Centroid of a Triangle
The centroid of a triangle is the intersection of the three medians, or the "average" of the three vertices. It has several important properties and relations with other parts of the triangle, including its circumcenter, orthocenter, incenter, area, and more.
The centroid is typically represented by the letter \(G\).
Contents
Finding the centroid
The centroid is easily found using coordinates: a triangle with vertices at \((x_1, y_1), (x_2, y_2)\), and \((x_3, y_3)\) has centroid at \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\).
Triangle \(ABC\) has \(A = (3,4)\), \(B=(5,12)\), and \(C=(8,15)\). What are the coordinates of the centroid of \(ABC\)?
The centroid lies at
\[\left(\frac{3+5+8}{3}, \frac{4+12+15}{3}\right)=\left(\frac{16}{3}, \frac{31}{3}\right)\]
Proof of existence
The simplest proof is a consequence of Ceva's theorem, which states that \(AD, BE, CF\) concur if and only if
\(\frac{AE}{EC} \cdot \frac{CD}{DB} \cdot \frac{BF}{FA} = 1\)
In this case, \(D,E,F\) are the midpoints of their respective sides. Therefore, \(AE=EC, CD=DB, BF=FA\), so the equality above is immediately true, demonstrating the existence of the centroid.
Properties
A median of a triangle is the line segment between a vertex of the triangle and the midpoint of the opposite side. Each median divides the triangle into two triangles of equal area.The centroid is the intersection of the three medians.
The three medians also divide the triangle into six triangles, each of which have the same area.
The centroid divides each median into two parts, which are always in the ratio 2:1.
The centroid also has the property that
\(AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2)\).
This is a consequence of the more general property that
\(PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3PG^2\) for any point \(P\)
(the above formula comes from taking \(P=A, B, C\) in turn, then adding the three results).
The sides of triangle \(ABC\) are \(5,6,\) and \(7\). \(P\) is a point in the plane of the triangle such that \(PA^2 + PB^2 + PC^2 = 70\). The locus of \(P\) is a circle of radius \(r\), where \(r\) can be expressed in the form \(\frac{m}{n}\) for some relatively prime positive integers \(m\) and \(n\). Find \(100m+n\).
A similar property is the following: any line through the centroid hits \(AB\) at a point \(D\) and \(AC\) at a point \(E\). Then
\(\frac{BD}{DA}+\frac{CE}{EA}=1\).
It is also possible to calculate the length of a median from the side lengths:
\(AD = \sqrt{\frac{2b^2+2c^2-a^2}{4}}\)
\(BE = \sqrt{\frac{2a^2+2c^2-b^2}{4}}\)
\(CF = \sqrt{\frac{2a^2+2b^2-c^2}{4}}\)
Note that this also gives the length of \(AG, BG,\) and \(CG\), since the median is divided in a 2:1 ratio by the centroid:
\(AG = \frac{\sqrt{2b^2+2c^2-a^2}}{3}\)
\(BG = \frac{\sqrt{2a^2+2c^2-b^2}}{3}\)
\(CG = \frac{\sqrt{2a^2+2b^2-c^2}}{3}\)
which is another way of showing that \(AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2)\).
Relations to other triangle centers
If \(O\) is the circumcenter of a triangle, then
\(OG = R^2-\frac{1}{9}(a^2+b^2+c^2)\)
where \(R\) is the circumradius of the triangle.
The centroid also lies on the Euler line of the triangle, so
\(GH = \frac{2}{3}OH, GO=\frac{1}{3}OH\)
where \(H\) is the orthocenter of the triangle.
If \(A', B', C'\) are the circumcenters of triangles \(BCG, ACG, ABG\) respectively, then
(O) is the centroid of triangle \(A'B'C'\). Furthermore, \(G\) is the symmedian point of \(\triangle A'B'C'\).
Finally, the medians of \(\triangle A'B'C'\) pass through the midpoints of \(AB, BC,\) and \(CA\), so the medians of \(\triangle A'B'C'\) and \(\triangle ABC\) intersect at the midpoints of the original triangle.
Other polygons
Other polygons have analogous interpretations of the centroid; it remains the center of mass of the vertices of the polygon.
However, the centroid is no longer (necessarily) the intersection of the medians; in fact, the medians do not necessarily intersect in larger polygons.