Take $\triangle A_1B_1C_1$ to be the reference triangle. Define $A'$ to be the reflection of $A$ over $\overline{B_1C_1}$, and define $B'$ and $C'$ similarly. Then \[ \angle B_1A'C_1=\angle B_1AC_1=\angle CAB = \angle BA_1C_1, \]so $A' \in (A_1B_1C_1)$. Similarly $B'$ and $C'$ lie on the circumcircle. We now use complex numbers, with $(A_1B_1C_1)$ the unit circle. All of $a_1,b_1,c_1,a',b',c'$ lie on the unit circle. By the reflection formula, $a=b_1+c_1-\tfrac{b_1c_1}{a'}$ and symmetric statements. We know $A_1\in \overline{BC}$ and symmetric statements. So $b,a_1,c$ are collinear. Hence \[ \mathbb{R} \ni \frac{b-a_1}{c-a_1}=\frac{c_1-\frac{a_1c_1}{b'}}{b_1-\frac{a_1b_1}{c'}} = \frac{c_1c'(b'-a_1)}{b_1b'(c'-a_1)}. \]So it is equal to its conjugate, so we have the equation \[ \frac{c_1c'(b'-a_1)}{b_1b'(c'-a_1)} = \frac{\frac{1}{c_1c'}(\frac{1}{b'}-\frac{1}{a_1})}{\frac{1}{b_1b'}(\frac{1}{c'}-\frac{1}{a_1})} = \frac{b_1(a_1-b')}{c_1(a_1-c')} \implies \frac{c_1c'}{b_1b'}=\frac{b_1}{c_1} \implies c'c_1^2=b'b_1^2. \]Hence by symmetry, $a_1^2a'=b_1^2b'=c_1^2c' := N$ for some complex number $N$ with magnitude 1. So now \[ \boxed{a'=\frac{N}{a_1^2}, \quad b'=\frac{N}{b_1^2}, \quad c'=\frac{N}{c_1^2}}. \]We now locate $A_2$. We have $A'A_2=AA_2=A_1A_2$. Hence \begin{align*} |a'-a_2|=|a_1-a_2| &\implies (a'-a_2)(\frac{1}{a'}-\overline{a_2}) = (a_1-a_2)(\frac{1}{a_1}-\overline{a_2}) \\ &\implies 1-a'\overline{a_2}-\frac{a_2}{a'} + a_2\overline{a_2} = 1-a_1\overline{a_2}-\frac{a_2}{a_1}+a_2\overline{a_2} \\ &\implies (a_1-a')\overline{a_2} = a_2(\frac{1}{a'}-\frac{1}{a_1}) \\ &\implies \overline{a_2} = \frac{a_2}{a'a_1}. \end{align*}The second condition is that $A_2\in \overline{B_1C_1}$. Unpacking this and using $\overline{a_2}$ we found above: \begin{align*} &\qquad b_1+c_1=a_2+b_1c_1\overline{a_2}=a_2+b_1c_1\cdot \frac{a_2}{a'a_1} = a_2 \left( \frac{a'a_1+b_1c_1}{a'a_1}\right) \\ &\implies a_2 = \frac{(b_1+c_1)a'a_1}{a'a_1+b_1c_1} = \frac{(b_1+c_1)\cdot \frac{N}{a_1}}{\frac{N}{a_1}+b_1c_1} = \frac{(b_1+c_1)N}{N+a_1b_1c_1}. \end{align*}So by symmetry, we have \[ \boxed{a_2 = \frac{(b_1+c_1)N}{N+a_1b_1c_1}, \quad b_2 = \frac{(c_1+a_1)N}{N+a_1b_1c_1}, \quad c_2 = \frac{(a_1+b_1)N}{N+a_1b_1c_1}}. \]This is super clean. Finally, we want to show $\angle B_2A_2C_2=\angle B_1A_2C_1$. So \[ \frac{c_2-a_2}{b_2-a_2} = \frac{\frac{(a_1+b_1)N}{N+a_1b_1c_1} - \frac{(b_1+c_1)N}{N+a_1b_1c_1} } {\frac{(c_1+a_1)N}{N+a_1b_1c_1} - \frac{(b_1+c_1)N}{N+a_1b_1c_1}} = \frac{a_1-c_1}{a_1-b_1}. \]Note $B_2\to C_2$ is a CCW rotation around $A_2$ and $B_1\to C_1$ is a CCW rotation around $A_1$. So from the above $\frac{c_2-a_2}{b_2-a_2} \div \frac{c_1-a_1}{b_1-a_1} = 1\in \mathbb{R}$, so $\angle C_2A_2B_2=\angle C_1A_1B_1$. By symmetry, all corresponding angles are equal, so $\triangle A_2B_2C_2\sim \triangle A_1B_1C_1\sim \triangle ABC$.

