ABC is an isosceles triangle with AB = AC. O is a variable point on the line BC such that the circle center O radius OA does not have the lines AB or AC as tangents. The lines AB, AC meet the circle again at M, N respectively. Find the locus of the orthocenter of the triangle AMN.
Solution
Let A' be the reflection of A in BC. Then OA = OA', so the circle also passes through A'. ∠MAA' = ∠NAA', so A' is the midpoint of the arc MN. Hence OA' is perpendicular to MN. Let X be midpoint of MN, so X lies on OA'.
Now ∠OA'M = ∠MA'N/2 = (180o-∠A)/2 = 90o - ∠A/2, so ∠MOX = 2(90o-∠OA'M) = ∠A. Hence OX/OA' = OM cos A/OA' = cos A, which is fixed. So the locus of X is line parallel to BC. Let G be centroid of AMN, then AG/AX = 2/3, so G also lies on a line parallel to BC. But H lies on ray OG with GH = 2OG (Euler line), so H also lies on a line parallel to BC.
Given any point H on the line take a line through A' parallel to AH. It meets the line BC at a point O, which is the required point to generate H. (Arguably, we do not generate the two points on the line corresponding to OA perpendicular to AB and AC, because then one of M, N coincides with A and AMN is degenerate.)
© John Scholes
jscholes@kalva.demon.co.uk
9 February 2004
Last updated/corrected 9 Feb 04