Let $ K$ be the circumcenter of the $ \triangle ABC.$ Let $ \triangle P_aP_bP_c, \triangle Q_aQ_bQ_c$ be the pedal triangles of $ P, Q$; the midpoint $ M$ of $ (PQ)$ is their common circumcenter. The midpoint $ N_a$ of $ AQ$ is the circumcenter of the $ \triangle AQ_bQ_c,$ on account of the rigth angles $ \angle QQ_bA, \angle QQ_cA.$ The isogonal $ AP$ of $ AQ \equiv AN_a$ with respect to the angle $ \angle CAB \equiv \angle Q_cAQ_b$ is its A-altitude $ \Longrightarrow$ $ Q_bQ_c \perp AP.$ But $ O_2O_3 \perp AP$ as well $ \Longrightarrow$ $ Q_bQ_c \parallel O_2O_3$ and similarly, $ Q_cQ_a \parallel O_3O_1, Q_aQ_b \parallel O_1O_2.$ The $ \triangle O_aQ_bO_c \sim \triangle O_1O_2O_3$ are centrally similar, having the correspnding sides parallel. $ M, O$ are their corresponding circumcenters and $ Q, K$ are their corresponding points, the concurrency points of their corresponding cevians $ (Q_aQ \parallel O_1K) \perp BC,$ $ (Q_bQ \parallel O_2K) \perp CA,$ $ (Q_cQ \parallel O_3K) \perp AB.$ Their corresponding lines $ QM \parallel KO$ are parallel. In ezactly the same way, $ PM \parallel KO'$ $ \Longrightarrow$ $ OO' \parallel PQ.$

yetti wrote: Let $ K$ be the circumcenter of the $ \triangle ABC.$ Let $ \triangle P_aP_bP_c, \triangle Q_aQ_bQ_c$ be the pedal triangles of $ P, Q$; the midpoint $ M$ of $ (PQ)$ is their common circumcenter. The midpoint $ N_a$ of $ AQ$ is the circumcenter of the $ \triangle AQ_bQ_c,$ on account of the rigth angles $ \angle QQ_bA, \angle QQ_cA.$ The isogonal $ AP$ of $ AQ \equiv AN_a$ with respect to the angle $ \angle CAB \equiv \angle Q_cAQ_b$ is its A-altitude $ \Longrightarrow$ $ Q_bQ_c \perp AP.$ But $ O_2O_3 \perp AP$ as well $ \Longrightarrow$ $ Q_bQ_c \parallel O_2O_3$ and similarly, $ Q_cQ_a \parallel O_3O_1, Q_aQ_b \parallel O_1O_2.$ The $ \triangle O_aQ_bO_c \sim \triangle O_1O_2O_3$ are centrally similar, having the correspnding sides parallel. $ M, O$ are their corresponding circumcenters and $ Q, K$ are their corresponding points, the concurrency points of their corresponding cevians $ (Q_aQ \parallel O_1K) \perp BC,$ $ (Q_bQ \parallel O_2K) \perp CA,$ $ (Q_cQ \parallel O_3K) \perp AB.$ Their corresponding lines $ QM \parallel KO$ are parallel. In ezactly the same way, $ PM \parallel KO'$ $ \Longrightarrow$ $ OO' \parallel PQ.$ 이 자의 풀이를 보지 말거라. 짐이 더 좋은 풀이를 올리리라.

Could you please use English instead of Korean here. Thanks.

$ \mbox{Prove that circumcircle of the } \triangle{ABC}, \triangle{O_1O_2O_3} \mbox{ and } \triangle{O'_1O'_2O'_3} \mbox{ coaxial circles.}$

My solution : Let $ T $ be the circumcenter of $ \triangle ABC $ . Let $ D \equiv O_2O_3 \cap O_2'O_3', E \equiv O_3O_1 \cap O_3'O_1', F \equiv O_1O_2 \cap O_1'O_2' $ . Easy to see $ T=O_1O_1' \cap O_2O_2' \cap O_3O_3' $. Since $ O_2O_3, O_2'O_3' $ is the perpendicular bisector of $ AP, AQ $, respectively , so $ \angle O_2O_3O_3' =180^{\circ} - \angle BAP =180^{\circ} - \angle QAC =\angle O_2O_2'O_3' $ . i.e. $ O_2, O_3, O_2', O_3' $ are concyclic Similarly, we can get $ O_3, O_1, O_3', O_1' $ are concyclic and $ O_1, O_2, O_1', O_2' $ are concyclic. Since $ O_2D \cdot DO_3=O_2'D \cdot DO_3', O_3E \cdot EO_1=O_3'E \cdot EO_1', O_1F \cdot FO_2=O_1'F \cdot FO_2' $ , so $ D, E, F $ are collinear at the radical axis of $ (O_1O_2O_3) $ and $ (O_1'O_2'O_3') $ . i.e. $ \overline{DEF} $ is perpendicular to $ OO' $ ... $ (1) $ Since $ O_2O_3, O_2'O_3' $ is the perpendicular bisector of $ AP, AQ $, respectively , so $ D $ is the circumcenter of $ \triangle APQ $ . i.e. $ D $ lie on the perpendicular bisector of $ PQ $ Similarly, we can get $ E, F $ lie on the perpendicular bisector of $ PQ $ . i.e. $ \overline{DEF} $ is the perpendicular bisector of $ PQ $ ... $ (2) $ From $ (1) $ and $ (2) $ we get $ OO' \parallel PQ $ . Q.E.D Remark : $ (1) $ Since $ TO_1 \cdot TO_1'=TO_2 \cdot TO_2'=TO_3 \cdot TO_3' $ , so $ (O_1'O_2'O_3') $ is the image of $ (O_1O_2O_3) $ under the inversion WRT $ (O) $ . i.e. $ (O), (O_1O_2O_3), (O_1'O_2'O_3') $ are coaxial $ (2) $ This problem can be generalized to the quadrilateral : Let $ P, Q $ be the isogonal conjugate of $ ABCD $ . Let $ O_1, O_2, O_3, O_4 $ be the circumcenter of $ \triangle ABP, \triangle BCP, \triangle CDP, \triangle DAP $ . Let $ O_1', O_2', O_3', O_4' $ be the circumcenter of $ \triangle ABQ, \triangle BCQ, \triangle CDQ, \triangle DAQ $ . Let $ O, O' $ be the circumcenter of $ O_1O_2O_3O_4, O_1'O_2'O_3'O_4' $ . Then $ OO' \parallel PQ $