Editing Centroid (section)

== Determination {{anchor|Locating}} ==
{{see also|Center of mass#Determination}}

=== Plumb line method ===
The centroid of a uniformly dense [[planar lamina]], such as in figure (a) below, may be determined experimentally by using a [[plumbline]] and a pin to find the collocated center of mass of a thin body of uniform density having the same shape.  The body is held by the pin, inserted at a point, off the presumed centroid in such a way that it can freely rotate around the pin; the plumb line is then dropped from the pin (figure b).  The position of the plumbline is traced on the surface, and the procedure is repeated with the pin inserted at any different point (or a number of points) off the centroid of the object. The unique intersection point of these lines will be the centroid (figure c).  Provided that the body is of uniform density, all lines made this way will include the centroid, and all lines will cross at exactly the same place.

{| cellpadding=3
|- valign=center
| [[File:Center gravity 0.svg|147px|center]]
| [[File:Center gravity 1.svg|183px|center]]
| [[File:Center gravity 2.svg|157px|center]]
|-
| align=center | (a)
| align=center | (b)
| align=center | (c)
|}

This method can be extended (in theory) to concave shapes where the centroid may lie outside the shape, and virtually to solids (again, of uniform density), where the centroid may lie within the body. The (virtual) positions of the plumb lines need to be recorded by means other than by drawing them along the shape.

=== Balancing method ===
For convex two-dimensional shapes, the centroid can be found by balancing the shape on a smaller shape, such as the top of a narrow cylinder.  The centroid occurs somewhere within the range of contact between the two shapes (and exactly at the point where the shape would balance on a pin).  In principle, progressively narrower cylinders can be used to find the centroid to arbitrary precision.  In practice air currents make this infeasible.  However, by marking the overlap range from multiple balances, one can achieve a considerable level of accuracy.

=== Of a finite set of points ===
The centroid of a finite set of <math>k</math> points <math>\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_k</math> in <math>\R^n</math> is<ref name = protter520/>
<math display=block>\mathbf{C} = \frac{\mathbf{x}_1 + \mathbf{x}_2 + \cdots + \mathbf{x}_k}{k} .</math>
This point minimizes the sum of squared Euclidean distances between itself and each point in the set.

=== By geometric decomposition ===
The centroid of a plane figure <math>X</math> can be computed by dividing it into a finite number of simpler figures <math>X_1, X_2, \dots, X_n,</math> computing the centroid <math>C_i</math> and area <math>A_i</math> of each part, and then computing

<math display=block>
C_x = \frac{\sum_i {C_i}_x A_i}{\sum_i A_i}, \quad
C_y = \frac{\sum_i {C_i}_y A_i}{\sum_i A_i}.
</math>

Holes in the figure <math>X,</math> overlaps between the parts, or parts that extend outside the figure can all be handled using negative areas <math>A_i.</math> Namely, the measures <math>A_i</math> should be taken with positive and negative signs in such a way that the sum of the signs of <math>A_i</math> for all parts that enclose a given point <math>p</math> is <math>1</math> if <math>p</math> belongs to <math>X,</math> and <math>0</math> otherwise.

For example, the figure below (a) is easily divided into a square and a triangle, both with positive area; and a circular hole, with negative area (b).
{{multiple images
|align=center
|height=200
|direction=horizontal
|image1=COG 1.png
|caption1=(a) 2D Object
|image2=COG 2.png
|caption2=(b) Object described using simpler elements
|image3=COG 3.png
|caption3=(c) Centroids of elements of the object
}}
The centroid of each part can be found in any [[list of centroids|list of centroids of simple shapes]] (c). Then the centroid of the figure is the weighted average of the three points.  The horizontal position of the centroid, from the left edge of the figure is
<math display=block>x = \frac{5 \times 10^2 + 13.33 \times \frac{1}{2}10^2 - 3 \times \pi2.5^2}{10^2 + \frac{1}{2}10^2 -\pi2.5^2} \approx 8.5 \text{ units}.</math>
The vertical position of the centroid is found in the same way.

The same formula holds for any three-dimensional objects, except that each <math>A_i</math> should be the volume of <math>X_i,</math> rather than its area.  It also holds for any subset of <math>\R^d,</math> for any dimension <math>d,</math> with the areas replaced by the <math>d</math>-dimensional [[measure (mathematics)|measure]]s of the parts.

