Talk:Triangle center

Useful film
Covering the first four centers: http://www.archive.org/details/journey_to_the_center_of_a_triangle 76.117.247.55 (talk) 00:38, 2 May 2011 (UTC)

Descriptions
I was just thinking, we really need a brief description of each centre, rather than merely the trilinears. I've just experimented, and this is what I've come up with for the classical ones:

(*) : actually the 1st isogonic center, but also the Fermat point whenever A,B,C ≤ 2π/3

I'm not sure if this is satisfactory. I can see that for some users it would cause the trilinears column to be squashed up. As such, I'm further wondering whether to reduce these to just the first trilinear (as the "Recent" section is doing, basically a triangle centre function but in some cases not truly so). Or even do away with them and leave this more technical detail to the individual articles. What do people think? — Smjg (talk) 23:13, 27 October 2017 (UTC)


 * Your table with descriptions has been moved to the article. — Rgdboer (talk) 01:30, 3 March 2019 (UTC)

Proof for orthocenter equation
Think $$H$$ be the orthocenter. According to table contents, $$H$$ should be given as


 * $$H=\frac{(\vec{BC},\vec{BA})(\vec{CA},\vec{CB})A+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})B+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})C}{(\vec{BC},\vec{BA})(\vec{CA},\vec{CB})+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})}.$$

In order to verify if $$H$$ be unmistakably the orthocenter or not, goal can be if $$(\vec{AH},\vec{BC})=0$$ can be said or not. To avoid chaos, let's apply $$m$$ as the denominator for above $$H$$ equation, like


 * $$m\equiv(\vec{BC},\vec{BA})(\vec{CA},\vec{CB})+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA}),$$

then


 * $$mH=(\vec{BC},\vec{BA})(\vec{CA},\vec{CB})A+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})B+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})C,$$


 * $$m\vec{AH}=((\vec{BC},\vec{BA})(\vec{CA},\vec{CB})A+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})B+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})C)$$
 * $$-((\vec{BC},\vec{BA})(\vec{CA},\vec{CB})A+(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})A+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})A)$$


 * $$=(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})\vec{AB}+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})\vec{AC},$$


 * $$(m\vec{AH},\vec{BC})=(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})(\vec{AB},\vec{BC})+(\vec{AB},\vec{AC})(\vec{BC},\vec{BA})(\vec{AC},\vec{BC})$$


 * $$=(\vec{CA},\vec{CB})(\vec{AB},\vec{AC})(\vec{AB},\vec{BC})-(\vec{AB},\vec{AC})(\vec{AB},\vec{BC})(\vec{CA},\vec{CB})$$
 * $$=0.$$

Usually $$m\neq 0$$. So,


 * $$(\vec{AH},\vec{BC})=0.$$

Tsukitakemochi (talk) 14:38, 2 March 2019 (UTC)

Relation between Circumcentre and Orthocentre
By amending the flawed proof at the following link and using the fact that if the tan of two angles are the same then they differ by a multiple of pi (the ‘proof’ here says the angles are identical), then we can easily prove that the circumcentre and orthocentre of a triangle coincide iff it is equilateral.Overlordnat1 (talk) 12:21, 10 May 2021 (UTC)

Oops!Forgot to post link:-https://www.toppr.com/ask/question/if-origin-is-the-orthocentre-of-a-triangle-formed-by-the-points-cos/Overlordnat1 (talk) 12:23, 10 May 2021 (UTC)

Higher Dimensional Triangle Centers
I am wondering if there are higher dimensional triangle centers. A triangle is a simplex which can be generalized to higher dimensions such as a tetrahedron in 3 dimensions. ScientistBuilder (talk) 18:03, 16 June 2022 (UTC)


 * This is mentioned very briefly in the "Non-Euclidean and other geometries" section, with a reference to the Ungar book "Barycentric Calculus in Euclidean and Hyperbolic Geometry". It doesn't give a page number, and from the table of contents online none of the chapters appear to solely be about handle higher-dimensional triangle centers. It'd be worth verifying the reference, at leas to add the page number. Apocheir (talk) 21:21, 16 June 2022 (UTC)