Let the reflection of $H_2 H_3$ in $T_2 T_3$ be $l_1$ etc. Let $M$, $N$ be the intersections of $BI$ and $CI$ respectively with $T_{2}T_{3}$. Let $M^\prime$ be the point on $H_{2}H_{3}$ such that $M^\prime \hat{M} H_2 = T_2 \hat{M} H_2$. Let $T_{2}^{\prime}$ be the reflection of $T_2$ in $BI$. Clearly, this lies on the incircle. $T_2 \hat{M} I = A\hat{T} _3 T_2 - A\hat{B}M = 90-\frac{A}{2}-\frac{B}{2}=\frac{C}{2}=T_2 \hat{C} I$ $\therefore IT_2 MC$ is cyclic. Since $I \hat{T_2}C=90$, $BMC = 90$. So $BH_3 N H_2 M C$ are concyclic. $\therefore M^\prime \hat{H_2} M = H_3 \hat{B} M = \frac{B}{2} = M \hat{B} C = M \hat{H_2} T_2$ $\therefore MM^\prime H_2 \equiv alent MT_2 H_2$ $\therefore MM^\prime = MT_2$ Also, $T_2 ^\prime \hat{M} T_2 = 2B\hat{M}N=C=2N\hat{M}H_2 = T_2 \hat{M} M^\prime$ This means that $T_2 ^\prime$ is the reflection of $M^\prime$ in $T_2 T_3$, so that $T_2$ is the intersection of $l_1$ with the incircle. Similarly, $T_2$ is the intersection of $l_3$ with the incircle. This implies the result.

Here's a different, probably less tricky approach. Let $A'B'C'$ be the image of $ABC$ under the negative homothety taking the circumcircle to the incircle. It suffices to show that $A'B'C'$ reflects across $T_{1}T_{2}, T_{2}T_{3}, T_{3}T_{1}$ to the orthic triangle. First, note that $B'C'$ is parallel to $BC$, and $AT_{3}= AT_{2}$, so the reflection of $B'C'$ across $T_{2}T_{3}$ is antiparallel to $BC$. Using analogous facts for the other sides, the three reflections are some homothety of the orthic triangle; we need only show that this homothety is identity. Now let $l_{B}$ and $l_{C}$ be respective reflections of $A'C'$, $A'B'$ across $T_{1}T_{3}$, $T_{1}T_{2}$. Note that $T_{1}$ is equidistant from $A'B'$ and $A'C'$ since it is the midpoint of the arc $B'C'$. Then it must also be equidistant from $l_{B}$ and $l_{C}$. Furthermore, since $l_{B}$ is antiparallel to $AC$, the angle $(l_{B}, BC)$ is $A$. Similarly, $(l_{C}, BC) =-A =-(l_{B}, BC)$. Hence $BC$ is parallel to an angle bisector of $l_{B}, l_{C}$, and at the same time $T_{1}$ lies on one of the angle bisectors of $l_{B}, l_{C}$. We quickly verify that $l_{B}$ and $l_{C}$ cut $BC$ on the same side of $T_{1}$, so it follows that $BC$ is an angle bisector of $l_{B}, l_{C}$, and so it contains $l_{B}\cap l_{C}$. Using the corresponding facts for the other two pairs of lines, we conclude that our triangle is a homothety of the orthic triangle which keeps its vertices on the same sides of $ABC$. It then follows that the homothety is centered around $A$, $B$, and $C$ at the same time, so it must be identity, finishing the proof.

Let $A'B'C'$ be the triangle such that $A'B' \| AB$, $B'C' \| BC$, $C'A' \| CA$ and $A'$, $B'$ and $C'$ lie on the incircle of $ABC$. We claim that $A'B'C'$ is the triangle that results from the described reflections. To show this result we first will prove the following lemma. Lemma: In an acute-angled triangle $ABC$ with circumcircle $\Gamma$, $D$, $E$ and $F$ are the midpoints of major arcs $\widehat{BC}$, $\widehat{AC}$ and $\widehat{AB}$, respectively. If $\omega$ is the tangent to $\Gamma$ at $A$, $\ell_C$ is the reflection of $AB$ across $DE$ and $\ell_B$ is the reflection of $AC$ across $DF$, then $\omega$, $\ell_B$ and $\ell_C$ are concurrent. Proof: Denote $\angle{A}=a$, $\angle{B}=b$ and $\angle{C}=c$. Let $P$ and $Q$ be the intersections of the perpendicular bisector of $AD$ with $AB$ and $AC$, respectively, and let $\ell_B$ and $\ell_C$ intersect at $R$. Now consider the reflection $P'$ of $P$ across $DE$. Since $E$ is the midpoint of major arc $\widehat{AC}$, $\angle{ADE}=\angle{ACE}=90^\circ - b/2$. Further, since $D$ is the midpoint of major arc $\widehat{BC}$, it follows that $AD$ is the external bisector of $\angle{A}$. This combined with the fact that $PA=PD$ implies that $\angle{PDA}=\angle{PAD}=90^\circ - a/2$. Therefore if $M$ is the intersection of $DE$ with $PP'$, $\angle{PDM}=180^\circ - \angle{PDA}-\angle{ADE}=90^\circ - c/2$. Since $P'$ is the reflection of $P$ across line $EDM$, $DM$ is the bisector of $\angle{PDP'}$ and hence $\angle{PDP'}=2\angle{PDM}=180^\circ - c$. By the same argument, if $Q'$ is the reflection of $Q$ across $DF$, it follows that $\angle{QDQ'}=180^\circ - b$. Since $P$ and $Q$ lie on the perpendicular bisector of $AD$ and $AD$ is the external bisector of $\angle{A}$, it follows that $\angle{PDQ}=180^\circ - a$ and that $PD=QD$. Now since $P'D=PD=QD=Q'D$ and $\angle{PDQ}+\angle{P'DP}+\angle{Q'DQ}=180^\circ - a + 180^\circ - b + 180^\circ - c = 360^\circ$, it follows that $P' = Q'$. Further, since $P$ and $Q$ lie on lines $AB$ and $AC$, respectively, $P'$ and $Q'$ lie on $\ell_C$ and $\ell_B$, respectively. Hence $P' = Q' = R$. As previously established, $\angle{MDR}=90^\circ - c/2=\angle{DFE}$ which implies that $DR$ is tangent to $\Gamma$ by tangent-angle theorem. Hence $\omega$, $\ell_B$ and $\ell_C$ concur at $R$. $\blacksquare$ Now note that since $B'C' \| BC$ and $BC$ is tangent to the circumcircle of $A'B'C'$ at $T_1$, it follows that $T_1$ is the midpoint of major arc $\widehat{B'C'}$. By the same argument $T_2$ and $T_3$ are the midpoints of major arcs $\widehat{A'C'}$ and $\widehat{A'B'}$, respectively. Let $k_A$, $k_B$ and $k_C$ denote the reflections of $B'C'$, $A'C'$ and $B'C'$ across $T_2 T_3$, $T_1 T_3$ and $T_2 T_3$, respectively. By the lemma, the pairwise intersections of $k_A$, $k_B$ and $k_C$ lie on $BC$, $AC$ and $AB$. Let these points be $X$, $Y$ and $Z$, respectively. Since $T_2 T_3$, $T_1 T_3$ and $T_1 T_2$ are perpendicular to the bisectors of $\angle{A}$, $\angle{B}$ and $\angle{C}$, $XY$, $XZ$ and $YZ$ are antiparallel to $AB$, $AC$ and $BC$ with respect to the bisectors of triangle $ABC$. This implies that $\angle{BXC}=\angle{BYC}$ and hence that $\angle{AZC}=\angle{AXC}=\angle{AYB}=\angle{BZC}$. Hence $\angle{BZC}=\angle{AZC}=90^\circ$ which implies that $Z = H_1$. By the same argument, $Y=H_2$ and $X=H_3$. This completes the proof of the claim.

SnowEverywhere wrote: Now note that since $B'C' \| BC$ and $BC$ is tangent to the circumcircle of $A'B'C'$ at $T_1$, I think this part is wrong.. $T_1$ may not be on the circumcircle of $A'B'C'$ at this step. But I think there is another way to solve this problem by that lemma...

Dear Mathlinkers, only a link http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=380345. Sincerely Jean-Louis

cadiTM wrote: SnowEverywhere wrote: Now note that since $B'C' \| BC$ and $BC$ is tangent to the circumcircle of $A'B'C'$ at $T_1$, I think this part is wrong.. $T_1$ may not be on the circumcircle of $A'B'C'$ at this step. But I think there is another way to solve this problem by that lemma... Sorry if I was unclear. I defined $A'B'C'$ to be the triangle inscribed in the incircle with sides parallel to those of $ABC$. By this definition, $T_1$ lies on the incircle and therefore the circumcircle of $A'B'C'$. I think it is right and have tried to make the definition clearer in the beginning.

Let $D,E,F$ be the points where the bisectors of $A$, $B$, and $C$ meet $BC$, $CA$, and $AB$, respectively. Let $U_1, U_2, U_3$ be the reflections of $T_1, T_2, T_3$ across $AD$, $BE$, and $CF$, respectively. We claim that the images of the reflections of $H_1H_2, H_2, H_3H_1$ across $T_1 T_2, T_2 T_3, T_3 T_1$ meet at $U_1, U_2, U_3$. To prove this claim, it suffices to show that the reflection of $U_1$ across $T_1 T_2$ lies on $H_1H_2$. Denote by $d(\ell, X)$ the distance from $X$ to $\ell$, for any line $\ell$ and point $X$. Let $M$ be the midpoint of $T_1 T_2$. Let $H_1', H_2', U_1'$ be the images of the reflections of $H_1,H_2,U_1$ across $MC$. Because $MC \perp T_1 T_2$, $T_1 T_2$ is fixed by this reflection. Thus, it suffices to show that the reflection of $U_1'$ over $T_1 T_2$ lies on $H_1' H_2'$. Since $MC \perp T_1 T_2$ and $M$ is the intersection of $MC$ and $T_1T_2$, the reflection of $U_1$ across $MC$, reflected across $T_1 T_2$, is simply the reflection of $U_1$ across $M$. Therefore, we need only show that the reflection of $U_1$ across $M$, which we shall call $U_1''$, lies on $H_1'H_2'$. Since $H_1H_2$ is antiparallel to $AB$ with respect to $\angle C$, $H_1'H_2'$ is parallel to $AB$, so it is enough to show that $d(AB, U_1'') = d(AB, H_1'H_2')$. Let $h_c = d(AB, C)$. We have $d(AB, H_1'H_2') = h_c - d(C, H_1'H_2') = h_c - d(C, H_1H_2) = h_c(1-\cos C)$, since $\triangle CH_1H_2$ and $\triangle CAB$ are similar with factor of similarity $\frac{H_2C}{CB} = \cos C$. Since $U_1''$ is the reflection of $U_1$ across $M$, $d(AB, U_1'') = 2d(AB, M) - d(AB, U_1)$. We have $2d(AB, M) = d(AB,T_1) + d(AB, T_2) = (s-b) \sin B + (s-a) \sin A$. To compute $d(AB, U_1)$, we extend $DU_1$ to meet $AB$ at $C'$ and $AC$ at $B'$. $\triangle AB'C'$ is the reflection of $\triangle ABC$ across the bisector of $\angle A$, so $d(AB, U_1) = d(AC', U_1) = d(AC, T_1) = (s-c) \sin C$. Thus, if we denote by $K$ the area of $\triangle ABC$, we have \begin{align*} d(AB, U_1'') &= 2d(AB, M) - d(AB, U_1) = (s-b) \sin B + (s-a) \sin A - (s-c) \sin C = (s-b) \frac{2K}{ac} + (s-a) \frac{2K}{bc} - (s-c) \frac{2K}{ab}\\ &= \frac{K}{abc} (b(a+c-b) + a(b+c-a) - c(a+b-c)) = \frac{2K}{c} \cdot \frac{2ab - a^2 - b^2 + c^2}{2ab} = h_c (1 - \cos C) = d(AB, H_1'H_2'), \end{align*} as desired.

