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For what natural numbers n is it true that whenever the midpoints of the sides of a convex n-sided polygon $K_n$ form a regular n-gon, then $K_n$ itself is also a regular n-gon?

I know it is true for $n=3$, and that it isn't true for $n=4$, but I'm stuck with $n>4$.

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  • $\begingroup$ The proof you know that it isn't true for $n=4$ should be able to be generalized to all even $n$. $\endgroup$ – Greg Martin Apr 10 '16 at 8:17
  • $\begingroup$ Yes you're right, I haven't noticed that before... $\endgroup$ – user327929 Apr 10 '16 at 8:26
  • $\begingroup$ As for odd $n$: if you think of the vertices as complex numbers, then the operation that takes the original vertices to the midpoints is the result of applying a certain matrix (with entries $\frac12$ on two adjacent diagonals and $0$s elsewhere) to the vector of original vertices. When $n$ is odd, this matrix is invertible, which means that the vector of midpoints uniquely determines the vector of original vertices. $\endgroup$ – Greg Martin Apr 10 '16 at 8:35
  • $\begingroup$ Greg Martin I disagree. The matrix should have additional entries of 1/2 at (1,n) and (n,1), no? How does this affect invertibility? $\endgroup$ – Oscar Lanzi Apr 10 '16 at 10:37
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Let us work in the plane identified with the set of complex numbers $\mathbb{C}$.

Let us prove it is false for an even number of points.

Let us call $v_k \in \mathbb{C} (k=1\cdots n)$ the vertices of the polygon and $m_k \in \mathbb{C}$ the midpoint of $[v_k,v_{k+1}] $ with $v_{n+1}=v_1$. The $m_k$ are known, on a regular polygon, and the $v_k$ are unknown.

Let us call $S_k$ the (central) symmetry wrt $m_k$. Its expression is $Z=S_k(z)=2m_k-z$

(explanation: $Z=2m_k-z \Leftrightarrow \dfrac{z+Z}{2}=m_k$)

If, for example, $S_1$ is followed by $S_2$, one has:

$S_2(S_1(z))=2m_2-(2m_1-z)=2(m_2-m_1)+z$

More generally, by composing $S_1$ followed by $S_2 \cdots S_n$, (a transformation that we will call $S$) when $n$ is even, one has, by an immediate recurrence:

$$S(z)=2(\sum(-1)^k m_k)+z=z \ \ (1) \ \ \text{which boils down to} \ \ S(z)=z$$

(last equality because the sum is zero for a regular polygon).

Thus, whatever the initial $z=v_1$, one will be able to construct a closed polygon. It is clear that if one takes $z_1$ very close from $m_1$, $v_2=S_1(v_1)$ will be as well close to $v_1$ but far away from $v_3$, thus generating a non-regular polygon (see image below).

enter image description here

Remark: there is a funny 3D interpretation of the above figure as a perspective view of a particular section of a cube by a plane passing by certain midpoints.

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  • $\begingroup$ I have tried to be a little more explicit. In particular, I have given the expression of a symmetry $S(z)=2m_k-z$ with respect to point $m_k$. $\endgroup$ – Jean Marie Apr 10 '16 at 21:46

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