Problem (2021GOWACAp3): In triangle $ABC$ with incircle $\omega$, let $I$ be the incenter and $D$ be the point where $\omega$ touches $\overline{BC}$. Let $S$ be the point on $(ABC)$ with $\angle ASI = 90^\circ$ and $H$ be the orthocenter of $\triangle BIC$, so that $Q \ne S$ on $\overline{HS}$ also satisfies $\angle AQI = 90^\circ$. Prove that $X$, the reflection of $I$ over the midpoint of $\overline{DQ}$, lies on $\omega$.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; d = foot i b c; o = circumcenter a b c; m1 = midpoint a i; s = on_circle o a, on_circle m1 a; h = orthocenter b i c; q = on_line h s, on_circle m1 a; m2 = midpoint d q; x = mirror i m2 ? cong x i i d
HAGeo formalization: A B C = triangle; I = incenter A B C; BC = line B C; D = foot I BC; O = circumcenter A B C; M1 = midpoint A I; (O) = circle_center_point O A; (AI) = circle_center_point M1 A; S = intersection (O) (AI); H = orthocenter B I C; HS = line H S; Q = intersection HS (AI); M2 = midpoint D Q; X = reflect_point_wrt_point I M2; Prove: cong X I I D
Difficulty score: 4.8