Geometry in an affine space

By John Mike on Monday, June 17, 2002 - 06:10 pm:

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

By David Loeffler on Monday, June 17, 2002 - 07:06 pm:

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

By David Loeffler on Monday, June 17, 2002 - 07:38 pm:

Let's fix some arbitrary origin in our affine space and make it into a vector space.

For odd $n$ , where the number of points is even, consider $M_{0} + M_{2} + \dots + M_{n - 1}$ ; this is $1 / 2 (a_{0} + a_{1}) + 1 / 2 (a_{1} + a_{2}) \dots + 1 / 2 (a_{n - 1} + a_{n})$ .

On the other hand

$M_{1} + M_{3} + \dots + M_{n} = 1 / 2 (a_{1} + a_{2}) + 1 / 2 (a_{3} + a_{4}) \dots + 1 / 2 (a_{n} + a_{0})$ .

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 $λ$ for $a_{0}$ . Then we canproceed to write down equations for each of the $a_{i}$ , giving them in terms of the $λ$ and the $M_{i}$ . Note that $a_{0}$ is defined as $λ + stuff$ , $a_{1}$ is $- λ + stuff$ , and so on; $a_{i}$ is $(- 1)^{i} λ$ plus some additional mess.

We eventually get to $a_{n + 1} = - λ + R$ where $R$ is some linear combination of the $M_{i}$ . We also know that $a_{n + 1} = a_{0} = 1$ . This is an equation which can easily be solved for $λ = R / 2$ . This is the solution (which is evidently unique).

(If $n + 1$ is even, we have $a_{n + 1} = λ + R$ , the $λ$ s 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