Construct incenter of tetrahedron

Question

How do I construct the incenter of a tetrahedron? Assuming that the tetrahedron in question is not regular/isoceles, of course.

Answer 1

The incenter is the intersection of the bisector planes of the dihedral angles formed by three tetrahedron faces which don't have a common vertex.

If $ABCD$ are your tetrahedron vertices, to obtain the bisector plane of planes $ABC$ and $ABD$, construct on those planes respectively lines $AC'$ and $AD'$, perpendicular to $AB$. The plane formed by the bisector of $\angle C'AD'$ with $AB$ is the bisector plane of faces $ABC$ and $ABD$. Repeat the same construction for faces $ABC$, $ACD$ and then for faces $ABC$, $DBC$.

See picture below for an example, showing perpendicular lines $AC'$ and $AD'$, their bisector (blue line) and the bisector plane of the dihedral angle formed by $ABC$ and $ABD$ (blue plane). Point $O$ is the incenter.

Answer 2

This question is computationally explored (not ruler&compass-explored) in the MathematicaSE posting, Insphere for Irregular Tetrahedron, which supplements @JeanMarie's link.

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