Just bash it Set $(ABC)$ as the unit circle and define: $$P=l_b \cap l_c~,~Q=l_c \cap l_a~,~R=l_a \cap l_b$$$G:$ the centroid of $\triangle PQR$ $O:$ the circumcenter of $\triangle PQR$ $B_1:$ the reflection of $C ~\text{w.r.t}~ AH$ $B_2:$ the reflection of $A ~\text{w.r.t}~ CH$ We will instead prove that $H,O$ and $G$ are collinear and then euler line implies the required result. Extend ray $AH$ to meet $(ABC)$ at $D$ $$h=a+b+c~ , ~d=-\frac{bc}{a}~ ,~ b_1=a+d-ad\overline{c}$$$$ \text{So} ~b_1=\frac{a^2+ab-bc}{a}~~\text{and similarly}~ b_2=\frac{c^2+bc-ab}{c}$$$$P\in l_b\iff\begin{vmatrix} p & \overline{p} & 1\\ b_1 & \overline{b_1} & 1\\ b_2 & \overline{b_2} & 1\\ \end{vmatrix}=0\iff (\overline{b_2}-\overline{b_1})p+(b_1-b_2)\overline{p}=b_1\overline{b_2}-\overline{b_1}b_2$$$$\iff ac(a+b+c)p+abc(ab+bc+ca)\overline{p}=2a^2bc+2ab^2c+2abc^2+2a^2c^2-a^3b-bc^3$$Similarly for $P\in l_c:$ $$ab(a+b+c)p+abc(ab+bc+ca)\overline{p}=2a^2bc+2ab^2c+2abc^2+2a^2b^2-a^3c-b^3c$$$$ \text{By subtraction}:~ p=\frac{a^3+2a^2b+2a^2c-b^2c-bc^2}{a(a+b+c)}$$$$ \text{Now consider the map}: ~\phi (z)=\frac{h(h-z)}{ab+bc+ca}$$For a point $z$ let $z'$ denote the image $\phi(z)$ Notice that $\phi$ is just a combination of a dilation and a translation, so it suffices to prove that $H',O'$ and $G'$ are collinear. We trivially see $h'=0$ and $p'=\frac{b+c}{a}$ similarly $q'=\frac{c+a}{b},r'=\frac{a+b}{c}$ Now notice: $$|p'+1|=|q'+1|=|r'+1|=|a+b+c|$$ So $o'=-1$ and because $h'=0$ it's left to prove $g'$ is a real number which is true since: $$p'+q'+r'=\frac{a^2b+a^2c+b^2a+b^2c+c^2a+c^2b}{abc}\in \mathbb{R}~~\blacksquare$$

Let $A_B$ be the reflection of $A$ over the $B$-altitude, and define $A_C,B_A,B_C,C_A,C_B$ similarly. Let $A_BB_A\cap A_CC_A=A'$ and define $B',C'$ similarly. Notice $\measuredangle BA_CC=-\measuredangle BAC=\measuredangle BHC,$ so $(BHC)$ passes through $A_C,$ and by symmetry $A_B.$ Next, since $AB\cdot AA_C=AC\cdot AA_B,$ we get $A$ lies on the radical axis of $(C_ABA_C)$ and $(B_ACA_B),$ but pairs $B,B_A$ and $C,C_A$ share a midpoint, which is the foot of the altitude from $A,$ and thus the radical axis is line $AH.$ In particular we have $(C_ABA_C)$ and $(B_ACA_B)$ are congruent, so $\measuredangle A'A_CA=\measuredangle C_AA_CB=\measuredangle B_AA_BC=\measuredangle A'A_BA,$ so $A,A',A_B,A_C$ are concyclic. By definition we have $AH=A_BH=A_CH,$ so $AH=A'H.$ Also $\measuredangle C'A'B'=\measuredangle A_CA'A_B=\measuredangle A_CAA_B=\measuredangle BAC,$ so by symmetry we have $\triangle ABC\sim\triangle A'B'C',$ oppositely oriented. Next reflect $\triangle A'B'C'$ over the perpendicular bisector of $AA',$ which passes thorugh $H.$ If $B',C'$ are sent to $D,E$ then $\triangle ADE\sim\triangle ABC,$ similarly oriented. We also have $HB=HB'=HD$ and $HC=HC'=HE.$ Now, letting $M,N$ be midpoints of $BD,CE$ we get $BD\perp MH,CE\perp NH.$ Now if $P=BD\cap CE,$ then by spiral similarity we have $(ADEP),(ABCP)$ cyclic, and by Gliding Principle, also $(AMNP).$ But the right angles give $(PMHN)$ is cyclic with diameter $HP.$ Since $AH\perp BC$ we must then have $AP\parallel BC,$ and since $APBC$ is cyclic we get that it is an isosceles trapezoid. Now we want to show that $H$ lies on the Euler line of $\triangle ADE,$ because reflecting back gives the desired result. Call $O,O'$ the circumcenters of $\triangle ABC$ and $\triangle ADE,$ and call $H'$ the orthocenter of $\triangle ADE.$ Now if $HO\cap H'O'=X\ne H,$ we would have $A,O,O',X$ cyclic by spiral similarity. Thus it suffices to show $A,O,O',H$ are cyclic. Since $O,O'$ lie on the perpendicular bisector of $AP,$ it suffices to show $AO'OH$ is an isosceles trapezoid. Now let $O''$ be the midpoint of $PH$ and thus the circumcenter of $\triangle AMN.$ There is a spiral similarity at $A$ sending $\triangle BCO$ to $\triangle MNO''$ to $\triangle DEO',$ so there is a spiral similarity at $A$ sending $BMD$ to $OO''O'$ to $CNE,$ so by Gliding Principle $O''$ is the midpoint of $OO'.$ Thus the perpendicular bisectors of $AH$ and $OO'$ are parallel and both pass through $O'',$ so we are done.

[asy][asy] size(9cm); defaultpen(fontsize(10pt)); import olympiad; import geometry; pair A = (1,3.6); pair B = (0,0); pair C = (4,0); pair H = orthocenter(A,B,C); pair B_c = reflect(C,H) * B; pair C_b = reflect(B,H) * C; pair C_a = reflect(A,H) * C; pair A_c = reflect(C,H) * A; pair A_b = reflect(B,H) * A; pair B_a = reflect(A,H) * B; pair A_1 = extension(A_b,B_a,A_c,C_a); pair B_1 = extension(B_c,C_b,A_b,B_a); pair C_1 = extension(B_c,C_b,A_c,C_a); pair O_1 = circumcenter(A_1,B_1,C_1); pair H_1 = orthocenter(A_1,B_1,C_1); pair O = circumcenter(A,B,C); fill(A--B--C--cycle, mediumgray); fill(A_1--B_1--C_1--cycle, mediumgray); draw(circumcircle(A,A_1,A_b), linewidth(0.7)); draw(B_c--H--C_b, gray); draw(B_1--C_b, dashed); draw(B_1--A_b, dashed); draw(A_1--C_a, dashed); draw(C_b--A, dashed); draw(C_a--B, dashed); draw(A--B--C--cycle, linewidth(1)); draw(A_1--B_1--C_1--cycle, linewidth(1)); draw(H--O_1, black); dot("$A$", A, dir(63)); dot("$B$", B, dir(-90)); dot("$C$", C, dir(-26)); dot("$H$", H, dir(-15)); dot("$B_c$", B_c, dir(170)); dot("$C_b$", C_b, dir(107)); dot("$C_a$", C_a, dir(157)); dot("$A_c$", A_c, dir(-129)); dot("$A_b$", A_b, dir(16)); dot("$B_a$", B_a, dir(-85)); dot("$A_1$", A_1, dir(-140)); dot("$B_1$", B_1, dir(-42)); dot("$C_1$", C_1, dir(-70)); dot("$O_1$", O_1, dir(135)); dot("$H_1$", H_1, dir(150)); dot("$O$", O, dir(90)); [/asy][/asy] Let $B_c$ denote the reflection of $B$ across $CH$. Define $C_b$, $A_c$, $C_a$, $A_b$, and $B_a$ similarly. Let $O$ be the circumcenter of $\triangle ABC$. Let $A_1 = \ell_b\cap \ell_c$, and define $B_1$ and $C_1$ similarly. Claim. $\triangle A_1B_1C_1$ and $\triangle ABC$ are inversely similar. Proof. By APMO 2010 P4, the circumcenter of $\triangle HB_cC_b$ lies on $OH$. Since $(HB_c, HC_b)$ are isogonal w.r.t. $\angle BHC$, it follows that $(\perp B_cC_b, OH)$ are isogonal w.r.t. $\angle BHC$ as well. Therefore, we have $$\measuredangle(B_cC_b, BH) = 90^\circ + \measuredangle(CH, OH).$$Similarly, we get that $\measuredangle(A_bB_a, BH) = 90^\circ + \measuredangle(AH, OH)$, so $$\measuredangle(B_cC_b, A_bB_a) = \measuredangle(CH, AH) = \measuredangle(AB, BC) = -\measuredangle CBA,$$implying the similarity. $\blacksquare$ The claim above also implies that$\measuredangle A_cA_1A_b = \measuredangle BAC = \measuredangle A_cAA_b$, so $A, A_1, A_b, A_c$ are concyclic. However, since $HA=HA_b=HA_c$, it follows that $HA=HA_1$. Similarly, $HB=HB_1$ and $HC=HC_1$. Therefore, we have $$\frac{A_1H_1}{A_1H} = \frac{B_1H_1}{B_1H} = \frac{C_1H_1}{C_1H}.$$To finish, note that $H$ is another concurrency point of $H$-Apollonius circles of $\triangle B_1H_1C_1$, $\triangle C_1H_1A_1$, and $\triangle A_1H_1B_1$. Since $B$ and $C$ are inverses w.r.t. the Apollonius circle of $\triangle B_1H_1C_1$, this circle must be orthogonal to $\odot(A_1B_1C_1)$ and hence passes through the inverse of $H_1$ w.r.t. $\odot(A_1B_1C_1)$. Repeating this logic on other two circles, we find that $H$ and $H_1$ are inverses w.r.t. $\odot(A_1B_1C_1)$, implying the conclusion.

