Existence of the orthocenter can be shown in a variety of ways. Here I would like to present an algebraic proof that invokes the representation of points in the plane as complex numbers. We applied complex numbers to construct similar triangles and concentric polygons, prove Napoleon's theorem, and investigated Morley's complex interpolation theory that led to a spectacular theorem, known as Morley's Miracle.

As the main tool in the proof, I choose the circular coordinates that served Morley in such good stead. In passing, these are also known as Gaussian or conjugate coordinates, the former for historical reasons, the latter in association with the manner in which they are defined.

If X and Y are usual Cartesian coordinates, define x = X + iY and y = X - iY. Conversely, a pair (x,y) (or, actually, either of the conjugate complex numbers x and y) of circular coordinates uniquely determines Cartesian coordinates X and Y. Indeed,

In the Cartesian coordinates a straight line is defined by a linear equation, say

m being the slope, b the y-intercept of the line. (I leave the exceptional case of vertical lines as an exercise.) We can express the same equation in the circular coordinates as

where M = (1 - mi)/(1 + mi) and B = 2bi/(1 + mi). M is the line's clinant, not the slope. The important point is that in the circular coordinates straight lines are again described by linear equations. (Be cautioned, though, that not every linear equation in the circular coordinates describes a line.) For a given point z and its conjugate , the straight line through z with the clinant M is defined by

y = M(x - z) + ,

In the Cartesian coordinates, the line through two points (X₁, Y₁) and (X₂, Y₂) has the slope (Y₂ - Y₁)/(X₂ - X₁). Similarly (the verification is quite easy), in the circular coordinates, the line through two points (x₁, y₁) and (x₂, y₂) has the clinant defined by the ratio (y₂ - y₁)/(x₂ - x₁).

The last important fact about the clinants that I should mention is that the clinant of a line perpendicular to a line with a clinant M, is -M. This is because, for m = tan(f),

by Euler's and de Moivre's formulas.

But cos(2f + p) = - cos(2f) and sin(2f + p) = - sin(2f), so that e^{-(2f + p)i} = - e^-2fi.

We are ready to tackle the problem of concurrency of the altitudes in a triangle. As we'll see, there is a bonus for approaching the problem in the framework of circular coordinates.

Let a triangle be defined by its three vertices x₁, x₂, and x₃. I shall assume existence of the circumcircle: there exists a circle that passes through all three given points. As neither location, nor the "size" of the triangle matter, let's translate and resize it so as to have its circumcenter at the origin and the unit circle xy = 1 as its circumcircle. In other words, let's place the vertices of a triangle on the unit circle. Please note that, with this choice of vertices, the circumcenter is given by x = 0.

The equation of the side x₂x₃ is

where M = (y₂ - y₃)/(x₂ - x₃). Since x₂ and x₃ lie on the unit circle, we can simplify the expression for the clinant M:

The equation of the side x₂x₃ is, therefore,

The clinant of lines perpendicular to x₂x₃ is then y₂y₃. The altitude through x₁ is one of those perpendiculars and its equation is simply:

From (1)-(2) we easily find the foot of the altitude through x₁. Just add up the two equations and pass to the conjugates:

• the circumcenter O is at the origin: O = 0,
• the orthocenter H is at the point x₁ + x₂ + x₃,
• the centroid M is known to be defined by (x₁ + x₂ + x₃)/3.

to be continued...

References

1. S.L.Greitzer, Conjugate Coordinate Geometry, Arbelos, v3, n1-5, MAA, 1997
2. Liang-shin Hahn, Complex Numbers & Geometry, MAA, 1994

Copyright © 1996-2008 Alexander Bogomolny