Show for the centroid $M$ of a triangle $ABC$ that $\vec{OM}=\frac13(\vec{OA}+\vec{OB}+\vec{OC})$

Question

Show for the centroid $M$ of a triangle $ABC$ that $$\vec{OM}=\dfrac13\left(\vec{OA}+\vec{OB}+\vec{OC}\right)$$ where $O$ is an arbitrary point. I haven't studied position vectors and am very new to vectors so I would like a simple solution. Any help would be appreciated. I don't even know how to start. Maybe we can use the fact that the centroid divides each median in a ratio of $2:1$? Thank you!

Answer 1

As $A_1$ is the midpoint of $BC$,

$2\vec{OA_1} = \vec{OB} + \vec{OC}$

As $M$ is the centroid of the triangle, it divides median $AA_1$ into ratio of $AM:MA_1 = 2:1$

$\vec{OM} = \frac{\vec{OA}+2\vec{OA_1}}{3} = \frac{\vec{OA}+\vec{OB} + \vec{OC}}{3}$

That is all the proof is.

But if you need to further show how $2\vec{OA_1} = \vec{OB} + \vec{OC}$,

$\vec{OA_1} = \vec{OB} + \vec{BA_1}$

But also, $\vec{OA_1} = \vec{OC} + \vec{CA_1} = \vec{OC} - \vec{A_1C}$

As $\vec{BA_1} = \vec{A_1C}$, adding both you have

$2\vec{OA_1} = \vec{OB} + \vec{OC}$

Similarly you can show $\vec{OM} = \frac{\vec{OA}+2\vec{OA_1}}{3}$.

$\vec{OM} = \vec{OA}+\vec{AM}$

As $\vec{AM} = \frac{2}{3} \vec{AA_1}$,

$\vec{OM} = \frac{3\vec{OA}+2\vec{AA_1}}{3} = \frac{\vec{OA}+2(\vec{OA}+\vec{AA_1})}{3} = \frac{\vec{OA}+2\vec{OA_1}}{3}$

Answer 2

If the coordinate of vertices A, B and C are $(x_1, y_1),( x_2, y_2)$ and $(x_3, y_3)$ then coordinate of centroid is $M(\frac{x_1+x_2+x_3)}3,\frac{y_1+y_2+y_3)}3)$.

The vector form is:

$\vec{OM}=(x_M i+y_Mj)=(\frac{x_1+x_2+x_3)}3 i+\frac{y_1+y_2+y_3)}3 j)$

$A(x_1,y_1)$ means vector $\vec{OA}=(x_1 i+y_1 j)$ , so is about B and C so we can write:

$(x_M i+ y_M j)=\frac 13[(x_1 i+y_1j)+(x_2 i+y_2 j)+(x_3 i+y_3 j)]$

Or in vector form:

$\vec{OM}=\frac 13(\vec{OA}+\vec{OB}+\vec{OC})$

Now if you transform coordinate to an arbitrary point like $O(x_O. y_O)$ the coordinates of vertices will be:

$(x_1-x_O, y_1-y_O)$ which is vector $\vec{OA}=(x_1-x_O)i+(y_1-y_O)j$ ,$(x_2-x_o, y_2-y_o)$ which is vector $\vec{OB}=(x_2-x_O)i+(y_2-y_O)j$ and $(x_3-x_O, y_3-y_O)$ which is vector $\vec{OC}=(x_3-x_O)i+(y_3-y_O)j$ .Hence the coordinate of centroid is:

$M(\frac{(x_1-x_o)+(x_2-x_o)+(x_3-x_o)}3),(\frac{(y_1-y_o)+(y_2-y_o)+(y_3-y_o)}3)$

Or :

$\vec{OM}=(\frac{(x_1-x_o)+(x_2-x_o)+(x_3-x_o)}3)i+(\frac{(y_1-y_o)+(y_2-y_o)+(y_3-y_o)}3 )j$

The vector representation of this is:

$\vec{OM}=\frac{\vec{OA}+\vec{OB}+\vec{OC}}3$

Answer 3

The median passing through $A$ has equation$$r=a+s((b+c)/2-a)=(1-s)a+(s/2)b+(s/2)c.$$The median through $B$ is$$r=(t/2)a+(1-t)b+(t/2)c.$$These meet at$$s=t=2/3,\,r=(a+b+c)/3.$$This point lies at $u=2/3$ on the third median,$$r=(u/2)a+(u/2)b+(1-u)c.$$

Answer 4

We take the triple $(A,B,C)$ as barycentric coordinate system for the plane. That means: any point $P$ of the plane is written

$$ P = \alpha A + \beta B + \gamma \,C \qquad \text{ with }\;\alpha+\beta+\gamma=1, \tag{1} $$

which is equivalent to

$$ P-O = \alpha(A-O) + \beta(B-O) + \gamma\,(C-O) \qquad \big[\alpha+\beta+\gamma=1\big] \tag{2} $$

for any point $\,O\,$ of the plane (see Note 1 below).