For points $P$ and $Q$, let $P_Q$ be the reflection of $P$ over $QH$. Then define $A' = \overline{C_AA_C} \cap \overline{B_AA_B}$ and define $B'$, $C'$ similarly. Clearly this is the triangle that we care about. Now, taking a homothety at $A$ with scale factor $\frac 12$, we find that $A_CA_B$ is antiparallel to $BC$, whence $A_CBA_BC$ is cyclic. More particularly this means that $\triangle AC_AB_A \sim^+ \triangle AA_CA_B$, so $A$ is the spiral center taking $C_AB_A$ to $A_CA_B$. In particularly it follows that $AA_CA'A_B$ is cyclic. Moreover since we have reflections over $BH$ and $CH$, we conclude that $A'H = AH$. Let $H'$ and $O'$ be the orthocenter and circumcenter respectively of $A'B'C'$. From similar triangles we have \[ \frac{A'H'}{B'H'} = \frac{AH}{BH} = \frac{A'H}{B'H} \]and similar, so $H'$ lies on the $H$-apollonian circles of $\triangle HA'B'$, $\triangle HB'C'$, and $\triangle HC'A'$. But then it is well known that these apollonian circles are orthogonal to $(A'B'C')$ by say, inversion. Particularly these three circles stay fixed, and $H$ must be the inverse of $H'$ with respect to $(A'B'C')$. We are done.

We will employ complex numbers, with $(ABC)$ as the unit circle. Let $F = \ell_a \cap \ell_b$ etc. Let $X$ be the reflection of $C$ over the $A$-altitude and $Y$ be the reflection of $A$ over the $C$-altitude. We will denote $\sigma_1 = a+b+c$ and $\sigma_2 = ab+ac+bc$. We easily get that $x = b+c-ab\bar c$ and $y = a+b-\bar abc$. Because we know that $F \in \ell_b$, then we have \begin{align*} \frac{f-x}{y-x} &\in \mathbb R \\ \frac{a(cf+ab-bc-c^2)}{(a-c)\sum ab} &\in \mathbb R \\ \frac{a(cf+ab-bc-c^2)}{(a-c)\sum ab} &= \frac{\frac 1a (\frac 1c\bar f+ \frac{1}{ab}-\frac{1}{bc}-\frac{1}{c^2})}{(\bar a - \bar c)\sum \overline{ab}} \\ \frac{cf+ab-bc-c^2}{\sum ab} &= \frac{- bc\bar f - \frac{c^2}{a} + b + c}{\sum a} \\ a\left(cf + ab - bc - c^2\right) \sigma_1 &= \left(-abc \bar f - c^2 + ab + ac\right) \sigma_2 \\ \left(ac \sigma_1 \right)f + (abc\sigma_2)\bar f &= \left(ab+ac-c^2\right) \sigma_2 + \left(ac^2+abc-a^2b\right)\sigma_1 \end{align*} Similarly, we also have $F \in \ell_a$, so we have \[\begin{cases} \left(ac \sigma_1 \right)f + (abc\sigma_2)\bar f = \left(ab+ac-c^2\right) \sigma_2 + \left(ac^2+abc-a^2b\right)\sigma_1 \\ \left(bc \sigma_1 \right)f + (abc\sigma_2)\bar f = \left(ab+bc-c^2\right) \sigma_2 + \left(bc^2+abc-ab^2\right)\sigma_1 \end{cases}\] Subtracting the second equation from the first equation gives \begin{align*} (a-b)c\sigma_1f &= (a-b)c \sigma_2 + \left(ac^2-bc^2+ab^2-a^2b\right)\sigma_1 \\ c\sigma_1 f &= c \sigma_2 + (c^2 -ab)\sigma_1 \\ f &= \frac{\sigma_2}{\sigma_1} + c - \frac{ab}{c} \\ &= \frac{\sigma_2}{\sigma_1} + \sigma_1 - \frac{\sigma_2}{c} \end{align*}Similarly, we can get that $d = \frac{\sigma_2}{\sigma_1} + \sigma_1 - \frac{\sigma_2}{a}$ and $e = \frac{\sigma_2}{\sigma_1} + \sigma_1 - \frac{\sigma_2}{b}$. Thus, we see that circumcenter of the the triangle $DEF$ is simply $\frac{\sigma_2}{\sigma_1}+\sigma_1$ and the orthocenter of the triangle $DEF$ is $\frac{\sigma_2}{\sigma_1}+\sigma_1-\sigma_2(\bar a+\bar b +\bar c) = \frac{\sigma_2}{\sigma_1} + \sigma_1 - \frac{\sigma_2^2}{abc}$. It suffices to show that $h = \sigma_1$ lies on this line. So it suffices to show that \[\frac{\sigma_2/\sigma_1-\sigma_2^2/abc}{\sigma_2/\sigma_1} \in \mathbb R \Longleftrightarrow \frac{\sigma_1\sigma_2}{abc} \in \mathbb R \] That is correct since \[\frac{\overline{(a+b+c)} \cdot \overline{(ab+ac+bc)}}{\overline{abc}} = \frac{(ab+ac+bc)(a+b+c)}{abc}\]

I had a terrible time solving this problem because I thought it was going to be harder than it was. The crux is to introduce $K$ the isogonal conjugate of the point of infinity along the Euler line; $T = KH \cap (ABC)$ actually lies on the circumcircle of $\mathcal{T}$. This can be proven with a terrible amount of angle chasing because $T$ lies on $(AHP)$, $(BHQ)$, $(CHR)$ as well (where $P$, $Q$, $R$ are the vertices of $\mathcal{T}$). This allows one to do inversion at $H$ and it turns out that the orthocenter of $\mathcal{T}$, which is $PH_A \cap QH_B \cap RH_C$, maps to the center of the image under inversion of the circumcircle of $\mathcal{T}$.