basically triangle $A_2B_2C_2$ wrt triangle $A_1B_1C_1$ is triangle $A_1B_1C_1$ wrt $ABC$ ^^in order to prove this, we can let $A'$ be on $B_1C_1$ such that $A'$ wrt $A_1B_1C_1$ is $A_1$ wrt $ABC$ and prove $A'A = A'A_1$.

i'm sorry Let $\triangle A'B'C'$ be the medial triangle of $\triangle A_1B_1C_1$, and let $P,Q,R$ be the midpoints of $AA_1, BB_1, CC_1$. So $A_2$ is the point on $B_1C_1$ with $A_1P \perp A_2P$, etc.

Claim: $A'B'C'PQR$ are concyclic. Proof: We show $P$ lies on $(A'B'C')$. Taking homothety at $A_1$ of ratio $2$, it suffices to show that if $T$ is the reflection of $A_1$ over $A'$, then $AB_1C_1T$ are concyclic. This follows from$$\measuredangle B_1TC_1 = -\measuredangle B_1AC_1 = -\measuredangle BAC = \measuredangle B_1AC_1.$$ Note that by gliding principle $\triangle PQR$ is similar to $\triangle ABC, \triangle A_1B_1C_1, \triangle A'B'C'$. Hence it suffices to demonstrate the following, where we relabel so that $\triangle A'B'C'$ is refrence triangle: Quote: Let $\triangle A_1B_1C_1$ be the antimedial triangle of $\triangle ABC$. The triangles $\triangle PQR$ and $\triangle ABC$ are directly similar and share a circumcircle. Point $A_2$ is the point on $B_1C_1$ such that $\angle A_1PA_2 = 90^\circ$, and points $B_2, C_2$ are defined similarly. Then the orthocenter $H$ of $\triangle ABC$ is the center of a spiral similarity sending $\triangle ABC$ to $\triangle A_2B_2C_2$.

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Lazy sol oops Let $A_X, B_X, C_X$ be reflections of $A_1, B_1, C_1$ over $B_1C_1, A_1C_1, A_1B_1$ respectively, and $H =$ orthocenter of $\triangle{A_1B_1C_1}$. Then clearly, $AA_XHB_1C_1$ is cyclic and same for $B, C$. Claim $1$: $AA_X \parallel BC$, and same for $B, C$ Proof: Suffices to show $\angle{A_XAC}=\angle{C}$; this is obvious by angle chasing since $\measuredangle{A_XAC} = \measuredangle{A_XHB_1} = \measuredangle{B_1C_1A_1}$. Now let $M_A, M_B, M_C$ be midpoints of $AA_X$, etc, and $\alpha =$ angle formed by $AB \& A_1B_1$, $AC \& A_1C_1$, and $BC \& B_1C_1$. Claim $2$: $\measuredangle{A_1AA_2} = \measuredangle{A_2A_1A} = \alpha$, and same for $B, C$ Proof: Clearly, since $A_2A = A_2A_X$, $\measuredangle{A_1AA_2} = 2\measuredangle{M_AA_2B_1} = 2(90^{\circ}-\alpha)$. $A_2A = A_2A_1$, hence proven. Thus, if we let point $Q$ be such that $\measuredangle{B_1A_1Q} = \alpha$, $QA_1 \parallel AB$ so $\measuredangle{A_2A_1A} = \measuredangle{B_1A_1Q} \implies \measuredangle{A_2A_1B_1} = \measuredangle{AA_1Q} = \measuredangle{A_1AB} \implies BA_1:A_1C = B_1A_2 : A_2C_1$ and same for $B_2, C_2$. $\triangle{A_1B_1C_1} \sim \triangle{ABC} \implies \triangle{A_2B_2C_2} \sim \triangle{A_1B_1C_1}$, and so we are done. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.060034895778351, xmax = 13.115131401339172, ymin = -10.732990198075434, ymax = 6.082974325206182; /* image dimensions */ pen qqzzff = rgb(0,0.6,1); pen wwzzff = rgb(0.4,0.6,1); pen cczzff = rgb(0.8,0.6,1); pen ccccff = rgb(0.8,0.8,1); 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geo lmao.