Dear Mathlinkers, you can see also http://perso.orange.fr/jl.ayme vol. 9 Le triangle réféchi Sincerely Jean-Louis

There is also a simple solution using complex numbers. WLOG let $ \omega $, the incircle of $ \triangle{ABC}, $ be the unit circle and let the complex coordinates of $ A, B, C, T_1, T_2, T_3, H_1, H_2, H_3 $ be denoted by $ a, b, c, t_1, t_2, t_3, h_1, h_2, h_3 $ respectively. Now, it is well-known that $ a = \frac{2t_{2}t_{3}}{t_2 + t_3}, b = \frac{2t_{1}t_{3}}{t_1 + t_3}, c = \frac{2t_{1}t_{2}}{t_1 + t_2} $. Since $ H_2 $ is the projection of $ B $ onto "chord" $ T_{2}T_{2} $ of $ \omega $ we have that $ h_2 = \frac{1}{2}\left(b + 2t_2 - t_{2}^2\overline{b}\right) = \frac{t_{1}t_{2} + t_{2}t_{3} + t_{1}t_{3} - t_{2}^2}{t_1 + t_3} $. Now, let $ P_2 $ be the reflection of $ H_2 $ over $ T_{2}T_{3} $ and denote its complex coordinate by $ p_2 $. Then $ p_2 = t_2 + t_2 - t_{2}t_{3}\overline{h_2} = \frac{t_1(t_{2}^2 + t_{3}^2)}{t_2(t_1 + t_3)} $. Letting $ P_3 $ be the reflection of $ H_3 $ over $ T_{2}T_{3} $ and denoting its complex coordinate by $ p_3 $ we similarly obtain that $ p_3 = \frac{t_1(t_{2}^2 + t_{3}^2)}{t_3(t_1 + t_2)} $. Now let $ Z $ be an intersection of line $ P_{2}P_{3} $ with $ \omega $. Let its complex coordinate be $ z $. It is clear that $ z $ satisfies the following equation: $ \frac{z - p_2}{\overline{z} - \overline{p_2}} = \frac{p_2 - p_3}{\overline{p_2} - \overline{p_3}} $. Now we can compute $ p_2 - p_3 = \frac{t_{1}^2(t_3 - t_2)(t_{2}^2 + t_{3}^2)}{t_{2}t_{3}(t_1 + t_3)(t_1 + t_2)} $ and that $ \overline{p_2} - \overline{p_3} = \frac{(t_2 - t_3)(t_{2}^2 + t_{3}^2)}{t_{2}t_{3}(t_1 + t_3)(t_1 + t_2)} $ so $ \frac{p_2 - p_3}{\overline{p_2} - \overline{p_3}} = -t_{1}^2 $. Moreover, since $ Z \in \omega $ we have that $ \overline{z} = \frac{1}{z} $. Therefore we have that $ \frac{z - p_2}{\frac{1}{z} - \overline{p_2}} = -t_{1}^2 \Longrightarrow z^2 - (p_2 + t_{1}^2\overline{p_2})z + t_{1}^2 = 0 $. We can compute that $ p_2 + t_{1}^2\overline{p_2} = \frac{t_1(t_{2}^2 + t_{3}^2)}{t_{2}t_{3}} $. By a quick application of the quadratic formula we find that the two possibilities for $ z $ are $ \frac{t_{1}t_{2}}{t_3} $ and $ \frac{t_{1}t_{3}}{t_2} $. This immediately implies that the vertices of the triangle formed by the lines $ P_{1}P_{2}, P_{2}P_{3}, P_{3}P_{1} $ have complex coordinates $ \frac{t_{1}t_{2}}{t_3}, \frac{t_{2}t_{3}}{t_1}, \frac{t_{3}t_{1}}{t_2} $ all of which clearly lie on $ \omega $ as desired. The motivation for using complex numbers is clear since, after taking the incircle to be the unit circle, all relevant points are projections or reflections with respect to chords of the circle, which are "nice" in complex coordinates.

Let $X_1$ be on the incircle such that $T_1X_1\parallel T_2T_3$, and define $X_2,X_3$ similarly. We claim that $X_1X_2X_3$ is the triangle formed. Note that \[ \angle X_3T_1B=\angle T_3T_1T_2=\angle X_2T_1C \] so the sides are indeed parallel to the sides of $ABC$. Let $T_1T_3$ meet $X_1T_2$ at $D$. By symmetry, $D$ lies on the angle bisector of $\angle BAC$, which is the perpendicular bisector of $T_3T_2$ and $T_1X_1$. So by a well known lemma, $D$ lies on the circle with diameter $AC$. Let $E$ be the reflection of $X_1$ over $T_1T_2$. It suffices to show that $E$ lies on $H_1H_3$. \[ \angle DT_3T_2=\angle DT_2T_3=\angle CT_1T_2=\angle CT_2T_1 \] so \[ \angle H_1DA=\angle H_1CA=\angle T_1DX_1 \] Thus $\angle X_1DA=\angle ADT_1=\angle T_1DH_1$, so $E$ is also the reflection of $T_1$ across $H_1D$. But $D$ is the midpoint of arc $H_3C$, so \[ \angle DH_1E=\angle T_1H_1D=\angle CH_1D=180-\angle H_3H_1D \] and so $E$ lies on $H_1H_3$, as desired.

Let Ω be the homothety with negative radius taking the circumcircle to the inradius. Construct the triangle A'B'C' that is the image of ABC under Ω, and reflect each side w.r.t the respective side of TaTbTc. It is enough to prove the resulting triangle, call it HaHbHc, is the orthic triangle. But Ta is equidistant A'B' and A'C', and so circle centered at Ta is tangent to lines A'B', A'C', and the reflections of these across TaTb and TaTc. From this TaHa is parallel to BC and so Ha lies on BC. Similarly Hb and Hc lie on the sides of ABC and we finish easily.

My solution: Let $I$ be the incenter of $\triangle{ABC}$, $AI, BI, CI\cap{BC, CA, AB} = D, E, F$, $X, Y, Z$ are the reflections of $T_1, T_2, T_3$ WRT $AD, BE, CF$ $\Rightarrow$ $DX, EY, FZ$ are tangent to $(I)$ $\Rightarrow$ $T_2T_3, EF, YZ$ are concurrent (familiar property of circumscribed quadrilateral) (1) Moreover: $T_2T_3 = T_1Z$ (reflect WRT $CI$) and $T_2T_3 = T_1Y$ (reflect WRT $BI$) so $T_1Z = T_1Y \Rightarrow YZ\parallel{BC}$ Analogously, we have: $ZX\parallel{CA}, XY\parallel{AB}$ On the other hand: $(EH_2AT_2) = \frac{\overline{AE}}{\overline{AH_2}}:\frac{\overline{T_2E}}{\overline{T_2H_2}} = \frac{\overline{AE}}{\overline{AH_2}}:\frac{\overline{IE}}{\overline{IB}} = \frac{\overline{AE}}{\overline{AH_2}}:\frac{\overline{AE}}{\overline{AB}} = \frac{\overline{AB}}{\overline{AH_2}} = \frac{\overline{AC}}{\overline{AH_3}} = \frac{\overline{AF}}{\overline{AH_3}}:\frac{\overline{AF}}{\overline{AC}} = \frac{\overline{AF}}{\overline{AH_3}}:\frac{\overline{IF}}{\overline{IC}} = \frac{\overline{AF}}{\overline{AH_3}}:\frac{\overline{T_3F}}{\overline{T_3H_3}} = (FH_3AT_3)$ $\Rightarrow$ $EF, H_2H_3, T_2T_3$ concur (2) (1), (2) $\Rightarrow$ $YZ, H_2H_3, T_2T_3$ concur (3) Moreover: $(H_2H_3, T_2T_3) \equiv (T_2T_3, YZ)$ (mod $\pi$) (because $YZ\parallel{BC}$ and $H_2H_3, BC$ are antiparallel in $\angle{BAC}$, and $T_2T_3$ is perpendicular to the bisector of $\angle{BAC}$) (4) (3), (4) $\Rightarrow$ $YZ$ is the reflection of $H_2H_3$ WRT $T_2T_3$ and the conclusion follows. Q.E.D.

We use complex numbers with $\omega$ the unit circle. Let $T_1 = a$, $T_2 = b$, $T_3 = c$. The main content of the problem is to show that the triangle in question has vertices $ab/c$, $bc/a$, $ca/b$ (which is evident from a good diagram). Since $A = \frac{2bc}{b+c}$, we have \[ H_1 = \frac{1}{2} \left( \frac{2bc}{b+c} + a + a - a^2 \cdot \frac{2bc}{b+c} \right) = \frac{ab+bc+ca-a^2}{b+c}. \]The reflection of $H_1$ over $\overline{T_1 T_2}$ is \begin{align*} H_1^C &= a + b - ab \overline{H_1} \\ &= a + b - b \cdot \frac{ac+ab+a^2-bc}{a(b+c)} \\ &= \frac{a(a+b)(b+c) - b(a^2+ab+ac-bc)}{a(b+c)} \\ &= \frac{c(a^2+b^2)}{a(b+c)}. \end{align*}Now, we claim that $H_1^C$ lies on the chord joining $\frac{ca}{b}$ and $\frac{cb}{a}$, which implies the problem (since by analogy $H_1^B$ does as well). To see this, it suffices to compute \begin{align*} H_1^C + \left( \frac{ca}{b} \right)\left( \frac{ca}{a} \right) \overline{H_1^C} &= \frac{c(a^2+b^2)}{a(b+c)} + c^2 \frac{\frac 1c \cdot \frac{a^2+b^2}{a^2b^2}} {\frac1a\left( \frac{b+c}{bc} \right)} \\ &= \frac{c(a^2+b^2)}{a(b+c)} + \frac{c(a^2+b^2)}{abc^{-1}(b+c)} \\ &= \frac{c(a^2+b^2)}{a(b+c)} \left( \frac{b+c}{b} \right) \\ &= \frac{c(a^2+b^2)}{ab}= \frac{ca}{b} + \frac{cb}{a} \end{align*}as desired.

