# Quod Erat Demonstrandum

## 2019/05/10

### 線長乘積

Filed under: Pure Mathematics — johnmayhk @ 11:52 下午
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$(8\sin^236^o\sin54^o)^2$

$|1-\omega||1-\omega^2|\dots |1-\omega^{n-1}|$

$= |(1-\omega)(1-\omega^2)\dots (1-\omega^{n-1})|$

$z=1,\omega, \omega^2, \dots , \omega^{n-1}$ 皆為下式的（不相同的）根

$z^n-1=0$

$z^n-1$ 可因式分解為

$z^n-1\equiv (z-1)(z-\omega)(z-\omega^2)\dots (z-\omega^{n-1})$

$z^n-1\equiv (z-1)(z^{n-1}+z^{n-2}+\dots +z+1)$

$(z-\omega)(z-\omega^2)\dots (z-\omega^{n-1})\equiv (z^{n-1}+z^{n-2}+\dots +z+1)$

$(1-\omega)(1-\omega^2)\dots (1-\omega^{n-1})=1+1+\dots +1=n$

$|1-\omega||1-\omega^2|\dots |1-\omega^{n-1}|=|n|=n$

Let $P_0,P_1,P_2,\dots ,P_{n-1}$ be a regular n-gon inscribed in a circle with radius r and centre O.

If P is a point on the plane of the circle such that $\angle POP_0=\theta$, show that

$PP_0\cdot PP_1 \cdot PP_2 \dots PP_{n-1}=\sqrt{r^{2n}-2r^n(OP)^n\cos(n\theta)+OP^{2n}}$

solution