ABC is a triangle and A, B, C are not collinear. Prove that the distance between the orthocenter and the circumcenter is less than three times the circumradius.
Solution
We use vectors. It is well-known that the circumcenter O, the centroid G and the orthocenter H lie on the Euler line and that OH = 3 OG. Hence taking vectors with origin O, OH = 3 OG = OA + OB + OC. Hence |OH| ≤ |OA| + |OB| + |OC| = 3 x circumradius. We could have equality only if ABC were collinear, but that is impossible, because ABC would not then be a triangle.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002