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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 800.
http://img15.imageshack.us/img15/3285/problem800.pngDenote S(XYZ)= area of triangle XYZDraw lines per attached sketchLet K is the projection of O over AB.AB ⊥to OK and AB orthogonal to OC => AB ⊥ to CKIn right triangle COK , draw altitude OH . Since AB ⊥ to plane CKO => OH orthogonal to ABOH ⊥ to KC and orthogonal to AB => OH ⊥ to plane ABC1. We have S(AOB)=1/2. AB.OK and S(AHB)=1/2.AB.KHSo S(AOB)=S(AHB)/cos(x) where x is the angle form by 2 planes AOB and ABCSimilarly 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)^2Note that S(ABC)=S(ABH)+S(AHC)+S(BHC)3. Calculation volume of tetrahedron OABCV=1/3.a.b.c=1/3.h.S(ABC) => a.b.c=S(ABC).h4. 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^2Combine 2 above expressions we get 1/h^2=1/a^2+1/b^2+1/c^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,
See below for the sketch of the problem 800https://s25.postimg.org/e47zco94v/Problem_800.jpg