Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We all know that all three altitudes of a triangle meets in the orthocenter of the triangle. It's a quite classical problem and is proven. However, what I really wanna know is what characteristic of the triangle is the profound for this to happen? E.g: Is this because of the sum of 3 internal angles equals 180? In Non-Euclidean geometry, where sum of 3 internal angles is greater or smaller than 180 degree, does the 3 altitudes meets in a single point? Or is it because of another reason?

share|improve this question
2  
Euler has quite a few things named after him, so there is no need to deprive Euclid of any honors! :D –  Mariano Suárez-Alvarez Feb 19 '13 at 7:43
    
In the hyperbolic plane, the perpendicular bisectors of the sides of a triangle can be concurrent, have a common perpendicular or be asymptotic. In general, then, there is no orthocenter. –  Mariano Suárez-Alvarez Feb 19 '13 at 7:53
    
+1 Damn intresting observation! –  Arjang Feb 19 '13 at 7:54
1  
Have you heard, that the orthocenter is isogonal conjugate of circumcentre? It does not explain everything about the orthocenter, but for me it does explain a lot (isogonal conjugation contains traces of $z \mapsto z^{-1}$ transformation). –  dtldarek Feb 19 '13 at 8:11
    
Something else that bothers me more is that the existence of the Euler line: the orthocenter, centroid, and circumcenter of any triangle are always collinear. It sounds interesting but doesn't it seem a little bit coincident? I mean I know how to prove it but it doesn't sound natural to me. –  Nhím Hổ Báo Feb 19 '13 at 15:37

3 Answers 3

It is perhaps interesting to note that the definition of altitudes is perfectly straightforward for simplexes in higher dimensions, but that already in dimension $3$ the altitudes of a general tetrahedron are not concurrent. For instance for the tetrahedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(1,0,1)$, two of the altitudes meet in the origin and two others meet in $(1,0,0)$, but there are no other points of intersection; in general position none of the altitudes will intersect.

share|improve this answer

The proofs I know all use Euklidean geometry (e.g. the orthocenter is the intersection of the middle orthogonals for a bigger triangle).

In synthetic geometry, one can consider translation planes with an orthogonality relation and the Fano axiom (diagonals of a nondegenerate parallelogram intersect), thus minimally allowing the proof above. One can show that this makes the geometry at least a Pappus plane.

share|improve this answer
    
What is synthetic geometry? –  Arjang Feb 20 '13 at 2:57

Vladimir Arnol'd often emphasized that it was because of the Jacobi identity in Lie algebras. I think he might have meant the algebra so(3) represented as $R^3$ with vector cross-product as the multiplication, but am not sure, and it would be nice to see his remark explained.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.