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$OABC$ is a tetrahedron where $O$ is the origin.Position vectors of its angular points $A,B$ and $C$ are $\vec{a},\vec{b}$ and $\vec{c}$ respectively. Segments joining each vertex with the centroid of the opposite face are concurrent at the point $P$.Find the position vector of $P$.


I know that the position vector of the centroid of a plane face $ABC$ is $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$.But i have no idea how to find the position vector of $P$ in this question.Please help me.

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The position vector of the centroid of the plane $OAB,OBC,OCA,ABC$ is $$\frac{\vec a+\vec b}{3},\quad\frac{\vec b+\vec c}{3},\quad\frac{\vec c+\vec a}{3},\quad\frac{\vec a+\vec b+\vec c}{3}$$ respectively.

Now, let $\vec p$ be the position vector of $P$. Then, there exist $s,t,u,v$ such that $$\vec p-\vec a=s\left(\frac{\vec b+\vec c}{3}-\vec a\right),\qquad \vec p-\vec b=t\left(\frac{\vec c+\vec a}{3}-\vec b\right)$$ $$\vec p-\vec c=u\left(\frac{\vec a+\vec b}{3}-\vec c\right),\qquad \vec p=v\left(\frac{\vec a+\vec b+\vec c}{3}\right),$$ i.e. $$\vec p=(1-s)\vec a+\frac s3\vec b+\frac s3\vec c,\qquad \vec p=\frac t3\vec a+(1-t)\vec b+\frac t3\vec c$$ $$\vec p=\frac u3\vec a+\frac u3\vec b+(1-u)\vec c,\qquad \vec p=\frac v3\vec a+\frac v3\vec b+\frac v3\vec c$$ So, solving $$1-s=\frac t3=\frac u3=\frac v3,\quad\frac s3=1-t=\frac u3=\frac v3,\quad\frac s3=\frac t3=1-u=\frac v3$$ gives $s=t=u=v=\frac 34$.

Thus, the position vector of $P$ is $\color{red}{\frac 14\vec a+\frac 14\vec b+\frac 14\vec c}.$

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