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    A wire frame is made by wrapping a wire around the 'equator' of a perfectly spherical balloon of unit radius and wrapping two wires at right angles around the sphere going through the North and South poles. The balloon is then deflated but remains perfectly spherical until it can just be moved out of the framework. What is the radius of the largest sphere that can pass through this frame, without distorting it, from inside to outside?
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    (Original post by atkelly)
    A wire frame is made by wrapping a wire around the 'equator' of a perfectly spherical balloon of unit radius and wrapping two wires at right angles around the sphere going through the North and South poles. The balloon is then deflated but remains perfectly spherical until it can just be moved out of the framework. What is the radius of the largest sphere that can pass through this frame, without distorting it, from inside to outside?
    I get,

    r=1/√3

    is that right?
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    Well, I think it's right. Here's the working.

    The small sphere has got to pass through a quadrant of a larger hemisphere.
    This smaller sphere must just touch the three"wires" making up the hemispherical quadrant and will just touch it at the mid-pts shown, viz. A, B and C.
    ABC is an equilateral triangle, and the radius of the smaller sphere can be found by finding out the circle which will circumscribe that equilateral triangle.

    To find the distance between A and B.
    ========================
    Take the dot product of two vectors, OA and OB

    In the 3-dimensional cartesian system with the axes x, y, z the vectors OA and OB are given as,

    OA = R(0, cos45, sin45)
    OB = R(cos45,0, sin45)

    OA.OB = 0 + 0 + (1/√2)(1/√2)
    OA.OB = ½

    OA.OB = |OA||OB|cosθ
    OA.OB = R.R.cosθ
    OA.OB = cosθ (R=1)

    cosθ = ½
    θ = 60°
    =====

    therefore AB² = R² + R² - 2.R.R.cosθ
    AB² = 2R²(1 - cos60°)
    AB² = 2*½ = 1
    AB = 1
    =====

    The circumscribing circle
    ================
    Looking at the second figure,

    Side of equilateral triangle is L.

    L = AB
    L = 1

    using cosine rule,

    L² = r² + r² - 2.r.r.cosø
    L² = 2r²(1 - cos120°)
    1 = 2r²(1 + 0.5)
    2r² = 2/3
    r = 1/√3
    =====
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Updated: June 16, 2004

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