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Given an arc PQ with curvature $\frac{1}{9}$ Three identical circles with radii 3 and centered at B,G,A respectively. The circumference of the circles pass through each other's centers. Find the area of the shaded region. Would solving angle DHF be possible in this figure? So I could just get the area of the sector DHF and remove excess area. Any other solutions are very much welcome!

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I am not really good with geometry and know little theorems. I would really appreciate if people could help me out here. Thank you for all your help!

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See this: Here is the link meta.math.stackexchange.com/questions/1803/… –  amWhy Nov 12 '12 at 22:33
    
Is the arc part of a circle? Sorry if this is a dumb question. –  Braindead Nov 13 '12 at 0:06
    
@Braindead, Hmm. The problem doesn't say so. So I think it doesn't imply that. :( This is just a bonus question in my class. –  John Chang Nov 13 '12 at 0:10
    
The statement that the curvature is $\frac 19$ implies it is part of a circle of radius $9$ –  Ross Millikan Nov 13 '12 at 0:33
    
@RossMillikan I thought that too. But does it imply that given an arc with curvature, that the arc is part of a circle? If it is, how do we go from there? –  John Chang Nov 13 '12 at 0:38

1 Answer 1

up vote 1 down vote accepted

If you know about polar coordinates (and especially equations of circles that are tangent to the origin) and you know about integration, then you could find these areas by placing the origin at $G$. You would make use of the radii ($3$ and $9$) to find the angles at which the $A$- and $B$-centered circles are oriented relative to your primary polar axis.

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Hi sir, could you show me how to set up the integral? If it is not too much to ask. Thank you! –  John Chang Nov 13 '12 at 1:28
    
Is this actually homework for an integral calculus class? –  alex.jordan Nov 13 '12 at 1:29
    
Yes sir. It is a bonus question. I already got the black part. (9/2 arccos(17/18) - 3pi - 9sqrt(3)/2) but i have problems setting up the red and the blue part –  John Chang Nov 13 '12 at 1:54
    
I think I managed solving the red part with just simple geometry. For the blue part, would it be possible to use polar integrationas well? How do i get angle DGH? –  John Chang Nov 13 '12 at 2:11

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