Let $ABC$ be an acute triangle and suppose points $A, A_b, B_a, B, B_c, C_b, C, C_a,$ and $A_c$ lie on its perimeter in this order. Let $A_1 \neq A$ be the second intersection point of the circumcircles of triangles $AA_bC_a$ and $AA_cB_a$. Analogously, $B_1 \neq B$ is the second intersection point of the circumcircles of triangles $BB_cA_b$ and $BB_aC_b$, and $C_1 \neq C$ is the second intersection point of the circumcircles of triangles $CC_aB_c$ and $CC_bA_c$. Suppose that the points $A_1, B_1,$ and $C_1$ are all distinct, lie inside the triangle $ABC$, and do not lie on a single line. Prove that lines $AA_1, BB_1, CC_1,$ and the circumcircle of triangle $A_1B_1C_1$ all pass through a common point. Josef Tkadlec (Czech Republic), Patrik Bak (Slovakia)
Problem
Source: 2021 Czech-Polish-Slovak Match, P6
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03.08.2021 22:49
By Miquel's theorem, let $X=(AA_bC_a)\cap (BB_cA_b)\cap (CC_aB_c)$ and similarly $Y=(AA_cB_a)\cap (BB_aC_b)\cap (CC_bA_c)$. Now, by cyclic quadrilaterals, we have $\measuredangle XA_1Y=\measuredangle XA_1A-\measuredangle YA_1A=\measuredangle XC_aA-\measuredangle YAA_c$. However this expression is cyclic, since it is well known (by cyclic quads), that $\measuredangle YA_cA=\measuredangle YB_aB=\measuredangle YC_bC$ and similarly $\measuredangle XC_aA=\measuredangle XA_bB=\measuredangle XB_cC$. Therefore the pentagon $XYA_1B_1C_1$ is cyclic with circumcenter $\omega$. Now let $Z_a=AA_1\cap \omega$, and cyclically define the points $Z_b$ and $Z_c$. Now we must have $\measuredangle XA_1Z_a=\measuredangle XA_1A=\measuredangle XC_aA$. Once again, this expression is cyclic, and therefore $\measuredangle XA_1Z_a=\measuredangle XB_1Z_b=\measuredangle XC_1Z_c$, implying $Z_a\equiv Z_b\equiv Z_c=: Z$. Thus the three lines $AA_1,BB_1,CC_1$ and the circle $\omega=(XYA_1B_1C_1)$ all concur in the point $Z$