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ABC circle sector turns on ground (x axis) as shown in the figure. A is the center of the circle. $\angle{OAB}=\angle{OAC}=\alpha $

$|AB|=r$

$\cfrac{|AP|}{|PC|}=k$

The corners meet on point $H$ in the limit points. What is the curve equation $f(x)$ that passing from $P',P,P''$ Thanks for answers

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How is it moving? You may look for the cycloid curve. en.wikipedia.org/wiki/Cycloid –  Berci Oct 3 '12 at 21:09
    
@Berci Thanks for the link.It is similiar to Cycloid but turning point in circle in my question, not on circle. –  Mathlover Oct 3 '12 at 21:13
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1 Answer

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This is a curtate cycloid. Refer to http://mathworld.wolfram.com/CurtateCycloid.html

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