This problem is an extension to the Isreal problem https://artofproblemsolving.com/community/u1035145h3212242p31098325. The triangle obtained by the reflections of circumcenters over altitudes shifted by the vector $\overrightarrow{OX}$ will give you the triangle formed by the three lines $\ell_a$, $\ell_b$, $\ell_c$. Noticing this will give you the desired synthetic solution.

This beautiful problem was proposed by Fedir Yudin from Ukraine. In my opinion, one of the best geometry problems that I've seen in my life (if not bashing it). The official solution with Apollonius circle is just insane. My solution could be find as Solution 2 in the official Shortlist booklet. Also, strangely enough that this problem was also proposed to IMO 2022, but PSC preferred known problem from 2006 than this.

Let $\mathcal{T} = \triangle A'B'C'$, $H'$ be the orthocenter of $\triangle A'B'C'$. Let $A_B$ be the reflection of $A$ in $BH$ and other points are defined similarly. Claim: $B$, $C$, $A_C$, $A_B$, $H$ are concyclic. Proof: Notice that $AC_A=AC=CA_C$ and $AB_A=BA=BA_B$. Hence $\measuredangle CA_CB=\measuredangle BAC=\measuredangle CA_BB$ and $\measuredangle BHC = -\measuredangle BAC$. Claim: $A$, $A'$, $B_A$, $C_A$ are concyclic. Proof: Since $\measuredangle B_AC_AA=\measuredangle ACB = \measuredangle A_BA_CA$ and $\measuredangle AB_AC_A=\measuredangle CBA = \measuredangle AA_BA_C$, so $A'=B_AA_B \cap C_AA_C$ must be the intersection of $(AB_AC_A)$ and $(AA_BA_C)$ by Miquel points. Notice that $H$ is the circumcenter of $AA_BA_C$ and so $HA'=HA=2r(ABC)\cos \angle A$. Since $H'A'=2r(A'B'C')\cos \angle A'$ we must have $\frac{HA'}{H'A'}=\frac{r(ABC)}{r(A'B'C')}$ as $\angle A = \angle A'$ by aforementioned concyclicity. This expression is symmetric in $A'$, $B'$, $C'$ then by Apollonius circles $H$ and $H'$ are inversive pairs in $(A'B'C')$, i.e. $HH'$ passes through the circumcenter $O'$ of $\mathcal{T}$.

squarc_rs3v2m wrote: I had a terrible time solving this problem because I thought it was going to be harder than it was. The crux is to introduce $K$ the isogonal conjugate of the point of infinity along the Euler line; $T = KH \cap (ABC)$ actually lies on the circumcircle of $\mathcal{T}$. This can be proven with a terrible amount of angle chasing because $T$ lies on $(AHP)$, $(BHQ)$, $(CHR)$ as well (where $P$, $Q$, $R$ are the vertices of $\mathcal{T}$). This allows one to do inversion at $H$ and it turns out that the orthocenter of $\mathcal{T}$, which is $PH_A \cap QH_B \cap RH_C$, maps to the center of the image under inversion of the circumcircle of $\mathcal{T}$. Actually, proved with a little bit of oriented-angle chasing, the isogonal conjugate of a point $P$ is a point at infinity if and only if $P$ lies on the circumcircle. Nice solution!

Mild generalization: If $P$ is any point moving on a line through $O$, the circumcenter of $ABC$, then $P$ lies on the corresponding line in $\mathcal{T}$. (Insted of reflection in $AP$, then reflect in perpendicular line through $P$ to $BC$). Also, if $P$ is on the Jerabek Hyperbola, then the Euler line of $T$ pass through the reflection of the Parry-reflection-point in the nine-point center of $ABC$.

Here are two possible hybrid solutions presented with motivation.

keglesnit wrote: Mild generalization: If $P$ is any point moving on a line through $O$, the circumcenter of $ABC$, then $P$ lies on the corresponding line in $\mathcal{T}$. (Insted of reflection in $AP$, then reflect in perpendicular line through $P$ to $BC$). Also, if $P$ is on the Jerabek Hyperbola, then the Euler line of $T$ pass through the reflection of the Parry-reflection-point in the nine-point center of $ABC$. Is there a solution?

Looks pretty similar to the configuration in this problem (and my solution to that problem works directly too). Perhaps a bit far-fetched, but does this constitute a leak of the problem before the ISL was released?

WizardMath wrote: Looks pretty similar to the configuration in this problem (and my solution to that problem works directly too). Perhaps a bit far-fetched, but does this constitute a leak of the problem before the ISL was released? It's in july, so even if it did I don't think it needs looking into

Nice problem btw, enjoyed it. Congrats to the proposer!

I totally forgot that Appollonian circles existed so here goes my long conclusion to something that could've been a one-liner. fml. Let $A_B$ and $A_C$ be the reflections of $A$ over $BH$ and $CH$, and define $B_A$, $B_C$, $C_A$, $C_B$ similarly. Then $\triangle AC_BB_C$ is the reflection of $\triangle ABC$ over $AH$, so they are inversely similar. Also, $\measuredangle CA_BA = \measuredangle ABC = \measuredangle AA_CB$, so $BA_CCA_B$ is cyclic, and hence $\triangle AA_CA_B$ and $\triangle ABC$ are also inversely similar. Therefore, $\triangle AC_BB_C \stackrel{+}{\sim} \triangle AA_BA_C$, so $A$ is the Miquel point of quadrilateral $A_BC_BB_CA_C$. In particular, $A, A', A_B, A_C$ are concyclic, and $\measuredangle B'A'C' = -\measuredangle BAC$. Since $H$ is the circumcenter of $\triangle A'A_BA_C$, it also follows that $HA = HA' = HA_B = HA_C$. Repeating this for other vertices, we see that $\triangle ABC$ and $\triangle A'B'C'$ are inversely similar. Let $H'$ be the orthocenter of $\triangle A'B'C'$. Let $M', N', P'$ be the midpoints, and let $D'$, $E'$, $F'$ (resp. $X$, $Y$, $Z$) be the feet of perpendiculars from $H'$ (resp. $H$) onto the sides $B'C', C'A'$ and $A'B'$. To show the collinearity, it then suffices to show that \[ \frac{M'X}{M'D'} = \frac{N'Y}{N'E'} = \frac{P'Z}{P'F'}, \]in the directed sense. We will show the second equality, using complex numbers to avoid dealing with config issues. Since $HC_B = HC'$, $Y$ is the midpoint of $C_BC'$, so $A'C_B = 2N'Y$, and similarly, $A'B_C = 2P'Z$. Therefore, \[ \frac{y - n'}{z - p'} = \frac{c_b - a'}{b_c - a'}. \]On the other hand, if we let $D$, $E$, $F$, and $M$, $N$, $P$ be the feet of perpendiculars and midpoints of $\triangle ABC$ instead, then \[ \frac{e' - n'}{f' - p'} = \overline{\bigg(\frac{e - n}{f - p}\bigg)} = \overline{\bigg(\frac{c_a - a}{b_a - a}\bigg)}, \]and so to show our identity, it suffices to show that $\triangle A'C_BB_C$ and $\triangle AC_AB_A$ are inversely similar. But this is clear since we have $\measuredangle AB_AC_A = \measuredangle BB_AB' = \measuredangle BB_CB' = -\measuredangle A'B_CC_B$ because $BB_AB_CB'$ is cyclic, and similarly $\measuredangle AC_AB_A = -\measuredangle A'C_BB_C$ because $CC_AC_BC'$ is cyclic.

Define $A_b$ to be the reflection of $A$ about $HB$, let $l_b$ and $l_c$ meet at $A_1$, and define similar points cyclically. Let $H_1$ and $O_1$ be the orthocenter and circumcenter of $\mathcal{T}$. Claim: $AA_bA_cA_1$ is cyclic with center $H$ Clearly $H$ is the center of $(AA_bA_c)$. Triangles $AA_bB_a$ and $AA_cC_a$ are similar as $\measuredangle A_bAB_a=\measuredangle A_cAC_a$ and $AA_c\cdot AB_a=AA_b\cdot AC_a$. Thus $\measuredangle (AC, l_c)=\measuredangle (AB,l_b)$, as desired. The cyclic claims also hold. Claim: $A_1B_1C_1$ and $ABC$ are similar Simply $\measuredangle (l_b,l_c)=\measuredangle(AB,AC)$ with cyclic equalities also holding. Now we have that $HA_1:HB_1:HC_1=HA:HB:HC=H_1A:H_1B:H_1C$. Let $\lambda$ be such that $$\lambda=\frac{HA_1}{H_1A_1}=\frac{HB_1}{H_1B_1}=\frac{HC_1}{H_1C_1}$$Thus $(A_1B_1C_1)$ is the Apollonian Circle of $HH_1$ with ratio $\lambda$, implying that $O_1$, $H_1$, and $H$ are collinear.

