Here is a solution with involution (originally submitted as a rough sketch by an unofficial TSTST participant and cleaned up by me). Let $\mathcal{C}_A$ be the conic given by the $A$-altitude and the line through $A$ parallel to $BC$, and define $\mathcal{C}_B,\mathcal{C}_C$ similarly. Let $\ell$ be the perpendicular bisector of $PQ$, and let $M$ be the midpoint of $PQ$. Let $H$ be the orthocenter of $\triangle ABC$ and $G$ be the centroid.

It then suffices to show that $\mathcal{C}^{**}$ contains the line at infinity. Let $B',C'$ be the reflections of $B,C$ over the midpoints of $AC,AB$. Then we know by DDIT on $H$ and the complete quadrilateral with opposite pairs of points $(B, C'), (C, B'), (G, \infty_{BC})$ that $(HB, HC'), (HC, HB'), (HG, H\infty_{BC})$ correspond under some involution. But if we intersect this pencil with $B'C'$, the first pair corresponds to $\ell_B\cap B'C'$ and the second pair corresponds to $\ell_C \cap B'C'$. It follows that since $\mathcal{C}^{**}$ contains $HG\cap B'C'$, it also contains $H\infty_{BC}\cap B'C' = \infty_{B'C'}$. Similarly $\mathcal{C}^{**}$ contains $\infty_{C'A'}, \infty_{A'B'}$, so it contains $\ell_{\infty}$, and we are done.

My problem! Here is the original problem statement and solution. Original problem statement wrote: Let distinct points $P$ and $Q$ lie inside scalene triangle $ABC$. Suppose that the angle bisectors of $\angle PAQ$, $\angle PBQ$, and $\angle PCQ$ are altitudes of triangle $ABC$. Prove that the midpoint of $\overline{PQ}$ lies on the Euler line of triangle $ABC$.

Lemma 1 (widely known): The triangle $\triangle ABC$ is given. $\mathcal{C}$ isogonal conjugate to the perpendicular bisector $BC$ WRT $\triangle ABC$. $l$ tangent to $\mathcal{C}$ at point $A$. Then the line $l$ coincides with the $A$-symedian of the triangle $\triangle ABC$ Let's get back to the main problem. Let's make an inversion with the center at the point $H$. We will keep the designations of the images. $=> H$ incenter $\triangle ABC$. Also circles $\odot (AHP)$ and $\odot (AHQ)$ symmetric with respect to the straight line $AH$. $=> P,Q$ antigonal conjugate WRT $\triangle BHC, \triangle AHB, \triangle AHC, \triangle ABC =>$ The middle of the segment $PQ$ is the Feuerbach point of the triangle $\triangle ABC$. Let $\mathcal{C}$ be the Feuerbach hyperbola of the triangle $\triangle ABC => P,Q \in \mathcal{C}$. It remains to apply Lemma 1 for the triangle $\triangle HPQ$

Is there a moving points solution to this - I am thinking of fixing $P_1$ on $BC$ and then animating $P$ on $PP_1$ (there might be a weird way of animating after an appropriate inversion).

Let \((ABC)\) be the unit circle. The condition that the \(A\)-altitude bisects \(\angle PAQ\) gives \begin{align*} \frac{(p-a)(q-a)}{(b+c)^2}&\in\mathbb R\\ \implies b^2c^2\left( \overline p-\frac1a \right)\left( \overline q-\frac1a \right) &=(p-a)(q-a)\\ \implies\left(b^2-c^2\right) \left( \overline{pq}-\frac1a(\overline p+\overline q)+\frac1{a^2} \right) &=\frac{b^2-c^2}{b^2c^2}\left(pq-a(p+q)+a^2\right). \end{align*}Note that when summing cyclically, most terms vanish, leaving \begin{align*} (\overline p+\overline q)\sum_\mathrm{cyc}\frac{b^2-c^2}a &=(p+q)\sum_\mathrm{cyc}\frac{a\left(b^2-c^2\right)}{b^2c^2}\\ \implies\frac{p+q}{\sum_\mathrm{cyc}\frac{b^2-c^2}a}&\in i\mathbb R. \end{align*} We finish by checking that \[\sum_\mathrm{cyc}\frac{b^2-c^2}a=\frac{(a-b)(b-c)(c-a)(a+b+c)}{abc}\]is indeed perpendicular to \(a+b+c\).