Wait lol this problem is so silly. Also, can we assume they're directly similar? (not that I took this competition but still) Let $B_1'$ be the reflection of $B_1$ over $C_1A_1$. Note that $(AB_1C_1)$ and its cyclic variants all pass through the circumcneter $O$ of $ABC$, which is also the orthocenter of $A_1B_1C_1$ (provable by angle chasing). Hence $B_1'$ lies on $(BA_1C_1)$. Moreover this gives us a nice description of $B_2$ as the circumcenter of $BB_1B_1'$. Now, note that: \begin{align*} \angle{C_1B_2B_1} &= \frac{1}{2}\angle{B_1'B_2B_1} \\ &= \angle{B_1'BB_1} \\ &= \angle{B_1'C_1A_1} + \angle{C_1BB_1} \\ &= \angle{ABB_1} + \angle{BAB_1} \\ &= \angle{BB_1C} \end{align*} This implies that $AB_1CA_1BC_1A$ and $A_1B_2C_1A_2B_1C_2A_1$ are similar figures; thus proving the problem.

We proceed similarly to this problem. Let $M_aM_bM_c$ be the medial triangle of $ABC$; let $H_a=M_bM_c\cap B_1C_1$ and similarly $H_b,H_c$; let $N_a=M_bM_c\cap AA_1$ and similarly $N_b,N_c$. Finally, let $O$ be the Miquel point of $A_1B_1C_1$ wrt $ABC$. By the cyclicities, we have $\measuredangle BOC=\measuredangle BOA_1+\measuredangle A_1OC=\measuredangle BC_1A_1+\measuredangle A_1B_1C=\measuredangle C_1AB_1+\measuredangle B_1A_1C_1=2\measuredangle A$ But the only point which satisfies all these three conditions is the circumcenter of $ABC$. Furthermore since $\measuredangle (A_1O,B_1C_1)=\measuredangle OA_1C_1+\measuredangle A_1C_1B_1=\measuredangle OBA+\measuredangle A_1BO+\measuredangle OAB_2=\frac{\pi-\measuredangle A+\pi-\measuredangle B+\pi-\measuredangle C}{2}=\frac{\pi}{2}$ $O$ is the orthocentre of $A_1B_1C_1$. This is true for all choices of $A_1,B_1,C_1$ so all these triangles are directly similar to each other, and in particular with $M_aM_bM_c$ (with center of similarity $O$). In particular this implies that the foot of the perpendicular from $A_1$ to $B_1C_1$ is $H_a$; indeed assume that this foot was $H_a'$, and $K_a$ is the foot of $M_a$ on $M_bM_c$. By the similarity of the two triangles, we must also have that of $OA_1M_a$ and $OH_a'K_a$, which implies $H_a'K_a\perp OK_a\implies H_a'\in M_bM_c\implies H_a'\equiv H_a$. Since $A_2$ lies on the axis of $AA_1$, it follows that $N_aA_2\perp AA_1$. So we now have, by the direct similarity and the cyclicity of $A_2N_aA_1H_a$ (because of the two right angles), that $$\measuredangle A_1AA_2=\measuredangle A_2A_1A=\measuredangle A_2A_1N_a$$$$=\measuredangle A_2H_aN_a=\measuredangle (B_1C_1,M_bM_c)$$$$=\measuredangle (C_1A_1,M_cM_a)=\cdots =\measuredangle B_1BB_2=\measuredangle B_2B_1B$$$$=\cdots =\measuredangle C_1CC_2=\measuredangle C_2C_1C$$So we in fact have $AA_1A_2\sim BB_1B_2\sim CC_1C_2$. Now the conclusion can be seen to follow through complex numbers. The last fact means there is a number $z$ such that $\frac{a_2-a}{a_1-a}=\frac{b_2-b}{b_1-b}=\frac{c_2-c}{c_1-c}=z$ (in fact we have $Im(z)=\frac 12$). Now, by the similarities, we have $$\frac{c-a}{b-a}=\frac{c_1-a_1}{b_1-a_1}=\frac{(1-z)(c-a)+z(c_1-a_1)}{(1-z)(b-a)+z(b_1-a_1)}=\frac{(c+z(c_1-c))-(a+z(a_1-a)}{(b+z(b_1-b))-(a+z(a_1-a))}=\frac{c_2-a_2}{b_2-a_2}$$which finally implies the thesis.