Angle chasing and spiral similarity Let $O$ be the circumcenter of $ABC$. Let $XYZ$ be the triangle $\mathcal{T}$, $DEF$ be the orthic triangle of $ABC$. Let $A_B$ and $A_C$ be the reflections of $B$ and $C$ across $F$ and $E$. Define $B_A$, $B_C$, $C_A$ and $C_B$ similarly. Claim: $X \in (AB_AC_A)$ and $H$ is the center of $(AB_AC_AX)$ Proof. $\angle B_CAF = \angle C_BAE$ and $\frac{B_CA}{AB_A} = \frac{B_CA}{2 \cdot AF} = \frac{AC}{2 \cdot AF} =\frac{AB}{ 2 \cdot AE} =\frac{AC_B}{2 \cdot AE} = \frac{AC_B}{AC_A}$ together implies that $AB_CB_A \sim AC_BC_A$. This gives $\angle XB_AA = \angle XC_AA$. Note that $HA = HB_A = HC_A$ so we are done. $\square$ By symmetry, $HY = HA_B$ and $HZ = HA_C$. Additionally, because $\frac{A_BF}{FB} = \frac{A_CE}{EC}$, circles $(AA_BA_C), (AFE)$ and $(ABC)$ are concurrent at a point other than $A$. Call it $Q$. Let $O_A$ and $H_A$ be the circumcenter and orthocenter of $AA_BA_C$. If $X'$ is the reflection of $X$ wrt the line passing through $H$ perpendicular to $A_BA_C$, then $XA_BA_C \cong XYZ$. Note that $XYZ \sim ABC$ by the main claim. Thus $X' \in (AA_BA_C)$. It is enough to prove that $H$ is on $O_AH_A$. The following angle chasing uses the facts that the spiral similarity centered at $Q$ takes $AA_BA_C$ to $ABC$ and $AO_AHO$ is a parallelogram: $\angle (O_AH_A,OH) = \angle (A_BA_C,BC) = \angle BQA_B = \angle OQO_A = \angle OAO_A = \angle OHO_A$ which implies $H \in O_AH_A$.

very cool problem!!!! Let $O$ be the circumcenter of $\triangle{ABC}$ and $H'$ the reflection of $H$ over $O$. Also, let $H_A$, $H_B$, $H_C \in (HH')$ such that $HH_A \parallel BC$, $HH_B \parallel AC$, $HH_C \parallel AB$. Then we can see by angle chasing that $\triangle{H_AH_BH_C} \sim \triangle{ABC}$. Claim 1: $H$ is the Anti-Steiner point of $\triangle{H_AH_BH_C}$. (The Anti-Steiner point of a triangle is the intersection of the reflections of the Euler Line across the three sides, and lies on the circumcircle.) Proof: Let $S$ be the Anti-Steiner point of $\triangle{ABC}$. Then by angle chasing, $\angle{OSA} = \angle(OH, BC) = \angle{OHH_A}$ so this is true. Claim 2: The line through $A$ parallel to $H_BH_C$, and the corresponding other two lines for $B$ and $C$, concur at $S$. Proof: Since $ABCS \sim H_AH_BH_CH$, $\angle(AS, BC) = \angle(H_BH_C, HH_A)$, which implies that this is true. Now note that $\ell_a$ and $H_BH_C$ are reflections across the line through the midpoint of $AH$ parallel to $H_BH_C$, and by a $\tfrac{1}{2}$ homothety at $H$, this line passes through the midpoint of $HS$, so $\triangle{H_AH_BH_C}$ and the triangle formed by $\ell_a$, $\ell_b$, $\ell_c$ are reflections across the midpoint of $HS$. Reflecting everything over this midpoint, it suffices to show that $S$ lies on the Euler Line of $\triangle{H_AH_BH_C}$. However, $\triangle{ABC} \sim \triangle{H_AH_BH_C}$ implies that $\angle{HOS} = \angle{HOH_1}$, where $H_1$ is the orthocenter of $\triangle{H_AH_BH_C}$, so $S$, $O$, $H_1$ are collinear. $\square$

Let $O$ be center of $(ABC);$ $AD, BE, CF$ be the altitudes of $\triangle ABC;$ $DE, DF$ intersect $AB, CA$ at $Z, Y;$ $B_c, C_b$ be reflections of $B, C$ in $HC, HB,$ respectively. Since $(B_cB, B_cH) \equiv (BH, BA) \equiv (CA, CH) \pmod \pi,$ we have $A, B_c, H, C$ lie on a circle. Similarly, we have $A, C_b, H, B$ lie on a circle. Consider the inversion $$\mathcal{I}^{k = \overline{AH} \cdot \overline{AD}}_A: H \longleftrightarrow D, B \longleftrightarrow F, C \longleftrightarrow E$$So $$\mathcal{I}^{k = \overline{AH} \cdot \overline{AD}}_A: (AHC) \longleftrightarrow DE, (AHB) \longleftrightarrow DF, B_c \longleftrightarrow Z, C_b \longleftrightarrow Y$$This means $B_c, C_b, Y, Z$ lie on a circle. From this, if we let $S$ be Anti - Steiner point of $HO$ WRT $\triangle ABC,$ $S_a, S_b \in (O)$ such as $AS_a \perp HO, BS_b \perp HO$ then $$(B_cC_b, AB) \equiv (YZ, CA) \equiv (AS_a, CA) \equiv (AB, AS) \pmod \pi$$or $\ell_a \parallel AS$. Similarly, we have $\ell_b \parallel BS, \ell_c \parallel CS$. We define $C_a, A_c, A_b, B_a$ with the same way as $B_c, C_b$. Note that $AH = A_bH = A_cH,$ we have $$(A_bA, A_bA_c) \equiv \dfrac{1}{2} (\overrightarrow{HA}, \overrightarrow{HA_c}) \equiv (HA, HC) \equiv (BC, BA) \equiv - (BA, BC) \pmod \pi$$But $(AA_b, AA_c) \equiv - (AB, AC) \pmod \pi$ then $\triangle ABC \stackrel{-}{\sim} \triangle AA_bA_c$. From the definition of $B_a, C_a,$ we have $\triangle ABC \cong \triangle AB_aC_a$ with the opposite direction, hence $\triangle AB_aC_a \stackrel{+}{\sim} \triangle AA_bA_c$. So there exists a spiral similarity $\mathcal{S}_A: B_aC_a \mapsto A_bA_c$. Then the intersection of $\ell_b, \ell_c,$ which we define $A'$ (similarly, we have $B', C'$), must lies on $(H, HA)$ and $(AB_aC_a)$. Therefore, $$(S_aS, S_aA') \equiv (S_aS, S_aA) + (S_aA, S_aA') \equiv (CS, CA) + (CA, \ell_c) \equiv (CS, \ell_c) \equiv 0 \pmod \pi$$or $S, S_a, A'$ are collinear. Similarly, we have $S, S_b, B'$ are collinear. Hence $$(SA', SB') \equiv (SS_a, SS_b) \equiv (CB, CA) \equiv (\ell_b, \ell_a) \pmod \pi$$or $S \in (A'B'C')$. It's easy to see that $\triangle ABC \stackrel{-}{\sim} \triangle A'B'C'$. Since $$(B'A', B'S) \equiv (\ell_c, B'S_b) \equiv (SC, SS_b) \equiv (SC, SA) + (SA, SS_b) \equiv (BC, BA) + (AB, AS_a) \equiv (BC, BA) + (AS, AC) \equiv (BC, BA) + (BS, BC) \equiv (BS, BA) \equiv - (BA, BS) \pmod \pi$$and $$(SA', SB') \equiv (CB, CA) \equiv (SB, SA) \equiv - (SA, SB) \pmod \pi$$we have $\triangle ABS \stackrel{-}{\sim} \triangle A'B'S$. From this, if we let $O', H'$ be circumcenter and orthocenter of $\triangle A'B'C'$ then $S$ is Anti - Steiner point of $H'O'$ WRT $\triangle A'B'C'$. Suppose that $AS$ intersects $(A'B'C')$ again at $S'_a$. Note that $AS \parallel \ell_a,$ we have $A'S, A'S'_a$ are isogonal conjugate in $\angle{B'A'C'}$ or $A'S'_a \perp O'H'$ and $(AS, AS_a) \equiv (A'S, A'S'_a) \pmod \pi$ or $S'_a \in (H, HA)$. Hence $HO' \perp A'S'_a$ or $H, H', O'$ are collinear

