Do a negative inversion around the radical center. (If radical center doesn't exist, then it is trivial)

Homothety also works. ayan.nmath wrote: If radical center doesn't exist, then it is trivial Oh no, forgot to mention this

Attachments:
Problem 17.pdf (111kb)

Desargue anyone? Problem wrote: Three circles $\omega_1,\omega_2,\omega_3$ are given.let $A_0$ and $A_1$ be the common points of $\omega_1$ and $\omega_2$,similarly define $B_0,B_1,C_0,C_1$.Let $O_{ijk}$ be the circumcenter of $A_iB_jC_k$.Prove that the four lines of the form $O_{ijk}O_{1-i,1-j,1-k}$ are either concurrent or parallel. WLOG let $A_0,B_0,C_0$ lie in the outer parts of the circles and $A_1,B_1,C_1$ in the inner.Let $D,E,F$ be the centers of $\omega_1,\omega_2,\omega_3$ respectively. We will use desargues theorm repeatedly. Firstly note that if $A,B$ are points on a circle with center $O$ and $C,D$ are an arbitary points then the circumcenter of $(ABC),(ABD)$ and $O$ are collinear as all lie on the perpendicular bisector of $AB$.This simple looking observation is really powerful.As this gives the following triples of collinearities- $$ O_{010}-O_{110}-F ; O_{010}-O_{011}-E ; O_{001}-O_{011}-D ; $$$$ O_{001}-O_{101}-F ; O_{110}-O_{111}-E ; O_{101}-O_{111}-D ; $$$$ O_{011}-O{111}-F ; O_{100}-O_{101}-E ; O_{100}-O_{110}-D ; $$Thus $$F=O_{010}O_{110}\cap O_{001}O_{101}\cap O_{011}O_{111}$$$$E=O_{010}O_{011}\cap O_{110}O_{111}\cap O_{100}O_{101}$$$$D=O_{001}O_{011}\cap O_{101}O_{111}\cap O_{100}O_{110}$$Thus the following are perspective $$O_{010}O_{001}O_{011} \to O_{110}O_{101}O_{111}$$$$O_{010}O_{110}O_{100} \to O_{011}O_{111}O_{101}$$$$O_{001}O_{101}O_{100} \to O_{011}O_{111}O_{110}$$Thus by desargues theorm on these triplets we get the following collinearities $$O_{010}O_{001}\cap O_{110}O_{101}-D-E$$$$O_{100}O_{100} \cap O_{011}O_{101}-F-D$$$$O_{100}O_{001}\cap O_{011}O_{110}-E-F$$But note that by desargues's theorm again on the last collinearity only implies that $$O_{011}O_{010}O_{110} \to O_{100}O_{101}O_{001} $$which last but not the least again by desargue implies $O_{011}O_{100}; O_{010}O_{101}; O_{110}O_{001}$ concur which is what we want!Similar working gives $O_{000}O_{111};O_{010}O_{101}; O_{110}O_{001}$ also concur and combined together give $$O_{000}O_{111};O_{001}O_{110};O_{010}O_{101};O_{011}O_{100}$$are concurrent so we are done. phew! $\blacksquare$ Someone pls check this I surely messed up something

ayan.nmath wrote: If radical center doesn't exist, then it is trivial Oops, forgot to mention this case.

This is similar to RMM 2009 P3

TheDarkPrince wrote: Homothety also works. ayan.nmath wrote: If radical center doesn't exist, then it is trivial Oh no, forgot to mention this did you use LaTeX for your solutions