Here is a different approach with complex numbers that does not even need to set $ABC$ on the unit circle. Note that $ABC$ must be acute as the condition implies that the altitudes of $ABC$ are in its interior. Using complex numbers, the angle bisector condition can be encoded as $\frac{(a-p)(a-q)}{(b-c)^2},\frac{(b-p)(b-q)}{(c-a)^2},\frac{(c-p)(c-q)}{(a-b)^2}$ being (negative) reals. The key idea is to take a real linear combination of these expressions to obtain that $\frac{p+q}2$ lies on a fixed line, noting that for fixed $u\ne0,v\in\mathbb C$ the locus of $z$ for which $uz+v\in\mathbb R$ is a line. This is done by finding $X,Y,Z\in\mathbb R$ that kills the $pq$ term in $X\cdot\frac{(a-p)(a-q)}{(b-c)^2}+Y\cdot\frac{(b-p)(b-q)}{(c-a)^2}+Z\cdot\frac{(c-p)(c-q)}{(a-b)^2}$ but not the $p+q$ term. Lemma: there exist real numbers $X,Y,Z$ such that $\frac X{(b-c)^2}+\frac Y{(c-a)^2}+\frac Z{(a-b)^2}=0$ and $\frac{aX}{(b-c)^2}+\frac{bY}{(c-a)^2}+\frac{cZ}{(a-b)^2}$ is nonzero. Proof: Since $\mathbb C$ has real dimension $2$, we can always satisfy the first equality with some $X,Y,Z$ not all $0$. Furthermore, $X,Y,Z$ are all nonzero as otherwise if say $X=0$ then $\frac{(a-b)^2}{(a-c)^2}$ would be real, which is impossible for an acute triangle. Now if $\frac{aX}{(b-c)^2}+\frac{bY}{(c-a)^2}+\frac{cZ}{(a-b)^2}=0$ as well, then $\frac{Y(b-a)}{(c-a)^2}+\frac{Z(c-a)}{(a-b)^2}=0$, so $\frac{(a-b)^3}{(a-c)^3}$ is real, which implies $\angle A=60^\circ$. Similarly, we have that $\angle B=\angle C=60^\circ$, contradicting $ABC$ being scalene. Taking these reals $X,Y,Z$ we have that $\frac{X(a-p)(a-q)}{(b-c)^2}+\frac{Y(b-p)(b-q)}{(c-a)^2}+\frac{Z(c-p)(c-q)}{(a-b)^2}$ is real, so \[\frac{p+q}2\left(\frac{aX}{(b-c)^2}+\frac{bY}{(c-a)^2}+\frac{cZ}{(a-b)^2}\right)-\frac12\left(\frac{a^2X}{(b-c)^2}+\frac{b^2Y}{(c-a)^2}+\frac{c^2Z}{(a-b)^2}\right)\]is real. Since $\frac{aX}{(b-c)^2}+\frac{bY}{(c-a)^2}+\frac{cZ}{(a-b)^2}$ and $\frac{a^2X}{(b-c)^2}+\frac{b^2Y}{(c-a)^2}+\frac{c^2Z}{(a-b)^2}$ are some fixed complex numbers with the former being nonzero, the midpoint $\frac{p+q}2$ of $PQ$ must lie on some fixed line. It remains to show that this line is the Euler line of $ABC$. Taking both $P$ and $Q$ to be the orthocenter of $ABC$, the midpoint of $PQ$ is the orthocenter of $ABC$. (Note here that the algebraic condition that $\frac{(a-p)(a-q)}{(b-c)^2}, \frac{(b-p)(b-q)}{(c-a)^2}, \frac{(c-p)(c-q)}{(a-b)^2}$ are real is still satisfied, even if the problem condition of distinctness is not.) Taking $P$ and $Q$ to be the foci of the Steiner inellipse of the anticomplementary triangle of $ABC$, the midpoint of $PQ$ is the centroid of $ABC$. (This can be verified by standard properties of inscribed conics, and even if $P$ and $Q$ are outside $ABC$ the algebraic condition is still satisfied.) Since $ABC$ is scalene, we have found two distinct positions for the midpoint of $PQ$, and thus the midpoint of $PQ$ must always lie on the line between them, i.e. the Euler line of $ABC$.

Very cool problem! My sol is pretty similar to nervy's but I'll post it anyway. Invert everything with center orthocenter $H$. Let $X$ be $X'$ after invertion where $X=A, B, C, P, Q$. Let the circumcenter and Feuerbach point of $A'B'C'$ be $O$ and $Fe$. By the angle conditions, since $\measuredangle HXP=- \measuredangle HXQ$, $\measuredangle HP'X=- \measuredangle HQ'X$ where $X=A', B', C'$, so there exists a rectangle hyperbola $\Gamma$ passing through $H, P', Q', A', B', C'$ and its center is the midpoint of $P'Q'$. $H$ is incenter of $\Delta A'B'C'$, so $\Gamma$ is Feuerbach hyperbola of $\Delta A'B'C'$, so the midpoint of $P'Q'$ is $Fe$. Well known that $OH$ is tangent to $\Gamma$, so $OH$ is the symmedian of $\Delta P'HQ'$, thus before invertion the Euler line of $\Delta ABC$ bisects $PQ$. Done.

Synthetic solution using Sondat's theorem, Miquel Circle, and Steiner Line . Let $H$ be the orthocenter, and let $O$ be the circumcenter of $\triangle ABC$ Let $P_A$, $P_B$, $P_C$ be the reflections of $P$ across $AH$, $BH$, and $CH$, so $H$ is the center of $\odot(P_AP_BP_C)$. Moreover, note that $\triangle ABC$ and $\triangle P_AP_BP_C$ are orthologic. In particular, the perpendiculars from $P_A$ to $BC$, from $P_B$ to $AC$, and from $P_C$ to $AB$ meet at the point $P' = 2H-P$. $\triangle ABC$ and $\triangle P_AP_BP_C$ are perspective at $Q$. Thus, by Sondat's theorem, line $P'Q$ is perpendicular to the line $\ell$ joining $D=BC\cap P_BP_C$, $E=AC\cap P_AP_C$, and $F=AB\cap P_AP_B$. Upon considering homothety at $P$, line $P'Q$ is parallel to the line joining $H$ and the midpoint of $PQ$, so it suffices to show that $\overline{DEF}\perp OH$. Now, we can forget about $P$, $Q$; what we will only need from now is $H$ is the center of $\odot(P_AP_BP_C)$ and that $\triangle P_AP_BP_C\stackrel -\sim \triangle ABC$. The following diagram sketches what remains. [asy][asy] size(12cm); import olympiad; import geometry; defaultpen(fontsize(10pt)); pair A = (1,3.2); pair B = (0,0); pair C = (4,0); pair O = circumcenter(A,B,C); pair H = orthocenter(A,B,C); pair D = 0.37*B + 0.63*C; pair E = extension(D, foot(D,O,H), A, C); pair F = extension(D, foot(D,O,H), A, B); pair M = reflect(circumcenter(A,E,F),O) * A; pair O_A = circumcenter(A,E,F); pair O_B = circumcenter(B,D,F); pair O_C = circumcenter(C,D,E); pair X = reflect(E,F) * orthocenter(A,E,F); pair O_prime = 2 * foot(circumcenter(M,O,O_A), X, M) - M; pair A_1 = reflect(O_prime,O_A) * M; pair B_1 = reflect(O_prime,O_B) * M; pair C_1 = reflect(O_prime,O_C) * M; pair P_A = reflect(E,F) * A_1; pair P_B = reflect(E,F) * B_1; pair P_C = reflect(E,F) * C_1; fill(A--B--C--cycle, palegray); fill(A_1--B_1--C_1--cycle, 0.7*paleblue+0.3*white); draw(A--B--C--cycle, linewidth(1)); draw(D--F, red+linewidth(1.0)); draw(A--F, gray); draw(circumcircle(A, E, F), red); draw(circumcircle(B, D, F), red); draw(circumcircle(C, D, E), red); draw(circumcircle(M, O, O_A), deepgreen+linewidth(1.0)); draw(A_1--B_1--C_1--cycle, blue+linewidth(1.2)); draw(F--A_1--E, blue+linewidth(0.9)); draw(B_1--D, blue+linewidth(0.9)); draw(P_A--P_B--P_C--cycle, dashed+blue); draw(E--P_A--F, dashed+blue); draw(P_B--D, dashed+blue); dot("$A$", A, dir(154)); dot("$B$", B, dir(-135)); dot("$C$", C, dir(-43)); dot("$O$", O, dir(57)); dot("$H$", H, dir(90)); dot("$D$", D, dir(-92),red); dot("$E$", E, 1.25*dir(20),red); dot("$F$", F, dir(106),red); dot("$M$", M, dir(7)); dot("$O_A$", O_A, dir(85),deepgreen); dot("$O_B$", O_B, dir(179),deepgreen); dot("$O_C$", O_C, dir(-60),deepgreen); dot("$X$", X, dir(6)); dot("$O'$", O_prime, dir(138),deepgreen); dot("$A_1$", A_1, dir(-127),blue); dot("$B_1$", B_1, dir(-50),blue); dot("$C_1$", C_1, dir(16),blue); dot("$P_A$", P_A, dir(124),blue); dot("$P_B$", P_B, dir(-105),blue); dot("$P_C$", P_C, dir(183),blue); [/asy][/asy] Now, let $A_1$, $B_1$, $C_1$ be the reflections of $P_A$, $P_B$, $P_C$ across $\ell$. The inverse similarity implies that $A_1\in\odot(AEF)$, and similar for $B_1$ and $C_1$. Let $O_A$, $O_B$, $O_C$ be the centers of $\odot(AEF)$, $\odot(BDF)$, and $\odot(CDE)$, and let $O'$ be the center of $\odot(A_1B_1C_1)$. Let $M$ be the Miquel point of $\triangle ABC\cup\ell$. We then claim that Claim: $O'$ lies on$\odot(MO_AO_BO_CO)$ (these five points are well-known to be concyclic). Proof. Follows from that $M$ is also Miquel point of $\triangle A_1B_1C_1\cup\ell$, so by that fact applied on this complete quadrilateral, we get $M, O_A, O_B, O_C, O'$ are concyclic. $\blacksquare$ Now, we construct point $X$ on $\odot(AEF)$ such that $AX\perp\ell$. We then claim the following. Claim: $M$, $X$, $O'$ are collinear. Proof. The reflection of $M$ across $\ell$ lies on the Steiner's line of this complete quadrilateral, and so are $H$ and the orthocenter of $\triangle AEF$. Reflecting these three points across $\ell$ gives the result. $\blacksquare$ Finally, the problem becomes just angle chasing. We have that \begin{align*} \measuredangle O'OO_A &= \measuredangle O'MO_A \\ &= \measuredangle XMO_A \\ &= 90^\circ + \measuredangle XMM \\ &= 90^\circ + \measuredangle XAM \\ &= \measuredangle(\ell, AM) \\ &= 90^\circ + \measuredangle(\ell, OO_A), \end{align*}so $OO'\perp\ell$. Thus, $O\in HO'$, implying also that $\ell\perp OH$, done.

