More fun in 2015, Part 20

We will use complex numbers. WLOG, let the vertex which you are drawing chords from be $1$ . Let $\omega = e^{\frac{2\pi}{2015}i}$ , that is, the primitive 2015th root of unity. We will define a polynomial $P(x) = x^{2014} + x^{2013} + ... + x + 1 = (x - \omega)(x - \omega^2)...(x - \omega^{2014})$ Thus, the product of the chords is just $|1 - \omega| \cdot |1 - \omega^2| \cdot ... \cdot |1 - \omega^{2014}| = |P(1)| = \boxed{2015}$

We can generalize this to an $n$ -gon and we will see that the product is just $n$ .

May be 2015 gon is too much to think of ,so let's look at the triangle first. The triangle that inscribe the circle that has radius = 1 has side's length = √3 so the product of chords that draw from one vertice = √3×√3 = 3 The square that inscribe the circle radius=1 has side's lenht =√2 , diagonal leght =2 so the product of chords that draw from one vertice = √2×√2×2 = 4 And for the Pentagon product = 1×√5×√5×1 = 5 So 2015 gon = 2015