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In Euclidean geometry, the centroid, orthocenter and circumcenter of a triangle lie on a line.

In which other geometries does this hold?

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Does Absolute Geometry + Euler collinearity $\ \Longrightarrow\ $ Euclid ? –  Włodzimierz Holsztyński Jan 10 at 17:23

3 Answers 3

In the paper "On some classical constructions extended to hyperbolic geometry", A. V. Akopyan proves an analogue of Feuerbach's theorem for hyperbolic geometry. Let $M_a,M_b,M_c$ be three points on sides of a triangle so that the corresponding cevians bisect the area of the triangle $ABC$. The intersection of these cevians, G, is the analogue of the centroid. Next if we take the circle passing through $M_a,M_b,M_c$, it will intersect each side a second time, in points $H_a,H_b,H_c$. The intersection of $AH_a,BH_b,CH_c$ is the analogue of the orthocenter, call it $H$.

The paper proves that there is a line passing through $G$, $H$, the circumcenter and the center of the circle passing through $M_a,M_b,M_c$. This is pretty much the most natural analogue for the Euler line in hyperbolic geometry.

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On the subject of A. V. Akopyan: He has a really cool little book called "geometry in pictures". –  Igor Rivin Sep 10 '12 at 2:13
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This non-Euclidean Euler line can be seen in my Mathematica Demonstration entitled Non-Euclidean Triangle Continuum at demonstrations.wolfram.com/NonEuclideanTriangleContinuum. –  Robert A. Russell Apr 6 '13 at 17:18

Another non-euclidean version of the Euler line, different from Akopyan's one appears in the paper "Non-euclidean shadows of classical projective theorems". This line is the line called "orthoaxis" by N. Wildberger in "Universal hyperbolic geometry II: a pictorial overview", and so it is called "Euler-Wildberger line".

As in Akopyan's paper, the Euler-Wildberger line uses alternative definitions for some centers, in this case for the centroid and the circumcenter of a triangle. These alternative definitions are equivalent to the standard ones in the euclidean case but not in the non-euclidean world. For a non-euclidean triangle $ABC$, its double triangle $A''B''C''$ is constructed by taking at each vertex the orthogonal line to the altitude through the vertex. The double triangle and the original triangle are allways perspective. In euclidean geometry, the center of perspectivity is the centroid of the triangle, and the lines $m_a,m_b,m_c$ joining $A,B,C$ with $A'',B'',C''$ respectively are the medians of the triangle, passing through the midpoints of the sides $a,b,c$ of $ABC$ respectively. In hyperbolic or elliptic geometry, this center of perspectivity is not necessarily the centroid, it is a "pseudo-centroid" of the triangle, and the lines $m_a,m_b,m_c$ do not pass through the midpoints of $a,b,c$ in general. The intersection points of $m_a,m_b,m_c$ with $a,b,c$ respectively are the "pseudo-midpoints" of the sides. The lines orthogonal to the sides through the pseudo-midpoints are concurrent and their intersection point is the "pseudo-circumcenter" of the triangle. The orthocenter, the pseudo-centroid and the pseudo-circumcenter are collinear and the Euler-Wildberger line is the line joining them. This construction is developed in Cayley-Klein projective models and because of this it is valid also for other hyperbolic figures such as right-angled hexagons, for example.

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Only in Euclidean geometry.

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Why do you say that? –  Igor Rivin Sep 9 '12 at 18:17
    
Igor, I say it because it is true :) BTW it is really easy to check. –  Anton Petrunin Sep 10 '12 at 0:56
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How does this fit with Gjergji Zaimi's answer? (Maybe you are using different definitions...) –  Henry Cohn Sep 10 '12 at 1:55
    
I use the standard defs and Akopyan modifies them. (BTW if you would read Gjergji's answer then it would be clear for you.) –  Anton Petrunin Sep 11 '12 at 14:52

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