Here's an alternative finish to the involution solution that's completely analytic: for any line $\ell$, let $d_{\ell}$ denote homogenous linear polynomial in $\mathbb C[x,y,z]$ equal to the signed distance to $\ell$ on the normalized affine plane $z=1$ (or $x+y+z=1$ if you prefer). Furthermore, define $d_{\ell^\infty}$ as a homogenous linear polynomial that is constant $1$ on the normalized affine plane. We show that \[ d_{A\infty_{BC}} d_{AH}, \quad d_{A\infty_{BC}} d_{AH}, \quad d_{A\infty_{BC}} d_{AH}, \quad \text{and} \quad d_{HG} d_{\ell^\infty} \]are linearly dependent over $\mathbb C$. Let $h_A$ denote the length of the $A$-altitude, and define $h_B$ and $h_C$ similarly. Observe that $A\infty_{BC}$ and $BC$ are parallel and are separated by $h_A$, so (after choosing an appropriate sign) we have $d_{A\infty_{BC}} = d_{BC} + h_A d_{\ell^\infty}$. Doing the same for $B$ and $C$, we have, for any constants $\alpha, \beta, \gamma$, \begin{align*} \alpha d_{A\infty_{BC}} d_{AH} + \beta d_{A\infty_{BC}} d_{AH} + \gamma d_{A\infty_{BC}} d_{AH} &= (\alpha h_A d_{AH} + \beta h_B d_{BH} + \gamma h_C d_{CH}) d_{\ell^\infty} \\ &\;+(\alpha d_{AH} d_{BC} + \beta d_{BH} d_{CA} + \gamma d_{CH} d_{AB}). \end{align*}But $d_{AH} d_{BC}$, $d_{BH} d_{CA}$, and $d_{CH} d_{AB}$ must be linearly dependent since they all lie on the pencil of conics through $\{A,B,C,H\}$, so we can find $\alpha$, $\beta$, and $\gamma$ not all zero such that the second term vanishes everywhere. Plugging in $G$ specifically, we have \[ \alpha d_{AH} (G) d_{BC} (G) + \beta d_{BH} (G) d_{CA} (G) + \gamma d_{CH} (G) d_{AB} (G) = 0.\]But it is well known that $d_{BC} (G) = \tfrac13 h_A$, etc., so we must have $\alpha h_A d_{AH} (G) + \beta h_B d_{BH} (G) + \gamma h_C d_{CH} (G) = 0$ as well. This means the variety of $\alpha h_A d_{AH} + \beta h_B d_{BH} + \gamma h_C d_{CH}$ passes through $G$, but clearly it also passes through $H$, so we are done. $\blacksquare$

ABCDE wrote: Here is a different approach with complex numbers that does not even need to set $ABC$ on the unit circle. In fact, this solution easily generalizes to the following: Let $ABC$ be a triangle and $X$ be any point that is not an incenter of excenter of $ABC$. If $P$ and $Q$ are points such that $AX$ bisects $\angle PAQ$, $BX$ bisects $\angle PBQ$, and $CX$ bisects $\angle PCQ$, then the midpoint of $PQ$ lies on a fixed line. The fixed line has the following explicit description: let $A'$ be the intersection of the reflection of $AB$ across $BX$ and the reflection of $AC$ across $CX$, and define $B'$ and $C'$ similarly. At least one of $A',B',C'$ is well-defined as $X$ is not an incenter or excenter of $ABC$, and all of the well-defined points among the midpoints of $AA',BB',CC'$ are collinear with $X$. It would be interesting to see a synthetic proof of the above.