Easy to see that the the symmetric images of $ H_2H_3$ wrt the lines $ T_2T_3$ ( called it $ l_1$) is parallel to $BC$ Let the Medial triangle of $ABC$ be $DEF$ Consider the Feuerbach hyperbola $f$ of $ABC$, Feuerbach point of $ABC$ be $V$, $H_2H_3$ cut $T_2T_3$ at $K_1$. Then $K_1$ is the pole of $BC$ wrt $f$, so $l_1$ is the polar of $D$ wrt $f$. So it's sufficient to prove that the pole of $EF$ wrt $f$ is on the incircle of $ABC$. Let $EF$ cut $T_2T_3$ at $P_1$. From Fontene theorem, $D,P_1,V$ are collinear. Note that the pole of $V$ pass through $T_1$ and parallel to $T_2T_3$, and the pole of the infinity point of $BC$ is $VD$, and they meet on the incircle of $ABC$ (by inversion of $D$), so the problem is done.

Nice problem with a very rich configuration Let $X_i$ be the point on the incircle $\omega$ so that $TX_i\parallel X_{i+1}X_{i+2}$. We claim that $X_1X_2X_3$ is the desired triangle. In particular, we will show that $X_iX_{i+1}$ is the reflection of $H_iH_{i+1}$ over $T_iT_{i+1}$, which will finish. Actually, we'll just show that $X_1$ is on the reflection of $H_1H_2$ over $T_1T_2$; the rest follows similarly. If $AB = AC$ the result is immediate from $H_1 = T_1 = X_1$, so assume now that $AB < AC$. If $\ell$ be the reflection of $H_1H_2$ over $T_1T_2$, then since $H_1H_2$ and $AB$ are antiparallel and $T_1T_2$ is perpendicular to an angle bisector of $\angle ACB$, we get $\ell \parallel AB$. Let $\ell'$ be the line through $X_1$ parallel to $\ell\parallel AB$, then we wish to show $\ell = \ell'$. Let $\ell$ meet $AI$ at $P$, where $I$ is the incenter of $ABC$ and $D = AI\cap BC$. We note that since $T_1P, X_1P$ are reflections over $AI$ (from $T_1X_1\perp AI$) we obtain $T_1P\parallel AC$ since $AC, AB$ are reflections over $AI$ and $AB\parallel X_1P$. Thus, $\triangle DTP \sim \triangle DCA$. Now, \[ \frac{DT_1}{H_1T_1} = \frac{DI}{IA} = \frac{CD}{DA} = \frac{DT_1}{T_1P}, \]where the equalities follow from $T_1I\parallel AH_1$, the Angle Bisector Theorem from $CI$ bisecting $\angle ACD$, and $\triangle DTP \sim \triangle DCA$. Thus $H_1T_1 = T_1P$. Note that since $AB < AC$, $A, B, H_1$ are on the same side of $T_1I$, so $CH_1 > CT_1$. Thus, the reflection of $H_1$ over $T_1T_2$ lies on the line through $T_1$ parallel to $AC$, lies the same distance from $T_1$ as $H_1$, and is on the opposite side of $BC$ as the incircle. Now, since $T_1, D, C$ lie on $BC$ in that order, it follows that $\frac{AD}{DP} = \frac{T_1D}{DC}$ where we direct segments, by considering the homothety centered at $D$ taking $T_1P\to CA$. Thus, $A,P$ lie on opposite sides of $BC$, so $P$ is the reflection of $H_1$ over $T_1T_2$. Now, the line through $P$ parallel to $AB$ is $\ell$, so $\ell = \ell'$ and $X_1\in \ell$ as desired. Similarly, $X_1$ is on the reflection of $H_1H_3$ over $T_1T_3$, and so $X_1$ is indeed one of the vertices of our triangle, as are $X_2, X_3$, which completes the proof. $\blacksquare$

I think this solution is new. Let $I,O$ be the incenter and circumcenter of $\triangle ABC$. First, since $H_2H_3\perp AO$, and the reflection of $AO$ over $T_2T_3$ gives a line perpendicular to $BC$, it follows that the reflection of $H_2H_3$ is parallel to $BC$. Now let $A_1$ be the incenter of $\triangle AH_2H_3$. By a well-known lemma, $A_1$ is also the orthocenter of $AT_2T_3$ (indeed, this is true because $AA_1 = AI\cos A$). Now when we reflect triangle $A_1H_2H_3$ about $T_2T_3$, our previous observation tells us that the image of $A_1$ is $I$, so we get some triangle $IXY$, where $XY||BC$ as noted earlier. Furthermore, the distance from $I$ to $XY$ is the distance from $A_1$ to $H_2H_3$, which is just $r\cos A$ (since $AH_2H_3, ABC$ are similar with ratio $\cos A$). Therefore, the three reflected lines in the problem statement determine some triangle $A_2B_2C_2$ which is homothetic and therefore similar to $ABC$, and which satisfies that there exists a point $I$ in the plane whose distances to the three sides of the triangle are $r\cos A, r\cos B, r\cos C$. The only point $P$ in the plane such that the ratios of its projections to the sides of $A_2B_2C_2$ are $\cos A:\cos B:\cos C$ is the circumcenter of $A_2B_2C_2$, so $I$ is the circumcenter and $A_2B_2C_2$ has circumradius $r$, meaning it lies on the incircle, and we're done. (There are some configuration issues, but since $ABC$ is acute, we can verify that $A_1$ lies on the "correct side" of $H_2H_3$, so $I$ lies on the "correct side" of $B_2C_2$ and is therefore in the interior of $A_2B_2C_2$, so this is fine.)

Let $Y$ lie on the incircle such that $\overline{T_2Y}\parallel\overline{T_3T_1}$. We will show that $Y$ lies on the reflection of $\overline{H_2H_3}$ over $\overline{T_2T_3}$, which is sufficient by symmetry. [asy][asy] size(9cm); defaultpen(fontsize(10pt)); pen pri=springgreen; pen sec=red; pen tri=blue; pen qua=purple; pen qui=orange; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pen qifil=invisible; pair A, B, C, H1, H2, H3, I, T1, T2, T3, X, Y, Z, P, Yp; A=dir(110); B=dir(200); C=dir(340); H1=foot(A, B, C); H2=foot(B, C, A); H3=foot(C, A, B); I=incenter(A, B, C); T1=foot(I, B, C); T2=foot(I, C, A); T3=foot(I, A, B); X=2*foot(T2+T3-T1, T2, T3)-(T2+T3-T1); Y=2*foot(T3+T1-T2, T3, T1)-(T3+T1-T2); Z=2*foot(T1+T2-T3, T1, T2)-(T1+T2-T3); P=extension(T1, Y, T2, T3); Yp=2*foot(Y, T2, T3)-Y; filldraw(A -- B -- C -- A -- cycle, fil, pri); filldraw(incircle(A, B, C), fil, pri); filldraw(T1 -- T2 -- T3 -- cycle, sfil, sec); filldraw(X -- Y -- Z -- cycle, tfil, tri); draw(T1 -- P -- T2, qua); draw(H1 -- H2 -- H3 -- H1, qui+dotted); draw(H2 -- Yp, qui+dotted); draw(Y -- Yp, tri+dashed); draw(H2 -- P, qui+dotted); draw(arc((B+C)/2, C, B), qui); fill(arc((B+C)/2, C, B) -- cycle, qifil); draw(B -- P, qui+dashed); dot("$X$", X, S); dot("$Y$", Y, SE); dot("$Z$", Z, dir(195)); dot("$Y'$", Yp, N); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$T_1$", T1, S); dot("$T_2$", T2, N/2); dot("$T_3$", T3, dir(150)); dot("$H_1$", H1, S); dot("$H_2$", H2, N); dot("$H_3$", H3, dir(120)); dot("$P$", P, E); [/asy][/asy] Let $P=\overline{T_1Y}\cap\overline{T_2T_3}$, which lies on $\overline{BI}$ by reflection, and let $Y'$ be the reflection of $Y$ over $\overline{T_2T_3}$. Then by the Iran Lemma, $P$ lies on $(BCH_2H_3)$. Since $\measuredangle H_2PB=\measuredangle H_2CB=\measuredangle T_2PY$, but $\overline{PB}$ bisects $\angle T_2PY$, $\overline{PH_2}$ bisects $\angle T_2PY'$, so we deduce by $PY'=PY=PT_2$ that $PT_2H_2Y'$ is a kite. Now \[\measuredangle PH_2Y'=\measuredangle T_2H_2P=\measuredangle CBP=\measuredangle PBH_3=\measuredangle PH_2H_3,\]completing the proof.

Pretty long but straightfoward by complex number. Anw first G8 Ive solved Set all of these at complex plane. Let the incircle touchpoint of $\triangle ABC$ opposite to $A,B,C$ be $a,b,c$ respectively in complex number. Therefore, $A = \frac{2bc}{b+c}, B = \frac{2ac}{a + c}, C = \frac{2ab}{a+b}$. Since $H_1$ is the projection of $A$ towards $BC$, which gives us \[ H_1 = \frac{ab + ac + bc - a^2}{b + c} \]Similarly, $H_2 = \frac{ab + ac + bc - b^2}{a +c} $ and $H_3 = \frac{ab + ac + bc - c^2}{a+b}$ Now, we'll find the reflection of $H_1H_2$ over $T_1T_2$. Reflection of $H_1$ across $T_1T_2$ is \begin{align*} & a + b - ab \cdot \frac{\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} - \frac{1}{a^2}}{ \frac{1}{b} + \frac{1}{c} } \\ &= a + b - \frac{b(ac + ab + a^2 - bc)}{a(b+c)} \\ &= \frac{a(a+b)(b+c) - b(ac + ab + a^2 - bc)}{a(b+c)} \\ &= \frac{c(a^2 + b^2)}{a(b+c)} \end{align*}Similarly, we will get the other point as $ \frac{c(a^2 + b^2)}{b(a+c)} $ Therefore, the three lines are determined by lines passing through \[ \left( \frac{c(a^2 + b^2)}{a(b+c)} , \frac{c(a^2 + b^2)}{b(a+c)} \right), \left( \frac{a(b^2 + c^2)}{b(a+c)} , \frac{a(b^2 + c^2)}{c(a+b)} \right), \left( \frac{b(a^2 + c^2)}{a(b+c)}, \frac{b(a^2 + c^2)}{c(a+b)} \right) \]Now, the conjugate are \[ \left( \frac{a^2 + b^2}{ab(b+c)}, \frac{a^2 +b^2}{ab(a+c)} \right), \left( \frac{b^2+c^2}{bc(a+c)}, \frac{b^2+c^2}{bc(a + b)} \right) , \left( \frac{a^2 + c^2}{ac(a+b)} , \frac{a^2 + c^2}{ac(b+c)} \right) \]Now, we want to determine the intersection of these three lines using the complex intersection formula. It suffices to find for one point as the others are analogously symmetrical. \begin{align*} \text{Numerator} &= \frac{a^2c(a^2 + b^2)^2 (b - a)(c - b)(b^2 + c^2)}{a^2 b^3c (a+c)(b+c)(a+b)(a+c)} - \frac{c^2a(a^2 + b^2)(b-a)(b^2+c^2)^2(c-b)}{ab^3 c^2 (a+c)^2 (b+c)(a+b)} \\ &= \frac{(b-a)(a^2 + b^2)^2 (c - b)(b^2 + c^2)}{b^3(a + c)^2(a+b)(b+c)} - \frac{(b^2+c^2)^2(c-b)(b-a)(a^2 + b^2)}{b^3(a+c)^2(b+c)(a+b)} \\ &= \frac{(c-b)(b-a)(a^2 + b^2)(b^2+c^2)(a^2 - c^2)}{b^3(a + c)^2 (b+c)(a+b)} \\ &= \frac{(c-b)(b-a)(a-c)(a^2+b^2)(b^2+c^2)}{b^3(a+c)(b+c)(a+b)} \end{align*}and furthermore, we have \begin{align*} \text{Denominator} &= a^2 (b^2 + c^2) \left( \frac{c-b}{bc(a+c)(a+b)} \right) \left( \frac{b^2 + c^2}{bc} \right) \frac{a - b}{(a+c)(b+c)} - c^2 (a^2 + b^2) \left( \frac{b - a}{ab(a+c)(b+c)} \right) \left( \frac{b^2+c^2}{bc} \right) \frac{b-c}{(a+b)(a+c)} \\ &= \frac{(a^2 + b^2)(b^2+c^2)(a-b)(c-b)a^2}{ab^2c(a+c)^2(a+b)(b+c)} - \frac{c^2(a^2 + b^2)(b^2+c^2)(a-b)(c-b)}{ab^2c (a+c)^2(a+b)(b+c)} \\ &= \frac{(a^2+b^2)(b^2+c^2)(a-b)(b-c)(c-a)}{ab^2c(a+c)(a+b)(b+c)} \end{align*}Divide the two of them, we get the intersection being \[ \frac{\frac{(c-b)(b-a)(a-c)(a^2+b^2)(b^2+c^2)}{b^3(a+c)(b+c)(a+b)} }{\frac{(a^2+b^2)(b^2+c^2)(a-b)(b-c)(c-a)}{ab^2c(a+c)(a+b)(b+c)} } = \frac{-ac}{b} \]Notice that $-\frac{ac}{b}$ lies on the unit circle since \[ -\frac{\overline{ac}}{\overline{b}} = -\frac{\frac{1}{ac}}{\frac{1}{b}} = -\frac{b}{ac} = \frac{1}{-\frac{ac}{b}}\]

