$ABCDE$ is a regular pentagon of center $O$. Let $$\vec{V}=\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}+\vec{OE}$$ Express $\vec{V}$ in terms of $\vec{OA}$, and then in terms of $\vec{OB}$. Conclude.
I tried to use Chasles relation and draw the vectors and add them up (4th vertex of the parm), but I was not able to reach a precise answer, and I feel like there is a much easier way to solve this.
I also tried giving each point a pair of coordinates but I couldn't reach the answer.
So this is my main question, finding $\vec{V}$ with simple math. (No matrices, maybe only vectors).
This question came as an activity before introducing the $n^{th}$ of a complex number, so I guess the conclusion is related to complex numbers, but finding $\vec{V}$ should be done algebraicly.
Any help is appreciated
Assuming the vectors are in $\mathbb{R}^2$, then we can use the rotation matrix $$M=\begin{bmatrix} \cos\frac{2\pi}{5} & -\sin\frac{2\pi}{5}\\\sin\frac{2\pi}{5}&\cos\frac{2\pi}{5}\end{bmatrix}$$ to express all the vectors in terms of $\overrightarrow{OA}$. The expression for $\overrightarrow{V}$ simplifies to $$\overrightarrow{V}=\left(M^0+M^1+M^2+M^3+M^4\right)\overrightarrow{OA}$$ $$\overrightarrow{V}=\begin{bmatrix} \sum_{k=0}^4 \cos\frac{2\pi k}{5}&\sum_{k=0}^4 -\sin\frac{2\pi k}{5}\\\sum_{k=0}^4 \sin\frac{2\pi k}{5}&\sum_{k=0}^4 \cos\frac{2\pi k}{5}\end{bmatrix}\overrightarrow{OA}$$ You can geometrically prove that this final matrix is the $2\times 2$ zero matrix by positioning a unit pentagon with two vertices on $(0,0)$ and $(1,0)$ and then find the horizontal/vertical displacement as you go around the pentagon (instead of this geometric approach, you can also use complex numbers, but that defeats the point). Hence, $$\overrightarrow{V}=\begin{bmatrix}0&0 \\ 0&0\end{bmatrix}\overrightarrow{OA}=\vec{0}$$