Denote $A_B$ the point obtained by reflecting $A$ across $BH$, and $B_A$ the point obtained by reflecting $B$ across $AH$, where $\ell_c = B_A A_B$. Define $A_C, C_A, B_C, C_B$ analogously. Let $A_1 = A_C C_A \cap B_A A_B$, define $B_1,C_1$ similarly. $\boxed{\text{Claim :} \triangle A_1 B_1 C_1 \text{ is similar to the triangle with the side length} B_A A_B, B_C C_B, A_C C_A}$ We will show that $B_A A_B, B_C C_B, A_C C_A$ form a triangle. Consider the Vector $$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{0}$$$$\overrightarrow{A_B C} = \overrightarrow{AC}( 1 - \frac{2\cos a \sin c}{\sin b})$$$$ \overrightarrow{C B_A} = \overrightarrow{CB}( 1 - \frac{2\cos b \sin c}{\sin a})$$$$\overrightarrow{A_B B_A} = \overrightarrow{AC}(\frac{\sin(c-a)}{\sin b}) + \overrightarrow{CB}(\frac{\sin(c-b)}{\sin a})$$$$\overrightarrow{A_B B_A} + \overrightarrow{B_C C_B} + \overrightarrow{C_A A_C} = \overrightarrow{0}$$So, $B_A A_B, B_C C_B, A_C C_A$ form a triangle. Next we will show that $$\frac{B_A A_B}{AB} = \frac{B_C C_B}{BC} = \frac{A_C C_A}{AC}$$Let $AB = \sin c, BC = \sin a, AC = \sin b$ By cosine law : $$(\frac{B_A A_B}{\sin c})^2 = 3 - 2\cos 2a - 2\cos 2b - 2\cos 2c$$Thus, $$\frac{B_A A_B}{AB} = \frac{B_C C_B}{BC} = \frac{A_C C_A}{AC} = \sqrt{3 - 2\cos 2a - 2\cos 2b - 2\cos 2c}$$Then, we have $\triangle ABC \sim \triangle A_1B_1C_1$ Since, the two triangle are similar, $$\angle A_1B_1C_1 = \angle B_C B_1 B_A$$So, $B B_1 B_C B_A$ are cyclic, with the center $H$. Thus, $$ BH = B_1H$$. Suppose the orthocenter and circumcenter of $A_1B_1C_1$ are $E$ and $D$, respectively. Let the angle bisector of $\angle HB_1E$ intersect $HE$ at $X$ ans $X_1$ be the point such that $$ (H, E ; X, X_1) = -1$$Consider the ratio $$ \frac{AH}{A_1E} = \frac{A_1H}{A_1E} = \frac{BH}{B_1E} = \frac{B_1H}{B_1E} = \frac{CH}{C_1E} = \frac{C_1H}{C_1E}$$We get that $$\angle XA_1X_1 = \angle XB_1X_1 = \angle XC_1X_1 = 90^{\circ}$$Thus, $XX_1$ is a diameter of the circumcircle of $A_1B_1C_1$. So, we can conclude that $H,E,D$ collinear, as desired.

Hopefully didn't make a typo WLOG let $AC>AB>BC$. Now we can use normal lenghts and angles. Let $A_C$ and $A_B$ be the reflections of $A$ with respect to $BH$ and $CH$, respectively. Define $B_A$, $B_C$, $C_A$ and $C_B$ similarly. Hence, $\ell_a = B_AC_A$, $\ell_b = A_BC_B$, $\ell_c = A_CB_C$. Let $\ell_b \cap \ell_c = \{A_1\}$. Define $B_1, C_1$ similarly. Claim: $\triangle{ABC} \sim \triangle A_1B_1C_1$. Proof: See second figure. We can bring $\triangle{C_AAB_A}, \triangle{A_BBC_B}, \triangle{A_CCB_C}$ together like that figure without rotating any of them thanks to the lenght equalities and angles among these triangles. Now just consider this quadrilateral $XYZT$. If we prove it is cyclic, $\angle A , \angle B , \angle C$ appears among the lines $\ell_a, \ell_b, \ell_c$. so we will be done. Just consider $XYZT$. We will prove it is cyclic via complex numbers. Now let $\triangle{ABC}$ be inscribed in unit circle. Hence: $b_a = a+c-\frac{ab}{c}$, $b_c = a+c-\frac{bc}{a}$, $c_a = a+b-\frac{ac}{b}$. Assume that $y = 0$, thus $x = c - b_c$, $z = c_a - a$ and $t = b_a - a$. $\implies x = \frac{a^2-bc}{a} , z = \frac{b^2-ac}{b} , t = \frac{c^2-ab}{c}$. $$XYZT \text{ is cyclic} \iff \frac{z}{t} \div \frac{z-x}{t-x} \in \mathbb{R}$$. $$\iff \frac{\frac{b^2-ab}{b}}{\frac{c^2-ab}{c}} \div \frac{\frac{(ab+ac+bc)(b-a)}{ab}}{\frac{(ab+ac+bc)(c-a)}{ac}} \in \mathbb{R}$$$$\iff \frac{b^2-ac}{c^2-ab}.\frac{c-a}{b-a} \in \mathbb{R}$$One can easily check that this is true. $\square$ Now by this claim, $\angle A_BAA_C = \angle A_BA_1A_C$ hence $(AA_BA_CA_1)$. And it is clear that $H$ is the center of $(AA_BA_CA_1)$. Let $H_1$ be the orthocenter of the $\triangle{A_1B_1C_1}$ Hence, $HA = HA_1$. Similarly, $HB = HB_1$, $HC = HC_1$. Also by similarity, $\frac{H_1A_1}{HA} = \frac{H_1B_1}{HB} =\frac{H_1C_1}{HC}$ hence $\frac{H_1A_1}{HA_1} = \frac{H_1B_1}{HB_1} =\frac{H_1C_1}{HC_1}$. Thus, $(A_1B_1C_1)$ is an apollonian circle of $[HH_1]$. This implies the line $HH_1$ is diameter of $(A_1B_1C_1)$. $\blacksquare$

