47205 等角差線定義修正與n倍等角差線(nE.D.L.)猜想

### 貳、等角差線--定義修正

$$\left\{\begin{array}{l} x=\cfrac{l\tan\theta}{\tan(\theta+\alpha)+\tan \theta}, \\ y=\cfrac{l\tan\theta \tan(\theta+\alpha)}{\tan (\theta+\alpha)+\tan \theta}. \end{array}\right.$$

### 參、$n$倍等角差線與$n$倍等角差線組(The Group of $n$E.D.L.)

$$\left\{\begin{array}{l} x=\cfrac{l\tan\theta}{\tan(n\theta+\alpha)+\tan \theta}, \\ y=\cfrac{l\tan\theta \tan(n\theta+\alpha)}{\tan (n\theta+\alpha)+\tan \theta}. \end{array}\right.$$

### 肆、 $n$ 倍等角差線的漸近線

\begin{align*} &\hskip -25pt\begin{bmatrix} \cos\beta & -\sin\beta\\ \sin\beta & \cos\beta \end{bmatrix} \begin{bmatrix} x^\prime\\ y^\prime \end{bmatrix}= \begin{bmatrix} \cos\beta & -\sin\beta\\ \sin\beta & \cos\beta \end{bmatrix} \begin{bmatrix} \cfrac{l\tan\theta}{\tan(n\theta+\alpha)+\tan\theta}\\ \cfrac{l\tan\theta \tan(n\theta+\alpha)}{\tan(n\theta+\alpha)+\tan\theta} \end{bmatrix}\\ &=\begin{bmatrix} \left(\cfrac{l\displaystyle\tan\theta}{\displaystyle\tan(n\theta+\alpha)+\tan\theta}\right)(\cos\beta-\sin\beta\tan(n\theta+\alpha))\\ \left(\cfrac{l\displaystyle\tan\theta}{\displaystyle\tan(n\theta+\alpha)+\tan\theta}\right)(\sin\beta+\cos\beta\tan(n\theta+\alpha)) \end{bmatrix}= \begin{bmatrix} x^{\prime\prime}\\ y^{\prime\prime} \end{bmatrix}. \end{align*}

\begin{align*} \lim_{\theta \to \beta}x^{\prime\prime} &= \lim_{\theta \to 0^-} \left[\frac{l\tan(\theta-\beta)(\cos\beta-\sin\beta\tan(n(\theta-\beta)+\alpha))}{\tan(\theta-\beta)+\tan(n(\theta-\beta)+\alpha)}\right]=-\infty,\\ \lim_{\theta \to \beta}y^{\prime\prime} &= \lim_{\theta \to \beta}\left[\frac{l\tan\theta(\sin\beta+\cos\beta\tan(n\theta+\alpha))}{\tan\theta+\tan(n\theta+\alpha)}\right]\\ &=\lim_{\theta \to \beta}\left[\frac{l\sec^2\theta(\sin\beta+\cos\beta\tan(n\theta+\alpha))+nl\cos\beta\tan\theta\sec^2(n\theta +\alpha)}{sec^2\theta+\sec^2(n\theta+\alpha)}\right]\\ &=\frac{nl\sin\beta}{n+1}. \end{align*}

\begin{align*} &\hskip -25pt\begin{bmatrix} \cos\beta & -\sin\beta\\ \sin\beta & \cos\beta \end{bmatrix} \begin{bmatrix} X\\ Y \end{bmatrix}= \begin{bmatrix} X^\prime\\ Y^\prime \end{bmatrix}= \begin{bmatrix} t\\ \cfrac{nl\sin\beta}{n+1} \end{bmatrix} \hbox{($t$ 為參數)},\\ X &= \left | \begin{array}{cccc} t & -\sin\beta\\ \cfrac{nl\sin\beta}{n+1} & \cos\beta \end{array}\right |=\cos\beta\cdot t+\cfrac{nl\sin^2\beta}{n+1},\\ Y &= \left | \begin{array}{cccc} \cos\beta & t\\ \sin\beta & \cfrac{nl\sin\beta}{n+1} \end{array}\right |=-\sin\beta\cdot t+\cfrac{nl\sin\beta\cos\beta}{n+1}, \end{align*} 消除參數$t$, 可得$X$, $Y$關係為： \begin{align*} Y&=-\tan\beta+\frac{nl}{n+1}(\frac{\sin^3\beta}{\cos\beta}+\sin\beta\cos\beta)\\ &= -\tan\beta X+\frac{nl}{n+1}(\frac{\sin^3\beta}{\cos\beta}+\frac{\sin\beta\cos^2\beta}{\cos\beta})=-\tan\beta X+\frac{nl}{n+1}\tan\beta, \end{align*}

### 伍、特殊$n$倍等角差線(Singular $n$E.D.L.)

\begin{align*} \lim_{\theta\to 0}x &=\lim_{\theta\to 0}\cfrac{l\tan\theta}{\tan(n\theta)+\tan \theta} = \lim_{\theta\to 0} \cfrac{l\sec^2\theta}{n\sec^2(n\theta) +\sec^2 \theta}=\cfrac{l}{n+1},\\ \lim_{\theta\to 0}y &=\lim_{\theta\to 0}\cfrac{l\tan\theta \tan(n\theta)}{\tan (n\theta)+\tan \theta}=\cfrac{l}{n+1}\cdot 0=0.\tag*{$\Box$} \end{align*}

