The Insphere of a Tetrahedron

A contiguous derivation of radius and center of the insphere of a general tetrahedron is given. Therefore a linear system is derived. After a transfor-mation of it the calculation of radius and center can be separated from each other. The remaining linear system for the center of the insphere can be solved after discovering the inverse of the corresponding coefficient matrix. This procedure can also be applied in the planar case to determine radius and center of the incircle of a triangle.


Introduction
A coherent derivation of radius and center of the insphere of a general tetrahedron is hard to find (see e.g. [1]). When setting up the linear system for their calculation the uniform orientation of the faces of the tetrahedron has to be observed. This is exemplified in the examples at the end. After transforming the original linear system into a system having a coefficient matrix of block upper triangular form the calculation of radius and center of the insphere of the tetrahedron can be separated from each other. The solution of the remaining linear system for the center of the insphere rests upon an exercise in [2]. for each triangle is chosen in such a way, that the normal to a face is directed into the interior of the tetrahedron according to the corkscrew rule: the direction of the normal of a face is defined such that the right thumb points in the direction of the normal and the fingers curl along the orientation of the bounding curve of a triangle.
x x y y z z x x y y z z x x y y z z Expanding determinant (1) in terms of the elements of the first row leads to: An illustration is given in Figure 1.
Repeating the above procedure for the plane through the vertices P 1 , P 4 , P 2 of the tetrahedron gives by exchanging the indices 2 4 → , 3 2 → :  Repeating the above procedure for the plane through the vertices P 1 , P 3 , P 4 of the tetrahedron gives by exchanging the indices 4 3 → , 2 4 → : .
x x y y z z C x x y y z z x x y y z z From determinant C one obtains (17) by subtracting the first row from the second and third row, exchanging rows two and three and expanding the resulting determinant with respect to the first row. This process is reversible.
Corollary 1 For determinant C holds 0 C = if and only if the four vertices P 1 , P 2 , P 3 , P 4 of the tetrahedron are lying in a plane.
Proof: Considering for instance the plane through P 1 , P 2 , P 3 according to (1) the coordinates of vertex P 4 are fulfilling (1) if and only if determinant C is zero after exchanging the third row with the second and afterwards with the first row, leaving the value of the determinant unchanged. 

A Linear System
From Equations (6), (7), (10) and (16) one obtains the following system of linear equations for the unknowns 123  123  123  1   142  142  142  142  1   134  134  134  134  1   243  243  243  243   0  , , ,  134  123  142   243  134  123  142   243  134  123  142   ,  ,  ,  , y y z z d y z y y z z y y y y z z z z y y y y z z z z y y z z y y z z y y z z y y z z y y z z y y z z y y z z y y z z y y z z y y z z y y z z y y z z -or what is equivalent, by adding the first three rows of (19) to the last row-yields with the result of Proposition 1 a decomposed linear system, whose coefficient matrix has block upper triangular form , The summation of the terms ijk w in the last row of (20) ranges over the in-  134  134  1  134   142  142  142  1  142   123  123  123  1  123 , , , The next Proposition is formulated for the coefficient matrix 134  134   142  142  142   123  123  123   ,  ,  ,  , , of the linear system (22). This Proposition rests upon an exercise in [2]. Proposition 2 For matrix C from (18) and matrix D from (23) the following statement is true: C denotes the transpose of C and E the identity matrix. This means that D is the cofactor matrix of C. Furthermore for 0 C ≠ holds: T 1 C C is the inverse of D and 2 D C = .
Proof: Expanding C in terms of the elements of the first, second and third column yields with denotations (11), (8), (3): 1  3  1  2  1  2  1  2  1  3  1  4  1  4  1  4  1  4  1   2  1  2  1  4  1  3  1  3  1   2  1  134  3  1  142  4  1  123 , , , y y z z y y z z C x x x x y y z z y y z z y y z z x x y y z z , , By multiplication of matrices T C and D one obtains with the above results for determinant C the elements on the main diagonal of the product matrix: 1  3  1  4  1  134  134  134  T  2  1  3  1  4  1  142  142  142   2  1  3  1  4  1  123  123  123 , , , The non-diagonal elements of the product matrix are zero because they can be written as 3 × 3 determinants with two equal columns. As an example this shall be shown for the element in row two and column one of the product matrix: y y z z y y z z y y y y y y z z y y z z y y z z y y y y z z y y y y z z y y y y z z y y y y z z Forming determinants in (24) yields Theorem 1 If the vertices P 1 , P 2 , P 3 , P 4 of the tetrahedron are not lying in a plane the center of the insphere is given by:

The Planar Case
Assume that the three vertices of a triangle ( ) , P x y = , ( ) Adding the first and the second row of the linear system (37) to the last row yields where the sum in the third row of (38) ranges over the index doublets 12, 31 and 23. From the third row of (38) one obtains Multiplying the linear system (40) from the left by T B yields because of ( ) ( )

Examples
The tetrahedrons considered in the sequel are having as vertex the origin, a vertex on the x-axis, a vertex in the x-y plane and a vertex in three-dimensional space: , , P x y z = . For determinant C according to (26) holds: The solution of the linear system is: The solution of the linear system is: ( ) 3.73968,1.73656,1.14508 m P = . The distance of m P to P 1 is 4.27926, the distance of m P to P 2 is 3.07179, the distance of m P to P 3 is 2.83245 and the distance of m P to P 4 is 6.23721. These numerical results have been obtained with Mathematica.

Conclusion
A coherent derivation of radius and center of the insphere of a general tetrahedron is given. By means of elementary techniques of linear algebra the original linear system for the determination of radius and center of the insphere of a tetrahedron can be broken up to calculate the radius and the coordinates of the center separately. The remaining linear system for the coordinates of the center of the insphere can be solved by multiplying with the inverse of the coefficient matrix, which is found in the course of the treatise. This procedure can readily be applied in the planar case.