The barycenter of a wire triangle is the "Spieker" point, namely the center of the inscribed circle of its medial triangle, that is the triangle whose vertices are the midpoints of the sides. Is there a nice caracterisation for the barycenter of a wire tetrahedron? It is easy to prove that it is on the line which joints the spieker point of any face to the barycenter of the triangle formed by the midpoints of the three remaining sides of the tetrahedron and that it divides that line in the ratio of the total weight of the triangle to the total weight of the three remaining sides. But, I cannot do anything with that. I can also prove that if masses are placed at the vertices proportional to the area of the opposite facet, then the barycenter is the center of the inscribed sphere. But...no farther.
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You mean a generic tetrahedron? If all faces are equal then it's easy, but in general I don't know. – Intelligenti pauca Sep 21 '15 at 16:25
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I mean a generic tetrahedron. The answer IS easy for a regular or equifacial tetrahedron...but the question is for a general tetrahedron. – mark Sep 21 '15 at 16:30
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Can't we just consider the Spieker points of the facets, give them a weight that equals the perimeter of the corresponding facet, then take the weigthed average of such points? – Jack D'Aurizio Sep 21 '15 at 17:09
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Since each edge belongs to two facets, you would have to give each Spieker point half the weight of the the perimeter. But, there is no obvious nice geometric interpretation of the resulting barycenter as there is for the Spieker point...at least I have not found one...for example as the center of the inscribed sphere of the tetrahedron formed by the Spieker points...which I do not think is true.. – mark Sep 21 '15 at 17:27