Let \(A'\) be the reflection of \(A_1\) over \(\overline{B_1C_1}\), and define \(B'\) and \(C'\) similarly. Then \(\measuredangle B_1A'C_1=\measuredangle C_1A_1B_1=\measuredangle CAB\), so \(A'\) lies on \((AB_1C_1)\), etc. Moreover \(\measuredangle A'AB=\measuredangle A'B_1C_1=\measuredangle CBA\), implying \(\overline{AA'}\parallel\overline{BC}\), etc. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair O,A,B,C,A1,B1,C1,Ap,Bp,Cp,A2,B2,C2; real t=30; O=origin; A=dir(110); B=dir(210); C=dir(330); A1=extension(B,C,O,rotate(t)*(B+C)); B1=extension(C,A,O,rotate(t)*(C+A)); C1=extension(A,B,O,rotate(t)*(A+B)); Ap=reflect(B1,C1)*A1; Bp=reflect(C1,A1)*B1; Cp=reflect(A1,B1)*C1; A2=circumcenter(A,A1,Ap); B2=circumcenter(B,B1,Bp); C2=circumcenter(C,C1,Cp); draw(circumcircle(A,A1,Ap),gray+dashed+linewidth(.3)); draw(A--Ap,gray+linewidth(.8)); draw(B--Bp,gray+linewidth(.8)); draw(C--Cp,gray+linewidth(.8)); draw(circumcircle(A,B1,C1),linewidth(.3)); draw(circumcircle(B,C1,A1),linewidth(.3)); draw(circumcircle(C,A1,B1),linewidth(.3)); draw(A2--B2--C2--cycle,gray+linewidth(.9)); draw(A1--B1--C1--cycle,linewidth(.9)); draw(A--B--C--cycle,linewidth(1.2)); dot("\(A\)",A,N); dot("\(B\)",B,W); dot("\(C\)",C,SE); dot("\(A_1\)",A1,SW); dot("\(B_1\)",B1,E); dot("\(C_1\)",C1,dir(190)); dot("\(A_2\)",A2,dir(120)); dot("\(B_2\)",B2,dir(220)); dot("\(C_2\)",C2,SE); dot("\(A'\)",Ap,N); dot("\(B'\)",Bp,S); dot("\(C'\)",Cp,E); [/asy][/asy] Note \(\measuredangle(\overline{BB'},\overline{CC'})=\measuredangle C_1A_1B_1\) implies \(\measuredangle(\overline{BB'},\overline{C_1A_1})=\measuredangle(\overline{CC'},\overline{A_1B_1})\). But since \(\overline{C_1A_1}\perp\overline{B_1B'}\) and \(\overline{A_1B_1}\perp\overline{C_1C'}\), we have \(\measuredangle BB'B_1=\measuredangle CC'C_1\). Since \(B_2\) and \(C_2\) are the circumcenters of \(\triangle BB'B_1\) and \(\triangle CC'C_1\), we have (by symmetry) that \(\triangle AA_2A_1\sim\triangle BB_2B_1\sim\triangle CC_2C_1\). Since \(\triangle ABC\sim\triangle A_1B_1C_1\), it follows from mean geometry that \(\triangle ABC\sim\triangle A_2B_2C_2\) as well. Remark: [Rigorous ``mean geometry''] We justify that the problem follows from \(\triangle AA_2A_1\sim\triangle BB_2B_1\sim\triangle CC_2C_1\) via complex numbers. Indeed there is \(\lambda\in\mathbb C\) such that \(A_2=\lambda A+(1-\lambda)A_1\), etc.; hence \[\frac{B_2-A_2}{C_2-A_2}=\frac{\lambda(B-A)+(1-\lambda)(B_1-A_1)}{\lambda(C-A)+(1-\lambda)(C_1-A_1)} =\frac{B-A}{C-A}\]since \(\frac{B_1-A_1}{C_1-A_1}=\frac{B-A}{C-A}\).

This problem was proposed by Fedir Yudin (Mindstormer on AoPS). Amazing configuration and a great problem, as always from Fedir. He is an extremely talented author! I am sure that a lot of his problems will appear in different competitions, including IMO, EGMO, RMM in the next few years. Can't wait to see them!

gives that $A'$, $B'$ and $C'$ lie on $\odot{(AB_1C_1)}$, $\odot{(BA_1C_1)}$, and $\odot{(CA_1B_1)}$, respectively. Claim.$\triangle AA_2A_1\sim\triangle BB_2B_1\sim\triangle CC_2C_1$.

