1st Brazil 1979 Problem 3 The vertex C of the triangle ABC is allowed to vary along a line parallel to AB. Find the locus of the orthocenter.
A parabola. Solution Take axes so that A is (-a,0), B is (a,0) and C is (k,b). Then the orthocenter lies on the line x = k. The line AC has gradient b/(k+a), so the perpendicular has gradient -(k+a)/b. Hence the altitude from B has equation y + (x-a)(k+a)/b = 0. So the intersection is x = k, y = -(k-a)(k+a)/b. So the locus is all or part of the parabola by = a 2 -x 2 . But we can get an orthocenter with any x-coordinate (by taking C to have the same x-coordinate), so we can get all points on the parabola. 1st Brazil 1979 © John Scholes jscholes@kalva.demon.co.uk 23 Oct 2003 Last corrected/updated 23 Oct 03