ABCDE wrote: ABCDE wrote: Here is a different approach with complex numbers that does not even need to set $ABC$ on the unit circle. In fact, this solution easily generalizes to the following: Let $ABC$ be a triangle and $X$ be any point that is not an incenter of excenter of $ABC$. If $P$ and $Q$ are points such that $AX$ bisects $\angle PAQ$, $BX$ bisects $\angle PBQ$, and $CX$ bisects $\angle PCQ$, then the midpoint of $PQ$ lies on a fixed line. The fixed line has the following explicit description: let $A'$ be the intersection of the reflection of $AB$ across $BX$ and the reflection of $AC$ across $CX$, and define $B'$ and $C'$ similarly. At least one of $A',B',C'$ is well-defined as $X$ is not an incenter or excenter of $ABC$, and all of the well-defined points among the midpoints of $AA',BB',CC'$ are collinear with $X$. It would be interesting to see a synthetic proof of the above. My solution still works

Oops, I was only looking at the official synthetic solution and missed that one. A further challenge that does not seem to follow from the same argument: Suppose now that $X,Y,Z$ are fixed points such that $AX,BY,CZ$ are not necessarily concurrent but do not concur at the incenter or an excenter. Then there exists a line $\ell$ such that whenever $P$ and $Q$ are such that $AX$ bisects $\angle PAQ$, $BY$ bisects $\angle PBQ$, and $CZ$ bisects $\angle PCQ$, the midpoint of $PQ$ lies on $\ell$.

Solved with AdhityaMV and Rg230403. Hyperbolas yay! Let $DEF$ be the orthic triangle and $H$ be the orthocenter. Let $M$ be the midpoint of $PQ$. Invert about $H$ swapping $A$ with $D$ and say points $P$ and $Q$ go to $P'$ and $Q'$, and $M$ goes to $M'$. Then the angle bisector condition becomes that $\angle HP'D = \angle HQ'D$ and analogously for $E$ and $F$. This means that the isogonal conjugate of the perpendicular bisector of $P'Q'$ in triangle $HP'Q'$ is a rectangular hyperbola passing through $H,P',Q'$ and $D,E,F$. Note that for any point $X$ on the perpendicular bisector and its isogonal conjugate $X'$, $HX$ and $HX'$ are isogonal, so taking $X$ limiting to $M'$, we get that the isogonal of $HM'$ is the tangent to the hyperbola at $H$. Now, shift reference triangle to $DEF$ with its incenter $H$. From its perspective, this hyperbola is rectangular, passes through vertices and incenter, and so must in fact be the feuerbach hyperbola of $\triangle DEF$. Let $N$ be the nine point center of $ABC$, and the circumcenter of $DEF$. Suppose $NH$ intersects the hyperbola at $H'$ with $H' \neq H$. Then since $NH$ isogonal conjugates to the hyperbola and $H$ remains fixed, $H'$ must also lie on the hyperbola but be distinct from $H'$ as it is not the incenter, but then the hyperbola intersects $NH$ at three distinct points, impossible. Therefore, $NH$ must in fact be tangent to the hyperbola, and from the previous paragraph, $NH$ is symmedian in $\triangle HP'Q'$. Since the symmedian of $\triangle HP'Q'$ is the Euler line of $ABC$, inverting back, we get that the median of $\triangle HPQ$ is the Euler line of $\triangle ABC$, and so $M$ lies on it, as desired, so we're done. $\blacksquare$

Unfortunately recently I have been relying on non-synthetic techniques... Identify $(ABC)$ as the unit circle, with $o=0, |a|=|b|=|c|=1, h=a+b+c.$ From the angle condition, \[\frac{\left(\frac{p-a}{h-1}\right)}{\left(\frac{q-a}{h-a}\right)} = \overline{\left(\frac{\left(\frac{p-a}{h-1}\right)}{\left(\frac{q-a}{h-a}\right)}\right)} \implies \frac{(p-a)(q-a)}{(h-a)^2} = \overline{\left(\frac{(p-a)(q-a)}{(h-a)^2}\right)}\]\[ \frac{(p-a)(q-a)}{(b+c)^2} = \frac{(a\overline{p}-1)(a\overline{q}-1)b^2c^2}{(b+c)^2a^2} \implies a^2(p-a)(q-a)=(a\overline{p}-1)(a\overline{q}-1)b^2c^2 \]\[a^2(pq-a(p+q)+a^2)=(a^2\overline{pq}-a(\overline{p}+\overline{q})+1)b^2c^2 \implies \]\begin{align*} a^4-b^2c^2+a^2pq - a^2b^2c^2\overline{pq} &= a^3(p+q) - ab^2c^2(\overline{p}+\overline{q}), \\ b^4-a^2c^2+b^2pq - b^2a^2c^2 \overline{pq} & = b^3(p+q) - ba^2c^2(\overline{p}+\overline{q}), \\ c^4-a^2b^2 + c^2pq - c^2a^2b^2 \overline{pq} & = c^3(p+q)-ca^2b^2(\overline{p}+\overline{q}). \end{align*}Sum $b^2-c^2, c^2-a^2, a^2-b^2$ the first, second, third equations respectively. The coefficients of $pq$ and $\overline{pq}$ on the left hand side cancel to zero, and $\sum_{\text{cyc}} (b^2-c^2)(a^4-b^2c^2)=0$ as well. Now \[(p+q) \sum_{\text{cyc}}(b^2-c^2)a^3 = (\overline{p}+\overline{q}) abc \sum_{\text{cyc}} bc(b^2-c^2).\]But now we find \[(p+q)(a-b)(b-c)(c-a)(ab+bc+ca) = (\overline{p}+\overline{q})(a-b)(b-c)(c-a)(a+b+c)abc\]from expansion and factorization. Now divide $(a-b)(b-c)(c-a)$ from both sides to obtain \[(p+q)(ab+bc+ca)= (\overline{p}+\overline{q})(a+b+c)abc.\]But $\overline{h} = \tfrac{ab+bc+ca}{abc},$ so we have $(p+q)\overline{h} = (\overline{p}+\overline{q})h.$ Let $m$ be the midpoint of $pq.$ We want to prove $o,m,h$ collinear. But this is true since \[\frac{\frac{p+q}{2}}{h} = \overline{\left(\frac{\frac{p+q}{2}}{h}\right)} \Longleftrightarrow \frac{p+q}{2h} = \frac{\overline{p}+\overline{q}}{2\overline{h}},\]which follows from above.

