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Thread: Geometry Puzzle

  1. #1

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    Geometry Puzzle

    Draw a right-angled triangle.

    3 ropes are attached to the midpoints of each side, and each rope is half the length of the side it is attached to. A goat is tied to the end of the rope on the hypotenuse, and two sheep are attached to the ropes on the shorter sides. The area of the triangle is 1 acre. Find the area that the goat can reach that the sheep cannot.


    ~
    I'm not sure how hard this should be - it was written by a university professor for my Uncle who has limited mathematical knowledge but enjoys puzzles. I can't think where to start with it though. Thanks for any help.

  2. #2
    Addicted Member Rassis's Avatar
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    Re: Geometry Puzzle

    If I understood it correctly, it seems that the sheep can reach everywhere inside the triangle or does its exterior count as well?
    ...este projecto dos Deuses que os homens teimam em arruinar...

  3. #3

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    Re: Geometry Puzzle

    I've drawn a diagram of it. We want the red area. The blue circles represent the outside of where the sheep can reach.
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  4. #4
    Addicted Member Rassis's Avatar
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    Re: Geometry Puzzle

    First consider a generic circle with radius R and a rope connecting any two points. The area A limited by the rope and the circle is given by:

    A = R^2*[a/2 – sin(a/2)*cos(a/2)]

    Where “a” is the angle formed by the two lines that connect the center of the circle and each of the two extreme points of the rope. Let us call it the inner-angle (perhaps the expression is not the most correct in English…).

    Second, consider a right-angled triangle with sides X and Y and hypotenuse Z.



    The inner-angle “a1” of the circle with diameter Y is given by:

    a1 = (Pi) – 2*arc.tan(X/Y)

    And the inner-angle “a2” of the circle with diameter X is given by:

    a2 = (Pi) – 2*arc.tan(Y/X)

    If the area of the triangle is S, then S = ½*(X*Y)

    Now, suppose X = 1. Because it is assumed that the area of the triangle S = 1, you have: Y = 2*1/1 = 2. And you get sequentially:

    a1 = 2.2142974
    a2 = 0.9272952
    Ax = 0.0159119
    Ay = 0.7071487

    And finally the area A we are after is:

    A = Az – Ax – Ay = (Pi)/8*(X^2 + Y^2) – Ax – Ay = 1.9634954 – 0.0159119 – 0.7071487 = 1.2404348

    If X = Y = 1.4142136, you get A = 1.2853982 which is a maximum.

    Rui
    Last edited by Rassis; May 1st, 2006 at 04:01 AM.
    ...este projecto dos Deuses que os homens teimam em arruinar...

  5. #5

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    Re: Geometry Puzzle

    Super! I'll send it to my uncle and see if that's what the answer was supposed to be. That all makes sense though. Thanks.

  6. #6
    Addicted Member Rassis's Avatar
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    Re: Geometry Puzzle

    It was a pleasure.

    Rui
    ...este projecto dos Deuses que os homens teimam em arruinar...

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