Lemma(well-known). In convex quadrilateral $ABCD$ $\angle A = \angle C$. Let perpendicular bisector of $\overline{AC}$ cuts $\overline{BD}$ at $X$. Then $\angle BCX = \angle DAC$ and $\angle ACD = \angle XAB$. Claim. $\triangle B_2C_1B_1\sim\triangle B_1CB$.

gives $\triangle A_2B_2C_2 \sim \triangle A_1B_1C_1 \sim \triangle ABC$. $\blacksquare$

nixon0630 wrote:

gives that $A'$, $B'$ and $C'$ lie on $\odot{(AB_1C_1)}$, $\odot{(BA_1C_1)}$, and $\odot{(CA_1B_1)}$, respectively. Claim.$\triangle AA_2A_1\sim\triangle BB_2B_1\sim\triangle CC_2C_1$.

Lemma(well-known). In convex quadrilateral $ABCD$ $\angle A = \angle C$. Let perpendicular bisector of $\overline{AC}$ cuts $\overline{BD}$ at $X$. Then $\angle BCX = \angle DAC$ and $\angle ACD = \angle XAB$. Claim. $\triangle B_2C_1B_1\sim\triangle B_1CB$.

gives $\triangle A_2B_2C_2 \sim \triangle A_1B_1C_1 \sim \triangle ABC$. $\blacksquare$ nice explanation!

Define a complex coordinate system in which the points $A$, $B$, $C$, $A_1$, $B_1$, $C_1$ have coordinates $a$, $b$, $c$, $az$, $bz$, $cz$ respectively for certain complex $a$, $b$, $c$ and $z$. If $z \in \mathbb{R}$, it is easy to see that the only possibility is $z = -\frac{1}{2}$. Now assume $z$ has a non-zero complex component. There exist real numbers $\lambda$, $\mu$, $\nu$ such that $az = b\lambda + c(1-\lambda)$, $bz = c\mu + a (1-\mu)$ and $cz = a\nu + b(1-\nu)$. This can be rewritten as $M \begin{bmatrix} a \\ b \\ c\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$, where $M = \begin{bmatrix} -z & \lambda & 1-\lambda \\ 1-\mu & -z & \mu \\ \nu & 1-\nu & -z \end{bmatrix} $. This implies $0 = -\det M = z^3+(r-1)z-r=(z-1)(z^2+z+r)$, where $r = \lambda\mu+\mu\nu+\nu\lambda-\lambda-\mu-\nu+1 \in \mathbb{R}$. Hence $z$ is a non-real root of $z^2+z+r=0$, meaning that just as in the real case, $\operatorname{Re} z = -\frac{1}{2}$ and $|z|$=$|z+1|$. It remains to notice that $|az^2-az| = |a||z||z-1| = |a||z+1||z-1| = |az^2-a|$, so $az^2$ is the only possible coordinate for $A_2$. The other two points are handled in the same exact way.