spent 2.5 hours doing nothing in contest We use complex numbers with $(ABC)$ as the unit circle. We wish to prove that $\tfrac{p+q}{2}$, or equivalently $p+q$, is a real multiple of the orthocenter $a+b+c$, i.e. $$\frac{p+q}{a+b+c}=\frac{\overline{p}+\overline{q}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \iff (ab+bc+ca)(p+q)=abc(a+b+c)(\overline{p}+\overline{q}).$$The conditions on the bisectors of $\angle PAQ$, etc. are equivalent to cyclic variants of $$\frac{\frac{p-a}{h-a}}{\frac{h-a}{q-a}} \in \mathbb{R} \iff \frac{(p-a)(q-a)}{(b+c)^2} \in \mathbb{R} \iff \frac{(p-a)(q-a)}{(b+c)^2}=\frac{(\overline{p}-\frac{1}{a})(\overline{q}-\frac{1}{a})}{(\frac{1}{b}+\frac{1}{c})^2} \iff (p-a)(q-a)=b^2c^2\left(\overline{p}-\frac{1}{a}\right)\left(\overline{q}-\frac{1}{a}\right)$$which expands to $$pq-a(p+q)+a^2=b^2c^2\overline{p}\overline{q}-\frac{b^2c^2}{a^2}(\overline{p}+\overline{q})+\frac{b^2c^2}{a^2}.$$Multiply this equation by $\tfrac{1}{b^2}-\tfrac{1}{c^2}$ and sum cyclically (after doing the same to the other equations). Miraculously, almost everything cancels and we are left with \begin{align*} (p+q)\left(\frac{a}{b^2}-\frac{a}{c^2}+\frac{b}{c^2}-\frac{b}{a^2}+\frac{c}{a^2}-\frac{c}{b^2}\right)&=(\overline{p}+\overline{q})\left(\frac{c^2}{a}-\frac{b^2}{a}+\frac{a^2}{b}-\frac{c^2}{b}+\frac{b^2}{c}-\frac{a^2}{c}\right) &\iff\\ (p+q)(a^3c^2-a^3b^2+a^2b^3-b^3c^2+b^2c^3-a^2c^3)&=(\overline{p}+\overline{q})(ab^2c^4-ab^4c^2+a^4bc^2-a^2bc^4+a^2b^4c-a^4b^2c). \end{align*}By inspection, $a-b$ and cyclic variants divide both sides, leading to the factorization of the LHS as $(p+q)(a-b)(b-c)(c-a)(ab+bc+ca)$ and the RHS as $(\overline{p}+\overline{q})(a-b)(b-c)(c-a)(abc(a+b+c))$, which finishes the problem. $\blacksquare$

We uploaded our solution https://calimath.org/pdf/USATSTST2023-6.pdf on youtube https://youtu.be/CMX3iHomSBI.

Amazing Problem! Sketch: Denote $P_A,P_B,P_C$ be the reflections of $P$ across $AH,BH,CH$. Obviously, $H$ is center of $({P_AP_BP_C})$, notice $ABC,P_AP_BP_C$ are perspective at $Q$, and also orthologic. So by Sondat's theorem and homothety it suffices to show that if $\overline{P_AP_B \cap AB} = F$ similarly define $D,E$ then $\overline{DEF} \perp \overline{OH}$, reflect $P_AP_BP_C$ about $\overline{DEF}$ to get $A_1B_1C_1$, let $M$ be the miquel point of $ABDECF$. Let $O_A,O_B,O_C,O'$ be the centers of $(AEF),(BDF),(DEC),(A_1B_1C_1)$. Let $X$ be the point on $(AXF)$ such that $AX \perp \overline{DEF}$. It is easy to see $MOO'O_AO_BO_C$ is cyclic, further $O'-M-X$ by a short angle chase. So $AX \parallel OO'$ and we're done

Where does the multiplying the equation by (b^2-c^2) get motivated from?

Extension: Let $H'$ be the reflection of $H$ over the Feuerbach point of the orthic triangle and let the image of $H'$ after a negative inversion about the polar circle of $ABC$ be $X$. If $Q'$ is the reflection of $Q$ over line $XH$, show that $XP \cdot XQ'=XH^2$.

Take $(ABC)$ to be the unit circle. Then it is clear that in terms of complex coordinates, \begin{align*} \angle PAH = \angle QAH \implies \frac{q - a}{h - a} \div \frac{h - a}{p - a} \in \mathbb{R} \end{align*}Thus we find, \begin{align*} \frac{(q - a)(p - a)}{(h - a)^2} &= \overline{\left(\frac{(q - a)(p - a)}{(h - a)^2} \right)}\\ \iff \frac{(q - a)(p - a)}{(b+c)^2} &= \frac{\left( \overline{q} - \frac{1}{a} \right)\left(\overline{p} - \frac{1}{a} \right)}{\left(\frac{1}{b} + \frac{1}{c} \right)^2}\\ \iff \frac{(q - a)(p - a)}{(b + c)^2} &= \frac{b^2c^2(\overline{q}a - 1)(\overline{p}a - 1)}{a^2(b + c)^2}\\ \iff a^2(q - a)(p - a) &= b^2c^2(\overline{q}a - 1)(\overline{p}a - 1) \end{align*}Similar cyclic statements give, \begin{align*} a^2(pq - a(p + q) + a^2) &= b^2c^2(a^2\overline{pq} - a(\overline{p} + \overline{q}) + 1)\\ b^2(pq - b(p + q) + b^2) &= a^2c^2(b^2\overline{pq} - b(\overline{p} + \overline{q}) + 1)\\ c^2(pq - c(p + q) + c^2) &= a^2b^2(c^2\overline{pq} - c(\overline{p} + \overline{q}) + 1) \end{align*}Multiply the first equation by $(b^2 - c^2)$ and so on; summing cyclically, \begin{align*} \sum (a^4(b^2 - c^2) - (b^2 - c^2)b^2c^2) + \sum a^2(b^2 - c^2)pq - \sum a^3(b^2 - c^2)(p + q) &= a^2b^2c^2\overline{pq} \sum (b^2 - c^2) - abc\sum bc(b^2 - c^2)(\overline{p} + \overline{q})\\ (p + q)\sum a^3(b^2 - c^2) &= abc(\overline{p} + \overline{q})\sum bc(b^2 - c^2)\\ \end{align*}Note that it suffices to show that, \begin{align*} \frac{\frac{p + q}{2}}{a + b + c} &\in \mathbb{R}\\ \iff \frac{p + q}{2(a + b + c)} &= \overline{\left(\frac{p + q}{2(a + b + c)} \right)}\\ \iff \frac{p + q}{2(a + b + c)} &= \frac{\overline{p} + \overline{q}}{2\left(\frac{ab + bc + ca}{abc} \right)}\\ \iff \frac{p + q}{\overline{p} + \overline{q}} &= \frac{abc(a + b + c)}{ab + bc + ca} \end{align*}Then it suffices to show that, \begin{align*} \frac{abc\sum bc(b^2 - c^2)}{\sum a^3(b^2 - c^2)} &= \frac{abc(a + b + c)}{ab + bc + ca}\\ \iff \frac{\sum bc(b^2 - c^2)}{\sum a^3(b^2 - c^2)} &= \frac{a + b + c}{ab + bc + ca}\\ \iff \frac{(a + b + c)(a^2b - a^2c + b^2c - b^2a + c^2a - c^2b)}{(ab + bc + ca)(a^2b - a^2c + b^2c - b^2a + c^2a - c^2b)} &= \frac{a + b + c}{ab + bc + ca}\\ \iff \frac{a + b + c}{ab + bc + ca} &= \frac{a + b + c}{ab + bc + ca} \end{align*}as desired. This finishes the problem.

