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Math Help - Sphere inscribed in pyramid

  1. #1
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    Sphere inscribed in pyramid

    I need help:
    Sphere is inscribed in right pyramid with equilateral triangle in base.We know :
    angle of inclination =
    \alpha.
    side of the triangle = a
    Is it possible express the radius of sphere using a and \alfa? Or do we have to know the \beta? Use trygonometric functions.



    Sphere inscribed in pyramid-xmjyoo.png

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  2. #2
    GJA
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    Re: Sphere inscribed in pyramid

    Hi, aleschio.

    I don't think we need to know \beta. I will use m to denote the measure of a length. My idea is something like this:

    1) Since the base is equilateral we know m(AD)=\frac{\sqrt{3}}{2}a (see Equilateral triangle - Wikipedia, the free encyclopedia). There are many ways to show that m(DE)=\frac{1}{3}m(AD) (for example see Viviani's Theorem Viviani's theorem - Wikipedia, the free encyclopedia). Hence

    m(DE)=\frac{\sqrt{3}}{6}a.

    2) Since the tangents to the circle DF and DA are drawn from the same point (i.e. D), we know DG\cong DE. Therefore, m(DG)=\frac{\sqrt{3}}{6}a and \Delta DGO\cong \Delta DEO. Since \Delta DGO\cong \Delta DEO we also know that m(\angle GDO)=\alpha/2.

    3) We now use the right triangle \Delta DGO and the tangent function to obtain the radius.

    Does this answer your question? Let me know if anything is unclear/incorrect.

    Good luck!
    Last edited by GJA; September 21st 2012 at 03:47 PM.
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  3. #3
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    Re: Sphere inscribed in pyramid

    Thanks GJA

    I have a problem that I have to count the ratio of this pyramid to sphere inscribed in it when \alpha and \beta are given. I wondered if it's possible to count it when only \alpha is known. Your solution says that it is possible and fairly easy. But I have one more question, Is obvious that sphere is tangent to the base in point which is the end of the height of this pyramid or do we have to proof it?
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