Problem (2014ARMOg9p4): Let $M$ be the midpoint of the side $AC$ of acute-angled triangle $ABC$ with $AB>BC$. Let $\Omega $ be the circumcircle of $ ABC$. The tangents to $ \Omega $ at the points $A$ and $C$ meet at $P$, and $BP$ and $AC$ intersect at $S$. Let $AD$ be the altitude of the triangle $ABP$ and $\omega$ the circumcircle of the triangle $CSD$. Suppose $ \omega$ and $ \Omega $ intersect at $K\not= C$. Prove that $ \angle CKM=90^\circ $.
AlphaGeometry formalization: a b c = triangle; m = midpoint a c; o = circumcenter a b c; p = on_tline a o a, on_tline c o c; s = on_line b p, on_line a c; d = foot a b p; o1 = circumcenter c s d; k = on_circle o1 c, on_circle o a ? perp c k k m
HAGeo formalization: A B C = acute_triangle; M = midpoint A C; (ABC) = circumcircle A B C; l1 = tangent A (ABC); l2 = tangent C (ABC); P = intersection l1 l2; BP = line B P; AC = line A C; S = intersection BP AC; D = foot A BP; (CSD) = circumcircle C S D; K = intersection (CSD) (ABC); CK = line C K; KM = line K M; Prove: perpendicular CK KM
Difficulty score: 2.7