In an acute triangle ABC with ¯AB>¯AC, let D,E,F be the feet of the altitudes from A,B,C, respectively. Let P be an intersection of lines EF and BC, and let Q be a point on the segment BD such that ∠QFD=∠EPC. Let O,H denote the circumcenter and the orthocenter of triangle ABC, respectively. Suppose that OH is perpendicular to AQ. Prove that P,O,H are collinear.
Problem
Source: 2020 Korea Junior Math Olympiad p4 KJMO
Tags: collinear, geometry
25.11.2021 10:47
Let Q′ be the midpoint of BC. Note that (B,C;D,P)=−1, so Q′F2=Q′B⋅Q′C=Q′D⋅Q′P. Since AB>AC, D is between Q′ and P, so Q′ is on segment BD and Q′F is tangent to (FDP)⟺∠Q′FD=∠EPC. There can only be one point on segment BD that satisfies the angle condition for Q, so Q′=Q. Q is the midpoint of BC. Applying Brocard's theorem on cyclic quadrilateral BFEC with center Q, Q is the orthocenter of triangle AHP. Hence, AQ⊥HP. Thus, if AQ⊥OH, then P,O,H must be collinear.
26.11.2021 17:10
Can't believe I got this, but a solution without projective geo!: Firstly, ∠BFD=C, since FACD is cyclic. Now ∠QFD=∠EPC=C−B, just using angle chasing in △BFP. So BFQ=B. Now since △BFC is right angled, Q, must be its circumcentre using some direct angle chasing. So Q is the midpoint of BC. So AQ∩OH =G, which is the centroid, because AQ is the median. Now ∠QGH=90, so GHDQ is cyclic. . Similarly, AFGH and AGHE are al;so cyclic. Combining the last 2 results we get that AFGHE is cyclic. Also F,E,D,Q all lie on the nine point circle, so EFDQ is cyclic as well. So by radical axes theorem, OH,FE,QD concur , and since the latter 2 meet at P, so clearly O,H,P are collinear. ◼