Imagine having a straightforward compleks bash on TSTST. This was more of a sympoly factoring exercise than a hard geometry problem. Set $(ABC)$ be the unit circle as usual. We first start off with the desired conclusion. Let $M$ be the midpoint of $PQ$. Then, \[m=\frac{p+q}{2}\]Since the desired result is that points $O$ , $M$ and $H$ are collinear, it suffices to show that $\frac{m-o}{h-o} \in \mathbb{R}$. We first compute $\frac{m-o}{h-o}$. \begin{align*} \frac{m}{h} &= \frac{\frac{p+q}{2}}{a+b+c} \\ &= \frac{p+q}{2(a+b+c)} \end{align*}Further, \begin{align*} \overline{\left(\frac{m}{h}\right)} &= \overline{\left(\frac{p+q}{2(a+b+c)}\right)}\\ &= \frac{abc(\overline{p}+\overline{q})}{2(ab+bc+ca)} \end{align*}Thus, the desired goal will be the following equality. \begin{align} \frac{p+q}{a+b+c} = \frac{abc(\overline{p}+\overline{q})}{ab+bc+ca} \end{align} We now deal with the conditions. Since the $A-$altitude is the angle bisector of $\angle PAQ$, we must have $\angle PAH = \angle HAQ$. Thus, \begin{align*} \angle PAH &= \angle HAQ\\ \text{arg} \left(\frac{p-a}{a-h}\right) &= \text{arg} \left(\frac{h-a}{a-q}\right)\\ \text{arg} \left(\frac{(p-a)(q-a)}{(h-a)^2}\right) &=0 \\ \frac{(p-a)(q-a)}{(h-a)^2} &\in \mathbb{R}\\ \frac{(p-a)(q-a)}{(b+c)^2} \end{align*}Now, note that \[\overline{\left(\frac{(p-a)(q-a)}{(h-a)^2}\right)}=\frac{\left(\overline{p} - \frac{1}{a}\right)\left(\overline{q}-\frac{1}{b}\right)}{\left(\frac{1}{b}+ \frac{1}{c}\right)^2} \]which is equivalent to \begin{align*} \frac{(p-a)(q-a)}{(b+c)^2} &= \frac{\left(\overline{p} - \frac{1}{a}\right)\left(\overline{q}-\frac{1}{b}\right)}{\left(\frac{1}{b}+ \frac{1}{c}\right)^2}\\ a^2(p-a)(q-a) &= (\overline{p}a-1)(\overline{q}a-1)b^2c^2\\ a^2pq - a^3(p+q) + a^4 &= a^2b^2c^2 \overline{p}\overline{q} - ab^2c^2(\overline{p}+\overline{q}) + b^2c^2 \end{align*}Similarly, \[b^2pq - b^3(p+q) + b^4 = a^2b^2c^2\overline{p}\overline{q} - a^2bc^2(\overline{p}+\overline{q}) + a^2c^2\]\[c^2pq - c^3(p+q)+c^4 = a^2b^2c^2\overline{p}\overline{q} - a^2b^2c(\overline{p}+\overline{q})+a^2b^2\]Setting two pairs of these equations equal via the common $a^2b^2c^2\overline{p}\overline{q}$ term, \[pq = \frac{(a^3-b^3)(p+q)+(a^2bc^2-ab^2c^2)(\overline{p}+\overline{q})+b^4-a^4+b^2c^2-a^2c^2}{a^2-b^2}\]and \[pq = \frac{(b^3-c^3)(p+q)+(a^2b^2c-a^2bc^2)(\overline{p}+\overline{q})+c^4-b^4+a^2c^2-a^2b^2}{b^2-c^2}\]Now, setting these two expressions equal and completely eliminating $pq$ terms, we obtain a certain equation. We take all terms to one side. We break this calculation into 3 parts. First, we deal with the coefficient of the $(p+q)$ term. \begin{align*} (a^3-b^3)(b^2-c^2)(p+q)-(a^2-b^2)(b^3-c^3)(p+q) &= (a^3b^2-a^3c^2-b^5+b^3c^2\\ &-a^2b^3+a^2c^3+b^5-b^2c^3)(p+q)\\ &= (b-a)(c-b)(a-c)(ab+bc+ca)(p+q) \end{align*}Next, we look at the coefficient of the $(\overline{p}+\overline{q})$ term. Here, \begin{align*} (a^2bc^2-ab^2c^2)(b^2-c^2)(\overline{p}+\overline{q}) &- (a^2b^2c-a^2bc^2)(a^2-b^2)(\overline{p}+\overline{q})\\ &= abc((ac-bc)(b^2-c^2)-(ab-ac)(a^2-b^2))(\overline{p}+\overline{q})\\ &= abc(\overline{p}+\overline{q})(ab^2c-ac^3-b^3c+bc^3-a^3b+ab^3+a^3c-ab^2c)\\ &= abc(a-b)(b-c)(c-a)(a+b+c)(\overline{p}+\overline{q}) \end{align*}Finally, we have the constant terms, \begin{align*} (b^4-a^4+b^2c^2-a^2c^2)(b^2-c^2) &- (c^4-b^4+a^2c^2-a^2b^2)(a^2-b^2)\\ &= (b^2-a^2)(b^2-c^2)(a^2+b^2+c^2) - (a^2-b^2)(c^2-b^2)(a^2+b^2+c^2)\\ &=0 \end{align*}So, we are left with, \[ (b-a)(c-b)(a-c)(ab+bc+ca)(p+q) + abc(a-b)(b-c)(c-a)(a+b+c)(\overline{p}+\overline{q})=0\]\begin{align*} abc(a-b)(b-c)(c-a)(a+b+c)(\overline{p}+\overline{q}) &= (a-b)(b-c)(c-a)(ab+bc+ca)(p+q)\\ abc(a+b+c)(\overline{p}+\overline{q}) &= (ab+bc+ca)(p+q)\\ \frac{p+q}{a+b+c} &= \frac{abc(\overline{p}+\overline{q})}{ab+bc+ca} \end{align*}which was the desired result.

