Regular polygon

Sorry, your browser is not Java compliant Given a regular polygon with an odd number of sides n (here 5 sides) A1 ... An, and any point P on the circumcircle, on arc A1An :

 PA1 - PA2 + PA3 ... + PAn = 0 

Let 2t = angle POA1
Then in triangle PA1A2, angle PA2A1 = t, angle A2PA1 = π/n and naming a the side of the n-gon : PA1 = a. sin(t)/sin(π/n)
Generally, considering triangle PAkAk+1, angle PAk+1Ak = t + kπ/n, angle Ak+1PAk = π/n and PAk = a. sin(t + kπ/n)/sin(π/n)
The wanted relation becomes then

 ∑0n-1 (-1)k sin(t + kπ/n) = 0 

for all 0 ≤ t ≤ π/n, n odd

From cos(t) + i.sin(t) = ei.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 ei.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)

 

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