Question

Is there a 3D analogue of angle?

Answer 1

To prove that:-
In any tetrahedron with a cubic vertex the numerical square of the area of the face opposite the cubic vertex is equal to the sum of the numerical squares of the areas of the other three faces.


Figure 3.

Let:-
A = Area of Triangle OXZ
B = Area of Triangle OXY
C = Area of Triangle OYZ
D = Area of Triangle XYZ

To prove that D² = A² + B² + C²


Figure 4.I.D = ½ah(given)II.NowPX = Ö(b² - h²)(Pythagoras)III.andPZ = Ö(c² - h²)(Pythagoras)IV.anda = PX + PZV.\a = Ö(b² - h²) + Ö(c² - h²)VI.\a² = b² + c² - 2h² + 2Ö(b²c² - b²h² - c²h² + h²)VII.\a² - b² - c² + 2h² = 2Ö(b²c² - b²h² - c²h² + h²)VIIIa.\(a² - b² - c² + 2h²)² = 4(b²c² - b²h² - c²h² + h²)VIIIb.i.e.(a² - b² - c²)² + (a² - b² - c²)4h² + 4h² = 4b²c² - 4b²h² - 4c²h² + 4h²IX.\(a² - b² - c²)² + 4a²h² = 4b²c²X.\4a²h² = 4b²c² - (a² - b² - c²)²XI.butD = ½ah(I.)XII.\16D² = 4b²c² - (a² - b² - c²)²XIII.buta² = x² + z²(Pythagoras)XIV.andb² = x² + y²(Pythagoras)XV.andc² = y² + z²(Pythagoras)XVI.\16D² = 4(x² + y²)(y² + z²) - 4y4XVII.\4D² = x²y² + x²z² + y²z²XVIII.NowA = ½xz(given)XIX.andB = ½xy(given)XX.andC = ½yz(given)XXI.\A + B + C = ½(xy + xz + yz)XXII.\A² + B² + C² = ¼(x²y² + x²z² + y²z²)XXIII.\4(A² + B² + C²) = x²y² + x²z² + y²z²XXIV.but4D² = x²y² + x²z² + y²z²(XVII.)XXV.\4D² = 4(A² + B² + C²)XXVI.\D² = A² + B² + C²Q.E.D.

There is another proof which, unlike that given above, is neither so abstruse nor of my own devising (1976). For this proof a tetrahedron OXYZ with cubic vertex at O is cut by a plane through edge OY such that edge XZ is normal to it. The plane intersects edge XZ at P and angles OPX, OPZ, YPX and YPZ are all right angles.


Figure 5.

Let:-
A = Area of Triangle OXZ
B = Area of Triangle OXY
C = Area of Triangle OYZ
D = Area of Triangle XYZ

To prove that A² + B² + C² = D²

Answer 2

Explanation:

1. In this case, the Universe is the 3D analogue of the surface of a saddle. The angles of a triangle add up to less than 180 degrees and the Universe is infinite