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    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #1

    Feb 17, 2008, 06:49 AM
    Area, find value of k.
    I am posing this problem for kicks. For those out there who like a nice math challenge problem. Since this site rarely gets them, I thought I would pose one if anyone would like a go.

    "Looking at the graph, find the value of k so that areas of region A and region B are equal"
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    xphelper's Avatar
    xphelper Posts: 220, Reputation: 29
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    #2

    Feb 17, 2008, 07:59 AM
    Quote Originally Posted by galactus
    I am posing this problem for kicks. For those out there who like a nice math challenge problem. Since this site rarely gets them, I thought I would pose one if anyone would like a go.

    "Looking at the graph, find the value of k so that areas of region A and region B are equal"
    0.2
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #3

    Feb 17, 2008, 01:03 PM
    No, I am sorry to say that is incorrect. I will leave it up until tomorrow evening to see if anyone else wants a go.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Feb 18, 2008, 01:02 PM
    How about: k=0.724611?
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #5

    Feb 18, 2008, 01:57 PM
    Yes, indeed, ebaines. I knew you could get it. :)
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    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Feb 18, 2008, 02:19 PM
    Quote Originally Posted by galactus
    Yes, indeed, ebaines. I knew you could get it. :)
    Thanks! However, I must admit that I was unable to find a closed-form solution for this problem, but rather used an approximation method to solve a nasty equation:

    1 = b*sin(b) + cos(b)

    where b = arcsin(k)-pi.

    Is there a more elegant way?
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #7

    Feb 18, 2008, 02:59 PM
    Yes, that's exactly what I done. I used Newton's method.
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    galactus Posts: 2,271, Reputation: 282
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    #8

    Feb 23, 2008, 10:17 AM
    SOLUTION:


    Let (a,k) be the point of intersection of y=k with y=sin(x).

    Then, k=sin(a) and if the areas are equal:



    Solve for 'a' using Newton's method or some other estimation techniques and get: , so

    The horizontal line which intersects sin(x) creating two equal regions is y=0.724611

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