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One might want to read the official solutions (in Russian): [url]https://olympiads.mccme.ru/vmo/2022/final/day2.pdf [/url]

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.1255177985478038, xmax = 15.041779807182593, ymin = -5.606884619441723, ymax = 3.9453927871822243; /* image dimensions */ /* draw figures */ draw(circle((7.82,-0.82), 4.585255823591103), linewidth(1)); draw(circle((5.449940468924523,0.17151047149406307), 2.0192354176690226), linewidth(1)); draw(circle((9.571319950354923,0.015044357846555069), 2.6504010635306923), linewidth(1) + linetype("4 4")); draw((5.82156001922194,2.1562549340901267)--(5.336737596504696,-0.4005092991546239), linewidth(1)); draw((8.436817354602653,-2.380269746356897)--(9.03797817195929,-1.3394640052566549), linewidth(1)); draw((9.03797817195929,-1.3394640052566549)--(6.997625234472924,0.6480698762940378), linewidth(1)); draw((9.03797817195929,-1.3394640052566549)--(11.21781861962519,-2.061892793262225), linewidth(1)); draw((5.336737596504696,-0.4005092991546239)--(6.069610096000828,-1.7502910857037934), linewidth(1)); draw((5.336737596504696,-0.4005092991546239)--(4.322139599228562,-1.5034154582956776), linewidth(1)); draw((3.7427006533581038,-2.917665608634915)--(11.21781861962519,-2.061892793262225), linewidth(1) + blue); draw((12.04781446277594,-2.594867836239848)--(6.997625234472924,0.6480698762940378), linewidth(1) + blue); draw((5.750728454955485,3.271782770461454)--(8.436817354602653,-2.380269746356897), linewidth(1) + blue); draw((5.750728454955485,3.271782770461454)--(6.069610096000828,-1.7502910857037934), linewidth(1) + red); draw((3.7427006533581038,-2.917665608634915)--(5.82156001922194,2.1562549340901267), linewidth(1) + red); draw((4.322139599228562,-1.5034154582956776)--(12.04781446277594,-2.594867836239848), linewidth(1) + red); draw((5.750728454955485,3.271782770461454)--(3.7427006533581038,-2.917665608634915), linewidth(1)); draw((3.7427006533581038,-2.917665608634915)--(12.04781446277594,-2.594867836239848), linewidth(1)); draw((12.04781446277594,-2.594867836239848)--(5.750728454955485,3.271782770461454), linewidth(1)); /* dots and labels */ dot((5.750728454955485,3.271782770461454),dotstyle); label("$A$", (5.804679066259501,3.387772532480826), NE * labelscalefactor); dot((3.7427006533581038,-2.917665608634915),dotstyle); label("$B$", (3.9136190720547734,-3.182448729435645), NE * labelscalefactor); dot((12.04781446277594,-2.594867836239848),dotstyle); label("$C$", (11.683936099524198,-2.8551498842848244), NE * labelscalefactor); dot((4.322139599228562,-1.5034154582956776),linewidth(4pt) + dotstyle); label("$A_1$", (4.531850224006319,-1.4610992475313298), NE * labelscalefactor); dot((6.069610096000828,-1.7502910857037934),linewidth(4pt) + dotstyle); label("$B_1$", (6.144100090860349,-1.5823210420316336), NE * labelscalefactor); dot((5.82156001922194,2.1562549340901267),linewidth(4pt) + dotstyle); label("$C_1$", (5.913778681309774,1.7755226656267844), NE * labelscalefactor); dot((11.21781861962519,-2.061892793262225),linewidth(4pt) + dotstyle); label("$A_2$", (11.053582768122622,-1.9096198871824541), NE * labelscalefactor); dot((6.997625234472924,0.6480698762940378),linewidth(4pt) + dotstyle); label("$B_2$", (6.603231627761622,0.4420829261234415), NE * labelscalefactor); dot((8.436817354602653,-2.380269746356897),linewidth(4pt) + dotstyle); label("$C_2$", (8.229114956265562,-2.7218059103344903), NE * labelscalefactor); dot((7.544175791074714,10.39525971767555),linewidth(4pt) + dotstyle); label("$T$", (1.1255177985478038,3.9453927871822243), NE * labelscalefactor); dot((9.03797817195929,-1.3394640052566549),linewidth(4pt) + dotstyle); label("$T_2$", (9.089789697217713,-1.2429000174307827), NE * labelscalefactor); dot((5.336737596504696,-0.4005092991546239),linewidth(4pt) + dotstyle); label("$T_1$", (5.428891503308561,-0.38222527647862503), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Denote $\bigtriangleup A_1B_1C_1$ by the red triangle, and $\bigtriangleup A_2B_2C_2$ by the blue triangle (the labelling of the vertices are symmetric, i.e. $A_i$ is the intersection of lines from $B$ and $C$, etc.). The circles $\odot(A_1BC), \odot(AB_1C), \odot(ABC_2)$ are concurrent at a point by simple angle chasing; denote $T_1$ by their concurrency point, and define $T_2$ similarly. It's easy to see that $(A_1,A_2), (B_1, B_2), (C_1, C_2)$ are pairs of isogonal conjugates. Let $\Phi$ be an inversion centered at $T_1$ with any radius. The pith is the following claim. Main Claim. $\triangle ABC \cup \triangle A_2B_2C_2 \sim \triangle \Phi(A_1)\Phi(B_1)\Phi(C_1) \cup \triangle \Phi(A)\Phi(B)\Phi(C)$. Proof. We divide the proof into two steps: Step 01. $\triangle ABC \sim \triangle \Phi(A_1)\Phi(B_1)\Phi(C_1)$ Proof. Angle chasing: \begin{align*} \measuredangle \Phi(C_1)\Phi(B_1)\Phi(A_1) &= \measuredangle \Phi(C_1)\Phi(B_1)T_1 + \measuredangle T_1\Phi(B_1)\Phi(A_1) \\ &= \measuredangle T_1C_1B_1 + \measuredangle B_1A_1T_1 \\ &= \measuredangle T_1C_1A + \measuredangle CA_1T_1 \\ &= \measuredangle T_1BA + \measuredangle CBT_1 = \measuredangle CBA. \end{align*}Similarly, we prove that $\measuredangle \Phi(B_1)\Phi(A_1)\Phi(C_1) = \measuredangle BAC$ and $\measuredangle \Phi(A_1)\Phi(C_1)\Phi(B_1) = \measuredangle ACB$. Therefore, the two triangles are indeed similar. Step 02. $\measuredangle \Phi(C)\Phi(B_1)\Phi(A_1) = \measuredangle C_2BA$ Again, angle chasing: \begin{align*} \measuredangle \Phi(C)\Phi(B_1)\Phi(A_1) &= \measuredangle \Phi(C)\Phi(B_1)T_1 + \measuredangle T_1\Phi(B_1)\Phi(A_1) \\ &= \measuredangle T_1CB_1 + \measuredangle B_1A_1T_1 \\ &= \measuredangle T_1CA_1 + \measuredangle CA_1T_1 \\ &= \measuredangle CT_1A_1 = \measuredangle CBA_1 = \measuredangle CBC_1 = \measuredangle C_2BA. \end{align*}From the last step above, we can prove similarly the other equality $\measuredangle \Phi(C)\Phi(A_1)\Phi(B_1) = \measuredangle C_2AB$. Therefore, $\triangle \Phi(C)\Phi(A_1)\Phi(B_1) \sim \triangle C_2AB \implies \triangle ABC \cup C_2 \sim \triangle \Phi(A_1)\Phi(B_1)\Phi(C_1) \cup \Phi(C)$. Similarly for the other two ($A_2$ and $B_2$). Therefore, the main claim is proven. $\blacksquare$ From the main claim above, we conclude that: \begin{align*} &\odot(ABC) \text{ is tangent to } \odot(A_1B_1C_1) \\ \iff &\odot(\Phi(A)\Phi(B)\Phi(C)) \text{ is tangent to } \odot(\Phi(A_1)\Phi(B_1)\Phi(C_1)) \\ \iff &\odot(A_2B_2C_2) \text{ is tangent to } \odot(ABC), \end{align*}as desired. $\square$ Remark: $T_1, T_2$ are isogonal conjugates w.r.t. $\bigtriangleup ABC$. They also satisfy: $\triangle ABC \cup \triangle A_2B_2C_2 \cup T_2 \sim \triangle \Phi(A_1)\Phi(B_1)\Phi(C_1) \cup \triangle \Phi(A)\Phi(B)\Phi(C) \cup T_1$.

Very beautifull problem!!! Surprisingly enough: https://artofproblemsolving.com/community/c6h1724021

If interested, I proved the generalization to this nice problem here