If ABC and PQR are two triangles in space, prove that their…

If $ABC$ and $PQR$ are two triangles in space, prove that their centroids coincide if and only if $\overrightarrow{AP}+\overrightarrow{BQ}+\overrightarrow{CR}=\vec 0$.

Explanation of Method

When do the centroids of two triangles coincide?

Step 1: Centroids

\[ \vec g_1=\frac{\vec a+\vec b+\vec c}{3},\qquad \vec g_2=\frac{\vec p+\vec q+\vec r}{3}. \]

Step 2: Key identity

\[ \overrightarrow{AP}+\overrightarrow{BQ}+\overrightarrow{CR}=(\vec p-\vec a)+(\vec q-\vec b)+(\vec r-\vec c)=3(\vec g_2-\vec g_1)=3\,\overrightarrow{G_1G_2}. \]

Step 3: Both directions

A vector $3\,\overrightarrow{G_1G_2}$ is zero exactly when $G_1=G_2$. Therefore

\[G_1=G_2\iff\overrightarrow{AP}+\overrightarrow{BQ}+\overrightarrow{CR}=\vec 0\]

The identity from step 2 holds for any choice of origin, because the displacements $\overrightarrow{AP},\overrightarrow{BQ},\overrightarrow{CR}$ do not depend on it. It also says more: in general the sum equals $3\,\overrightarrow{G_1G_2}$.

$A$ $B$ $C$ $P$ $Q$ $R$ $G$

$\overrightarrow{AP},\overrightarrow{BQ},\overrightarrow{CR}$

Two triangles with the common centroid $G$.

$\overrightarrow{AP}$ $\overrightarrow{BQ}$ $\overrightarrow{CR}$

The same three vectors placed head to tail close up: $\overrightarrow{AP}+\overrightarrow{BQ}+\overrightarrow{CR}=\vec 0$.

Result

\[\boxed{\,G_1=G_2\iff\overrightarrow{AP}+\overrightarrow{BQ}+\overrightarrow{CR}=\vec 0\,}\]