2024 IMO Shortlist G7

Let $ABC$ be a triangle with incentre $I$ such that $AB<AC<BC$. The second intersections of $AI$, $BI$, and $CI$ with the circumcircle of triangle $ABC$ are $M_A$, $M_B$, and $M_C$, respectively. Lines $AI$ and $BC$ intersect at $D$, and lines $BM_C$ and $CM_B$ intersect at $X$. Suppose the circumcircles of triangles $XM_BM_C$ and $XBC$ intersect again at $S\ne X$. Lines $BX$ and $CX$ intersect the circumcircle of triangle $SXM_A$ again at $P\ne X$ and $Q\ne X$, respectively.

Prove that the circumcentre of triangle $SID$ lies on $PQ$.

设 $ABC$ 为三角形，内心为 $I$，且 $AB<AC<BC$。直线 $AI,BI,CI$ 与三角形 $ABC$ 外接圆的第二个交点分别为 $M_A,M_B,M_C$。直线 $AI$ 与 $BC$ 交于 $D$，直线 $BM_C$ 与 $CM_B$ 交于 $X$。假设三角形 $XM_BM_C$ 与 $XBC$ 的外接圆再次交于 $S\ne X$。直线 $BX,CX$ 分别与三角形 $SXM_A$ 的外接圆再次交于 $P\ne X,Q\ne X$。

证明三角形 $SID$ 的外心位于直线 $PQ$ 上。