Sum of arbitrary vertex to midpoint vectors

[Joffan]

I was looking at a homework question posted here requiring proof that the vectors from the vertices of a triangle to the midpoint of the opposite edge sum to zero, and it struck me that there is a more general property:

Consider a set of points, $\{A_0, A_1, \ldots A_n\}$. The midpoint of $A_k$ and $A_{k+1}$ is denoted by $a_k$, with $a_n$ as the midpoint of $A_n$ and $A_0$.

Now form a set of $n$ vectors defined by $\stackrel{\rightarrow}{A_ra_s}$ such that each vertex and each midpoint is used once by a vector in the set.

Now that set of vectors will sum to zero.

--------------------------

Sketch of messy proof
First step - the midpoints are a distraction. $\stackrel{\rightarrow}{A_ra_s} = \stackrel{\rightarrow}{A_rA_s} + \stackrel{\rightarrow}{A_sa_s} = \stackrel{\rightarrow}{A_rA_s }+ \frac{1}2\stackrel{\rightarrow}{A_sA_{s+1}}$, so the vector sum contains a scaled loop from the midpoint steps which sums to zero.

Next step - The remaining vectors form a set of closed loops which also sum to zero.

Improvements on this welcome...

 

[disregardthat]

The set of points A_k and a_k may be considered vectors from origon. So the n vectors you describe are on the form a_s-A_r, where s and r ranges from 0 to n with r depending on s via some bijection (<-- I assume this is what you mean).

a_k may be written as $\frac{1}{2}(A_k+A_{k+1})$, where indices are considered modulo n+1.

By writing out entire the sum, we get exactly 0.

 

[Joffan]

Nice and neat. Thanks.