Problem (2024AsiaPacificMOp1): Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $BQX$ and $CPX$ intersect at a point $Y \neq X$. Prove that the points $A, X$, and $Y$ are collinear.
AlphaGeometry formalization: a b c = triangle; d = on_line a b; e = on_line a c, on_pline d b c; x = free; p = on_line b c, on_line d x; q = on_line b c, on_line e x; y = on_circum b q x, on_circum c p x ? coll a x y
HAGeo formalization: A B C = acute_triangle; AB = line A B; D = on_line AB; BC = line B C; l1 = parallel_line D BC; AC = line A C; E = intersection AC l1; X = point; DX = line D X; P = intersection DX BC; EX = line E X; Q = intersection EX BC; (BQX) = circle B Q X; (CPX) = circle C P X; Y = intersection (BQX) (CPX); Prove: collinear A X Y
Difficulty score: 1.7