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Find elegant proof of 3D generalization of Pythagorean theorem?
A great many clever proofs have been devised for the classic a² + b² = c² Pythagorean theorem. Given a tri-rectangular pyramid, composed of 3 right triangles A, B, C meeting at a common orthogonal vertex, with a triangular base D, it is also generally true that A² + B² + C² = D², where A, B, C, D are the areas of those triangles. Can you find a clever proof for this?
BA goes to best effort.
Check out graph of tri-rectangular pyramid, orthogonal vertex is on left bottom:
Impressive work, guys. It's going to be hard to decide who deserves BA.
5 Antworten
- IndicaLv 7vor 7 JahrenBeste Antwort
Let (cosα,cosβ,cosγ) be direction cosines of a plane that slices off a solid like this in O1 and meets axes at X,Y,Z. Let area of triangle XYZ be Δ.
cos²α+cos²β+cos²γ = 1 → Δ² = (Δcosα)² + (Δcosβ)² + (Δcosγ)²
But angle between yz plane and XYZ is α and so projection of Δ onto yz plane is Δcosα.
∴ area OYZ = Δcosα with similar results for other faces.
- Let'squestionLv 7vor 7 Jahren
Without loss of generality, let the four vertices of the given tetrahedron be, D (0, 0, 0), A(a, 0, 0), B(0, b,0) and C(0, 0, c)
We have A = (1/2)bc, B = (1/2)ac an C= (1/2)ab and it can be proved that
D = (1/2)√(a²b² + b²c² c²a²). So it follows that
D² = A² + B² + C²
_____________________________________________
Above A, B and C meant areas and also points A, B C and D
Proof of area D of the triangle ABC is as follows, a, b c are line segments, and A, B and C are angles.
The triangle ABC has three sides of lengths:
√(a² + b²), √(b² + c²) and √(c² + a²)
D = Area of triangle ABC = (1/2)*{√(b² + c²)}*{√(c² + a²)}* sin C ----------- (1)
Now Cos C = [{b² + c² + c² + a² -(a² + b²)}/{2*{√(b² + c²)}*{√(c² + a²)}}]
so sin C = √(1 - cos² C) = [{√(a²b² + b²c² c²a²)}/{√(b² + c²)}*{√(c² + a²)} ----(2) from (1) and (2), we do get
D = (1/2)*√(a²b² + b²c² c²a²)}/
- GlippLv 7vor 7 Jahren
imagine a rectangular block with sides A, B, C and find the side D that is the straight line between two diagonally apposed corners.
First you look at the face made by A and B and compute the diagonal E by E^2 = A^2 + B^2
Then you notice that the line E also makes a right angle with C so that D^2 = C^2 + E^2
Therefore D^2 = C^2 + A^2 + B^2
- DukeLv 7vor 7 Jahren
May I offer a very belated proof, relying on "known facts", on a question CLOSED3 weeks ago? Oddly enough Y!A will let me do it NOW!
Let a→, b→, c→, d→ are outer normal vectors to the faces of the tetrahedron;
let as usual |a→| = A, |b→| = B, |c→| = C, |d→| = D; then
Known fact #1: (w→)² = |w→|² for every vector w→;
Known fact #2: a→ b→ = b→c→ = c→a→ = 0 (orthogonality);
Known fact #3: a→ + b→ + c→ + d→ = 0→.
Squaring both sides of a→ + b→ + c→ = -d→ we get what we need and similarly we can generalize the result for higher dimensions.
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- IanLv 7vor 7 Jahren
Without loss of generality, let the vertices be located at (0,0,0), (a,0,0), (0,b,0), (0,0,c), with a, b, and c positive.
Let A = area of right triangle with vertices (0,0,0), (0,b,0), (0,0,c).
Let B = area of right triangle with vertices (0,0,0), (a,0,0), (0,0,c).
Let C = area of right triangle with vertices (0,0,0), (a,0,0), (0,b,0).
Let D = area of triangle with vertices (a,0,0), (0,b,0), (0,0,c).
Clearly A = bc/2, B = ac/2, and C = ab/2.
To find D, we can use the fact that the area of a triangle, with two of its sides formed by vectors u and v, is half the magnitude of the cross product of u and v.
Two of the sides of the triangle with vertices (a,0,0), (0,b,0), (0,0,c) are formed by the vectors u = <-a,b,0> and v = <-a,0,c>. Therefore,
D = (1/2)|u cross v|
= (1/2)|<-a,b,0> cross <-a,0,c>|
= (1/2)|<det[b 0; 0 c], -det[-a 0; -a c], det[-a b; -a 0]>|
= (1/2)|<bc, ac, ab>|
= (1/2)sqrt((bc)^2 + (ac)^2 + (ab)^2).
Therefore,
D^2 = (1/2)^2 ((bc)^2 + (ac)^2 + (ab)^2)
= (bc/2)^2 + (ac/2)^2 + (ab/2)^2
= A^2 + B^2 + C^2.
(Note: the first answerer misinterpreted A, B, C, D as lengths instead of areas.)
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