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 $G$.

The centroid is easily found using coordinates: a triangle with vertices at $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ has centroid at $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right).$

Triangle $ABC$ has vertices $A = (3,4)$, $B=(5,12)$, and $C=(8,15)$. What are the coordinates of the centroid of triangle $ABC$?

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The centroid lies at

$\left(\frac{3+5+8}{3}, \frac{4+12+15}{3}\right)=\left(\frac{16}{3}, \frac{31}{3}\right).\ _\square$

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, $AE=EC, CD=DB,$ and $BF=FA,$ so the equality above is immediately true, demonstrating the existence of the centroid.

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

$AB^2+BC^2+CA^2=3\big(GA^2+GB^2+GC^2\big).$

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 $P.$

We are going to use Apollonius' theorem.

Let $D$ be the point where $AG$ and $BC$ meet, $E$ the point where $BG$ and $CA$ meet, and $F$ the point where $CG$ and $AB$ meet.

Use the formula on $\triangle ADP,~\triangle BEP,~\triangle CFP$ and add them together:

$PA^2+PB^2+PC^2+2(PD^2+PE^2+PF^2)=3\left(3PG^2+\dfrac{GA^2+GB^2+GC^2}{2}\right).\qquad (1)$

Use the formula on $\triangle ABP,~\triangle BCP,~\triangle CAP$ and add them together:

$2(PA^2+PB^2+PC^2)=2\left(PD^2+PE^2+PF^2+\dfrac{AB^2+BC^2+CA^2}{4}\right).\qquad (2)$

Use the formula on $\triangle ABD,~\triangle BCE,~\triangle CAF$ and add them together:

${\color{#D61F06}{AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2)}}.\qquad (3)$

$\big($Note that we accidentally proved $AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2)$ in the way.$\big)$

Substitute $(3)$ into $(2)$ and move some stuff around:

$2(PA^2+PB^2+PC^2)-2(PD^2+PE^2+PF^2)=\dfrac{3}{2}(GA^2+GB^2+GC^2).$

Add the above equation with $(1):$

$\begin{aligned} 3(PA^2+PB^2+PC^2)&=3(GA^2+GB^2+GC^2+3PG^2) \\ PA^2+PB^2+PC^2&=GA^2+GB^2+GC^2+3PG^2.\ _\square \end{aligned}$

$\big($The formula $AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2)$ can also be obtained from taking $P=A, B, C$ in turn, and then adding the three results.$\big)$

A similar property is the following: if 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:

$\begin{aligned} AD &= \sqrt{\frac{2b^2+2c^2-a^2}{4}}\\\\ BE &= \sqrt{\frac{2a^2+2c^2-b^2}{4}}\\\\ CF &= \sqrt{\frac{2a^2+2b^2-c^2}{4}}. \end{aligned}$

Note that this also gives the lengths of $AG, BG,$ and $CG$, since the median is divided in a 2:1 ratio by the centroid:

$\begin{aligned} AG &= \frac{\sqrt{2b^2+2c^2-a^2}}{3}\\\\ BG &= \frac{\sqrt{2a^2+2c^2-b^2}}{3}\\\\ CG &= \frac{\sqrt{2a^2+2b^2-c^2}}{3}, \end{aligned}$

which is another way of showing that $AB^2+BC^2+CA^2=3\big(GA^2+GB^2+GC^2\big)$.

Other centers of the triangle include the

• orthocenter
• incenter
• circumcenter.

The orthocenter is the point where the three altitudes of a triangle meet. The altitude is a line segment drawn from one vertex to the opposite side, and it is perpendicular to the opposite side.

The incenter is the center of the triangle's incircle. The incircle is the circle subscribed inside the triangle and it is tangent to each of its sides.

The circumcenter is the center of the circumcircle, the circle that passes through all three vertices of the triangle.

If $O$ is the circumcenter of a triangle, $R$ is the circumradius of the triangle, and $a,b,c$ are the lengths of $BC,CA,AB,$ respectively, then $OG^2 = R^2-\frac{1}{9}\big(a^2+b^2+c^2\big).$

Substitute $P=O$ into the formula $PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+PG^2,$ and we have

$\begin{aligned} OA^2+OB^2+OC^2 &=GA^2+GB^2+GC^2+3OG^2 \\ 9R^2&=3(GA^2+GB^2+GC^2)+9OG^2 &&\qquad [\text{since } OA=OB=OC=R]\\ &=a^2+b^2+c^2+9OG^2 &&\qquad \big[\text{since } AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2)\big] \\ \Rightarrow OG^2 &= R^2-\frac{1}{9}\big(a^2+b^2+c^2\big).\ _\square \end{aligned}$

The centroid also lies on the Euler line of the triangle, so

$GH = \frac{2}{3}OH,\quad GO=\frac{1}{3}OH,$

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 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.

• Circumcenter
• Incenter
• Orthocenter