# The altitudes and the Euler Line

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,

X = (x + y)/2 and Y = (X - y)/2i.

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

Y = mX + b,

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

y = Mx + B,

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 (X_{1}, Y_{1}) and (X_{2}, Y_{2}) has the slope (Y_{2} - Y_{1})/(X_{2} - X_{1}). Similarly (the verification is quite easy), in the circular coordinates, the line through two points (x_{1}, y_{1}) and (x_{2}, y_{2}) has the clinant defined by the ratio (y_{2} - y_{1})/(x_{2} - x_{1}).

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),

M | = (1 - mi)/(1 + mi) |

= (1 - i tan(f))/(1 + i tan(f)) | |

= (cos(f) - i sin(f))/(cos(f) + i sin(f)) | |

= e^{-2fi}
| |

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 x_{1}, x_{2}, and x_{3}. 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 x_{2}x_{3} is

y = M(x - x_{2}) + y_{2},

where M = (y_{2} - y_{3})/(x_{2} - x_{3}). Since x_{2} and x_{3} lie on the unit circle, we can simplify the expression for the clinant M:

M | = (y_{2} - y_{3})/(x_{2} - x_{3}) |

= (1/x_{2} - 1/x_{3})/(x_{2} - x_{3}) | |

= (x_{3} - x_{2})/(x_{2} - x_{3})/(x_{2}x_{3}) | |

= - 1/(x_{2}x_{3}) | |

= - y_{2}y_{3} |

The equation of the side x_{2}x_{3} is, therefore,

(1)

y = - y_{2}y_{3}(x - x_{2}) + y_{2},

The clinant of lines perpendicular to x_{2}x_{3} is then y_{2}y_{3}. The altitude through x_{1} is one of those perpendiculars and its equation is simply:

(2)

y = y_{2}y_{3}(x - x_{1}) + y_{1}

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

x | = (x_{1} + x_{2} + x_{2}x_{3}(y_{2} - y_{1}))/2 |

= (x_{1} + x_{2} + x_{3} - x_{2}x_{3}/x_{1})/2
| |

= (H - x_{2}x_{3}/x_{1})/2, |

where H = x_{1} + x_{2} + x_{3}. Similarly to (2), the altitudes through x_{2} and x_{3} are given by

y = y_{1}y_{3}(x - x_{2}) + y_{2} and

y = y_{1}y_{2}(x - x_{3}) + y_{3}

Multiply the first equation by y_{2} and the second by y_{3}. Subtracting the results eliminates x. But solving for y and then changing to conjugates will resurrect x in an unexpectedly simple form: x = H! Since H depends symmetrically on the vertices x_{1}, x_{2}, and x_{3}, all three altitudes meet at the point x = H, which proves existence of the orthocenter as promised.

It's time for the bonus. Note that, in our triangle,

- the circumcenter O is at the origin: O = 0,
- the orthocenter H is at the point x
_{1}+ x_{2}+ x_{3}, - the centroid M is known to be defined by (x
_{1}+ x_{2}+ x_{3})/3.

This implies two things. First, the three centers - O, H, and M - lie on a straight line. And second, M is one third way from O to H. In other words,

2 OM = MH, |

The line is known as the *Euler line* or *Euler's line*.

### References

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

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

67011094