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[几何] 小球依次撞击正n边形的每条边,求发射角度的范围

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abababa posted 2015-7-11 15:56 |Read mode
正$n$边形$A_1A_2A_3\cdots A_n$的边$A_1A_2$上的中点$P$处有一小球,以角度$\theta$出发撞击边$A_2A_3$,反射后撞击$A_3A_4$,依次下去,最后撞击边$A_1A_2$,求$\theta$的范围

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kuing posted 2015-7-11 16:45
不断反射就行了吧,以正五边形为例,如图
QQ截图20150711164644.gif
然后就计算吧……应该不难,就看怎么计算会简单点……

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kuing posted 2015-7-11 17:51
不妨设正 $n$ 边形的边长为 $2$,将其放在坐标系中,使 $A_1(-1,0)$, $A_2(1,0)$,其余顶点均在 $x$ 轴的上方,则经 $n-1$ 次那样的反射后,最后那条边的中点坐标为
\[\left( n+n\cos\frac{2\pi}n, n\sin\frac{2\pi}n \right).\]

(1)当 $n$ 为奇数,则最后那条边的两个端点坐标分别为
\[\left( n+(n-1)\cos\frac{2\pi}n, (n-1)\sin\frac{2\pi}n \right), \left( n+(n+1)\cos\frac{2\pi}n, (n+1)\sin\frac{2\pi}n \right),\]
所以此时发射的角度范围为
\[\left[ \arctan\frac{(n-1)\sin\frac{2\pi}n}{n+(n-1)\cos\frac{2\pi}n}, \arctan\frac{(n+1)\sin\frac{2\pi}n}{n+(n+1)\cos\frac{2\pi}n} \right];\]

(2)当 $n$ 为偶数,则最后那条边的两个端点坐标分别为
\[\left( n-1+n\cos\frac{2\pi}n, n\sin\frac{2\pi}n \right), \left( n+1+n\cos\frac{2\pi}n, n\sin\frac{2\pi}n \right),\]
所以此时发射的角度范围为
\[\left[ \arctan\frac{n\sin\frac{2\pi}n}{n+1+n\cos\frac{2\pi}n}, \arctan\frac{n\sin\frac{2\pi}n}{n-1+n\cos\frac{2\pi}n} \right].\]

素不素酱紫……

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青青子衿 posted 2015-7-11 19:07
回复 1# abababa
1.jpg 2.jpg
《斯坦因豪斯问题:从1道25省市自治区中学数学竞赛试题谈起》

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乌贼 posted 2015-7-12 01:18
与这贴相似
  bbs.pep.com.cn/forum.php?mod=viewthread&t … ine&orderby=dateline
  就是二楼的方法

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kuing posted 2015-7-12 02:07
回复 4# 青青子衿

不会吧,那公式对 n=3 都似乎不符合啊

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original poster abababa posted 2015-7-12 11:03
回复 3# kuing

方法理解了,计算最后撞击的边的端点坐标还是没看懂怎么算的

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kuing posted 2015-7-12 16:05
回复 7# abababa

$A_1A_2$ 的中点为原点,$A_2A_3$ 的中点 $(1+\cos\frac{2\pi}n,\sin\frac{2\pi}n)$,因为反射出来的那些中点都共线,所以最后那边的中点就是 $(n+n\cos\frac{2\pi}n,n\sin\frac{2\pi}n)$。
然后当奇数边时最后那边与 $A_2A_3$ 平行,所以两端就是中点分别加减向量 $(\cos\frac{2\pi}n,\sin\frac{2\pi}n)$;当偶数边时最后是水平的,所以横坐标分别加减$1$。

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original poster abababa posted 2015-7-12 19:56
回复 8# kuing
谢谢,这下理解了。

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其妙 posted 2015-7-12 21:34
回复 3# kuing
强厉害.gif
妙不可言,不明其妙,不着一字,各释其妙!

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caijinzhi posted 2015-7-13 17:39
厉害!

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其妙 posted 2015-7-25 10:31
回复 11# caijinzhi
图也画的蛮棒的!

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