Monday, November 16, 2009

Fox 186

Submitted by: http://geometri-problemleri.blogspot.com
You can find a more generalized version here in Newzad's blog.

11 comments:

  1. This comment has been removed by a blog administrator.

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  2. The bottom side of the bold area is 1/sin(tetha)
    where tetha between asin(1/3)=19.47 and 90

    and the bold area A= 1/sin(tetha)-B
    B is zero where asin(2/3)=41.8 < tetha < 90

    so if tetha=46.1622
    A=1/sin(46.1622)=1.3863

    if tetha=asin(2/3)=41.8
    A=1.5 > 1.4649

    so 1.4649 may not be the answer

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  3. I found A = 1,712 for theta = 29,4, by doing A = [3 - (upper triangle) - (lower trapezium)]. Sorry, I'm new here, and don't know the best way to show details.

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  4. Oscar, your answer checks out.
    theta = 29,4 gives A=1.712

    If you send us your solution we can publish it here. Thank you.

    8foxes.com

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  5. Yes you are right Oscar.
    I have to work on it again, thank you:))

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  6. For the name of Humanity, I deleted my comments. They don't have the rigth that made people make mistakes

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  7. I noticed that i made a mistake while writing eq. before

    max of the equation below gives the answer
    and it is nearly 1.712 as Oscar said

    1/sin(theta)-(2tan(theta)+2/tan(theta)-3sin(theta)tan(theta)-3cos(theta))*(2-3sin(theta))*1/2

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  8. Here goes the solution
    There are 3 situations of interest:


    a. 0 ≤ θ ≤ sin–1 (1/3), or 0 ≤ θ ≤ 19,47°  the intersection will be just a triangle, starting with area S = 0, going up to S = (√8)/2 = √2 = 1,414
    b. sin–1 (2/3) ≤ θ ≤ π/2, or 41,81° ≤ θ ≤ 90°  the intersection will be a rhombus and will end up as a square, with area going from S = 3/2 = 1,5 down to S = 1
    c. sin–1 (1/3) ≤ θ ≤ sin–1 (2/3)  the intersection will be the small rectangle ABCD, minus the triangle BDE, minus the trapezium OCFG.
    We have SBDE = EB*BD/2
    BD = 1 – DA = 1 – (2 – 3*sinθ) /cosθ
    EB = ED*cosθ = (BD/sinθ)*cosθ
    then, SBDE = [(BD/sinθ)*cosθ] * BD/2 = ½*[1 – (2 – 3*sin θ) /cosθ]2 * cosθ/sinθ
    = ½*(cosθ – 2 + 3*sinθ)2 /(cosθ * sinθ)
    and SOCFG = ½*(CF + OG) * 1
    OG = 3/(2 – 3*sinθ)
    CF = 3 – FE – EB = 3 – OG – EB
    then, SOCFG = ½*(3 – EB) = ½*[3 – (BD/sinθ)*cosθ] = ½*[3 – (cosθ /sinθ) – (2 – 3*sinθ) /sinθ]
    = 1/sinθ – cosθ/(2*sinθ)
    Therefore, S = 3 – SBDE – SOCFG
    = 3 – ½*(cosθ – 2 + 3*sinθ)2 /(cosθ * sinθ) – 1/sinθ – cosθ/(2*sinθ), or
    S = 1/sinθ – 2/(sinθ*cosθ) + (12 – 9*sinθ) /(2*cosθ), for sin–1 (1/3) ≤ θ ≤ sin–1 (2/3)
    Searching for the maximum of this function, we get
    Smax = 1,7120 for θmax = 29,40°

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  9. Updated solution

    http://i48.tinypic.com/2mhwqbp.gif

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  10. PS
    on updated solution, 3. case cannot be drawn becasue 3sin(34.16)-1+cos(34.16) > 1
    (forgot to say)

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