Pretty hard.

Another complex numbers solution, I'm posting because it uses a slightly different approach. Let $P, Q, R$ be the points on $\omega$ such that $T_1$ is the midpoint of arc $RPQ$, and symmetric results hold. It suffices to prove that reflection of $R$ in $T_2T_3$ lies on $H_2H_3$, since then other symmetric results hold. We use complex numbers with $\omega$ as unit circle. We may let $p = x^2, q = y^2, r = z^2, t_1 = yz, t_2 = zx, t_3 = xy$. Then by the intersection of tangents formula, $a = 2xyz/(y+z)$, and other symmetric results hold. By foot of perpendicular formula (on chord $T_3T_3$), $h_3 = xyz/(x+y) + xy - x^2y^2/((x+y)z)$, and other symmetric results hold. Let $R'$ be the reflection of $R$ in $T_2T_3$, then by reflection formula, $r' = xy + xz - x^2y/z$. Now, it suffices to prove that $(r'-h_2)/(r'-h_3)$ is real. The expression simplifies to $y(xy-z^2)(x+z)/((xy^2-z^3)(x+y))$, which is equal to its conjugate, so we're done.

here is my solution for this nice problem (we assume WLOG $AB\ge BC\ge AC$) let $C'$ be the reflection of $C$ over $T_1T_2$, we have $CT_1=CT_2$ so $CT_1C'T_2$ must be a rhombus, hence $T_1H'_1$ is parallel to $AC$ Let $X=(T_1T_2) \cap (AI)$ (this point lies on the circle with diameter $AB$) We have $\widehat{T_1XA}=\widehat{T_2XA}=\frac{\widehat{B}}2$ and $\widehat{T_1XH_1}=\widehat{H_1XA}-\widehat{T_2XA}=\widehat{B}-\frac{\widehat{B}}2$ so $H'_1$ lies on $(AI)$ ( the case where $H_1$ is the one that lies on $(AI)$ gives also $H'_1$ lies on $(AI)$ because in that case $H'_1=H_1$) by symmetry we can prove the same claim for the other points. We also have $\widehat{IH'_1H'_2}=\widehat{IH'_1T_1}-\widehat{H'_2H'_1T_1}=180-\frac{\widehat{A}}2-\widehat{T_1H_1H_2}=180-\frac{\widehat{A}}2-(180-\widehat{A})=\frac{\widehat{A}}2$ so $H'_1H'_2$ is parallel to $AB$ So $H'_1$ and $H''_1$ lie on $AI$ with $H''_1$ the reflection of $H_1$ over $T_1T_3$ I claim that the intersection of $H'1H'2$ and $H''_3H''_1$ is the reflection of $T_1$ over $AI$ $Y$ , this claim clearly solves the problem because the reflection of $T_1$ over $AI$ lies on the incircle since $T_2$ and $T_3$ and $I$ are fixed by this reflcection and so is the incircle We have $\widehat{YH'_1A}=\widehat{T_1H'_1A}=180-\frac{\widehat{A}}2$ so it lies on $H'_1H'_2$ by symmetry we prove that it lies on $H''_1H''_3$ QED

We use complex numbers. Set $\omega$ as the unit circle, and let $t_1, t_2, t_3$ be the free variables. From the tangent intersection formula, we have $a=\frac{2t_Bt_C}{t_B+t_C}$, $b=\frac{2t_At_C}{t_A+t_C}$, and $c=\frac{2t_At_B}{t_A+t_B}$. Next, using the foot of altitude formula, we have $h_A=\frac{-t_A^2+\sum_{cyc} t_At_B}{t_B+t_C}$, and $h_B=\frac{-t_C^2+\sum_{cyc} t_At_B}{t_A+t_C}$. Now, let $Z_A$ be the reflection of $H_A$ over $T_AT_B$, and $Z_B$ be the reflection of $H_B$ over $T_AT_B$. Using the complex reflection formula, we find $z_A=t_A+t_B-t_At_B\bar{h_A}=\frac{t_C(t_A^2+t_B^2)}{t_A(t_B+t_C)}$, and $z_B=\frac{t_c(t_A^2+t_B^2)}{t_B(t_A+t_C)}$. Now we compute the intersections of $Z_AZ_B$ with $\omega$. We have the equation $x(\overline{z_A-Z_B})+\bar{x}(z_B-z_A)+(z_A\bar{z_B}-\bar{z_A}z_B)=0$, which, since $|x|=1$, becomes $x^2(\overline{z_A-z_B})+(z_A\bar{z_B}-\bar{z_A}z_B)x+(z_B-z_A)=0$. We can compute $\overline{z_A-z_B}=\frac{(t_A^2+t_B^2)(t_A-t_B)}{t_At_B(t_B+t_C)(t_A+t_C)}$, $z_A\bar{z_B}-\bar{z_Az_B}=\frac{(t_A^2+t_B^2)^2t_C(t_A-t_B)}{t_A^2t_B^2(t_B+t_C)(t_A+t_C)}$, and $z_B-z_A=\frac{(t_A-t_B)t_C^2(t_A^2+t_B^2)}{t_At_B(t_B+t_C)(t_A+t_B)}$. Thus, after dividing out common factors and multiplying by $t_At_B$, the quadratic becomes $t_At_Bx^2+(t_A^2+t_B^2)t_Cx+t_At_Bt_C^2=0$. We can now apply quadratic formula to get $x=-\frac{t_Bt_C}{t_A}, -\frac{t_At_C}{t_B}$, so $l_C\cap \omega=-\frac{t_Bt_C}{t_A}, -\frac{t_At_C}{t_B}$. By symmetry (swapping variables), we get $l_B\cap \omega=-\frac{t_At_B}{t_C}, -\frac{t_Bt_C}{t_A}$, and $l_A\cap \omega=-\frac{t_At_B}{t_C}, -\frac{t_At_C}{t_B}$. Therefore, the three vertices of the triangle formed by $l_A, l_B, l_C$ are $-\frac{t_At_B}{t_C}, -\frac{t_At_C}{t_B}, -\frac{t_Bt_C}{t_A}$, which are all on the unit circle, so we are done.

@above is this how you treat a very very nice problem

complex numbers ftw!

mathlogician wrote: @above is this how you treat a very very nice problem You gotta do what you gotta do

Nothing new here. Let $DEF$ be the triangle determined by the three lines. A well-drawn diagram will inspire us to guess that $DT_1 \parallel T_2T_3$ and likewise equivalents. Thus it is sufficient (bearing in mind symmetry) to show that the reflection of $H_1$ and $H_2$ over $T_1T_2$ lie on the chord $DE$, where we redefine $D$ and $E$ as the points of intersection of the line through $T_1$ parallel to $T_2T_3$ and the line through $T_2$ parallel to $T_3T_1$, respectively. We use complex numbers with the incircle of $\triangle ABC$ as the unit circle. As usual, let lowercase letters denote the complex coordinates of the point represented by their respective uppercase letter unless otherwise stated. For convenience's sake, let $t_1 = x, t_2 = y, t_3 = z$. Then \[ a = \frac{2yz}{y+z}, b = \frac{2zx}{z+x}, c = \frac{2xy}{x+y}. \]Because $H_1$ is the foot from $A$ to $T_1T_1$, we obtain \[ h_1 = \frac{1}{2}( 2x - a + x^2 \overline{a}) = \frac{xy + yz + zx - x^2}{y+z} . \] Let $H_1'$ denote the reflection of $H_1$ over $T_1T_2$. \begin{align*} h_{1}' =& x + y - xy \overline{h_1} \\ &= x + y - \frac{xy(x(x+y+z) - yz)}{x^2(y+z)} \\ &= \frac{ x(x+y)(y+z) - xy(x+y+z) + y^2z}{x(y+z)} \\ &= \frac{x^2(y+z) - x^2y + y^2z}{x(y+z)} = \frac{z(x^2 + y^2)}{x(y+z)} \end{align*}Moreover, a straightforward computation gives $\overline{h_1'} = \frac{x^2 + y^2}{xy(y+z)}$. Now we check that $H_{1}' \in DE$. It suffices to show taht \[ h_1' + de \overline{h_{1}'} = d + e . \]Since $DT_1 \parallel T_2T_3$, we get $d = \frac{yz}{x}$. Likewise, $e = \frac{zx}{y}$. Thus \begin{align*} h_1' + de \overline{h_1'} &= \frac{z(x^2+y^2)}{x(y+z)} + \frac{z^2(x^2+y^2)}{xy(y+z)} \\ &= \frac{ z(x^2+y^2)(y+z))}{xy(y+z)} \\ &= \frac{z(x^2+y^2)}{xy} = \frac{zx}{y} + \frac{zy}{x} \\ &= d + e, \end{align*}as desired. By symmetry, the reflection of $H_2$ over $T_1T_2$ also lies on chord $DE$, so we are done.