We will use directed angles $\pmod{180^\circ}$. Furthermore, fix a line $\ell_0$ and let $\angle\ell=\angle\left(\ell_0,\ell\right)$. For $P,Q\in\{A,B,C\}$ with $P\ne{}Q$, let $P_Q$ denote the reflection of $P$ over $\overline{QH}$ and let $D=\overline{C_AA_C}\cap\overline{A_BB_A},E=\overline{A_BB_A}\cap\overline{B_CC_B},$ and $F=\overline{B_CC_B}\cap\overline{C_AA_C}$. Claim $1$. We have that $\angle{}B_ADC_A=\angle{}CAB$. Proof. Let $B'_A$ be the reflection of $B_A$ over $\overline{BH}$ and let $C'_A$ be the reflection of $C_A$ over $\overline{CH}$. It suffices that $\angle{}B'_AAC'_A=\angle{}CAB$ with $\angle{}BHC=\angle{}CAB$ and \begin{align*} \angle{}B'_AAC'_A&=\angle\overline{C'_AA}-\angle\overline{B'_AA}\\ &=\left(2\angle\overline{CH}-\angle\overline{C_AA_C}\right)-\left(2\angle\overline{BH}-\angle\overline{B_AA_B}\right)\\ &=\angle\overline{B_AA_B}-\angle\overline{C_AA_C}+2\left(\angle\overline{CH}-\angle\overline{BH}\right)\\ &=-\angle{}B_ADC_A+2\angle{}BHC. \end{align*}Note that \begin{align*} \angle{}BHB'_A&=\angle{}B_AHB\\ &=2\angle{}AHB\\ &=2\angle{}BCA \end{align*}and similarly $\angle{}CHC'_A=2\angle{}CBA$. We claim that applying a $2\angle{}BCA$ rotation around $H$, then a spiral similarity centered at $A$ oppositely oriented from the one sending $B$ to $C$, then a $2\angle{}ABC$ rotation around $H$, then a spiral similarity centered at $A$ centered $C$ to $B$ is the identity, which would prove the claim by applying this to $B$. It suffices to check $2$ cases. When this is applied to $H$ it is true and when it is applied to the point $P$ for which $\triangle{}APH\sim\triangle{}ABC$ positively oriented it is also true, proving the claim. Claim $1$ implies that $\triangle{}DEF\sim\triangle{}ABC$ negatively oriented. Claim $2$. We have that $AH=HD$. Proof. Note that $\angle{}A_BDA_C=\angle{}A_BAA_C$, so $A,A_B,A_C,$ and $D$ are concyclic. Then, note that $\overline{BH}$ and $\overline{CH}$ are the perpendicular bisectors of $\overline{AA_B}$ and $\overline{AA_C}$, respectively, so $H$ is the circumcenter of $AA_BA_CD$, implying the claim. Claim $3$. We have that $\overline{AD},\overline{BE},$ and $\overline{CF}$ are tangent to $(DEF)$. Proof. Note that \begin{align*} \angle{}ADE&=\angle{}AA_CA_B\\ &=-\angle{}ACB\\ &=\angle{}DFE, \end{align*}so $\overline{AD}$ is tangent to $(DEF)$. Similarly, we see that $\overline{BE}$ and $\overline{CF}$ are tangent to $(DEF)$. Let the tangents to $(DEF)$ at $E$ and $F$ meet at $X$, let the tangents to $(DEF)$ at $F$ and $D$ meet at $Y$, and let the tangents to $(DEF)$ at $D$ and $E$ meet at $Z$. Claim $4$. Let $\triangle{}ABC$ have inscribed triangles $\triangle{}DEF$ and $\triangle{}XYZ$. Then, the Miquel points of $DCEAFB$ and $XCYAZB$ are inverses in $(ABC)$ if and only if $\triangle{}DEF\sim\triangle{}XYZ$ negatively oriented. Proof. Assume that $\triangle{}DEF\sim\triangle{}XYZ$ negatively oriented. Let $P$ and $Q$ be the Miquel points of $DCEAFB$ and $XCYAZB$ and let $R$ and $S$ be their isogonal conjugates in $\triangle{}ABC$, respectively. It suffices that $R$ and $S$ are antigonal conjugates in $(ABC)$. This is true since \begin{align*} \angle{}BRC&=-\angle{}RCB-\angle{}CBR\\ &=\angle{}PCA+\angle{}ABP\\ &=\angle{}PDE+\angle{}FDP\\ &=\angle{}FDE \end{align*}and similarly $\angle{}BSC=\angle{}ZXY$, so $\angle{}BRC=-\angle{}BSC$, and similarly $\angle{}CRA=-\angle{}CSA$ and $\angle{}ARB=-\angle{}ASB$. All steps are reversible, proving the claim. Now, let $M$ be the Miquel point of $AZBXCY$. By claim $4$ we see that $M$ is the inverse of the circumcenter $O$ of $\triangle{}DEF$ in $(XYZ)$, which lies on the Euler line of $\triangle{}DEF$ since $O$ lies on the Euler line of $\triangle{}DEF$. We claim that $H$ is the midpoint of $\overline{OM}$, implying that $H$ lies on the Euler line of $\triangle{}DEF$. It suffices to prove the following restatement of the problem in $\triangle{}XYZ$. Let $\triangle{}ABC$ have intouch triangle $\triangle{}DEF$, circumcenter $O$, and incenter $I$. Let $I'$ be the inverse of $I$ in $(ABC)$ and let $H$ be the midpoint of $\overline{II'}$. Let the circles centered at $H$ through $D,E,$ and $F$ intersect $\overline{BC},\overline{CA},$ and $\overline{AB}$ at $D$ and $X$, at $E$ and $Y$, and at $F$ and $Z$, respectively. Show that $\triangle{}XYZ\sim\triangle{}DEF$ negatively oriented and that $H$ is the orthocenter of $\triangle{}XYZ$. First, note that $\triangle{}XYZ$ is the pedal triangles of $I'$ in $\triangle{}ABC$ since $\overline{ID}\perp\overline{BC},\overline{IE}\perp\overline{CA},$ and $\overline{IF}\perp\overline{AB}$. Therefore, we see that $\triangle{}XYZ\sim\triangle{}DEF$ negatively oriented since the Miquel points of $DCEAFB$ and $XCYAZB$ are $I$ and $I'$, respectively, so it suffices that the orthocenter of $\triangle{}XYZ$ is $H$. This is equivalent to the following problem by a homothety with factor $2$ centered at $I'$. Let $\triangle{}ABC$ have circumcenter $O$ and incenter $I$ and let $I'$ be the inverse of $I$ in $(ABC)$. Let $\triangle{}DEF$ be the reflection triangle of $I'$ in $\triangle{}ABC$. Prove that the orthocenter of $\triangle{}DEF$ is $I$. Let $I_1$ be the antigonal conjugate of $I$ in $\triangle{}ABC$ and let $H$ be the orthocenter of $\triangle{}ABC$. Note that $\overline{AI_1}\perp\overline{EF}$ with $I_1$ and $I'$ are isogonal conjugates in $\triangle{}ABC$. Also, note that if $I_A$ is the reflection of $I$ over $\overline{BC}$, then $I_A$ is the isogonal conjugate of $A$ in $\triangle{}IBC$, so by isogonal center we see that $\overline{I'I_A}$ goes through the circumcenter $M_A$ of $\triangle{}IBC$. Therefore, we see that if the conic through $A,B,C,H,$ and $I$ intersects $(ABC)$ at $A,B,C,$ and $I_2$, then \begin{align*} \angle\overline{DI}&=2\angle\overline{BC}-\angle\overline{I'M_A}\\ &=2\angle\overline{BC}-\angle{}OI'M_A-\angle\overline{OI}\\ &=2\angle\overline{AH}-\angle{}IM_AO-\angle\overline{OI}\\ &=2\angle\overline{AH}-\left(\angle\overline{OM_A}-\angle\overline{AI}\right)-\angle\overline{OI}\\ &=2\angle\overline{AH}-\left(\angle\overline{AH}-\angle\overline{AI}\right)-\angle\overline{OI}\\ &=\angle\overline{AH}+\angle\overline{AI}-\angle\overline{OI}\\ &=\angle\overline{AB}+\angle\overline{AC}-\angle\overline{OI}+\angle\overline{AH}-\angle\overline{AI}\\ &=\angle\overline{AI_2}+\angle\overline{AH}-\angle\overline{AI}\\ &=\angle\overline{AI_1}+\angle\overline{AI}-\angle\overline{AI}\\ &=\angle\overline{AI_1} \end{align*}where $\angle\overline{AI_2}+\angle\overline{AH}=\angle\overline{AI_1}+\angle\overline{AI}$ follows from the fact that $I_2$ and $H$ and $I_1$ and $I$ are antigonal conjugates in $\triangle{}ABC$ on the conic through $A,B,C,H,$ and $I$, so $\overline{DI}\parallel\overline{AI_1}$, so $\overline{DI}\perp\overline{EF}$. Similarly, we see that $\overline{EI}\perp\overline{FD}$ and $\overline{FI}\perp\overline{DE}$, so $I$ is the orthocenter of $\triangle{}DEF$, so we are done.

