Please, please, i need some hint about, tetrahedron

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Commandino’s Theorem asserts that the four medians of a tetrahedron intersect at a point that divides each median in a 1:3 ratio, with the longer segment adjacent to the vertex. The discussion emphasizes the use of vector algebra and linear transformations to prove this theorem, particularly focusing on the regular tetrahedron as a simplified case. Participants suggest that understanding dihedral angles and general geometry is essential for grasping the proof. Additionally, the conversation touches on the conditions necessary for linear transformations to preserve distance ratios.

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Commandino’s Theorem states that

The four medians of a tetrahedron
concur in a point that divides each of
them in the ratio 1:3, the longer
segment being on the side of the vertex
of the tetrahedron.

can someone put links below where about proof of this theorem

thx so much
 
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wowolala said:
Commandino’s Theorem states that

The four medians of a tetrahedron
concur in a point that divides each of
them in the ratio 1:3, the longer
segment being on the side of the vertex
of the tetrahedron.

can someone put links below where about proof of this theorem

thx so much

Why not just prove it yourself. If you know anything about dihedral angles, or even general geometry, this should be a piece of cake.
 
i think i would use vector algebra.
 
heres another point of view. since linear transformations take medians to medians, and all tetrahedra are equivalent by linear transformations it suffices to prove it for a regular tetrahedron.

but since the medians of the base triangle meet at a point, and the three medians originating from the base arise by lifting one end of those medians the same distance, it follows at least that those three medians meet in a point directly above the center of the base. hence dropping a median from the top vertex passes through this same point.

it remains to see where the point is.
 
mathwonk said:
since linear transformations take medians to medians, and all tetrahedra are equivalent by linear transformations it suffices to prove it for a regular tetrahedron.
Question: What condition should a linear transformation hold, in order to preserve ratios of distances? And does that condition will apply to a transformation which is custom-made to map a regular tetrahedron to a non-regular one (a "nice" transformation, since it won't scramble the vertices or something)?

At the bottom, degenerate case, I can imagine a transformation that always sets the Z coordinate to 0, mapping the regular tetrahedron to the X-Y plane and showing a perspective view of it. Obviously distance ratios are not preserved. Thus there must be a condition that a "nice" transformation satisfies.
 

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