Editing Centroid (section)

=== Of a triangle ===
{{main|Triangle center}}
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The centroid of a [[triangle]] is the point of intersection of its [[median (geometry)|medians]] (the lines joining each [[vertex (geometry)|vertex]] with the midpoint of the opposite side).<ref name="altshiller66" /> The centroid divides each of the medians in the [[ratio]] <math>2:1,</math> which is to say it is located <math>\tfrac13</math> of the distance from each side to the opposite vertex (see figures at right).<ref>{{harvtxt |Altshiller-Court|1925|p=65}}</ref><ref>{{harvtxt|Kay|1969|p=184}}</ref>  Its [[Cartesian coordinates]] are the [[arithmetic mean|means]] of the coordinates of the three vertices. That is, if the three vertices are <math>L = (x_L, y_L),</math> <math>M= (x_M, y_M),</math> and <math>N= (x_N, y_N),</math> then the centroid (denoted <math>C</math> here but most commonly denoted <math>G</math> in [[triangle geometry]]) is

<math display=block>
C = \tfrac13(L+M+N)
= \bigl( \tfrac13 (x_L+x_M+x_N), \tfrac13 (y_L+y_M+y_N) \bigr).
</math>

The centroid is therefore at <math>\tfrac13:\tfrac13:\tfrac13</math> in [[Barycentric coordinates (mathematics)|barycentric coordinates]].

In [[trilinear coordinates]] the centroid can be expressed in any of these equivalent ways in terms of the side lengths <math>a, b, c</math> and vertex angles {{nobr|<math>L, M, N</math>:<ref>Clark Kimberling's Encyclopedia of Triangles {{cite web |url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |title=Encyclopedia of Triangle Centers |access-date=2012-06-02 |url-status=dead |archive-url=https://web.archive.org/web/20120419171900/http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |archive-date=2012-04-19 }}</ref>}}

<math display=block>\begin{align}
C &= \frac{1}{a}:\frac{1}{b}:\frac{1}{c}
   = bc:ca:ab = \csc L :\csc M:\csc N \\[6pt]
  &= \cos L + \cos M \cdot \cos N :
     \cos M + \cos N \cdot \cos L :
     \cos N + \cos L \cdot \cos M \\[6pt]
  &= \sec L + \sec M \cdot \sec N :
     \sec M + \sec N \cdot \sec L :
     \sec N + \sec L \cdot\sec M.
\end{align}</math>

The centroid is also the physical center of mass if the triangle is made from a uniform sheet of material; or if all the mass is concentrated at the three vertices, and evenly divided among them.  On the other hand, if the mass is distributed along the triangle's perimeter, with uniform [[linear density]], then the center of mass lies at the [[Spieker center]] (the [[incenter]] of the [[medial triangle]]), which does not (in general) coincide with the geometric centroid of the full triangle.

The area of the triangle is <math>\tfrac32</math> times the length of any side times the perpendicular distance from the side to the centroid.<ref>{{harvtxt|Johnson|2007|p=173}}</ref>

A triangle's centroid lies on its [[Euler line]] between its [[orthocenter]] <math>H</math> and its [[circumcenter]] <math>O,</math> exactly twice as close to the latter as to the former:<ref>{{harvtxt|Altshiller-Court|1925|p=101}}</ref><ref>{{harvtxt|Kay|1969|pp=18,189,225–226}}</ref>

<math display=block>\overline{CH}=2\overline{CO}.</math>

In addition, for the [[incenter]] <math>I</math> and [[nine-point center]] <math>N,</math> we have
<math display=block>\begin{align}
\overline{CH} &=4\overline{CN}, \\[5pt]
\overline{CO} &=2\overline{CN}, \\[5pt]
\overline{IC} &< \overline{HC}, \\[5pt]
\overline{IH} &< \overline{HC}, \\[5pt]
\overline{IC} &< \overline{IO}.
\end{align}</math>

If <math>G</math> is the centroid of the triangle <math>ABC,</math> then

<math display=block>
(\text{Area of }\triangle ABG)
= (\text{Area of }\triangle ACG)
= (\text{Area of }\triangle BCG)
= \tfrac13(\text{Area of }\triangle ABC).
</math>

The [[isogonal conjugate]] of a triangle's centroid is its [[symmedian|symmedian point]].

Any of the three medians through the centroid divides the triangle's area in half. This is not true for other lines through the centroid; the greatest departure from the equal-area division occurs when a line through the centroid is parallel to a side of the triangle, creating a smaller triangle and a [[trapezoid]]; in this case the trapezoid's area is <math>\tfrac59</math> that of the original triangle.<ref name=Bottomley2002>{{cite web|last=Bottomley|first=Henry|title=Medians and Area Bisectors of a Triangle| url=http://www.se16.info/js/halfarea.htm|access-date=27 September 2013}}</ref>

Let <math>P</math> be any point in the plane of a triangle with vertices <math>A, B, C</math> and centroid <math>G.</math> Then the sum of the squared distances of <math>P</math> from the three vertices exceeds the sum of the squared distances of the centroid <math>G</math> from the vertices by three times the squared distance between <math>P</math> and {{nobr|<math>G</math>:<ref name = altshiller70>{{harvtxt|Altshiller-Court|1925|pp=70–71}}</ref>}}

<math display=block>
PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^2 + 3PG^2.
</math>

The sum of the squares of the triangle's sides equals three times the sum of the squared distances of the centroid from the vertices:<ref name = altshiller70/>

<math display=block>
AB^2 + BC^2 + CA^2 = 3(GA^2 + GB^2 + GC^2).
</math>

A triangle's centroid is the point that maximizes the product of the directed distances of a point from the triangle's sidelines.<ref>{{cite journal|first =Clark|last= Kimberling|title=Trilinear distance inequalities for the symmedian point, the centroid, and other triangle centers|journal =Forum Geometricorum|volume= 10|date=201|pages = 135–139|url = http://forumgeom.fau.edu/FG2010volume10/FG201015index.html }}</ref>

Let <math>ABC</math> be a triangle, let <math>G</math> be its centroid, and let <math>D, E, F</math> be the midpoints of segments <math>BC, CA, AB,</math> respectively. For any point <math>P</math> in the plane of <math>ABC,</math><ref>Gerald A. Edgar, Daniel H. Ullman & Douglas B. West (2018) Problems and Solutions, The American Mathematical Monthly, 125:1, 81-89, DOI: 10.1080/00029890.2018.1397465</ref>

<math display=block>
PA + PB + PC \le 2(PD + PE + PF) + 3PG.
</math>