Not a hard problem by bashing, first ISL G7 anyway

Solved with HoRI_DA_GRe8. S.Ragnork1729 also started with us but kicked himself out of his own meet 2 minutes after we started out. (His data pack got exhausted .) [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (9.15982,17.71344); pair B = (2.70665,-14.29283); pair C = (45.12867,-14.68292); pair X = (-2.10282,27.85747); pair Y = (-27.39869,-14.01600); pair Bp = (5.86979,1.39564); pair Ap = (-0.58337,-30.61063); pair D = (19.20189,-42.85472); pair T = (32.14381,21.33431); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle((24.03305,-1.93901), 24.64614), linewidth(0.6) + linetype("4 4") + red); draw(circle((-12.16657,5.35962), 24.64614), linewidth(0.6) + linetype("4 4") + blue); draw(circle((8.92903,-7.38428), 36.92808), linewidth(0.6)); draw(circle((20.74302,-18.25681), 24.64614), linewidth(0.6) + linetype("4 4") + red); draw(X--D, linewidth(0.6)); draw(D--Y, linewidth(0.6)); draw(Y--B, linewidth(0.6)); draw(X--A, linewidth(0.6)); dot("$A$", A, N); dot("$B$", B, dir(270)); dot("$C$", C, SE); dot("$X$", X, N); dot("$Y$", Y, SW); dot("$B'$", Bp, NW); dot("$A'$", Ap, SW); dot("$D$", D, dir(270)); dot("$T$", T, NE); [/asy][/asy] Let $X$ and $Y$ be the reflection of $C$ over the $B$-altitude and $A$-altitude respectively. Now let $T = \odot(ABC)\cap \odot(XCY)$. Let $A'$ and $B'$ be the reflection of $A$ and $B$ over the $C$-altitude respectively. Also, let $D=XB'\cap YA'$. Note that $\ell_{a}\equiv XB'$ and $\ell_{b}\equiv YA'$. Claim: $XABY$ is cyclic. Proof. Temporarily let $P$ and $Q$ denote the foot of the $A$-altitude and $B$-altitude respectively. By POP, we have that, \[ CP\cdot CB = CQ\cdot CA \implies 2\cdot CP\cdot CB =2\cdot CQ\cdot CA \implies CY\cdot CB = CX \cdot CA \]which gives us that $XABY$ is cyclic. $\blacksquare$ Note that as $C = XA\cap YB$ and $T=\odot(XCY)\cap \odot(ABC)$, $T$ becomes the center of spiral similarity mapping $AB\mapsto XY$. Now note that, \[ \measuredangle A'CB'=\measuredangle BCA =\measuredangle YCA. \]Also note that, \[ \measuredangle CB'A'=\measuredangle ABC =\measuredangle ABY=\measuredangle AXY=\measuredangle CXY .\] These two observations combined give us that $\triangle CB'A' \stackrel{+}{\sim} \triangle CXY$. This means that $C$ is the center of spiral similarity mapping $A'B'\mapsto YX$. This means that $D=XB'\cap YA'$ lies on $\odot(A'B'C)$ and $\odot(XCY)$. Now temporarily let $H$ denote the orthocenter of $\triangle ABC$. Note that $HX=HC=HY$ which gives us that $H$ is the center of $\odot(XCY)$. This means $\odot(XCY)$ remains fixed on reflecting about the line $CH$. Also note that $\odot(ABC)\mapsto \odot(A'B'C)$ on reflecting about $CH$. Thus on this reflection, $D = \odot(XCY)\cap \odot(A'B'C)\mapsto \odot(XCY)\cap \odot(ABC)=T$. Thus we also have that $D$ is the reflection of $T$ over $CH$. Now we start our complecks bash. Let $a$, $b$ and $c$ denote the affixes of $A$, $B$ and $C$ in the complex plane. We first find the affix for $Y$ by reflecting $C$ over $AH$. To do this, we have, \begin{align*} y &= \frac{((a+b+c)-a)\overline{c}+\overline{(a+b+c)}a- (a+b+c)\overline{a}}{\overline{a+b+c}-\overline{a}} \\ &= \frac{\frac{b+c}{c}+\left( \frac{1}{a}+\frac{1}{b} +\frac{1}{c}\right)a - \frac{a+b+c}{a}}{\frac{1}{a} +\frac{1}{b}+\frac{1}{c}-\frac{1}{a}} \\ &= b+a - \frac{bc}{a} .\end{align*} Similarly we also get that $x = a+b-\frac{ac}{b}$. Now using the formula for spiral similarity, we find the affix of $T$ which is the center of the spiral similarity mapping $AB\mapsto XY$. We have, \begin{align*} t&=\frac{a\left( b+a-\frac{bc}{a} \right)-b\left( a+b-\frac{ac}{b}\right)}{a+\left( b+a-\frac{bc}{a} \right) -b - \left( a+b-\frac{ac}{b} \right) } \\ &= \frac{(a+b)(a-b)+c(a-b)}{(a-b)+\frac{c(a-b)(a+b)}{ab}} \\ &= \frac{(a+b+c)ab}{ab+bc+ca} .\end{align*} Now we find the affix of $D$ by reflecting $T$ over $CH$. So, we have, \begin{align*} d &= \frac{(a+b)\overline{\left( \frac{(a+b+c)ab}{ ab+bc+ca } \right) +\overline{(a+b+c)}c}-(a+b+c)\overline{c}}{ \overline{(a+b+c)-c}} \\ &= \frac{(a+b)\left( \frac{ab+bc+ca}{(a+b+c)ab} \right) c\left(\frac{a+b}{ab}\right)-\frac{a+b}{c}}{\frac{a+b}{ab}} \\ &= \frac{ab+bc+ca}{a+b+c}+c-\frac{ab}{c}. \end{align*} Similarly we also get the vertices of the other vertices of the triangle formed by $\ell_{a}$, $\ell_{b}$ and $\ell_{c}$. The affixes of their vertices are, \[ \frac{ab+bc+ca}{a+b+c}+c-\frac{ab}{c},\qquad \frac{ab+bc+ca}{a+b+c} + a - \frac{bc}{a}\qquad, \frac{ab+bc+ca}{a+b+c}+b-\frac{ca}{b}. \] Now note that we just need to show that the Euler line of this triangle passes through the orthocenter of $\triangle ABC$. So we show that the line joining the centroid of this triangle and its circumcenter passes through orthocenter of $\triangle ABC$. Let $O$ and $G$ denote the circumcenter and centroid of this triangle respectively. The affix of $G$ is given by, \[ g=\frac{ab+bc+ca}{a+b+c}+\frac{a+b+c}{3}-\frac{\frac{ab}{c} +\frac{bc}{a}+\frac{ca}{b}}{3} .\] Now to find the circumcenter of the triangle, we shift one of the vertices to the origin and then find the affix of the shifted triangle and then shift the center back to its original position. We do this by shifting $\frac{ab+bc+ca}{a+b+c}+c-\frac{ab}{c} \mapsto 0$. Thus after shifting, we get \[ \frac{ab+bc+ca}{a+b+c}+a-\frac{bc}{a}\mapsto a-c+\frac{ab}{c} -\frac{bc}{a},\qquad \frac{ab+bc+ca}{a+b+c}+b-\frac{ca}{b} \mapsto b-c+\frac{ab}{c}-\frac{ca}{b}. \] We find the center using the formula which states that the circumcenter of the triangle formed by the complex numbers $x$, $y$ and $0$ is $\frac{xy(\overline{x}-\overline{y})} {\overline{x}y-\overline{x}y}$. Thus we have, \begin{align*} o &= \frac{(a-c)\left( \frac{b}{c}+\frac{b}{a}+1 \right)(b-c) \left( \frac{a}{c}+\frac{a}{b}+1 \right) \left( \left( \frac{1}{a}-\frac{1}{c}\right) \left( \frac{c}{b} +\frac{a}{b}+1\right) -\left( \frac{1}{b}-\frac{1}{c} \right) \left( \frac{c}{a}+\frac{b}{a}+1 \right) \right)}{ \left( \frac{1}{a}-\frac{1}{c} \right) \left( \frac{c}{b} +\frac{a}{b}+1\right) (b-c)\left( \frac{a}{c}+1+\frac{a}{b} \right)-(a-c)\left( \frac{b}{c}+\frac{b}{a}+1 \right) (\frac{1}{b}-\frac{1}{c})\left( \frac{c}{a}+\frac{b}{a}+1 \right) } \\ &=\frac{\left( \frac{(a-c)(b-c)(ab+bc+ca)^2(a+b+c)}{abc^2} \right)\cdot \left( \frac{c-a}{abc}-\frac{c-b}{abc} \right)}{\frac{(c-a)(a+b+c)}{abc}\cdot \frac{(b-c)(ab+bc+ca)}{bc} -\frac{(a-c)(ab+bc+ca)}{ca}}\cdot \frac{(c-b)(a+b+c)}{abc} \\ &=\frac{\frac{(a-b)(b-c)(c-a)(ab+bc+ca)^2(a+b+c)} {a^2b^2c^3}}{\frac{(ab+bc+ca)(a+b+c)(a-c)(c-b)}{abc^2} \left( \frac{1}{b}-\frac{1}{a} \right)} \\ &= \frac{\frac{(a-b)(b-c)(c-a)(ab+bc+ca)^2(a+b+c)} {a^2b^2c^3}}{\frac{(ab+bc+ca)(a+b+c)(a-b)(b-c)(c-a)} {a^2b^2c^2}} \\ &=\frac{ab+bc+ca}{c}\\ &= \frac{ab}{c}+b+a .\end{align*} Thus now shifting this center to its original position, we get that the original affix was, \[ \frac{ab}{c}+b+a\mapsto a+b+c+\frac{ab+bc+ca}{a+b+c} .\]Thus to prove that $O$, $G$ and $H$ are collinear, we need to show that, \[ a+b+c+\frac{ab+bc+ca}{a+b+c},\qquad a+b+c,\qquad \frac{ab +bc+ca}{a+b+c}+\frac{a+b+c}{3}-\frac{\frac{ab}{c} +\frac{bc}{a}+\frac{ca}{b}}{3} \]are collinear. To show this, we need to check that the following quantity is real. \begin{align*} \frac{a+b+c+\frac{ab+bc+ca}{a+b+c}-(a+b+c)} {a+b+c+\frac{ab+bc+ca}{a+b+c}-\frac{ab +bc+ca}{a+b+c}+\frac{a+b+c}{3}-\frac{\frac{ab}{c} +\frac{bc}{a}+\frac{ca}{b}}{3}} &= \frac{\frac{ab+bc+ca}{a+b+c}}{\frac{2(a+b+c)}{3} +\frac{\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}{3}}\\ &= \frac{3(ab+bc+ca)}{(a+b+c)\left( 2(a+b+c) +\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right) }\\ &= \frac{3(ab+bc+ca)(abc)}{(a+b+c)(2(a+b+c)abc +a^2b^2+b^2c^2+c^2a^2)}\\ &= \frac{3(ab+bc+ca)(abc)}{(a+b+c)(ab+bc+ca)^2}\\ &= \frac{3abc}{(a+b+c)(ab+bc+ca)}\\ &= \frac{3}{(a+b+c)\left( \frac{1}{a}+ \frac{1}{b}+\frac{1}{c}\right) } \end{align*}which is clearly real and we are done!

