|
|
isee
发表于 2014-5-19 21:25
本帖最后由 isee 于 2014-5-20 15:20 编辑 回复 2# kuing
个人以为
令$\angle AOP=\alpha<\dfrac \pi3$，直接由向量定义计算：
\begin{align*}
\vv {OP}\cdot \vv {PC}&=1\times PC \cos \left(\dfrac {\pi}{2}-\dfrac {\alpha}{2}\right)\\
&=PC \sin \dfrac {\alpha}{2}\\
&=\require{cancel}\xcancel{CQ}PQ
\end{align*}
即点C到直线OP的距离，于是，结果只能有最小值零了。如果由正弦定理，可全部化成角，不过，会出现3倍角关系。
当然，这里（默认）猜测让这个距离是简单的线性关系了，P与A无限靠近，当成了PC平行于OB，不过，似乎是相切更合理，才是。
于是，有上界：\[\require{cancel}\xcancel{0<\vv {OP}\cdot \vv {PC}<\sqrt 3}\]
若P与A重合时，个人没法理解，所以，这里P与A，C均不重合，即未考虑P在两个端点的情形.
硬算吧，全部化三角，继续，在三角形$POC$中由正弦定理，有
\begin{align*}
PC&=\dfrac {\sin\left(\dfrac \pi3-\alpha\right)}{\sin \left(\dfrac \pi6+\dfrac \alpha2\right)}\\[2em]
\therefore \vv {OP}\cdot{PC}&=\dfrac {\sin\left(\dfrac \pi3-\alpha\right)}{\sin \left(\dfrac \pi6+\dfrac \alpha2\right)}\cdot \sin\left(\dfrac \alpha2\right)\\
&=\dfrac {\sin\left(\dfrac \alpha2\right) \sin\left(\dfrac \pi3-\alpha\right)}{\sin \left(\dfrac \pi6+\dfrac \alpha2\right)}
\end{align*}
记$\beta=\dfrac \pi6+\dfrac \alpha2\in\left(\dfrac \pi6,\dfrac \pi3\right)$，则
\begin{align*}
\vv {OP}\cdot{PC}&=\dfrac {\sin\left(\beta -\dfrac \pi6\right) \sin\left(\dfrac {2\pi}3-2\beta\right)}{\sin \beta}\\
&=-\dfrac {\cos\left(\dfrac {\pi}2-\beta\right) -\cos\left(3\beta-\dfrac {5\pi}6\right)}{2\sin \beta}\\
&=-\dfrac {\sin\beta +\cos\left(3\beta+\dfrac {\pi}6\right)}{2\sin \beta}\\
&=-\dfrac12 - \dfrac {\cos\left(3\beta+\dfrac {\pi}6\right)}{2\sin \beta}\\
\end{align*}
考察函数
\[g(x)=- \dfrac {\cos\left(3x+\dfrac {\pi}6\right)}{\sin x},x\in\left(\dfrac \pi6,\dfrac \pi3\right) \]
对其求导，有
\begin{align*}
g'(x)&= \dfrac {3\sin\left(3x+\dfrac {\pi}6\right)\sin x+\cos \left(3x+\dfrac {\pi}6\right)\cos x}{\sin^2 x}\\
&=\dfrac {2\sin\left(3x+\dfrac {\pi}6\right)\sin x+\cos \left(2x+\dfrac {\pi}6\right)}{\sin^2 x}\\
&=\dfrac {-\cos\left(4x+\dfrac {\pi}6\right)+2\cos \left(2x+\dfrac {\pi}6\right)}{\sin^2 x}
\end{align*}
再单独考察分子，有
\begin{align*}
h(x)&= -\cos\left(4x+\dfrac {\pi}6\right)+2\cos \left(2x+\dfrac {\pi}6\right),x\in\left(\dfrac \pi6,\dfrac \pi3\right)\\
h'(x)&= 4\sin\left(4x+\dfrac {\pi}6\right)-4\sin \left(2x+\dfrac {\pi}6\right)\\[2ex]
&\left(4x+\dfrac {\pi}6\right)\in\left(\dfrac {5\pi}6,\dfrac {3\pi}2\right)\\
&\left(2x+\dfrac {\pi}6\right)\in\left(\dfrac \pi2,\dfrac {5\pi}6\right)\\
\therefore 4x+\dfrac {\pi}6&>2x+\dfrac {\pi}6\\
\Rightarrow \sin\left(4x+\dfrac {\pi}6\right)&<\sin \left(2x+\dfrac {\pi}6\right)\\[2em]
\end{align*}
这表明，$h(x)= -\cos\left(4x+\dfrac {\pi}6\right)+2\cos \left(2x+\dfrac {\pi}6\right)$在$x\in\left(\dfrac \pi6,\dfrac \pi3\right)$上单调递减，而
\begin{align*}
h\left(\dfrac \pi6\right)&= -\cos\left(\dfrac {4\pi}6+\dfrac {\pi}6\right)+2\cos \left(\dfrac {2\pi}6+\dfrac {\pi}6\right)=\frac {\sqrt 3}2>0\\
h\left(\dfrac \pi3\right)&= -\cos\left(\dfrac {4\pi}3+\dfrac {\pi}6\right)+2\cos \left(\dfrac {2\pi}3+\dfrac {\pi}6\right)=-\sqrt 3<0
\end{align*}
则$h(x)$在$x\in\left(\dfrac \pi6,\dfrac \pi3\right)$有惟一零点
\[h\left(x_0\right)= -\cos\left(4x_0+\dfrac {\pi}6\right)+2\cos \left(2x_0+\dfrac {\pi}6\right)=0=g'(x_0)\]
进一步知
\[g(x)_{\max}=- \dfrac {\cos\left(3x_0+\dfrac {\pi}6\right)}{\sin x_0} =??\]
硬算也没结果，不过，可以知道的是P在弧中点的时间并不是最大的 |
|
kuing
发表于 2014-5-19 21:42
，你莫非还把点$Q$的轨迹都搞出来？
回复 
