Problem (2019IranGOAp1): Circles $\omega_1$ and $\omega_2$ intersect each other at points $A$ and $B$. Point $C$ lies on the tangent line from $A$ to $\omega_1$ such that $\angle ABC = 90^\circ$. Arbitrary line $\ell$ passes through $C$ and cuts $\omega_2$ at points $P$ and $Q$. Lines $AP$ and $AQ$ cut $\omega_1$ for the second time at points $X$ and $Z$ respectively. Let $Y$ be the foot of altitude from $A$ to $\ell$. Prove that points $X, Y$ and $Z$ are collinear.
AlphaGeometry formalization: a b = segment; o1 = on_bline a b; o2 = on_bline a b; c = on_tline a o1 a, on_tline b a b; p = on_circle o2 a; q = on_line c p, on_circle o2 a; x = on_line a p, on_circle o1 a; z = on_line a q, on_circle o1 a; y = foot a c p ? coll x y z
HAGeo formalization: A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; O1A = line O1 A; l1 = perpendicular_line A O1A; AB = line A B; l2 = perpendicular_line B AB; C = intersection l1 l2; P = on_circle (O2); CP = line C P; Q = intersection CP (O2); AP = line A P; X = intersection AP (O1); AQ = line A Q; Z = intersection AQ (O1); Y = foot A CP; Prove: collinear X Y Z
Difficulty score: 2.1