The key observation is that the triangle is formed from isosceles-trapezoiding the intouch. More specifically, $T_1T_2 [\text{insertvertex}] T_3$ is an isosceles trapezoid, and this is cyclic. Set $T_1T_2T_3$ unit circle, at $x$, $y$, and $z$ respectively because subscripts are hard to type. Let DEF be the vertices, such that $T_1DT_2T_3$ is an isosceles trapezoid, etc. Then by isosceles trapezoid trick, $d = \frac{yz}{x}$, cyclic. Now, we need to calculate the reflections of the feet of altitudes. $a = \frac{2yz}{y + z}$ by ice cream cone; then foot of $a$ onto $xx$ gives $h_1 = x + \frac{yz - x^2}{y + z}$, which has conjugate $\frac 1 x + \frac{x^2 - yz}{x^2(y + z)}$. Now, reflect over $xy$; we get $h_1' = x - \frac{y(x^2 - yz)}{x(y + z)} = \frac{z(x^2 + y^2)}{x(y + z)}$. Now, determinants. We have \begin{align*} \begin{vmatrix} \frac{yz}{x} & \frac{x}{yz} & 1 \\ \frac{xz}{y} & \frac{y}{xz} & 1 \\ \frac{z(x^2 + y^2)}{x(y + z)} & \frac{x^2 + y^2}{xy(y + z)} & 1 \end{vmatrix} &\propto \begin{vmatrix} \frac y x & \frac{x}{yz} & 1 \\ \frac x y & \frac{y}{xz} & 1 \\ \frac{x^2 + y^2}{x(y + z)} & \frac{x^2 + y^2}{xy(y + z)} & 1 \end{vmatrix} \\ &= \frac{x^2 + y^2}{xy(y + z)} \cdot \left(\frac{x}{y} - \frac{y^2}{xz} - \frac{y}{x} + \frac{x}{z} \right) + \frac{y}{xz} \cdot \left( \frac y x - \frac x z \right) - \frac{x}{yz} \left( \frac x y - \frac{y^2}{xz} \right) \\ &= \frac{x^2 + y^2}{y^2z} - \frac{x^2 + y^2}{x^2z} + \frac{y^2}{x^2z} - \frac{x^2}{y^2z} \\ &= 0, \end{align*}as desired. The others are similar, so we are done.

We work in complex coordinates. Let the incircle be the unit circle, $t_{1} = x, t_{2} = y, t_{3} = z$. Then, by ice cream cone, $a = \frac{2yz}{y+z}$. By complex foot from $A$ to $T_{1}T_{1}$, we hae \[h_{1} = \frac{1}{2}(x + x +a - x^{2}a) = x + \frac{yz}{y+z} - \frac{x^{2}}{y+z} = \frac{xy + xz + yz-x^{2}}{y+z}\]By symmetry $h_{2} = \frac{xy+xz+yz-y^{2}}{x+z}, h_{3} = \frac{xy+xz+yz-z^{2}}{x+z}$ Now, for any point $R$ on the incircle, the reflection of $R$ over $T_{1}T_{2}$ is $x+y-\frac{xy}{r}$. If this lies on $H_{1}H_{2}$ (which would mean $R$ lies on the image of $H_{1}H_{2}$ wrt $T_{1}T_{2}$), then $\frac{h_{1} - (x+y-\frac{xy}{r})}{h_{1} - h_{2}}$ is real. We have \[\frac{h_{1} - x - y + \frac{xy}{r}}{h_{1} - h_{2}} = \frac{\frac{xy+xz+yz-x^{2}}{y+z} - x - y = \frac{xy}{r}}{\frac{xy+xz+yz-x^{2}}{y+z} - \frac{xy+xz+yz-y^{2}}{x+z}} = \frac{\frac{xy + xz + yz-x^{2}-xy-xz-y^{2}-yz + (y+z)\frac{xy}{r}}{y+z}}{\frac{(xy+xz+yz)(x-y) + y^{2}(y+z)-x^{2}(x+z)}{(y+z)(x+z)}}\]\[= \frac{-x^{2}-y^{2} + (y+z)\frac{xy}{r}}{\frac{(x-y)(xy + xz +yz - y^{2}-x^{2}-xy-zy-zx}{x+z}} = \frac{(x^{2}+y^{2}-(y+z)\frac{xy}{r})(x+z)}{(x-y)(x^{2}+y^{2})}\]Since this is real, this is equal to its conjugate. It's conjugate is \[\frac{(\frac{1}{x^{2}} + \frac{1}{y^{2}} - (\frac{1}{y} + \frac{1}{z})\frac{r}{xy})(\frac{1}{x} + \frac{1}{z})}{(\frac{1}{x} - \frac{1}{y})(\frac{1}{x^{2}} + \frac{1}{y^{2}}}\cdot \frac{x^{3}y^{3}z^{2}}{x^{3}y^{3}z^{2}} = \frac{(x+z)(y)(y^{2}z+x^{2}z-(y+z)rx)}{(y-x)z^{2}(y^{2}+x^{2})}\]Simplifying, we have \[x^{2}+y^{2}-(y+z)\frac{xy}{r} = \frac{y(y^{2}z + x^{2}z - (y+z)rx)}{z^{2}}\]\[x^{2}z^{2} + y^{2}z^{2} - z^{2}(y+z)\frac{xy}{r} = rxy(y+z) - y^{3}z-x^{2}yz\]\[xy(y+z)(r + \frac{z^{2}}{r}) = (y+z)(x^{2}z + y^{2}z) \Rightarrow xy(r+\frac{z^{2}}{r}) = y^{2}z + x^{2}z\]\[xyr^{2} - r(y^{2}z + x^{2}z) + xyz^{2} = 0 \Rightarrow (xr - yz)(yr - xz) = 0\]This means $r = \frac{yz}{x}$ or $r = \frac{xz}{y}$. By symmetry, the image of $H_{2}H_{3}$ intersect the incircle at $\frac{xz}{y}$ and $\frac{xy}{z}$, and the image of $H_{1}H_{3}$ intersects the incircle at $\frac{xy}{z}$ and $\frac{yz}{x}$. Therefore, the triangle they form have vertices at $\frac{xy}{z}, \frac{xz}{y}, \frac{yz}{x}$, which all lie on the incircle.

Let $I$ be the incenter of $\triangle ABC$, and $D$, $E$ the reflections of $H_1$ over $T_3T_1$ and $T_1T_2$, respectively. Let $K$ be the meeting point of the reflections of $H_3H_1$ over $T_3T_1$ and $H_1H_2$ over $T_1T_2$. Then by symmetry it suffices to prove that $KT_1T_2T_3$ is cyclic.

Lemma 1: $D$, $E$ lie on $AI$. Proof: Let $X$, $Y$, $M$ be the midpoints of $AH_1$, $IH_1$, $AI$, respectively and let $Z$ be the foot of $H_1$ to $T_3T_1$. Then let the perpendicular bisectors of $AH_1$ and $IH_1$ meet $T_3T_1$ at $P$, $Q$ respectively. Let $O = XP \cap YQ$ and $L = AI \cap T_3T_1$. By the Iran Lemma, we see that $\angle CLA = 90^\circ$. Since $X$ is the midpoint of $AH_1$ and $OX \perp AH_1$, $OX$ passes through the midpoints of $AB$ and $CA$ and so again by the Iran Lemma we see that $P$ lies on $CI$ and that $\angle CPA = 90^\circ$. Thus $CLH_1PA$ is cyclic. We have $\angle H_1OY = \angle IOY = \angle IMY = \angle LAH_1 = \angle QPH_1$, so $OPQH_1$ is cyclic. Then by Simson Line from $H_1$, we see that $X$, $Y$, $Z$ are collinear and by a homothety of scale $2$ from $H_1$, $D$ lies on $AI$. Similarly, we can find that $E$ also lies on $AI$. Lemma 2: $KD \parallel ET_1$ and $KE \parallel DT_1$ Proof: Define $F = KD \cap CI$, $G = T_3T_1 \cap KF$, and $S = H_1H_3 \cap CI$. $H_1E \parallel CI$ so $\angle T_1EH_1 = \angle T_1CI = \frac{\angle C}{2}$. We have: $$\angle KFC + \angle FGS + \angle ICH_1 = \angle H_1SC + \angle SCH_1 = \angle H_3H_1B = \angle A.$$Notice that: $$\angle FGS = 2\angle H_1GT_1 = 2(\angle H_3H_1B - \angle T_3T_1B) = \angle A - \angle C.$$Then we have: $$\angle KFC + (\angle A - \angle C) + \frac{\angle C}{2} = \angle A \Rightarrow$$$$\angle KFC = \angle T_1EH_1 = \frac{\angle C}{2}.$$Thus, $KD \parallel ET_1$ and similarly $KE \parallel DT_1$. $KDT_1E$ is a parallelogram, but $T_1D = T_1H_1 = T_1E$, so $KDT_1E$ is also a rhombus. Then $AI$ is the perpendicular bisector of $KT_1$ and of $T_2T_3$, so $KT_1T_2T_3$ is cyclic as desired. $\square$

12 hours [asy][asy] size(250); pair T1 = dir(47.26-137.26), T2 = dir(47.26), T3 = dir(47.26+119.74); pair A = (2*T2*T3)/(T2+T3), B = (2*T1*T3)/(T1+T3), C = (2*T1*T2)/(T1+T2); pair D = (T2*T3)/T1, EE = (T1*T3)/T2, F = (T1*T2)/T3; pair H1 = foot(A, B, C), H2 = foot(B, A, C), H3 = foot(C, A, B); pair X1 = extension(H2, H3, T2, T3), X2 = extension(H1, H3, T1, T3), X3 = extension(H1, H2, T1, T2); pair Y = extension(T2, D, T3, T1); pair D1 = 2*foot(D, T1, Y)-D; draw(A--B--C--cycle, fuchsia); draw(incircle(A, B, C), red); draw(circumcircle(A, H1, H3), magenta); draw(circumcircle(C, Y, T1), yellow); draw(A--Y--C, brown); draw(X2--D, orange); draw(X2--Y, heavyred); draw(X2--D1, orange); draw(H2--H3, orange+dashed); draw(T1--T3, heavyred+dashed); draw(EE--F, orange+dashed); draw(H1--X3, orange+dashed); draw(T1--X3, heavyred+dashed); draw(D--X3, orange+dashed); draw(T1--D1, red+dashed); draw(T3--T2--Y, brown); draw(T1--D, brown); draw(D1--Y, orange); draw(rightanglemark(A, Y, C, 4), brown); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D'$", D, SE); dot("$E'$", EE, NE); dot("$F'$", F, NW); dot("$H_1$", H1, S); dot("$H_2$", H2, N); dot("$H_3$", H3, NW); dot("$T_1$", T1, S); dot("$T_2$", T2, NE); dot("$T_3$", T3, NW); dot("$Y$", Y, SW); dot("$X_1$", X1, dir(60)); dot("$X_2$", X2, NW); dot("$X_3$", X3, NE); dot("$P$", D1, SW); [/asy][/asy] WLOG $AC>AB$ (to avoid annoying configuration issues). Denote by $D', E', F'$ points on the incircle $\omega$ that satisfy $\overline{T_1D'} \parallel \overline{T_2T_3}$ and cyclic permutations. We will show that $D' = D, E' = E$ and $F'=F$. Notice that it suffices to show that the reflection of $D'$ over the line $\overline{T_1T_3}$ lies on the line $\overline{H_1H_3}$. Let $P$ be the reflection of $D'$ over $\overline{T_1T_3}$, $X_1 = \overline{H_2H_3} \cap \overline{T_2T_3}$ (and cyclic permutations), and introduce $Y = \overline{T_1T_3} \cap \overline{D'T_2}$. Now by the Right Angles on Incircle Chord Lemma, $\angle AYC = 90^{\circ}$. This implies that both $AH_3H_1YC$ and $T_2IT_1YC$ are cyclic. The crux of the problem lies in the following: Claim. $T_1$ and $P$ are symmetric around $H_1Y$. Proof. Notice that $YT_1 = YD' = YP$ by symmetry and definition of $P$. Therefore, it suffices to show that $YH_1$ bisects the angle $PYT_1$. Notice that \begin{align*} \angle H_1YT_1 &= \angle H_1YA - \angle T_1YA \\ &= \angle H_1CA - \angle T_1CI \\ &=\frac C2, \end{align*}while \begin{align*} \angle PYT_1 &= \angle T_1YD' \\ &= 180^{\circ} - 2\angle T_1T_3T_2 \\ &= 180^{\circ} - (180^{\circ} - C) \\ &= C, \end{align*}proving the claim. $\blacksquare$ This means that $H_1T_1P$ is isosceles with $H_1P = T_1P$. Observe that \begin{align*} \angle H_1T_1P &= 360^{\circ} - 2\angle D'T_1Y - \angle H_1T_1D' \\ &=360^{\circ} - (180^{\circ} - C) - \left(180^{\circ} - \frac 12 \widehat{DT_1}\right) \\ &= C + \frac 12(360^{\circ} - 2\widehat{T_1T_3} - \widehat{T_2T_3}) \\ &= C+\frac 12(360^{\circ} - 2(180^{\circ} - B) - (180^{\circ} - A)) \\ &= C + \frac 12(2B+A-180^{\circ}) \\ &= \frac B2 + \frac C2 \\ &= 90^{\circ} - \frac A2, \end{align*}implying that $\angle T_1H_1P = A$. But $\angle H_3H_1T_1 = 180^{\circ} - A$ by cyclic $AH_3H_1C$ implies $\angle H_3H_1T_1 + \angle PH_1T_1 = 180^{\circ}$, so $P, H_1, H_3$ are collinear, as required. $\square$ Remark. By no means is this specific to this problem only, but I discovered that a lot of motivation for constructing points comes from completing the configuration -- in this problem, $Y$ completes the incircle configuration. Now, we were able to use the "dual" properties of $Y$ (that it lies on the two lines from which it is defined, and the two right angles) to help solve the problem.

