Problem (2022FakeCMOp3): In $\triangle ABC$ with excenters $I_A,I_B,I_C$, let $P$ be any point. Lines $AP,BP,CP$ meet the circumcircles of $\triangle BPC,\triangle CPA,\triangle APB$ again at $P_A,P_B,P_C$. The circumcircles of $\triangle I_BI_CP_A$ and $\triangle I_AI_BP_C$ meet again at $Q\ne I_B$. Lines $AQ,BQ,CQ$ meet the circumcircles of $\triangle BQC,\triangle CQA,\triangle AQB$ again at $Q_A,Q_B,Q_C$. Prove that $P_AQ_A$, $P_BQ_B$, $P_CQ_C$ are concurrent.
AlphaGeometry formalization: a b c = triangle; i_a = excenter a b c; i_b = excenter b c a; i_c = excenter c a b; p = free; p_a = on_line a p, on_circum b p c; p_b = on_line b p, on_circum c p a; p_c = on_line c p, on_circum a p b; q = on_circum i_b i_c p_a, on_circum i_a i_b p_c; q_a = on_line a q, on_circum b q c; q_b = on_line b q, on_circum a q c; q_c = on_line c q, on_circum a q b; k = on_line p_a q_a, on_line p_b q_b ? coll k p_c q_c
HAGeo formalization: A B C = triangle; Ia = excenter A B C; Ib = excenter B C A; Ic = excenter C A B; P = point; AP = line A P; (BPC) = circumcircle B P C; Pa = intersection AP (BPC); BP = line B P; (CPA) = circumcircle C P A; Pb = intersection BP (CPA); CP = line C P; (APB) = circumcircle A P B; Pc = intersection CP (APB); (IbIcPa) = circumcircle Ib Ic Pa; (IaIbPc) = circumcircle Ia Ib Pc; Q = intersection (IbIcPa) (IaIbPc); AQ = line A Q; (BQC) = circumcircle B Q C; Qa = intersection AQ (BQC); BQ = line B Q; (AQC) = circumcircle A Q C; Qb = intersection BQ (AQC); CQ = line C Q; (AQB) = circumcircle A Q B; Qc = intersection CQ (AQB); PaQa = line Pa Qa; PbQb = line Pb Qb; K = intersection PaQa PbQb; Prove: collinear K Pc Qc
Difficulty score: 6.4