Can you help me for this geometry problem :

Let M₀ , M₁ , ..., M_n , be n+1 points of an affine space, can we find A₀ , A₁ , ..., A_n , so that for i from 0 to n, M_i is the middle of (A_i ,A_i+1 ) (with A_n+1 =A₀ ) ?

Thanks

Is this actually true? Let's take the case n=1. Then we must have M₀ = 1/2(a₀ + a₁ ). However
M₁ = 1/2(a₁ + a₀ , so M₁ = M₀ . Similar contradictions occur when n = 3. It is, however, true if n is even.

David

For odd n, where the number of points is even, consider M₀+M₂+¼+M_n-1; this is 1/2(a₀+a₁)+1/2(a₁+a₂)¼+1/2(a_n-1+a_n).

On the other hand

M₁+M₃+¼+M_n=1/2(a₁+a₂)+1/2(a₃+a₄)¼+1/2(a_n+a₀).

These expressions are the same. Hence if such points exist for an (n+1)-gon, it mustsatisfy this property,which most (n+1)-gons do not. For example, in the case n=3 (a quadrilateral) wewould have to have its diagonals bisecting each other (constraining it to being a planar parallelogram).

In the case where n is even, let's pick some arbitrary value l for a₀. Then we canproceed to write down equations for each of the a_i, giving them in terms of the l and the M_i. Note that a₀ is defined as l+stuff, a₁ is -l+stuff, and so on; a_i is (-1)ⁱl plus some additional mess.

We eventually get to a_n+1=-l+R where R is some linear combination of the M_i. We also know that a_n+1=a₀=1. This is an equation which can easily be solved for l = R/2. This is the solution (which is evidently unique).

(If n+1 is even, we have a_n+1=l+R, the ls drop out and we have a condition on the M_i from the equation R=0, which is why the problem is not always solvable in this case.)

David