=== By integral formula ===
The centroid of a subset <math>X</math> of <math>\R^n</math> can also be computed by the vector formula

<math display=block>C = \frac{\int x g(x) \ dx}{\int g(x) \ dx} = \frac{\int_X x \ dx}{\int_X dx}</math>

where the [[integral]]s are taken over the whole space <math>\R^n,</math> and <math>g</math> is the [[Indicator function|characteristic function]] of the subset <math>X</math> of <math>\R^n \! : \ g(x) = 1</math> if <math>x \in X</math> and <math>g(x) = 0</math> otherwise.<ref name = protter526>{{harvtxt|Protter|Morrey|1970|p=526}}</ref>  Note that the denominator is simply the [[Measure (mathematics)|measure]] of the set <math>X.</math> This formula cannot be applied if the set <math>X</math> has zero measure, or if either integral diverges.

Alternatively, the coordinate-wise formula for the centroid is defined as

<math display=block>C_k = \frac{\int z S_k(z) \ dz}{\int g(x) \ dx},</math>

where <math>C_k</math> is the <math>k</math>th coordinate of <math>C,</math> and <math>S_k(z)</math> is the measure of the intersection of <math>X</math> with the [[hyperplane]] defined by the equation <math>x_k = z.</math>  Again, the denominator is simply the measure of <math>X.</math>

For a plane figure, in particular, the barycentric coordinates are

<math display=block>
C_{\mathrm x} = \frac{\int x S_{\mathrm y}(x) \ dx}{A}, \quad
C_{\mathrm y} = \frac{\int y S_{\mathrm x}(y) \ dy}{A},
</math>

where <math>A</math> is the area of the figure <math>X,</math> <math>S_{\mathrm y}(x)</math> is the length of the intersection of <math>X</math> with the vertical line at [[abscissa]] <math>x,</math> and <math>S_{\mathrm x}(y)</math> is the length of the intersection of <math>X</math> with the horizontal line at [[ordinate]] <math>y.</math>

==== Of a bounded region ====
The centroid <math>(\bar{x},\;\bar{y})</math> of a region bounded by the graphs of the [[continuous function]]s <math>f</math> and <math>g</math> such that <math>f(x) \geq g(x)</math> on the interval <math>[a, b],</math> <math>a \leq x \leq b</math> is given by<ref name="protter526" /><ref>{{harvtxt|Protter|Morrey|1970|p=527}}</ref>

<math display=block>\begin{align}
\bar{x} &= \frac{1}{A}\int_a^b x\bigl(f(x) - g(x)\bigr)\,dx, \\[5mu]
\bar{y} &= \frac{1}{A}\int_a^b \tfrac12\bigl(f(x) + g(x)\bigr)\bigl(f(x) - g(x)\bigr)\,dx,
\end{align}</math>

where <math>A</math> is the area of the region (given by <math display=inline>\int_a^b \bigl(f(x) - g(x)\bigr) dx</math>).<ref>{{harvtxt|Protter|Morrey|1970|p=528}}</ref><ref>{{harvtxt|Larson|1998|pp=458–460}}</ref>

=== With an integraph ===
An [[integraph]] (a relative of the [[planimeter]]) can be used to find the centroid of an object of irregular shape with smooth (or piecewise smooth) boundary.  The mathematical principle involved is a special case of [[Green's theorem]].<ref>{{harvtxt|Sangwin}}</ref>

=== Of an L-shaped object ===
This is a method of determining the centroid of an L-shaped object.

[[Image:CoG of L shape.svg|300px]]

#Divide the shape into two rectangles, as shown in fig 2. Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids. The centroid of the shape must lie on this line <math>AB.</math>
#Divide the shape into two other rectangles, as shown in fig 3. Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids. The centroid of the L-shape must lie on this line <math>CD.</math>
#As the centroid of the shape must lie along <math>AB</math> and also along <math>CD,</math> it must be at the intersection of these two lines, at <math>O.</math> The point <math>O</math> might lie inside or outside the L-shaped object.