oops heinous. We show an equivalent result that clearly suffices: if $A_2$ is chosen on segment $B_1C_1$ such that $A_2B_1/B_1C_1 = A_1B/BC$, then it lies on the perpendicular bisector of $AA_1$. This and symmetric results would imply the problem. Let $M$ be the spiral center mapping $\triangle ABC\to \triangle A_1B_1C_1$. The same similarity about $M$ maps $\triangle A_1B_1C_1\to \triangle A_2B_2C_2$. We work with complex numbers momentarily; let $M = 0$ and $A=1$ by translation, rotation, and scaling, and let $A_1=z$. Then we seek to show \[|z-1||z|=|z^2-z| =A_1A_2 = AA_2=|z^2-1|=|z-1||z+1|.\]As $|z-1|\ne 0$ because otherwise $z=1$ and $A=A_1,B=B_1,C=C_1$, absurd, we get $|z|=|z+1|$. This is equivalent to the assertion that $z$ has real part $-\frac 12$. That is, the foot $A'$ from $A_1$ to line $AM$ has $MA' = \frac 12 AM$. We now discard $A_2,B_2,C_2$ to focus on proving $MA' = \frac 12AM$. Relabel $A_1$ as $D$, $B_1$ as $E$, $C_1$ as $F$. Let $X_A$ denote the point along segment $AD$ such that $AX_A = \frac 12 DX_A$. Define $X_B,X_C$ symmetrically. Let $\omega_A$ be the circle about $X_A$ passing through $A$ and define $\omega_B,\omega_C$ symmetrically. Claim: These three circles concur at a point $M'$. Solution: To do so, by Miquel's Theorem, it suffices to show that $\omega_B$ and $\omega_C$ intersect along line $BC$ and symmetric variations. By symmetry, we only show $\omega_B$ and $\omega_C$ intersect along line $BC$. Let $\omega_B$ intersect $BC$ at $Z_1\ne B$ and $\omega_C$ intersect $BC$ at $Z_2\ne C$. The goal is to show that $Z_1=Z_2$. Let $Y_B$ be the point on segment $BE$ such that $BY_B = EX_B$. Define $Y_C$ similarly. It is clear that the foot from $Y_B$ to $BC$ must be $Z_1$ because $BY_B = 2BX_B$. Similarly, $Y_CZ_2\perp BC$. Thus it suffices to demonstrate $Y_BY_C\perp BC$. Shift the triangle so $B=(0,0)$ and $C=(1,0)$. For $P=(x,y)$ in the plane, let $v(P)=x$ be a function. The desired result $Y_BY_C\perp BC$ is equivalent to \[\frac 23 v(E) = \frac 23 v(E)+\frac 13 v(B) = v(Y_B)=v(Y_C)=\frac 23 v(F) + \frac 13 v(C) = \frac 23 v(F)+\frac 13.\]Thus we wish to show that $v(E)-v(F) = \frac 12$, so the projection of $EF$ onto line $BC$ has length half of $BC$.

After this bash, the claim is resolved. $\fbox{}$ We now argue that $M'=M$. This suffices because the foot $A''$ from $D$ to line $AM'$ must satisfy $A''M = \frac 12 AM'$ by definition. Note that $M'$ is the center of the spiral similarity mapping $AY_A$ to $BY_B$ by definition, as $Y_AY_B$ intersects $AB$ at the same place as the intersection of $\omega_A$ and $\omega_B$ that is not $M'$. Thus, by symmetry, if we define $B''$ and $C''$ symmetrically, $M'$ is the center of spiral similarity mapping $\triangle M'DA''\to \triangle M'EB''\to \triangle M'FC''$. As this implies a spiral similarity about $M'$ in the form of $\triangle DEF\to \triangle A''B''C''\to \triangle ABC$, we are done.

Why didn't I realize the two (degenerate) hexagons are actually similar ? Let the circumcircle of $A_1B_1C_1$ be $\omega$, the medial triangle of $ABC$ be $DEF$, the circumcenter of $ABC$ be $O$, $X = AA_1 \cap EF$, $Y = BB_1 \cap FD$, $Z = CC_1 \cap DE$, the reflection of $A$ over $B_1C_1$ be $A_3$, and the points $B_3$ and $C_3$ be defined similarly. Now, we can also define $M = AA_3 \cap B_1C_1$, $N = BB_3 \cap C_1A_1$, and $P = CC_3 \cap A_1B_1$. By homothety, we know $X, Y, Z$ are the midpoints of $AA_1, BB_1, CC_1$ respectively, which means $A_2X \perp AA_1$, $B_2Y \perp BB_1$, and $C_2Z \perp CC_1$. It's clear that $ABC$ and $A_1B_1C_1$ are directly similar, so we have $$DEF \overset{+}{\sim} ABC \overset{+}{\sim} A_1B_1C_1.$$Now, if $K$ is the Spiral Center mapping $DEF$ to $A_1B_1C_1$, then we know $K$ lies on $(AEF)$, $(BFD)$, and $(CDE)$ by the Spiral Center Lemma. Thus, $K$ must coincide with $O$, so the aforementioned lemma implies $(AB_1C_1)$, $(BC_1A_1)$, and $(CA_1B_1)$ all contain $O$. It's easy to see $$\measuredangle B_1A_1C_1 = \measuredangle BAC = \measuredangle C_1AB_1 = -\measuredangle C_1A_3B_1 = \measuredangle B_1A_3C_1$$so $A_3$ lies on $\omega$. Analogous angle chasing gives $B_3 \in \omega$ and $C_3 \in \omega$. It's well-known that $O$ is the orthocenter of $DEF$, so $O$ is also the orthocenter of $A_1B_1C_1$. This implies $A_1O \perp B_1C_1 \perp AA_3$, which yields $A_1O \parallel AA_3$. Similarly, we find $B_1O \parallel BB_3$ and $C_1O \parallel CC_3$. Thales' implies $AA_2MX$ and $BB_2YN$ are cyclic. In addition, properties of midlines gives $MX \parallel A_3A_1$, so we can compute $$\measuredangle AA_2X = \measuredangle AMX = \measuredangle AA_3A_1 = \measuredangle OA_1A_3$$$$= \measuredangle B_1A_1A_3 - \measuredangle B_1A_1O = \measuredangle B_1C_1A_3 - \measuredangle B_1CO$$$$= -\measuredangle B_1C_1A - \measuredangle ACO = \measuredangle AC_1B_1 - \measuredangle OAC$$$$= \measuredangle AOB_1 + \measuredangle CAO = \measuredangle AOB_1 + \measuredangle B_1AO = \measuredangle AB_1O.$$An analogous angle chase yields $\measuredangle BB_2Y = \measuredangle BC_1O$, implying $$\measuredangle AA_2X = \measuredangle AB_1O = \measuredangle AC_1O = \measuredangle BC_1O = \measuredangle BB_2Y.$$Now, we have $AA_2X \overset{+}{\sim} BB_2Y$, which easily gives $AA_1A_2 \overset{+}{\sim} BB_1B_2$. Hence, analogous angle chasing and transitivity of similarity yields $$AA_1A_2 \overset{+}{\sim} BB_1B_2 \overset{+}{\sim} CC_1C_2.$$Let $T$ be the unique point such that $TABC \overset{+}{\sim} TA_1B_1C_1$. Then, the Gliding Principle implies $TAA_1 \overset{+}{\sim} TBB_1 \overset{+}{\sim} TCC_1$, so $$TAA_1A_2 \overset{+}{\sim} TBB_1B_2 \overset{+}{\sim} TCC_1C_2.$$Now, applying the Gliding Principle again yields $$TABC \overset{+}{\sim} TA_1B_1C_1 \overset{+}{\sim} TA_2B_2C_2$$which finishes. $\blacksquare$

