Sunday, August 26, 2012

Problem 800: Gua's Theorem, Pythagorean theorem in 3-D, Tetrahedron, Cubic Vertex, Triangular Pyramid, Apex, Height, Right Triangle Area, Base Area, Projected Area

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 800.

Online Geometry Problem 800: de Gua's Theorem, Pythagorean theorem in 3-D, Tetrahedron, Cubic Vertex, Triangular Pyramid, Apex, Height, Right Triangle Area, Base Area, Projected Area.

3 comments:

  1. http://img15.imageshack.us/img15/3285/problem800.png

    Denote S(XYZ)= area of triangle XYZ
    Draw lines per attached sketch
    Let K is the projection of O over AB.
    AB ⊥to OK and AB orthogonal to OC => AB ⊥ to CK
    In right triangle COK , draw altitude OH .
    Since AB ⊥ to plane CKO => OH orthogonal to AB
    OH ⊥ to KC and orthogonal to AB => OH ⊥ to plane ABC
    1. We have S(AOB)=1/2. AB.OK and S(AHB)=1/2.AB.KH
    So S(AOB)=S(AHB)/cos(x) where x is the angle form by 2 planes AOB and ABC
    Similarly we have S(AOB)=S(ABC).cos(x)… ( O is the projection of C over plane AOB )
    From above expression we have S(AOB)^2=S(ABC).S(ABH)
    2. S(AOB)^2=S(ABC).S(ABH)
    S(AOC)^2=S(ABC).S(AHC)
    S(BOC)^2=S(ABC).S(BHC)
    Add above expression side by side we have S(AOB)^2+S(AOC)^2+S(BOC)^2=S(ABC)^2
    Note that S(ABC)=S(ABH)+S(AHC)+S(BHC)
    3. Calculation volume of tetrahedron OABC
    V=1/3.a.b.c=1/3.h.S(ABC) => a.b.c=S(ABC).h
    4. In right triangle KOC we have 1/h^2=1/c^2+1/k^2 … (relationship in a right triangle)
    In right triangle AOB we have 1/k^2=1/a^2+1/b^2
    Combine 2 above expressions we get 1/h^2=1/a^2+1/b^2+1/c^2

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  2. hai, can you give me the answer correctly about this problem please? or will you give me the picture about solution of this problem?
    I need it, please give you answer,

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  3. See below for the sketch of the problem 800

    https://s25.postimg.org/e47zco94v/Problem_800.jpg

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