existence of a convex polygon

definition
For n > 1, let
Σ(n)(x(1),...,x(n)) = x(1)+...+x(n), and
B(n)(x(1),...,x(n)) = Π(Σ(n)(x(1),...,x(n))-2x(i)).

Note that each factor of B(n) are obtained by replacing one + in Σ(n) by -.
Hence, B(4) = B, and Heron's Polynomial is given by H = Σ(3)B(3).
It is clear that a necessary condition for the existence of a convex polygon
with sides of length x(1),...,x(n) is B(n)(x(1),...,x(n)) > 0. In fact, this
condition is also sufficient.

a theorem in euclidean and hyperbolic geometry
Suppose that n≥3, and that x(1),...,x(n) are positive numbers. Then
there is a convex polygon with sides of length x(1),...,x(n) in order
if and only if B(n)(x(1),...,x(n)) > 0.
The condition is equivalent to 2max{x(i)} < Σ(n)(x(1),...,x(n)).

proof
We have only to prove sufficiency, so assume B(n)(x(1),...,x(n)) > 0.
In the course of the proof, it is convenient to write Σ(r) =Σ(r)(x(1),..,x(r)),
and B(r)=B(r)(x(1),...,x(r)).
Observe that, if x(k) is the maximum length, then, for i ≠ k, x(k) ≥ x(i)
Since Σ(n) includes x(i) and x(k), Σ(n)-2x(i)>0, provided n ≥ 3.
It follows that the sign of B(n) is that of Σ(n)-2x(k). This shows the
equivalence mentioned in the second sentence.

The proof is inductive. We know the result for n=3 (and 4), so assume
that n≥5, and that the result holds for n-1.

The starting point and direction of labelling do not affect the result,
so we may assume that x(n) = min{x(i)}, and x(n-1)≤x(1). For the
latter, note that reversing the direction swaps x(1) and x(n-1).
Note that x(1) ≥ x(n-1),x(n), so the maximum is x(k) with 1≤k≤n-2.

We want to choose x so there exist convex polygons with sides
x(n-1),x(n),x and x(1),...,x(n-2),x.
The former requires
(a) x < x(n-1)+x(n) and
(b) x > x(n-1)-x(n).
The latter requires
(c) x < Σ(n-2) (in case x >x(k)), and
(d) 2x(k) < Σ(n-2)+x, i.e. x > 2x(k)-Σ(n-2).
Note that Σ(n-2) includes x(1)(≥x(n-1), and x(2)(≥x(n)), so
x(n-1)+x(n) ≤ Σ(n-2). It follows that (a) implies (c).
Now (a), (d) will have solutions if and only if x(n-1)+x(n)>2x(k)-Σ(n-2).
i.e. Σ(n) > 2x(k). As noted, this is equivalent to B(n) > 0.
If we take account of (b), we see that the possible x are those between
max{x(n-1-x(n),2x(k)-Σ(n-2)} and x(n-1)+x(n).
Since we are free to choose x, we may even insist that x is in the
range X to Y=x(n-1)+x(n) for any choice of X < Y.

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