Let $O$ be the circumcenter of $\triangle A_1B_1C_1$, and let $O_A$, $O_B$, $O_C$ be the reflections of $O$ over the sides of $\triangle A_1B_1C_1$. Let $A'$, $B'$ and $C'$ be the reflections of $A_1$, $B_1$ and $C_1$ over the opposite sides. Note that $A_1, B_1, C, C'$ are concyclic with $O_C$ as their circumcenter, and that $C_2$ is the circumcenter of $\triangle C_1CC'$. Therefore, $\overline{C_2O_C}$ is perpendicular to $\overline{CC'}$. Similarly, $\overline{B_2O_B}$ is perpendicular to $\overline{BB'}$. Hence, \[ \measuredangle OC_2A_1 = \measuredangle A_1C_2O_C = \measuredangle C_1C'C = \measuredangle HC'C = \measuredangle HA_1C, \]where the last equality follows since the orthocenter $H$ of $\triangle A_1B_1C_1$ lies on $(A_1B_1C)$. Similarly, $\measuredangle OB_2A = \measuredangle HA_1B = \measuredangle HA_1C$, so $(A_1B_2C_2)$ passes through $O$. Consequently, we get $\triangle OA_2O_A \sim \triangle OB_2O_B \sim \triangle OC_2O_C$, and hence $O$ is the spiral similarity center mapping $\triangle A_2B_2C_2$ to $\triangle O_AO_BO_C$. Since the latter is similar to $\triangle ABC$, we are done.

