Problem (2014ARMOg10p4): Given a triangle $ABC$ with $AB>BC$, let $ \Omega $ be the circumcircle. Let $M$, $N$ lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. Let $K$ be the intersection of $MN$ and $AC$. Let $P$ be the incentre of the triangle $AMK$ and $Q$ be the $K$-excentre of the triangle $CNK$. If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; m = on_line a b; n = on_line b c, eqdistance c a m; k = on_line m n, on_line a c; p = incenter a m k; q = excenter k c n; x = angle_bisector a b c; r = on_circum a b c, on_tline b b x ? cong r p r q
HAGeo formalization: A B C = triangle; O = circumcenter A B C; AB = line A B; M = on_line AB; (C,AM) = circle_center_radius C A M; BC = line B C; N = intersection (C,AM) BC; MN = line M N; AC = line A C; K = intersection MN AC; P = incenter A M K; Q = excenter K C N; R = midarc A B C; Prove: cong R P R Q
Difficulty score: 2.8