## 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.

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