complex bash haunts me. not sure what to think. It is fairly clear that we can assume they are directly similar, else no such triangle exists. Let $A'$ be the reflection of $A_1$ over $\overline{B_1C_1}$, so $AA'B_1C_1$ is cyclic, and define $B',C'$ similarly. Since $A_2$ lies on the perpendicular bisector of $A'A_1$ and $AA_1$, it follows that is it the center of $(AA'A_1)$, hence it also lies on the perpendicular bisector of $\overline{AA'}$. Symmetric statements are true for $B_2,C_2$. Furthermore note that $(AA'A_1),(BB'B_1),(CC'C_1)$ concur, and this concurrency point is the orthocenter of $\triangle A_1B_1C_1$ by orthocenter reflections. Now fix $\triangle A_1B_1C_1$ and let $A$ move along $(A'B_1C_1)$, defining $B=\overline{AC_1} \cap (B'A_1C_1)$ and $C=\overline{AB_1} \cap (C'A_1B_1)$. I claim that $B,C,A_1$ are always collinear. To see this, note that this is true when triangle $A_1B_1C_1$ is the medial triangle of $ABC$, and that as $A$ moves at constant speed around $(A'B_1C_1)$ then $B$ and $C$ clearly also move at the same speed and direction (CW/CCW) around their respective circles, but so does $C^*:=\overline{BA_1} \cap (C'A_1B_1)$, hence $C^*=C$ always holds. We now use complex numbers. Let $A_1B_1C_1$ lie on the unit circle with $A_1=a,B_1=b,C_1=c$. The center of $(A'B_1C_1)$ is clearly $b+c$, and symmetric variants hold too, so by our above claim let $z$ be a complex number on the unit circle and set $A=b+c-az,B=c+a-bz,C=a-b+cz$. The $z=1$ case corresponds to $A_1B_1C_1$ being the medial triangle. The reflection of $A$ over $\overline{B_1C_1}$ is then $-\tfrac{bc}{az}$, hence the reflection of the two arc midpoints of $\widehat{A'A}$ in $(A'B_1C_1)$ are $\pm \sqrt{\tfrac{bc}{z}}$. Then by the complex intersection formula, we find $$A_2=\frac{\frac{bc}{z}(b+c)}{bc+\frac{bc}{z}}=\frac{b+c}{z+1},$$and similarly $B_2=\tfrac{c+a}{z+1},C_2=\tfrac{a+b}{z+1}$. Thus $\triangle A_2B_2C_2$ is similar to the triangle with vertices $\tfrac{b+c}{2},\tfrac{c+a}{2},\tfrac{a+b}{2}$, which is in turn similar to $\triangle ABC$. $\blacksquare$ EDIT: wait bruh you can actually just angle chase to get $BA_1CB_1AC_1 \sim B_1A_2C_1B_2A_1C_2$ after noting circumcenter i just cant do anything that requires more than one step on my 2 inch diagram rip

This is really good. Reflect $A_1$ over $B_1C_1$ to $A'$. Then, $AA'B_1C_1$ is cyclic so \[\measuredangle B_1AA' = \measuredangle B_1C_1A' = \measuredangle A_1C_1B_1 = \measuredangle ACB\]so $AA' \parallel BC$ and also $A_2A = A_2A_1=A_2A_1$. Now, \[\angle A_1A_2B_1 = \frac 12 \angle A_1A_2A' = \angle A_1AA' = \angle AA_1B\]and we conclude since $A_2$ and $A_1$ are in the same position on $B_1C_1$ and $BC$ which finishes by symmetry. Remark: The motivation for something like this is by noticing that it's really hard to compute angles of $A_2B_2C_2$. Thus we want to show the siilarity some other way, and this way leads to a clean solution i guess.

This lemma is important: Quote: Take two similar polygons $\mathcal P = P_1 P_2 \dots P_K$ and $\mathcal Q = Q_1 Q_2 \dots Q_k$. If $\mathcal R = R_1 R_2 \dots R_k$ is such that $P_1 Q_1 R_1 \sim P_2 Q_2 R_2 \sim \dots \sim P_k Q_k R_k$ (with the all orientations the same), then $\mathcal P \sim \mathcal Q \sim \mathcal R$. It is easy to show with complex numbers. Because of this, it suffices to show that $\triangle A_2 A_1 A \sim \triangle B_2 B_1 B \sim \triangle C_2 C_1 C$, or equivalently, that $\angle A_2 A_1 A = \angle B_2 B_1 B = \angle C_2 C_1 C$. Since $\angle B_1 A_1 C_1 = \angle B_1 A C_1$, the midpoint $A_3$ of $\overline{AA_1}$ lies on the nine-point circle of $\triangle A_1 B_1 C_1$. Similarly, $B_3$ and $C_3$ lie on the nine-point circle, and due to our lemma, it follows that if $\triangle KLM$ midtriangle of $\triangle A_1 B_1 C_1$, traingles $KLM$ and $A_3 B_3 C_3$ are rotations about the nine-point circle. So, letting $DEF$ be the orthic triangle of $\triangle A_1 B_1 C_1$, it follows that \[\angle A_3 DK = \angle B_3 E L = \angle C_3 F M.\]And since $A_1 DKA_3$ is cyclic, we have $\angle A_3 DK = \angle A_2 A_1 A_3 = \angle A_2 A_1 A$, so we're done.