Beautiful problem Let $H_1'$ and $H_2'$ be the reflections of $H_1$ and $H_2$ over $\overline{T_1T_2}$, respectively, and let $T_1'$ be the reflection of $T_1$ over the perpendicular bisector of $\overline{T_2T_3}$. It's sufficient to prove that $H_1'$, $H_2'$, and $T_1'$ are collinear. Claim 1: $H_1'$ lies on the angle bisector of $\angle BAC$. Proof: Let $X$ be the concurrency point of $\overline{T_1T_2}$, $\overline{T_1'T_3}$, and the angle bisector of $\angle BAC$. By the Iran Lemma, $\angle AXB=90^\circ$, so $AH_2XH_1B$ is cyclic. Thus, we have \[\measuredangle AXT_2=\measuredangle AT_2X+\measuredangle XAT_2=\measuredangle XT_1H_1+\measuredangle BAX=\measuredangle XT_1H_1+\measuredangle T_1H_1X=\measuredangle T_1XH_1H_1'XT_1,\]so $A$, $X$, and $H_1'$ are collinear. Claim 2: $\overline{AB} \parallel \overline{H_1'T_1'}$ Proof: We have \[\measuredangle XH_1'T_1'=\measuredangle T_1H_1'X=\measuredangle XH_1T_1=\measuredangle XAB,\]so $\overline{AB} \parallel \overline{H_1'T_1'}$. Claim 3: $\overline{AB} \parallel \overline{H_1'H_2'}$ Proof: Notice that $\overline{H_1H_2}$ and $\overline{H_1'H_2'}$ are antiparallel with respect to $\angle ACB$. Since $AH_2H_1B$ is cyclic, we know that $\overline{AB}$ and $\overline{H_1H_2}$ are antiparallel with respect to $\angle ACB$, so our claim is proven. Combining claims 2 and 3 gives the desired result.

Let $G_i$ be the reflection of $H_i$ over $\overline{T_iT_{i+1}}$ for $i=1,2,3$ where $T_4=T_1.$ Similarly, let $J_i$ be the reflection of $H_i$ over $\overline{T_{i-1}T_i}$ for $i=1,2,3$ where $T_0=T_3.$ We use complex numbers with unit circle $(T_1T_2T_3).$ Let $d=t_1,e=t_2,$ and $f=t_3.$ Notice that $H_1$ is the foot from $A=\frac{2}{\overline{e}+\overline{f}}=\frac{2ef}{e+f}$ to $\overline{T_1T_1}$ so $$h_1=\frac{1}{2}(2d+a-d^2\overline{a})=\frac{de+ef+fd-d^2}{e+f}.$$Also, $G_1$ is the reflection of $H_1$ over $\overline{DE}$ so \begin{align*}g_1&=\frac{(d-e)\overline{h_1}+\overline{d}e-d\overline{e}}{\overline{d}-\overline{e}}\\&=d+e-ed\overline{h_1}\\&=d+e-ed\left(\frac{\frac{1}{de}+\frac{1}{ef}+\frac{1}{fd}-\frac{1}{d^2}}{\frac{1}{e}+\frac{1}{f}}\right)\\&=\frac{(d+e)(e+f)d-de(d+e+f)-ef}{(e+f)d}\\&=\frac{(d^2+e^2)f}{d(e+f)}.\end{align*}Similarly, $j_2=\frac{(d^2+e^2)f}{e(f+d)}.$ We claim that $G_1,J_2,$ and $P=\frac{ef}{d}$ are collinear. Indeed, $$p=\lambda g_1+(1-\lambda)j_2$$where $\lambda=\frac{(e+f)(d^3-e^2f)}{(d-e)(d^2+e^2)f}$ and $\lambda\in\mathbb{R}$ as $$\overline{\lambda}=\frac{\left(\frac{1}{e}+\frac{1}{f}\right)\left(\frac{1}{d^3}-\frac{1}{e^2f}\right)}{\left(\frac{1}{d}-\frac{1}{e}\right)\left(\frac{1}{d^2}+\frac{1}{e^2}\right)\frac{1}{f}}=\frac{\frac{e+f}{ef}\cdot\frac{e^2f-d^3}{d^3e^2f}}{\frac{e-d}{de}\cdot\frac{d^2+e^2}{d^2e^2f}}=\lambda.$$Similarly, $P,G_2,$ and $J_3$ are collinear. Hence, $P=\overline{G_1J_2}\cap\overline{G_2J_3}$ and note that it lies on the unit circle. Similarly, the other intersections of the reflections of $\overline{H_1H_2},\overline{H_2H_3},$ and $\overline{H_3H_1}$ lie on $(T_1T_2T_3).$ $\square$

Let $I$ be the incenter. Throughout the solution we denote by $\mathcal {S}_i$ reflection over line $T_{i-1}T_{i+1}$ (indicates are taken modulo $3$). Our goal is to prove that $\mathcal {S}_{3} (H_1H_2),\mathcal {S}_{2} (H_1H_3)$ concur on incircle; then by analogous assertions we are done. We claim, that the concurrence point is reflection $R$ of $T_1$ over $AI.$ Claim. $\mathcal{S}_3 (H_1)\in AI.$ Proof. Let $AI\cap T_1T_2=P, BI\cap T_1T_2=Q.$ By the Iran lemma $A,B,H_1,H_2,P,Q$ are concyclic and $$\measuredangle BP\mathcal{S}_3 (H_1)=\measuredangle BPH_1+ 2\measuredangle H_1PT_1 =\measuredangle BAH_1+ 2\measuredangle H_1BQ=$$$$=\measuredangle BH_1A=\measuredangle BPA\implies \mathcal{S}_3 (H_1)\in AI.$$ Now observe that $\mathcal{S}_3 (T_1H_1)\parallel AC,$ so by reflection over $AI$ we deduce $R\mathcal{S}_3 (H_1)\parallel AB.$ But by reflection over $T_1T_2$ we deduce $\mathcal{S}_3 (H_1H_2)\parallel AB\implies R\in \mathcal{S}_3 (H_1H_2).$ Analogously $R\in \mathcal{S}_3 (H_1H_3),$ so done.

Let $E$ be the point on the incircle such that $ET_1\parallel T_2T_3$ and $F$ be the point on the incircle such that $FT_3\parallel T_1T_2.$ We claim that $EF$ is the reflection of $H_1H_3$ across $T_1T_3.$ Note that if we prove this, the analogous is also true, which solves the problem. Let the incircle be the $|z|=1$ on the complex plane, and let $T_3,T_1$ be $x=\text{cis}(\alpha),\overline{x}=\text{cis}(-\alpha)$ respectively for some complex number $x$. Clearly, $IB$ becomes simply the real axis. Let $T_2$ be $y=\text{cis}(\beta)$. $~$ First, let us determine $E$ and $F$. By the parallel condition, arcs $T_2E$ and $T_2F$ must both be congruent to arc $T_1T_3$. Thus, $\{E,F\}=\text{cis}(\beta\pm 2\alpha).$ We'll compute the intersection of $EF$ with the real axis. If $(a,b)$ and $(c,d)$ are points, then the real-intercept is \[\frac{bc-ad}{b-d}=\frac{\sin(\beta+2\alpha)\cos(\beta-2\alpha)-\cos(\beta+2\alpha)\sin(\beta-2\alpha)}{\sin(\beta+2\alpha)-\sin(\beta-2\alpha)}=\frac{\sin(4\alpha)}{2\sin(2\alpha)\cos(\beta)}=\frac{\cos(2\alpha)}{\cos(\beta)}\]Now, the real-intercept of $T_1T_3$ is just $\cos(\alpha).$ We need to calculate the real-intercept of $H_1H_3$, and to do so we shall first calculate $IB$, which is equal to $\tfrac{1}{\cos(\alpha)}.$ Next, let $H_1H_3$ and $IB$ intersect at $X$, then \begin{align*}BX &= BH_1\cdot \frac{\sin(\angle A)}{\sin(\angle C+\angle B/2)}\\ &= BA\cdot \cos(2\alpha)\cdot \frac{\sin(\beta-\alpha)}{\cos(\beta)}\\ &= \left(\tan(\alpha)+\tan\left(\frac{\beta-\alpha}{2}\right)\right)\cdot \frac{\cos(2\alpha)}{\cos{\beta}}\cdot \sin(\beta-\alpha)\\ &= \sin(\beta-\alpha)\left(\frac{\sin(\alpha)}{\cos(\alpha)}+\frac{\sin(\beta-\alpha)}{1+\cos(\beta-\alpha)}\right)\cdot \frac{\cos(2\alpha)}{\cos{\beta}} \\ &= \frac{\cos(2\alpha)}{\cos{\beta}}\cdot \sin(\beta-\alpha)\cdot \frac{\sin(\alpha)+\sin(\beta)}{\cos(\alpha)(1+\cos(\beta-\alpha))} \\ &= \cos(2\alpha)\left(\frac{1}{\cos(\beta)}-\frac{1}{\cos(\alpha)}\right) \end{align*}So we have \[IX=\frac{1}{\cos(\alpha)}-\frac{\cos(2\alpha)}{\cos(\beta)}+\frac{\cos(2\alpha)}{\cos(\alpha)}=2\cos(\alpha)-\frac{\cos(2\alpha)}{\cos(\beta)}.\]Now, let's put it all together. Let $H_1H_3$, $T_1T_3$ and $H_1H_3$ intersect $IB$ at $X,Y,Z$ respectively. We have $X=2Y-Z$ which means that $X$ and $Z$ are reflections across $Y.$ Since $T_1T_3$ is perpendicular to $IB$, it just remains to show that $\angle FZE=\angle H_3XI.$ That is just simple angle chasing. We are done.