Let $H$ be the orthocenter and let $P_A,P_B,P_C$ be the reflections of $P$ over $AH,BH,CH.$ Suppose $AP_A,BP_B,CP_C$ concur at $Q.$ Thus $\triangle ABC$ and $\triangle P_AP_BP_C$ are perspective at $Q,$ so the intersections $D=BC\cap P_BP_C,E=AC\cap P_AP_C,F=AB\cap P_AP_B$ are collinear along some line $\ell$ by Desargues's theorem. Let $A',B',C'$ be the reflections of $A,B,C$ across $\ell,$ respectively. First notice that $\measuredangle P_BP_AP_C=\measuredangle P_BPP_C$ since they all lie on the circle centered at $H$ through $P,$ and this is equal to $\measuredangle CAB$ since $P_BP\parallel AC$ and $P_CP\parallel AB.$ Thus $\triangle P_AP_BP_C$ is oppositely similar to $\triangle ABC.$ In particular, one consequence of this is that $\triangle P_AP_BP_C$ is not homothetic to $\triangle ABC,$ since it is scalene. Thus $\ell$ is not the line at infinity so $A',B',C'$ are defined. Claim: Lines $A'P_A,B'P_B,C'P_C$ are concurrent at some point $R$ such that $QR\perp\ell.$ Proof: Define $R$ to be the intersection of $A'P_A$ and the line through $Q$ perpendicular to $\ell.$ We want to show $QR=BB'\cdot\frac{QP_B}{BP_B}$ (in directed lengths) which would imply $\triangle P_BQR\sim\triangle P_BBB',$ which would mean $B',R,P_B$ collinear. But we know $QR=AA'\cdot\frac{QP_A}{AP_A}$ from $\triangle P_AQR\sim\triangle P_AAA',$ so we want to show $\frac{AP_A\cdot QP_B}{BP_B\cdot QP_A}=\frac{AA'}{BB'}.$ Then letting $\theta$ be the angle between lines $AP_A$ and $BP_B,$ we get $AP_A\cdot QP_B=\frac2{\sin\theta}[AP_AP_B]$ and $BP_B\cdot QP_A=\frac2{\sin\theta}[BP_AP_B],$ so we want $\frac{AA'}{BB'}=\frac{[AP_AP_B]}{[BP_AP_B]},$ (directed areas) which is equivalent to saying the ratio between the (directed) distances from $A,B$ to line $P_AP_B$ is the same as the ratio between the distances from $A,B$ to line $\ell,$ and this is by similar triangles since the three lines $P_AP_B,AB,\ell$ concur at $F.$ We prove the following lemma: Lemma: (F.E.Wood's theorem) Given two non-homothetic similar (similarly oriented) perspective triangles, both their perspector and their spiral center lie on the circumcircles of both triangles. Proof: Let one triangle be $ABC$ and the other $A'B'C'$ and let their perspector be $X$ and their spiral center be $Y.$ By the construction of the spiral center of two segments, we see that $Y$ must be the second intersection of $(ABX)$ and $(A'B'X),$ since $Y$ is the center of spiral similarity taking $AB$ to $A'B'.$ Thus $(ABXY),(BCXY),(ACXY)$ are concyclic, and thus $(ABCXY)$ is concyclic and similarly for $(A'B'C'XY).$ Now we apply our lemma to the triangles $P_AP_BP_C$ and $A'B'C'$ in our problem, since they are perspective at $R$ and they are both oppositely similar to $\triangle ABC.$ Letting the spiral center of $\triangle P_AP_BP_C$ and $\triangle A'B'C'$ be point $P^*,$ we have that $P^*$ lies on circles $P_AP_BP_CP$ and $A'B'C'.$ Claim: Points $P_A,A',E,F,P^*$ are concyclic. Proof: Since $P^*$ is the spiral center of $P_AP_B$ and $A'B',$ it is also the spiral centerof $P_AA'$ and $P_BB',$ and since $P_AP_B$ intersects $A'B'$ at $F$ we get $P_A,A',F,P^*$ are concyclic. Similarly $P_A,A',E,P^*$ are concyclic so the claim is proven. We thus also get that $P_B,B',D,F,P^*$ and $P_C,C',D,E,P^*$ are concyclic. We now consider the Steiner line of the complete quadrilateral given by lines $AB,BC,CA$ and $\ell.$ It passes through $H$ and the orthocenters $H_A,H_B,H_C$ of $\triangle AEF,\triangle BDF,\triangle CDE$ respectively. Claim: $P^*$ lies on this line. Proof: Notice $H_B$ lies on line $BB'$ since it is the perpendicular through $B$ to $\ell.$ Similarly for $H_C,$ and notice $BB'\parallel CC'.$ Then \[\measuredangle DP^*H_B=\measuredangle DB'H_B=\measuredangle DB'B=\measuredangle DC'C=\measuredangle DC'H_C=\measuredangle DP^*H_C.\] Now we let $P'$ be the reflection of $P^*$ over $\ell.$ Claim: $P'$ is the Anti-Steiner point of line $HP^*$ with respect to $\triangle ABC.$ Proof: Let $K_B$ and $K_C$ be the reflections of $H_B$ and $H_C$ over line $BC.$ First notice $P'$ lies on the circumcircles of $BDF$ and $CDE,$ by reflection over $\ell.$ Then $K_B$ and $K_C$ lie on these circles, respectively, by reflection over $BC.$ But then \[\measuredangle DP'K_B=\measuredangle DFK_B=\measuredangle(\ell,AH)=\measuredangle DEK_C=\measuredangle DP'K_C,\]so $P'$ lies on the reflection of $HP^*$ over $BC,$ and by symmetry it is the Anti-Steiner point. Now let $P^*_A,P^*_B,P^*_C$ be the reflections of $P^*$ over $AH,BH,CH$ respectively. Notice \[\measuredangle P^*P^*_AP^*_B=\measuredangle(HP^*,AC)+90=\measuredangle(AC,H'P')+90=\measuredangle BH'P'=\measuredangle BAP'=\measuredangle P^*A'B',\]where $H'$ is the reflection of $H$ over $AC.$ By symmetry this implies $P^*$ is the spiral center of $\triangle P^*_AP^*_BP^*_C$ and $\triangle A'B'C'.$ But then $P^*$ is the spiral center of $\triangle P_AP_BP_C$ and $\triangle P^*_AP^*_BP^*_C.$ But these triangles share a circumcircle, which is the circle centered at $H$ through $P$ which we have shown passes through $P^*.$ Thus their spiral center is either $H,$ which is impossible, or these two triangles are in fact identical. Thus, we must have $P^*=P.$ Next, we have \[\measuredangle(AC,PR)=\measuredangle P_BPR=\measuredangle P_BP_AR=\measuredangle FP_AA'=\measuredangle FEA'=\measuredangle(\ell,A'C')=\measuredangle(AC,\ell)\]so $PR\parallel\ell.$ Now consider the radical axes of pairs of circles $(ABC),(P_AP_BP_C),(A'B'C').$ Two of these radical axes are $\ell$ and $PR,$ and the other one is perpendicular to the line through the centers of the circles, which is the Euler line. Thus $\ell$ and $PR$ must both be perpendicular to the Euler line as well, by Radical Axis theorem. Now the line through the three circumcenters of $\triangle ABC,\triangle P_AP_BP_C$ and $\triangle A'B'C'$ is the Euler line, and thus $P,R$ are reflections over it. But we have shown $QR\perp\ell,$ so $QR$ is parallel to the Euler line, and thus it is the $P$-midline of $\triangle PQR$ which finishes.

