Question 5
Let \(ABC\) be an acute scalene triangle and let \(P\) be an interior point such that \(\angle APB=\angle APC=\alpha>180^\circ-\angle BAC\). The circumcircle of \(APB\) meets \(AC\) again at \(E\), and the circumcircle of \(APC\) meets \(AB\) again at \(F\). Let \(Q\) lie inside \(AEF\) with \(\angle AQE=\angle AQF=\alpha\). Reflect \(Q\) in \(EF\) to obtain \(D\). The internal bisector of \(\angle EDF\) meets \(AP\) at \(T\).
(a) Prove \(\angle DET=\angle ABC\) and \(\angle DFT=\angle ACB\).
(b) The line \(PA\) meets the lines \(DE,DF\) at \(M,N\), respectively. Let \(I,J\) be the incenters of \(PEM,PFN\), and let \(K\) be the circumcenter of \(DIJ\). The line \(DT\) meets this circumcircle again at \(H\). Prove that \(HK\) passes through the incenter of \(DMN\).
Editorial warning: the source statement is retained, but part (b) is false without an additional configuration restriction. The solutions distinguish the valid part (a), the source's restricted-configuration argument, and an exact counterexample to the printed part (b). Here \(AC,AB\) must mean their supporting lines: the given angle inequality forces \(E,F\) onto the opposite extensions beyond \(A\).
(a) Prove \(\angle DET=\angle ABC\) and \(\angle DFT=\angle ACB\).
(b) The line \(PA\) meets the lines \(DE,DF\) at \(M,N\), respectively. Let \(I,J\) be the incenters of \(PEM,PFN\), and let \(K\) be the circumcenter of \(DIJ\). The line \(DT\) meets this circumcircle again at \(H\). Prove that \(HK\) passes through the incenter of \(DMN\).
Editorial warning: the source statement is retained, but part (b) is false without an additional configuration restriction. The solutions distinguish the valid part (a), the source's restricted-configuration argument, and an exact counterexample to the printed part (b). Here \(AC,AB\) must mean their supporting lines: the given angle inequality forces \(E,F\) onto the opposite extensions beyond \(A\).