Let $C_B$ and $C_A$ be the reflections of $B$ and $A$ across $HC$ and define the other points similarly. Let $\ell_b$ meet $\ell_c$ at $A'$ and define $B'$ and $C'$ similarly and let $H'$ and $O'$ be the orthocenter and circumcenter of $\mathcal T$. Notice that $\measuredangle CC_AA=\measuredangle CB_AB=\measuredangle BAC$ so $C_ABHB_AC$ is cyclic. Similarly, $B_CAHBA_C$ and $CA_BHC_BA$ are cyclic. Claim: $\triangle ABC\sim \triangle A'B'C'$. Proof: Note that \[\measuredangle CC_AB_A=\measuredangle ACB=\measuredangle A_BA_CA\]And clearly $\measuredangle A_CAA_B=\measuredangle C_AAB_A$ so \[\triangle A_CAA_B\sim \triangle C_AAB_A\implies\frac{AA_C}{AC_A}=\frac{AA_B}{AB_A}\]This, together with $\measuredangle A_CAC_A=\measuredangle A_BAB_A$ gives $\triangle A_CAC_A\sim \triangle A_BAB_A$. Therefore, $\measuredangle A'C_AA=\measuredangle A'B_AA$ so $A'C_AAB_A$ is cyclic. In conclusion, $\measuredangle B'A'C'=\measuredangle BAC$ which implies the desired claim by symmetry. $\square$ We also have $AH=HC_A=HB_A$ so $H$ is the circumcenter of $\triangle AC_AB_A$. This implies that $AH=HA'$ and similarly $BH=HB'$, $CH=HC'$. Using the above claim yields \[\frac{A’H}{A'H'}=\frac{B’H}{B'H'}=\frac{C’H}{C'H'}\] We employ complex numbers. Let $A'=x$, $B'=y$, $C=z$ and $\vert x\vert=\vert y\vert=\vert z\vert=1$. We have $H'=x+y+z$ and \[k=\frac{\vert h-x\vert }{\vert y+z\vert}=\frac{\vert h-y\vert}{\vert x+z\vert}=\frac{\vert h-z\vert}{\vert x+y\vert}\]Where $H=h$ and $k$ is a real number. Squaring gives \[h\overline{h}-\frac{h}{x}-\overline{h}x+1=k^2\left(2+\frac{y}{z}+\frac{z}{y}\right)\]\[h\overline{h}-\frac{h}{y}-\overline{h}y+1=k^2\left(2+\frac{z}{x}+\frac{x}{z}\right)\]\[h\overline{h}-\frac{h}{z}-\overline{h}z+1=k^2\left(2+\frac{x}{y}+\frac{y}{x}\right)\]Subtracting the first two gives \[h\frac{x-y}{xy}+\overline{h}(y-x)=k^2\left(z\frac{x-y}{xy}+\frac{1}{z}(y-x)\right)\]\[\iff \frac{h}{xy}-\overline{h}=k^2\left(\frac{z}{xy}-\frac{1}{z}\right)\]From symmetry, we also have\[\frac{h}{yz}-\overline{h}=k^2\left(\frac{x}{yz}-\frac{1}{x}\right)\]Subtracting these yields \[\frac{h}{y}\cdot\frac{z-x}{zx}=k^2\left(\frac{1}{y}\cdot\frac{(z-x)(z+x)}{zx}+\frac{z-x}{zx}\right)\]\[\iff \frac{h}{xyz}=k^2\left(\frac{z+x}{xyz}+\frac{1}{zx}\right)\]\[\iff h=k^2(x+y+z)\]And since $k^2$ is real, we get that $H$ lies on $O'H'$. $\blacksquare$

Let $P_Q$ be the reflection of $P$ over $QH$ for $P \neq Q \in \{A, B, C\}$, let $X = \ell_b \cap \ell_c$, $Y = \ell_c \cap \ell_a$, $Z = \ell_a \cap \ell_b$, and let $H_{\mathcal T}$ be the orthocentre of triangle $\mathcal T$. Claim: $\triangle XYZ \stackrel{-}{\sim} \triangle ABC$. Proof. Complex bash. $\blacksquare$ Now $\measuredangle A_BXA_C = \measuredangle(\ell_c, \ell_b) = \measuredangle CAB = \measuredangle A_BAA_C$, so it follows that $AA_BXA_C$ is cyclic. In particular, the circumcentre of $\triangle AA_BA_C$ is $H$, so $AH = XH$; similarly, $BH = YH$ and $CH = ZH$. Therefore \[XH : YH : ZH = AH : BH : CH = XH_{\mathcal T} : YH_{\mathcal T} : ZH_{\mathcal T}\]which implies that $H$ and $H_{\mathcal T}$ are inverses (!) in $(XYZ)$.