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 Around the ring (Posted on 2010-12-20)
Take the line segment whose endpoints are the points (1,0) and (-1,0) and a point of rotation (x,y).

If the segment is rotated all the way around the point it will trace out an annulus.

Find simplified formula for the area of this annulus in terms of x and y.

Redo this problem for a rectangle with corners at (1,1), (-1,1), (-1,-1), and (1,-1).

 No Solution Yet Submitted by Jer Rating: 3.0000 (1 votes)

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 part 1 spoiler | Comment 1 of 9

Part 1:

One point will trace a circle of radius sqrt((1-x)^2+y^2), while the other will trace a circle of radius sqrt((1+x)^2+y^2). The areas will be these squared times pi, and the absolute values of the difference is the area of the annulus.

|pi((1-x)^2 + y^2 - (1+x)^2 - y^2)|

= |4*pi*x|

Part 2 requires a way of getting the largest and smallest distances from those to the four corners and subtracting the corresponding circle areas. Sounds like it needs some creative use of absolute values.

 Posted by Charlie on 2010-12-20 16:47:39

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