The numbers $\;\alpha,\beta,\gamma\;$ in $(1)$ are the barycentric coordinates of $\;P\;$ with respect to the given coordinate system.

We have $\;A_1-C=B-A_1\;$ and hence $A_1=\frac12(B+C)$. Similarly for others, so, in barycentric coordinates:

$$ A_1=\frac12(B+C)\qquad B_1=\frac12(A+C)\qquad C_1=\frac12(A+B). \tag{3} $$

The median $AA_1$ is therefore described by equation $\;tA+(1-t)A_1\quad[t\in\Bbb R],\;$ i.e. $\;tA+(1-t)\frac12(B+C)$, and similarly for others medians:

$$ AA_1:\qquad tA+\frac12(1-t)(B+C) \qquad \big[t\in\Bbb R\big] \tag{4}$$ $$ BB_1:\qquad sB+\frac12(1-s)(A+C) \qquad \big[s\in\Bbb R\big] \tag{5}$$ $$ CC_1:\qquad rC+\frac12(1-r)(A+B) \qquad \big[r\in\Bbb R\big]. \tag{6}$$

The lines $(4), (5)$ intersect in a point $M$ iff there exist $t,s\in\Bbb R\;$ such that

$$ M=tA+\frac12(1-t)(B+C) = sB+\frac12(1-s)(A+C) \tag{7}$$

i.e. (see Note 2 below) iff:

$$ t=\frac12(1-s) \qquad\quad \frac12(1-t)=s \qquad\quad \frac12(1-t)=\frac12(1-s) $$

which gives $\;t=s=\frac13$. A similar work on the other medians give $\;t=s=r=\frac13,$ therefore the medians are concurrent in $\;M$, and from $(7)$ we get

$$ M=\frac13A+\frac13B+\frac13C, \tag{8} $$

i.e. the barycentric coordinates of the centroid $M$ are $(\frac13,\frac13,\frac13).$

The equation $(8)$ immediately give our thesis

$$ M-O = \frac13\Big((A-O) + (B-O) + (C-O)\Big). $$

$\quad$

Note 1.$\quad$ We prove that $(2)$ is independent of $\,O$. If $\;O'\;$ is any other point, we have

\begin{align} P-O' &= (P-O) +(O-O') =\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\[1ex] &= \alpha(A-O) + \beta(B-O) + \gamma\,(C-O) + (O-O') =\\[1ex] &= \alpha\Big((A-O')+(O'-O)\Big) + \beta\Big((B-O')+(O'-O)\Big) +\\[1ex] &\qquad\qquad +\gamma\Big((C-O')+(O'-O)\Big) + (O-O')=\\[1ex] &= \alpha(A-O')+\beta(B-O')+\gamma\,(C-O')+(\alpha+\beta+\gamma)(O'-O)+(O-O') =\\[1ex] &= \alpha(A-O')+\beta(B-O')+\gamma\,(C-O'),\\ \end{align} being $\;\alpha+\beta+\gamma=1$.

The point $P$ in $(2)$, independent of $O$, is indicated as in $(1)$.

$\quad$

Note 2.$\quad$ If $\;\alpha+\beta+\gamma = \alpha'+\beta'+\gamma' = 1$, we have $$ \alpha A+\beta B+\gamma\,C = \alpha'A+\beta'B+\gamma'C \quad\implies\quad \alpha=\alpha', \;\;\beta=\beta', \;\;\gamma=\gamma' $$

In fact, by choosing $\;O=A\;$, the preceding equality give

$$ A+\alpha(A-A)+\beta(B-A)+\gamma\,(C-A) = A+\alpha'(A-A)+\beta'(B-A)+\gamma'(C-A) $$

hence

$$ \beta(B-A)+\gamma\,(C-A) = \beta'(B-A)+\gamma'(C-A). $$

As $\;B-A,\;C-A\;$ are linearly independent, we deduce $\beta=\beta'\;$ and $\;\gamma=\gamma'$; and finally $\;\alpha=1-\beta-\gamma=1-\beta'-\gamma'=\alpha'.$