Algebra question from Australia national olympiad 2013 Find all positive integers $n$ for which there are real numbers $x_1, \; x_2, \cdots,\; x_n$ satisfying $$(i) \; \; -1<x_i<1 \; for \; i=1,2, \cdots n$$ $$(ii) \; \; x_1+x_2+ \cdots +x_n=0 \; and$$ $$(iii) \;\; \sqrt{1-x_1^2}+\sqrt{1-x_2^2}+ \cdots +\sqrt{1-x_n^2}=1$$
Source: AMO 2013
My thoughts: It is easy to see that this works for all even $n$ and for $n=1$. I believe that all other odd numbers are impossible. To prove this I tried applying Cauchy-Schwarz to condition (iii). So that $ \sqrt{1-x_1^2}+\sqrt{1-x_2^2}+ \cdots +\sqrt{1-x_n^2} \leq 1$ with equality iff $x_1=x_2= \cdots =x_n$ which is not possible if $n$ is odd and greater than $1$.
However this didn't seem to work. So any help would be greatly appreciated.
 A: There may be some sign condition to be clarified about the square roots, but basically it is asking (in increasing order of relevance to solving the problem)

  for $n$ angles whose sines sum to $0$ and cosines add up to $1$, 

or better,

 $n$ unit complex numbers that add to $i$.

If the sign condition is that the square roots are positive, then you want

 $n$ vectors from the upper half of a unit circle, centered at $0$, excluding the endpoints, that sum to the unit vector on the midpoint of the same half circle.

which is equivalent to 

 for which $m$ is there an infinite rectangle with finite side of length $1$, and a convex $m$-sided polygon with sides all of length $1$, that fits inside the rectangle and has the finite side of the rectangle as one of its sides.

The problem seems tricky for odd $n$ and the above reformulations do not solve that case by themselves.
A: You can rephrase it as an optimization problem (taking into consideration that we already know what happens when $n$ is even):

Suppose that $n \geq 3$ is odd, $\delta > 0$ (sufficiently  small wrt $n$) and that $x_i \in [-1+\delta,1-\delta]$ are such that $\sum_i x_i = 0$. Then $F(x_1,\dots,x_n) := \sum_i \sqrt{1-x_i^2} > 1$. 

Because $B:=[-1+\delta,1-\delta]^n$ is a compact set, $F$ takes a minimal value at some point $z = (z_1,\dots,z_n)$; and without loss of generality we can assume that $z_1 \leq z_2\dots \leq z_n$. Let $z_1 = \dots = z_k = -1+\delta \neq z_{k+1}$, and $z_n = z_{n-1} = \dots = z_{n-l+1} = 1 -\delta \neq z_{n-l}$, so that $-1+\delta$ appears $k$ times, $1-\delta$ appears $l$ times, and $z_{k+1},\dots,z_{n-l} \in (-1+\delta,1-\delta)$.
I claim that $z_{k+1} = \dots = z_{n-l}$ (allowing for the possibility that this list is just one element, or empty). Indeed, if not, then you can select $k < i < j \leq n-l$. Consider the point $z'$ with $z_i' = z_i - \varepsilon$, $z_j' = z_j + \varepsilon$ and $z_k' = z_k$ otherwise. By basic analysis:
$$ F(z') = F(z) + \varepsilon (\frac{\partial F}{\partial x_j}(z) - \frac{\partial F}{\partial x_i}(z)) + O( \varepsilon^2) \\
F(z) + \varepsilon (f(z_j) - f(z_i)) + O( \varepsilon^2)
 $$
where $f(x) = \frac{-x}{\sqrt{1-x^2}}$ is the derivative of $\sqrt{1-x^2}$. If $\varepsilon$ is small enough, this we will have $F(z') < F(z) $, contradicting the choice of $z$. (In fact, this follows more easily from just noticing that $$\sqrt{1-x^2}$ is concave.)
So, the "optimal" sequence is of the form 
$$-1+\delta, -1+\delta,\dots,-1+\delta,z,z,z,\dots,z,1-\delta,1-\delta,\dots,1-\delta.$$
Next, I claim that $k+l \geq n-1$, i.e. in the sequence above there is at most one $z$. Else, we can replace a pair $z,z$ by $z-\varepsilon, z + \varepsilon$, again decreasing the value of $F$ because of concavity of $\sqrt{1-x^2}$.
Once we know that there is at most one $z$, it is easy see that the condition $\sum_i z_i = 0$ forces $k = l = \frac{n-1}{2}$ and $z = 0$. However, at this point $F$ takes value strictly greater than $1$, which finishes the proof.
