PA_{1} - PA_{2} + PA_{3} ... + PA_{n} = 0

Let 2t = angle POA_{1}

Then in triangle PA_{1}A_{2}, angle PA_{2}A_{1} = t,
angle A_{2}PA_{1} = π/n and naming a the side of the n-gon :
PA_{1} = a. sin(t)/sin(π/n)

Generally, considering triangle PA_{k}A_{k+1},
angle PA_{k+1}A_{k} = t + kπ/n, angle A_{k+1}PA_{k} = π/n and
PA_{k} = a. sin(t + kπ/n)/sin(π/n)

The wanted relation becomes then

∑_{0}^{n-1} (-1)^{k} sin(t + kπ/n) = 0

for all 0 ≤ t ≤ π/n, n odd

From cos(t) + i.sin(t) = e^{i.t}, the sum of trigonometric values of angles in
arithmetic progression with ratio α becomes a sum of exp functions in geometric progression,
with ratio e^{i.α} and results into the general relations :

sin(t) + sin(t+α) + sin(t+2α) + ... sin(t+nα) = sin((n+1)α/2).sin(t + nα/2) / sin(α/2)
cos(t) + cos(t+α) + cos(t+2α) + ... cos(t+nα) = sin((n+1)α/2).cos(t + nα/2) / sin(α/2) |

Here, with α = π + π/n, we get

∑ sin(t + kπ + kπ/n) = ∑ (-1)^{k}.sin(t + kπ/n)
and stopping at n-1 in the general relation, we get :

sin(n.α/2).sin(t + (n-1)α/2) / sin(α/2)
= sin((n+1)π/2).sin(t + ...) / sin(...) = 0 if n odd (n+1 even)