Distance between the Orthocenter and Circumcenter

Problem 2 from the 1994 Asian Pacific Mathematical Olympiad asked to

Prove that, for any triangle Δ, the inequality

HO ≤ 3R,

where H is the orthocenter, O the circumcenter and R the circumradius of Δ.

A stronger result can be easily proved with complex numbers: for any triangle Δ, with side lengths a, b, c the following identity holds:

(1) HO ² = 9R ² - (a² + b² + c²).

Solution

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Set the origin of the complex plane at the circumcenter O of triangle Δ and let x, y, z be the complex number corresponding to the vertices of Δ opposite sides a, b, c. Then, as we know, the orthocenter H can be determined with

  1. | x | = | y | = | z | = R,
  2. H = x + y + z,
  3. a = | z - y |, b = | x - z |, c = | y - x |.

With these, (1) becomes

(2) | x + y + z |² = 9R ² - (| z - y |² + | x - z |² + | y - x |²).

With a little insight and an appeal to symmetry, we are led to establishing an identity from which (2) follows immediately.

| x + y + z |² = 3(| x |² + | y |² + | z |²) - (| z - y |² + | x - z |² + | y - x |²).

The left-hand side of (3) equals

| x + y + z |² = | x |² + | y |² + | z |² + (xy* + yz* + zx* + x*y + y*z + z*x).

The second term in the right-hand side of (3) equals

| z - y |² + | x - z |² + | y - x |² = 2(| x |² + | y |² + | z |²) - (xy* + yz* + zx* + x*y + y*z + z*x).

Together they make (3) obvious.

(The same identity is derived elsewhere from the relation between the medians and the side lengths of a triangle.)

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Copyright © 1996-2018 Alexander Bogomolny

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