$r_m$ 之值都不同直到 $m\ge p$, 所以我們可知 $\{\Gamma^l_{\alpha,n}\}=\{\Gamma^l_{\gamma_m,n}\}_{m=0}^{p-1}$, 故必定存在 $0\le m\lt p$ 使得 $\gamma_m=0$, 於此, 滿足定理敘述的 $n$ 倍等角差線組有特殊 $n$ 倍等角差線。

$\Box$

### 陸、$n$倍等角差線組的對偶(The Dual of $n$E.D.L.)與輻射點(Radiation Point)

\begin{align*} &\hskip -25pt \cfrac{l}{2}\left(\frac{\tan p\theta}{\tan p\theta+\tan (q\theta+\alpha)}+\frac{\tan q\phi}{\tan q\phi+\tan (p\phi+\alpha^\prime)}\right)\\ =&\cfrac{l}{2}\left(\frac{\tan\Big( p\phi+\cfrac{p\alpha}{q}\Big)}{\tan \Big( p\phi+\cfrac{p\alpha}{q}\Big)+\tan q\phi} +\frac{\tan q\phi}{\tan q\phi+\tan (p\phi+\alpha^\prime)}\right)\\ =&\cfrac{l}{2}\left(\frac{\tan(p\phi+\alpha^\prime)}{\tan (p\phi+\alpha^\prime)+\tan q\phi}+\frac{\tan q\phi}{\tan q\phi+\tan (p\phi+\alpha^\prime)}\right) =\cfrac{l}{2}, \end{align*} \begin{align*} &\hskip -25pt \frac{l\tan p\theta \tan(q\theta+\alpha)}{\tan p\theta+\tan(q\theta+\alpha)}+\frac{l\tan q\phi \tan(p\phi+\alpha^\prime)} {\tan q\phi+\tan(p\phi+\alpha^\prime)}\\ =&\frac{-l\tan\Big( p\phi+\cfrac{p\alpha}{q}\Big)\tan q\phi}{\tan \Big( p\phi+\cfrac{p\alpha}{q}\Big)+\tan q\phi}+\frac{l\tan q\phi \tan(p\phi+\alpha^\prime)}{\tan q\phi+\tan(p\phi+\alpha^\prime)}\\ =&\frac{-l\tan(p\phi+\alpha^\prime)\tan q\phi}{\tan (p\phi+\alpha^\prime)+\tan q\phi}+\frac{l\tan q\phi \tan(p\phi+\alpha^\prime)} {\tan q\phi+\tan(p\phi+\alpha^\prime)}=0. \end{align*}

$\Box$

\begin{align*} &\hskip -25pt \cfrac{l}{2}\left(\frac{\tan p\theta}{\tan p\theta+\tan (q\theta+\alpha)}+\frac{\tan q\phi}{\tan q\phi+\tan (p\phi+\alpha^\prime)}\right)\\ =&\cfrac{l}{2}\left(\frac{\tan\Big( p\phi-\cfrac{p\alpha}{q}\Big)}{\tan \Big( p\phi-\cfrac{p\alpha}{q}\Big)+\tan q\phi}+\frac{\tan q\phi}{\tan q\phi+\tan (p\phi+\alpha^\prime)}\right)\\ =&\cfrac{l}{2}\left(\frac{\tan(p\phi+\alpha^\prime)}{\tan (p\phi+\alpha^\prime)+\tan q\phi}+\frac{\tan q\phi}{\tan q\phi +\tan (p\phi+\alpha^\prime)}\right)=\cfrac{l}{2}, \end{align*} \begin{align*} &\hskip -25pt \frac{l\tan p\theta \tan(q\theta+\alpha)}{\tan p\theta+\tan(q\theta+\alpha)}-\frac{l\tan q\phi \tan(p\phi+\alpha^\prime)} {\tan q\phi+\tan(p\phi+\alpha^\prime)}\\ =&\frac{l\tan\Big( p\phi-\cfrac{p\alpha}{q}\Big)\tan q\phi}{\tan \Big( p\phi-\cfrac{p\alpha}{q}\Big)+\tan q\phi} -\frac{l\tan q\phi \tan(p\phi+\alpha^\prime)}{\tan q\phi+\tan(p\phi+\alpha^\prime)}\\ =&\frac{l\tan(p\phi+\alpha^\prime)\tan q\phi}{\tan (p\phi+\alpha^\prime)+\tan q\phi} -\frac{l\tan q\phi \tan(p\phi+\alpha^\prime)}{\tan q\phi+\tan(p\phi+\alpha^\prime)}=0. \end{align*}

$\Box$

$\{\Gamma^l_{\alpha,n}\}$ 與其對偶之類對偶 $\{\Gamma^l_{\alpha_1,n}\}$ 中所有 $n$ 倍等角差線依定理1.可知在 $0\lt\theta\lt\beta^\prime$ 有漸近線 $$y=-\tan \beta^\prime x+\cfrac{\displaystyle nl}{\displaystyle n+1}\tan \beta^\prime.$$ 可以得到所有利用定理1.所得之漸近線皆交於點 $\Big(\cfrac{nl}{n+1},0\Big)$。

$\Box$

### 柒、$n$倍等角差線猜想

「任意在 $\mathbb{R}^2$ 上的一點, 先證明是否存在一 $n$ 倍等角差線 $\Gamma^l_{\omega,n}$ 通過此點, 而後取任意小的 $\epsilon\gt0$, 證明在 $\{\Gamma^l_{\alpha,n}\}$ 中是否存在一 $n$ 倍等角差線 $\Gamma^l_{\omega\pm\delta,n}$ ($\delta\gt0$) 與此點距離小於 $\epsilon$。」