Prove by using vectors $(\vec {AP} +\vec {CP}+\vec {BP})=\vec 0$

Question

Let A,B,C be the points of triangle and P inside of triangle. How can I prove using vectors that: $$(\vec {AP} +\vec {CP}+\vec {BP})=\vec 0$$ only if P is centroid

Answer 1

I assume that you are using the definition of the centroid of a triangle as the intersection of its medians. The essential idea is to use the following important property of vectors: for any points $X$, $Y$ and $Z$, $$ \vec{XZ} = \vec{XY} + \vec{YZ}. $$ (This is easy to prove by the definitions.) Now, note that \begin{align*} \vec{AP} + \vec{CP} + \vec{BP} = 0 &\iff \vec{AP} + \vec{AP} - \vec{AC} + \vec{AP} - \vec{AB} = 0 \\ &\iff \vec{AP} = \frac{\vec{AB} + \vec{AC}}{3} \\ &\iff \vec{AP} = \frac{2\vec{AD}}{3}, \end{align*} where $D$ is the midpoint of side $BC$. This proves that $P$ lies on the median $AD$; by symmetry, we know that $P$ also lies on the medians $BE$ and $CF$ (where $E$, $F$ are the midpoints of $CA$, $AB$ respectively), so $P$ is at the intersection of medians of the triangle.

In other words, your statement implies that $P$ is the centroid. I'll leave it as an exercise to you to prove the other direction: that is, if $P$ is the intersection of medians of the triangle, the equation is satisfied.

Answer 2

Let $\vec a,\vec b,\vec c$ be position vectors of points $A,B,C$. By definition, the position vector of the centroid $$\vec p =\frac{\vec a+\vec b+\vec c}{3}$$ But then $$\vec {AP}+\vec {BP} +\vec{CP} =(\vec p-\vec a) + (\vec p-\vec b)+(\vec p -\vec c) \\ = 3\vec p -\vec a-\vec b-\vec c \\ =0 $$