Stereometry

dwatna

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The faces of tetrahedron ABCD are acute-angled triangles and the dihedral angles by sides AB and CD are right. Prove that the orthocenters of its faces all lie on one plane.

I have tried many techniques, but gotten nowhere. I feel like the theorem on three perpendiculars might be useful as well as expressing the tetrahedron's volume in multiple ways using the right dihedral angles.
 

dwatna

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Any ideas, progress? Anyone?
 

Subhotosh Khan

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The faces of tetrahedron ABCD are acute-angled triangles and the dihedral angles by sides AB and CD are right. Prove that the orthocenters of its faces all lie on one plane.

I have tried many techniques, but gotten nowhere. I feel like the theorem on three perpendiculars might be useful as well as expressing the tetrahedron's volume in multiple ways using the right dihedral angles.
We have 3 faces and 3 points as orthocenters.

3 points will always lie on one plane!

What am I missing?
 

dwatna

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Subhotosh Khan

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A tetrahedron has 4 faces, I think.
Right - but:

When I took crystallography - one of those triangles was called base and the other three were designated as faces - at any given orientation.

Another point:

Dihedral angle is defined between two planes - here you say "the dihedral angles by sides AB and CD" - can you explain what is actually meant here?
 

dwatna

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Right - but:

When I took crystallography - one of those triangles was called base and the other three as designated as faces - at any given orientation.
Sorry then, there is no such distinction in my native language so I got confused.
A side is part of 2 faces. It is the dihedral angle between those two faces.
 
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Subhotosh Khan

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Sorry then, there is no such distinction in my native language so I got confused.
A side is part of 2 faces. It is the dihedral angle between those two faces.
AB and CD are lines in tetrahedron. To which two faces are you referring now?

The faces are triangles - so you need three points to define each.
 

dwatna

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AB and CD are lines in tetrahedron. To which two faces are you referring now?

The faces are triangles - so you need three points to define each.
The dihedral angle between faces ABC anc ABD is right, and so is the dihedral angle between faces ACD and BCD. Is it clear now?

We are to prove that the orthocenters of ABC, ABD, ACD and BCD all lie on one plane.
 

Dr.Peterson

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Dwatna means the two faces that meet at a given line.

The dihedral angle at the edge AB is between faces ABC and ABD.

@dwatna, it may help if you can tell us what "tools" you have available for the proof -- do you need to use synthetic geometry theorems, or can you uses analytic geometry (coordinates), or anything else? What sort of theorems or formulas have you learned that this exercise might be intended to use (or is the problem more like a contest problem with no specific context)?

I'm assuming that what you call stereometry is essentially what I would call "solid geometry".
 

dwatna

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Dwatna means the two faces that meet at a given line.

The dihedral angle at the edge AB is between faces ABC and ABD.

@dwatna, it may help if you can tell us what "tools" you have available for the proof -- do you need to use synthetic geometry theorems, or can you uses analytic geometry (coordinates), or anything else? What sort of theorems or formulas have you learned that this exercise might be intended to use (or is the problem more like a contest problem with no specific context)?

I'm assuming that what you call stereometry is essentially what I would call "solid geometry".
There are no restraints on tools, but I think that an analytical proof would be quite messy. There is no context to this problem, i.e. the intended technique is not given (it is "classified" as a hard problem). About tools: yourstandard planimetry+some stereometry facts and lemmas. Also 3D trigonometry and vectors.
 

dwatna

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@Dr.Peterson, I've come back to this problem after a few days and I wonder whether a coordinate solution would be feasible. There is a criterion that all faces are acute triangles, that might make the calculations tough.
 
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