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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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2 Answers 2

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
    
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

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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