=== Of a triangle ===
{{main|Triangle center}}
{| align="right" border="0" cellspacing="0" cellpadding="0"
|-
| [[Image:Triangle centroid 1.svg]] [[Image:Triangle centroid 2.svg|200px]]
|}

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>

=== Of a polygon ===
The centroid of a non-self-intersecting closed [[polygon]] defined by <math>n</math> vertices <math>(x_0, y_0),\;</math><math>(x_1, y_1),\; \ldots,\;</math><math>(x_{n-1}, y_{n-1}),</math> is the point <math>(C_x, C_y),</math><ref name=bourke /> where

<math display=block>
C_{\mathrm x} = \frac{1}{6A}\sum_{i=0}^{n-1}(x_i+x_{i+1})(x_i\ y_{i+1} - x_{i+1}\ y_i),</math>

and

<math display=block>
C_{\mathrm y} = \frac{1}{6A}\sum_{i=0}^{n-1}(y_i+y_{i+1})(x_i\ y_{i+1} - x_{i+1}\ y_i),
</math>

and where <math>A</math> is the polygon's signed area,<ref name=bourke>{{harvtxt|Bourke|1997}}</ref> as described by the [[shoelace formula]]:

<math display=block>
A = \frac{1}{2}\sum_{i=0}^{n-1} (x_i\ y_{i+1} - x_{i+1}\ y_i).
</math>

In these formulae, the vertices are assumed to be numbered in order of their occurrence along the polygon's perimeter; furthermore, the vertex <math>(x_n, y_n)</math> is assumed to be the same as <math>(x_0, y_0),</math> meaning <math>i+1</math> on the last case must loop around to <math>i=0.</math> (If the points are numbered in clockwise order, the area <math>A,</math> computed as above, will be negative; however, the centroid coordinates will be correct even in this case.)

{{anchor|Of a vertex set}}The centroid of a non-triangular polygon is not the same as its ''vertex centroid'', considering only its [[Vertex (geometry)|vertex]] set (as the [[#Of a finite set of points|centroid of a finite set of points]]; {{xref|see also: [[Polygon#Centroid]]}}).

=== Of a cone or pyramid ===
The centroid of a [[Cone (geometry)|cone]] or [[Pyramid (geometry)|pyramid]] is located on the line segment that connects the [[Apex (geometry)|apex]] to the centroid of the base. For a solid cone or pyramid, the centroid is <math>\tfrac14</math> the distance from the base to the apex. For a cone or pyramid that is just a shell (hollow) with no base, the centroid is <math>\tfrac13</math> the distance from the base plane to the apex.

===Of a tetrahedron and {{mvar|n}}-dimensional simplex===
A [[tetrahedron]] is an object in [[three-dimensional space]] having four triangles as its [[face (geometry)|faces]]. A line segment joining a vertex of a tetrahedron with the centroid of the opposite face is called a ''median'', and a line segment joining the midpoints of two opposite edges is called a ''bimedian''. Hence there are four medians and three bimedians. These seven line segments all meet at the ''centroid'' of the tetrahedron.<ref>Leung, Kam-tim; and Suen, Suk-nam; "Vectors, matrices and geometry", Hong Kong University Press, 1994, pp. 53–54</ref> The medians are divided by the centroid in the ratio <math>3:1.</math> The centroid of a tetrahedron is the midpoint between its [[Monge point]] and circumcenter (center of the circumscribed sphere). These three points define the ''Euler line'' of the tetrahedron that is analogous to the [[Euler line]] of a triangle.

These results generalize to any <math>n</math>-dimensional [[simplex]] in the following way. If the set of vertices of a simplex is <math>{v_0,\ldots,v_n},</math> then considering the vertices as [[vector (geometry)|vectors]], the centroid is

<math display=block>C = \frac{1}{n+1}\sum_{i=0}^n v_i.</math>

The geometric centroid coincides with the center of mass if the mass is uniformly distributed over the whole simplex, or concentrated at the vertices as <math>n+1</math> equal masses.

=== Of a hemisphere ===
The centroid of a solid hemisphere (i.e. half of a solid ball) divides the line segment connecting the sphere's center to the hemisphere's pole in the ratio <math>3:5</math> (i.e. it lies <math>\tfrac38</math> of the way from the center to the pole).
The centroid of a hollow hemisphere (i.e. half of a hollow sphere) divides the line segment connecting the sphere's center to the hemisphere's pole in half.