $AR$ is symmedian and $\angle BQ_AR=\angle RQ_AC=\angle BAC$. Note that $M,P,Q$ are collinear, so $(PFQE)$ is a harmonic bundle. Hence, $PQ$ is the $P-$symmedian of $\triangle PEF$. So, $K$ is the $P-$Dumpty Point. Hence, \[\angle FKM=\angle FKQ=\angle FPE=\angle FAE=\angle BAC=\angle FEM\]and it follows that $K,F,M,E$ are concyclic. Hence, $K$ lies on the nine-point circle of $ABC$. Take homothety at $H$ with factor 2 to get that $J$ lies on circumcircle, as required. $\blacksquare$

Surprisingly clean. Let $K$ be the midpoint of $\overline{AH}$, $N$ be the midpoint of $\overline{PQ}$, and $H_1$ be the desired orthocenter. Then $\angle KNM = 90^\circ$, thus $N$ lies on the nine-point circle. However, note that $N = \overline{HH_1} \cap \overline{PQ}$ is the midpoint of $\overline{HH_1}$; thus, the homothety sending the nine-point circle to $(ABC)$ sends $N$ to $H_1$, and $H_1$ lies on $(ABC)$.

Define $J$ as the orthocenter of $\triangle APQ$ and $K = AJ \cap PQ$. Notice the circles $(AH)$, $(AM)$, $(ABC)$ are coaxial, so if we aim to prove $J \in (ABC)$, our Coaxiality Lemma tells us this is equivalent to \[\frac{\operatorname{pow}(J,(AH))}{\operatorname{pow}(J,(AM))} = \frac{\operatorname{pow}((B,(AH))}{\operatorname{pow}(B,(AM))} \iff \frac{-AJ \cdot 2JK}{-AJ \cdot JK} = \frac{BF \cdot BA}{BD \cdot BM} \iff 2 = 2. \quad \blacksquare\]

ermmmm what the hyperbola Claim: $P,Q$ are antigonal conjugates. Proof. Let $P'$ be the reflection of $P$ over $M.$ Then $MP'\cdot MQ=MP\cdot MQ=MB\cdot MC$ by three tangents lemma, so $-\measuredangle BPC=\measuredangle BP'C=\measuredangle BQC.$ Also we have $\measuredangle APH=90^\circ=-\measuredangle AQH,$ so the antigonal conjugate of $P$ is given by one of the intersection points of two circles, one of which is $Q.$ We can show it must be $Q$ by considering the special cases $P=Q$ and using continuity. Now the midpoint of $PQ$ lies on the nine point circle. Since $H$ is the antipode of $A$ on $(APQ)$ we get the orthocenter of $APQ$ is the reflection of $H$ over the midpoint of $PQ,$ which lies on the circumcircle by homothety.

Since $H$ is the $A$ antipode in $(APQ)$, it lies on the reflection of orthocenter $N$ of $\Delta APQ$ through midpoint of $PQ$. Taking a homothety through $H$ with factor of $\frac{1}{2}$, it suffices to prove that midpoint $K$ of $PQ$ lies on the nine point circle which is the circle with diameter $(OM)$ where $O$ is center of $(AH)$/midpoint of $AH$. This is obvious because $OK \perp PQ \equiv MK$ due to $PQ$ being a chord on a circle centered at $O$.

Let $D$ be the orthocenter of $\triangle APQ$ and $H'$ the reflection of $H$ in $M$. Obviously $D$ is reflection of $H$ in midpoint of $PQ$, since $A$, $H$ are antipodes on $(APQ)$ and also $H'$ is antipode of $A$ on $(ABC)$. So the midpoint of $HD$ is on $PQ$. But $M$ is on $PQ$ so by midparallel we have $PQ\parallel DH'$. Also $AD\perp PQ$. Combining gives $AD\perp DH'$ and therefore $\angle ADH'=90^\circ$ which implies $H'$ on $(ABC)$.

Let $N$ be the midpoint of $AH$, let $X$ be the $A$-queue point, and define $E,F$ as normal. Note that there exists a fixed linear transformation that takes the midpoint of $PQ$ to the orthocenter of $\triangle APQ$ by first taking a $\frac23$-homothety from $A$ to get to the centroid, then taking a $3$-homothety from $N$. Since the midpoint of $PQ$ lies on the circle with diameter $NM$, it suffices to check three distinct such midpoints to show that the circle with diameter $NM$ maps to $(ABC)$ under the linear transformation. If $P=Q=F$, then the midpoint is $F$, so the image after the $\frac23$-homothety is $\frac23 F+\frac13 A$, so the image after the $3$-homothety is $2F+A-2N$. Indeed, this point is the reflection of $H$ over $AB$, say $H'$, because $H'=2F-H$ and $A=2N-H$. Similarly, $P=Q=E$ also works. Finally, $P=X$, $Q=H$ works since then the orthocenter is just $H$ itself. $\blacksquare$