Show how to construct the triangle ABC given the distance between the circumcenter O and the orthocenter H, the fact that OH is parallel to the side AB, and the length of the side AB.
Solution
Solution by Demetres Christofides
Let A be the point (-1, 0), B the point (1, 0), O the point (0, b) and H the point (3a, b). Then G is the point (a, b) (using the Euler line) and the midpoint M of AB is (0, 0). Hence C is the point (3a, 3b). We are given a but not b. Since OC = OA, we have 1 + b2 = 9a2 + 4b2, so 3b2 = 1 - 9a2.
Construct A, M, B. Then construct D (1, 1), E (1, 3a) and F(1-3a, 0). Take the line through E parallel to DF. Suppose it meets AB at K. Then BK/BF = BE/BD = 3a, so BK = 9a2, so K is (3b2, 0). Take L as (9b2, 0). Take the circle diameter AL. If it meets the y-axis at (0, ±k), then k2 = AM.ML = 9b2, so the have the point (0, 3b). The rest is now trivial.
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002