This was a failure even after 5 months of try, finally secured it with a hint. SL2000 G8 wrote: Let $ AH_1, BH_2, CH_3$ be the altitudes of an acute angled triangle $ ABC$. Its incircle touches the sides $ BC, AC$ and $ AB$ at $ T_1, T_2$ and $ T_3$ respectively. Consider the symmetric images of the lines $ H_1H_2, H_2H_3$ and $ H_3H_1$ with respect to the lines $ T_1T_2, T_2T_3$ and $ T_3T_1$. Prove that these images form a triangle whose vertices lie on the incircle of $ ABC$. Define $D,E,F \in \omega$ where $\omega$ is the incircle such that $T_1D \parallel T_2T_3$ and other variations hold. Claim : $DE || AB$ and other cyclical variations hold. Proof Kinda note that $D,E$ are reflections of $T_1,T_2$ over perpendicular bisectors of $T_2T_3$ and $T_1T3$ and other trivial things which hold. So, $$\angle DET_3=\angle DT_2T_3=\angle T_1T_2T_3=\angle ET_1T_3=\angle EDT_3 \cdots (1)$$Now we have , $$\angle BT_3D=\angle T_3T_1D=\angle T_3ED \stackrel{\text{By (1)}}=\angle T_3DE$$Which proves our claim $\blacksquare$ Claim : The reflections of $H_1$ over $T_1T_2$ and $T_1T_3$ lie on $AI$ where $I$ is the incentre and other variations hold too . Proof : Say $G$ be the point such that $BG \perp AI$ by Iran lemma $G \in T_1T_2$.Now $GT_2=GT_3$ and $BH_1GA$ cyclic.A little bit of angle chasing with this facts give that if $X$ is a point on $AG$ extended,$GT_1(T_1T_2)$ bisects $\angle H_1GX$.This gives that the reflection of $H_1$ over $T_1T_2$ lies on $AI$ and similarly the other reflection lies on $AI$ as well and hence the claim is proved $\blacksquare$ Claim :Let $H_1',H_2'$ be reflections of $H_1,H_2$ over $T_1T_2$, we have $H_1',H_2' \in DE$ Proof : Note that $T_3D \cap T_1T_2=G$ since $AG$ is the perpendicular biseector of $T_2T_3$. Now $$\angle H_1'DT_3=\angle H_1'T_1G=\angle H_1T_1G=\angle BT_1T_2=\angle AT_2T_1=\angle AT_3D $$The above observation implies $H_1'D \parallel AB \implies H_1' \in DE$.Similarly $H_2'E \parallel AB \implies H_2' \in DE $ $\blacksquare$ So the reflection of $H_1H_2$ over $T_1T_2$ is nothing but the line $DE$,similarly $EF$ and $FD$ are the other reflections.Since all the points $D,E,F$ lie on the incircle of $\triangle ABC$ we're done $\blacksquare$

Notice how reflecting $H_1H_2$ over $T_1T_2$ gives a line with the same direction as if we reclected it over $CI.$ But this just gives the direction of $AB.$ So this tells us that the desired triangle will be similar to $ABC.$ This motivates us to deffine $W_B$ and $W_C$ as the midpoint of arcs $AC$ and $BC$ containing $B$ and $A$ respectively, and $X_1=H_1H_2 \cap T_1T_2,$ $Y_1=AB\cap W_AW_B.$ Now as $W_BW_C \parallel EF$ then by homothety the problem is equivalent to showing that $\frac{Y_1W_A}{Y_1W_B}=\frac{X_1T_1}{X_1T_2},$ which follows easily (say by Menelaus on $\triangle AT_3T_2,$ and by the similarity $\triangle Y_1AW_A \sim \triangle Y_1W_BB$).

Let $\omega$ be the unit circle. Let $W$ be on $\omega$ such that $WT_1 \parallel T_2T_3$. Define $X$ and $Y$ similarly. The following claim finishes since $W$ is on $\omega$. Claim: $W$ is where the reflections of $\overline{H_1H_2}$ and $H_3H_1$ intersect. Similar goes for $X$ and $Y$. Proof: First note that $w = \frac{t_2t_3}{t_1}$ and $a = \frac{2t_2t_3}{t_2+t_3}$ and similar for $b$ and $c$. Therefore \[h_1 = \frac{1}{2}\left(t_1+t_1+\frac{2t_2t_3}{t_2+t_3}-\frac{2t_1^2}{t_2+t_3}\right) = \frac{t_1t_2+t_1t_3 + t_2t_3-t_1^2}{t_2+t_3}\]Next, reflecting $H_1$ over $T_1T_2$ yields \[h_1' = t_1+t_2-t_1t_2 \overline{h_1} = t_1+t_2 - \frac{t_1^2t_2+t_2^2t_1+t_1t_2t_3 - t_2^2t_3}{t_1(t_2+t_3)} = \frac{t_3(t_1^2+t_2^2)}{t_1(t_2+t_3)}\]Similar computation yields \[h_2' = \frac{t_3(t_1^2+t_2^2)}{t_2(t_1+t_3)}\]Now it suffices to show that $W,H_1',H_2'$ are collinear. This is equivalent to the following being real: \[\frac{w-h_1'}{w-h_2'} = \frac{\frac{t_3(t_2t_3-t_1^2)}{t_1(t_2+t_3)}}{\frac{t_3(t_2^2t_3-t_1^3)}{t_1t_2(t_1+t_3)}} = \frac{(t_2t_3-t_1^2)(t_2)(t_1+t_3)}{(t_2+t_3)(t_2^2t_3-t_1^3)}\]which is clearly real upon taking the conjugate, so we're done.

Too lazy to actually include all the details. Here's a decent sketch. We set the incircle be the unit circle. Let the intouch points have complex numbers $x$ , $y$ and $z$. We immediately have by Icecream cone that $a=\frac{2yz}{y+z}$ , $b= \frac{2xz}{x+z}$ and $c= \frac{2xy}{x+y}$. Now, let $D$ be the intersection of the lines parallel to $T_2T_3$ through $T_1$ and the incircle. Define $E$ and $F$ similarly. The claim is that these points are the vertices of the desired triangle. We first compute these points. Note that since $ET_2 \parallel T_1T_3$, we have \begin{align*} \overline{\left(\frac{e-y}{x-z}\right)} &= \frac{e-y}{x-z}\\ \frac{e-y}{x-z} &= \frac{\frac{1}{e}-\frac{1}{y}}{\frac{1}{x}-\frac{1}{z}}\\ \frac{e-y}{x-z} &= \frac{xz(y-e)}{ey(z-x)}\\ e &= \frac{xz}{y} \end{align*}Similarly, we also obtain that $d=\frac{yz}{x}$ and $f= \frac{xy}{z}$. Now, using the foot formula, we compute $h_1$. This turns out to be \[h_1=\frac{xy+xz+yz-x^2}{y+z}\]Similarly, we also have \[h_2 = \frac{xy+yz+zx-y^2}{x+z} \text{ and }h_3 = \frac{xy+yz+zx-z^2}{x+y}\]Now, let $H_1'$ be the reflection of $H_1$ across $T_1T_3$. It is clear that $h_1' = x+z-xz\overline{h}$. Further, we can compute \begin{align*} \frac{h_1'-d}{d-f} &= \frac{x+z-xz\overline{h_1}-\frac{yz}{x}}{\frac{yz}{x}-\frac{xy}{z}}\\ &= \\frac{\frac{x^2+xz-x^2z\overline{h_1}-yz}{x}}{\frac{yz^2-x^2y}{xz}}\\ &= \frac{zx^2+xz^2-x^2z^2\overline{h_1}-yz^2}{y(z^2-x^2)}\\ &= \frac{x^2yz-y^2z^2}{y(y+z)(z^2-x^2)}\\ &= \frac{x^2z-yz^2}{(y+z)(z^2-x^2)} \end{align*}And also, \begin{align*} \overline{\left(\frac{h_1'-d}{d-f}\right)} &= \overline{\left(\frac{x^2z-yz^2}{(y+z)(z^2-x^2)}\right)}\\ &= \frac{\frac{1}{x^2z}-\frac{1}{yz^2}}{\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{z^2}-\frac{1}{x^2}\right)}\\ &= \frac{z(yz-x^2)}{(y+z)(x^2-z^2)}\\ &= \frac{x^2z-yz^2}{(y+z)(z^2-x^2)}\\ &= \frac{h_1'-d}{d-f} \end{align*}which shows that $\frac{h_1'-d}{d-f} \in \mathbb{R}$, and thus $H_1'$ must lie on $\overline{DF}$. Similarly, we also obtain that $H_3'$ lies on $DF$. Thus, $\overline{DF}$ is the reflection of $\overline{H_1H_3}$ over $\overline{T_1T_3}$ as desired. After obtaining similar results for the other sides, the desired result follows.

We use complex numbers, setting $\omega$ to be the unit circle. We claim that the desired triangle is that formed by the points $$V_1 \colon = \frac{t_2 t_3}{t_1},$$$$V_2 \colon = \frac{t_3 t_1}{t_2},$$and $$V_3 \colon = \frac{t_1 t_2}{t_3}.$$To do so, we will show that the reflection of $H_1$ over $T_1 T_2$ lies on $V_1 V_2;$ this implies the result by symmetry. First, we have that $a = \frac{2t_2 t_3}{t_2 + t_3}.$ Then, since $H_1$ is the foot of the altitude from $A$ to $T_1 T_1,$ we have \begin{align*} h_1 &= \frac{1}{2} (a + t_1 + t_1 - t_1^2 \overline{a}) \\ &= \frac{1}{2} \left(\frac{2t_2 t_3}{t_2 + t_3} + 2t_1 - \frac{2t_1^2}{t_2 + t_3}\right) \\ &= \frac{t_2 t_3 + t_3 t_1 + t_1 t_2 - t_1^2}{t_2 + t_3}. \end{align*}Thus the reflection of $h_1$ over $t_1 t_2$ is \[ h_1' = t_1 + t_2 - t_1 t_2 \overline{h_1}. \]We now must verify that $H_1' \in V_1 V_2,$ or \[ v_1 + v_2 = h_1' + v_1 v_2 \overline{h_1'}. \]Plugging in, this becomes \[ \frac{t_2 t_3}{t_1} + \frac{t_3 t_1}{t_2} = t_! + t_2 + \frac{t_3^2}{t_1} + \frac{t_3^2}{t_2} - t_1 t_2 \overline{h_1} - \frac{t_3^2 h_1}{t_1 t_2}. \]Rearranging, this becomes \[ t_1 t_2 \overline{h_1} + \frac{t_3^2 h_1}{t_1 t_2} = t_1 + t_2 + \frac{t_3^2}{t_1} + \frac{t_3^2}{t_2} - \frac{t_2 t_3}{t_1} - \frac{t_3 t_1}{t_2}. \]Multiplying both sides by $t_1 t_2,$ this becomes \[ t_1^2 t_2^2 \overline{h_1} + t_3^2 h_1 = t_1^2 t_2 + t_2^2 t_1 + t_3^2 t_1 + t_3^2 t_2 - t_3 t_1^2 - t_3 t_2^2. \]Now, since \[ h_1 = \frac{t_2 t_3 + t_3 t_1 + t_1 t_2 - t_1^2}{t_2 + t_3}, \]we have \[ \overline{h_1} = \frac{t_1^2 + t_1 t_2 + t_1 t_3 - t_2 t_3}{t_1^2 (t_2 + t_3)}, \]so \begin{align*} t_1^2 t_2^2 \overline{h_1} + t_3^2 h_1 &= t_2^2 \cdot \frac{t_1^2 + t_1 t_2 + t_1 t_3 - t_2 t_3}{t_2 + t_3} + t_3^2 \cdot \frac{t_2 t_3 + t_3 t_1 + t_1 t_2 - t_1^2}{t_2 + t_3} \\ &= \frac{t_1^2 t_2^2 + t_1 t_2^3 + t_1 t_2^2 t_3 - t_2^3 t_3 + t_1 t_2 t_3^2 + t_2 t_3^3 + t_3^3 t_1 - t_3^2 t_1}{t_2 + t_3}. \end{align*}Therefore, since all of our steps are reversible, we must verify that \begin{align*} &t_1^2 t_2^2 + t_1 t_2^3 + t_1 t_2^2 t_3 - t_2^3 t_3 + t_1 t_2 t_3^2 + t_2 t_3^3 + t_3^3 t_1 - t_3^2 t_1^2 \\ &= (t_2 + t_3)(t_1^2 t_2 + t_2^2 t_1 + t_3^2 t_1 + t_3^2 t_2 - t_3 t_1^2 - t_3 t_2^2), \end{align*}which follows upon expansion.

