Ceva and POP , a similar definiton like this problem was in the plane geometry book which they considerd it to be cyclocevian conjugate.

This problem was proposed by me. One can prove similar results if the points are in any Conic Sections. I never seen this problem before and hopefully I discovered this first.

By Ceva's theorem, $AA_1, BB_1, CC_1$ are concurrent if and only if $$ \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A} \cdot \frac{AC_1}{C_1B} = 1 $$Similarly, $AA_2, BB_2, CC_2$ are concurrent if and only if $$ \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A} \cdot \frac{AC_2}{C_2B} = 1 $$Since $A_1, A_2, B_1, B_2, C_1, C_2$ lie on a circle, by the intersecting chords theorem, we have $$ BA_1 \cdot BA_2 = BC_1 \cdot BC_2 $$$$ CA_1 \cdot CA_2 = CB_1 \cdot CB_2 $$$$ AB_1 \cdot AB_2 = AC_1 \cdot AC_2 $$Dividing the first equation by the second, we have $$ \frac{BA_1 \cdot BA_2}{CA_1 \cdot CA_2} = \frac{BC_1 \cdot BC_2}{CB_1 \cdot CB_2} \implies \frac{BA_1}{CA_1} \cdot \frac{BA_2}{CA_2} = \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2} $$Dividing the third equation by the first, we have $$ \frac{AB_1 \cdot AB_2}{BA_1 \cdot BA_2} = \frac{AC_1 \cdot AC_2}{BC_1 \cdot BC_2} \implies \frac{AB_1}{BA_1} \cdot \frac{AB_2}{BA_2} = \frac{AC_1}{BC_1} \cdot \frac{AC_2}{BC_2} $$Dividing the second equation by the third, we have $$ \frac{CA_1 \cdot CA_2}{AB_1 \cdot AB_2} = \frac{CB_1 \cdot CB_2}{AC_1 \cdot AC_2} \implies \frac{CA_1}{AB_1} \cdot \frac{CA_2}{AB_2} = \frac{CB_1}{AC_1} \cdot \frac{CB_2}{AC_2} $$Assume that $AA_1, BB_1, CC_1$ are concurrent. Then, we have $$ \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A} \cdot \frac{AC_1}{C_1B} = 1 $$We want to show that $AA_2, BB_2, CC_2$ are concurrent, which is equivalent to $$ \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A} \cdot \frac{AC_2}{C_2B} = 1 $$From $\frac{BA_1}{CA_1} \cdot \frac{BA_2}{CA_2} = \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2}$, we have $$ \frac{BA_2}{CA_2} = \frac{CA_1}{BA_1} \cdot \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2} $$Similarly, from $\frac{AB_1}{BA_1} \cdot \frac{AB_2}{BA_2} = \frac{AC_1}{BC_1} \cdot \frac{AC_2}{BC_2}$, $$ \frac{AB_2}{BA_2} = \frac{BA_1}{AB_1} \cdot \frac{AC_1}{BC_1} \cdot \frac{AC_2}{BC_2} $$From $\frac{CA_1}{AB_1} \cdot \frac{CA_2}{AB_2} = \frac{CB_1}{AC_1} \cdot \frac{CB_2}{AC_2}$, $$ \frac{CA_2}{AB_2} = \frac{AB_1}{CA_1} \cdot \frac{CB_1}{AC_1} \cdot \frac{CB_2}{AC_2} $$Then, $$ \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A} \cdot \frac{AC_2}{C_2B} = \left( \frac{CA_1}{BA_1} \cdot \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2} \right) \left( \frac{AB_1}{CB_1} \cdot \frac{BA_2}{AB_2} \right) \left( \frac{AC_1}{CB_2} \cdot \frac{AB_2}{CA_2} \right) $$$$ = \left( \frac{CA_1}{BA_1} \cdot \frac{BC_1}{CB_1} \cdot \frac{BC_2}{CB_2} \right) \left( \frac{AB_1}{CB_1} \cdot \frac{BA_1}{AB_1} \cdot \frac{AC_1}{BC_1} \cdot \frac{AC_2}{BC_2} \right) \left( \frac{AC_1}{CB_2} \cdot \frac{AB_1}{CA_1} \cdot \frac{AC_1}{CB_1} \cdot \frac{AC_2}{CB_2} \right) $$Therefore, $$ \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A} \cdot \frac{AC_2}{C_2B} = \frac{CA_1}{BA_1} \cdot \frac{AB_1}{CB_1} \cdot \frac{AC_1}{CB_2} \cdot \frac{AC_1}{BC_1} \cdot \frac{BC_2}{CB_2} \cdot \frac{BC_1}{CB_1} \cdot \frac{AC_2}{BC_2} $$The same argument holds for the converse. So $AA_1, BB_1, CC_1 \text{ are concurrent} \iff AA_2, BB_2, CC_2 \text{ are concurrent}$

I found the exact same solution as #2 and #4 during the contest. Very nice problem btw, really enjoyed solving it Here is how I wrote it in the contest: Let $\Omega$ be the circle that points $A_1,A_2,B_1,B_2,C_1,C_2$ pass through From Power of the Point Theorem (POP) we get: $AC_1 \cdot AC_2=Pow(A,\Omega)=AB_2 \cdot AB_1 \implies AC_1 \cdot AC_2= AB_2 \cdot AB_1 \implies \boxed{\frac{AC_1}{AB_1}=\frac{AB_2}{AC_2}}$ $...(1)$ Simmilarly we find: $BA_2 \cdot BA_1=Pow(B, \Omega)=BC_2 \cdot BC_1 \implies BA_2 \cdot BA_1=BC_2 \cdot BC_1 \implies \boxed{\frac{BA_1}{BC_1}=\frac{BC_2}{BA_2} }$ $...(2)$ Finally we get: $CB_1 \cdot CB_2 = Pow(C,\Omega)=CA_1 \cdot CA_2 \implies CB_1 \cdot CB_2=CA_1 \cdot CA_2 \implies \boxed{\frac{CB_1}{CA_1}=\frac{CA_2}{CB_2} }$ $...(3)$

Ceva's theorem with power of a point works.

Okay, apparently this problem exists from before. (I found it in my notes for 2024 indian national olympiad camp.) Nonetheless, all the problems were quite nice from kosovo mo, and organising for 6 grades is a daunting task. So great job to the proposers and organisers.

What the sigma? Blud just stated the cyclocevian conjugate. Now that's rizz! https://mathworld.wolfram.com/CyclocevianConjugate.html