compleck

ABCDE wrote: ABCDE wrote: Here is a different approach with complex numbers that does not even need to set $ABC$ on the unit circle. In fact, this solution easily generalizes to the following: Let $ABC$ be a triangle and $X$ be any point that is not an incenter of excenter of $ABC$. If $P$ and $Q$ are points such that $AX$ bisects $\angle PAQ$, $BX$ bisects $\angle PBQ$, and $CX$ bisects $\angle PCQ$, then the midpoint of $PQ$ lies on a fixed line. The fixed line has the following explicit description: let $A'$ be the intersection of the reflection of $AB$ across $BX$ and the reflection of $AC$ across $CX$, and define $B'$ and $C'$ similarly. At least one of $A',B',C'$ is well-defined as $X$ is not an incenter or excenter of $ABC$, and all of the well-defined points among the midpoints of $AA',BB',CC'$ are collinear with $X$. It would be interesting to see a synthetic proof of the above. This might be approachable with isoptic cubic theory: consider the unique circumstrophoid of ABC with node at X (where X is not an incenter or excenter; as then the strophoid is not unique). Main claim which I can't prove: The only points P such that the reflections of AP,BP,CP across AX,BX,CX concur is all P on this strophoid. This might just be doable with finding 10 points and bezouting as the concurrency condition is deg3 by complex However once we have this it's quite easy. Note that by the fact that for two conjugate points P,Q(i.e Clawson-Schmidt conjugates in some quadrilateral that has this strophoid as its isoptic cubic) and any point N on the strophoid, we have that the reflection of NP over NX is NQ. This is true since NX must be isogonal with itself in some quadrilateral containing N as a vertex that has this strophoid as its isoptic cubic, as for a quadrilateral to have a nodal isoptic cubic, it must be a tangential quadrilateral with the node=X at its incenter, so isogonal conjugation in this quadrilateral at N must be reflection over NX. Thus we get that P,Q are conjugates on this strophoid. Note the Newton line of this strophoid is a line parallel to its asymptote through the node. By Newton II theorem, the inconic with foci P,Q has center lying on the Newton line, which is also the midpoint of P and Q, so we are done.

We will use directed angles. For any point $P$, reflect $\overline{AP},\overline{BP},$ and $\overline{CP}$ across $\overline{AH},\overline{BH},$ and $\overline{CH}$, respectively. The condition that these reflected lines concur at a point $Q$ is degree $3$ in $P$, so the locus of points for which $Q$ exists is a cubic $\mathcal{C}$. Let $f$ be the involution from $\mathcal{C}$ to itself map $P$ to $Q$ for all $P$ on $\mathcal{C}$. Let the circle points be $I$ and $J$. Since $f(I)=J$ and $f(J)=I$, we see that $I$ and $J$ lie on $\mathcal{C}$.

, so $\mathcal{C}=\mathcal{C}'$. Since $f$ swaps $P_1$ and $Q_1$ and swaps $P_2$ and $Q_2$, it must be isogonal conjugation in $P_1P_2Q_1Q_2$. Now, since $\mathcal{C}$ is some isoptic cubic and $f$ is an isogonal conjugation on it, we see that for any point $X$ on $\mathcal{C}$, there exists some reflection swapping $\overline{XP}$ and $\overline{Xf(P)}$ for all $P$ on $\mathcal{C}$. Since $f$ fixes $H$, this must be a reflection over $\overline{XH}$. Now, it suffices that $\mathcal{C}$ contains the point $\infty$ at infinity on the Euler line of $\triangle{}ABC$ by taking $X=\infty$. For this it suffices that $f\left(\infty\right)$ exists. In fact, we claim that $f\left(\infty\right)$ is the antigonal conjugate $O_1$ of the circumcenter $O$ of $\triangle{}ABC$ in $\triangle{}ABC$. This is true since if the antigonal conjugate of $H$ in $\triangle{}ABC$ on its Jerabek hyperbola is $H_1$, which is the isogonal conjugate of $\infty$ in $\triangle{}ABC$, then $$\angle{}O_1VH=\angle{}O_1VH_1+\angle{}H_1VH=\angle{}HVO+\angle{}OV\infty=\angle{}HV\infty,$$implying that $\overline{V\infty}$ and $\overline{VO_1}$ are reflections in $\overline{VH}$ for all $V\in\{A,B,C\}$, so we are done.

Complex FTW