Problem
GEO-B2-M09-P019 Reflection of the Orthocenter
In triangle \(A(0,0)\), \(B(1,0)\), \(C(u,v)\), where \(v\ne 0\), the orthocenter is \(H\left(u,\frac{u(1-u)}{v}\right)\). Prove that the point \(H'\), the reflection of \(H\) across \(AB\), lies on the circumcircle \((ABC)\).
Write the equation of the circle through \(A,B,C\), then substitute the coordinates of the reflected point.
The circle through \(A(0,0)\), \(B(1,0)\) has form \(x^2+y^2-x+\lambda y=0\). Substituting \(C(u,v)\), we get \(\lambda=\frac{u-u^2-v^2}{v}\). The reflected point is \(H'\left(u,-\frac{u(1-u)}{v}\right)\). Substitution gives \(u^2+\frac{u^2(1-u)^2}{v^2}-u-\frac{u(1-u)(u-u^2-v^2)}{v^2}=0\). After cancellation, this becomes \(0=0\). Hence \(H'\in(ABC)\).
The algebra looks longer, but the structure is standard: circle equation plus substitution.