The most beautiful configuration I've seen!! Rename the intouch triangle to $DEF,$ the orthic triangle to $GHI,$ the reflection of the orthic triangle in the intouch triangle to $JKL$ and the medial triangle $MNO.$ Let $T$ be the Feuerbach point and let $X$ be the intersection of $EF,HI,KL.$ We claim $X$ is the pole of $BC$ with respect to the Feuerbach hyperbola $\mathcal H.$ This is by Brokard's on first the triangle vertices and the orthocenter, and then on the triangle vertices and the Gergonne point. Thus the polar of $M$ is the line through $X$ parallel to $BC,$ which is $KL.$ By symmetry this implies that the polar of $J$ is $NO.$ Since this is parallel to the polar of $M$ we have $T,J,M$ collinear. Now simple angle chasing gives that $JKL$ is homothetic to $ABC$ and thus $MNO.$ From what we know the center of homothety must be $T.$ Now define $S$ to be the intersection of $EF$ and $NO.$ We know its polar is $DJ.$ We want to show $DJ\parallel EF,$ but this is equivalent to their poles $S,D$ lying on a line with $T,$ but this is by First Fontené theorem. Finally, consider the bottom point $Y$ on the nine-point circle. By homothety at the orthocenter with scale factor $2$ we see that $YM$ is perpendicular to the interior bisector of $\angle A,$ so it is parallel to $EF$ and thus to $DJ.$ But then the homothety at $T$ taking $MNO$ to $JKL$ takes $Y$ to $D,$ so it takes the nine-point circle to the incircle. Thus $J,K,L$ lie on the incircle, done.

The complex bash is clean. Define $\mathfrak A(z)=\frac z{\overline z}$. Rename the intouch points to $D$, $E$, $F$ and the feet to $P$, $Q$, $R$. Set the incircle to the unit circle. $P$ is the foot from $A$ to chord $DD$ of the unit circle, or $\frac{a+2d-d^2\overline a}2=\frac{-d^2de+df+ef}{e+f}$. Similarly, $q=\frac{-e^2+de+df+ef}{d+f}$ and $r=\frac{-f^2+de+df+ef}{d+e}$. The product of the intersections of the reflection of $PQ$ over $EF$ is $-\frac{\mathfrak A(e-f)^2}{\mathfrak A(p-q)}=-\frac{e^2f^2}{\frac{\mathfrak A(a-b)\mathfrak A(a-c)}{\mathfrak A(b-c)}}=-\frac{e^2f^2}{\frac{(-f^2)(-e^2)}{-d^2}}=d^2$. By the problem statement, the desired triangle has three vertices on the unit circle whose pairwise products are $d^2$, $e^2$, and $f^2$. So, the vertices are $\pm(\frac{ef}d,\frac{df}e,\frac{de}f)$; and setting $DEF$ to be equilateral confirms that the sign is $+$. So, it suffices to plug in $z=\frac{de}f$ and confirm that the reflection of $Z$ over $EF$ lies on $QR$. The reflection is $z'=e+f-f^2/d$. We have $z'-q=e+f-f^2/d-\frac{-e^2+de+df+ef}{d+f}=\frac{e^2d-f^3}{d(d+f)}$, so $\mathfrak A(z'-q)=-\frac{e^2f^2}{d^2}=\mathfrak A(q-r)$, as desired.

[asy][asy] import geometry; unitsize(5cm); point A = (-0.5, 1); point B = (-1, 0); point C = (1.1, 0); circle Omega = circumcircle(A, B, C); point O = Omega.C; line a1 = bisector(line(B, A), line(C, A)); line b1 = bisector(line(C, B), line(A, B)); point I = intersectionpoint(a1, b1); point T1 = intersectionpoint(perpendicular(I, line(B, C)), line(B, C)); point T2 = intersectionpoint(perpendicular(I, line(A, C)), line(A, C)); point T3 = intersectionpoint(perpendicular(I, line(A, B)), line(A, B)); circle omega = circumcircle(T1, T2, T3); point H1 = intersectionpoint(perpendicular(A, line(B, C)), line(B, C)); point H2 = intersectionpoint(perpendicular(B, line(A, C)), line(A, C)); point H3 = intersectionpoint(perpendicular(C, line(B, A)), line(B, A)); point R = intersectionpoint(line(H2, H3), line(T2, T3)); line l1 = parallel(R, line(B, C)); point V = intersectionpoint(l1, line(A, C)); point W1 = intersectionpoint(l1, line(A, B)); point U = intersectionpoint(l1, line(A, H1)); point Ip = intersectionpoint(perpendicular(I, line(A, H1)), line(A, H1)); draw(triangle(A, B, C)); draw(omega); draw(A--H1, dotted); draw(B--H2, dotted); draw(C--H3, dotted); draw(H2--R, red); draw(T2--R, blue); draw(R--V, green); draw(I--Ip, dotted); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C); dot("$I$", I, SW); dot("$T_2$", T2, NE); dot("$T_3$", T3, NW); dot("$O$", O); dot("$H_1$", H1, S); dot("$H_2$", H2, NE); dot("$H_3$", H3, NW); dot("$R$", R, W); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W1, NW); dot("$I'$", Ip, SW); label("$\omega$", (0.12, 0.3)); label("$\ell_1$", (-1, 0.4), green); [/asy][/asy] Let $\ell_1$ be the reflection of line $H_2H_3$ over line $T_2T_3$, and define $\ell_2$ and $\ell_3$ similarly. Quadrilateral $BCH_2H_3$ is cyclic, so we have $$\measuredangle(\ell_1, H_2H_3)=\measuredangle VRH_2=2\measuredangle T_2RH_2 = 2\measuredangle T_3T_2A + 2\measuredangle CH_2H_3=\measuredangle BAC + 2\measuredangle CBA=\measuredangle CBA-\measuredangle ACB=\measuredangle CH_2H_3+\measuredangle BCH_2=\measuredangle(BC, H_2H_3),$$therefore $\ell_1 \parallel BC$. Let $\ell_1$ intersect $AH_1$, $AC$, $AB$ at $U$, $V$, $W$ respectively. Now $\triangle AH_2H_3 \sim \triangle AWV \sim \triangle ABC$. Taking a perpectivity at $R$, we see that: $$(A, T_2; H_2, V)=(A, T_3; H_3, W)$$$$-\frac{AH_2}{AV}/\frac{T_2H_2}{T_2V}=-\frac{AH_3}{AW}/\frac{T_3H_3}{T_3W}$$$$\frac{T_2V}{T_2H_2}\cdot\frac{T_3H_3}{T_3W}=\frac{AH_3}{AH_2}\cdot\frac{AV}{AW}=\frac{b^2}{c^2}.$$Furthermore, $\triangle RVH_2 \sim \triangle RH_3W$, so by the angle bisector theorem: $$\frac{T_3H_3}{T_3W}=\frac{RH_3}{RW}=\frac{RV}{RH_2}=\frac{T_2V}{T_2H_2}.$$By the previous two equalities, $\frac{T_2V}{T_2H_2}=\frac{b}{c}$. From this, we have: $$AV-AT_2=\frac{b}{c}(AT_2-AH_2)$$$$AV=\frac{1}{c}((b+c)AT_2-bAH_2)=\frac{1}{c}((b+c)\frac{-a+b+c}{2}-b\cos\alpha)=\frac{1}{c}{(b+c)\frac{-a+b+c}{2}-b\frac{b^2+c^2-a^2}{2bc}}=\frac{a^2-ab-ac+2bc}{2c}$$$$AU=\frac{AH_1}{AC}\cdot AV=\frac{2[ABC]}{ab}\cdot \frac{a^2-ab-ac+2bc}{2c}=\frac{[ABC](a^2-ab-ac+2bc)}{abc}.$$ Let $I'$ be the foot of perpendicular from $I$ to $AH_1$. Now $d(I, \ell_1)=UI'$. Next: $$AI'=AH_1-r=\frac{2[ABC]}{a}-\frac{2[ABC]}{a+b+c}=\frac{2[ABC](b+c)}{a(a+b+c)}$$$$AI'-AU=\frac{2[ABC](b+c)}{a(a+b+c)}-\frac{[ABC](a^2-ab-ac+2bc)}{abc}=\frac{[ABC](b^2+c^2-a^2)}{bc(a+b+c)}.$$ As $b^2+c^2-a^2>0$ for acute-angled triangles, $AI'-AU$ is positive, so $I$ lies on the opposite side of $\ell_1$ as A, and $UI'=AI'-AU$. Finally, $$\frac{d(I, \ell_1)}{r}=\frac{UI'}{r}=\frac{2[ABC]}{r(a+b+c)}\cdot\frac{b^2+c^2-a^2}{2bc}=\cos\alpha=\frac{d(O, BC)}{R},$$and similar equalities hold for $\ell_2$ and $\ell_3$. Consider the negative homothety sending $\Omega$, the circumcircle of $\triangle ABC$, to $\omega$, its incircle. Let $A',B',C'$ be the images of $A,B,C$ respectively under this homothety. Now $\frac{d(I,B'C')}{r}=\frac{d(O, BC)}{R}$ and $I$ lies on the opposite side of $B'C'$ as $A$. Therefore, $\ell_1=B'C'$. Similary, $\ell_2=C'A'$ and $\ell_3=A'B'$. This means that the points $A',B',C'$, which lie on $\omega$, are precisely the vertices of the triangle formed by $\ell_1, \ell_2, \ell_3$, as desired.