Let I be the center of the inscribed sphere of a tetrahedron A B C...
Let be the center of the inscribed sphere of a tetrahedron and let be the length of the line passing through from the vertex to the opposite face. I am looking for a formula for which is analogous to the formula for the angle bisector in a triangle
I assume that if such a formula exist, it involves areas of the faces of the tetrahedron. Any comments are appreciated.
Answer & Explanation
Write for the area of the face opposite vertex .
We can find points , , on , , defining bisector planes of dihedral angles along , , .
The insphere's center, , lies on the line common to all three bisecting planes. therefore meets at the point, , where the cevians , , coincide. One readily shows this point to be
How long is ? Situate vertex at the origin. Then
Writing , , for the lengths of the vectors, and , , for appropriate face angles between them, we have