Let I be the center of the inscribed sphere of a tetrahedron A B C D and let I A be the length of the line passing through I from the vertex A to the opposite face. I am looking for a formula for I A 2 which is analogous to the formula for the angle bisector in a triangle I a 2 = b c ( b + c ) 2 [ ( b + c ) 2 − a 2 ] .I assume that if such a formula exist, it involves areas of the faces of the tetrahedron. Any comments are appreciated.

Question

Let   I be the center of the inscribed sphere of a tetrahedron   A  B  C  D and let       I    A   be the length of the line passing through   I from the vertex   A to the opposite face. I am looking for a formula for       I    A    2   which is analogous to the formula for the angle bisector in a triangle      I    a    2    =            b      c              (      b      +      c              )        2              [  (  b  +  c      )    2    −      a    2    ]  .I assume that if such a formula exist, it involves areas of the faces of the tetrahedron. Any comments are appreciated.

Marlie Frazier · Accepted Answer

Write             P      ^       for the area of the face opposite vertex   P.We can find points   X,   Y,   Z on             B      C        ¯  ,             A      C        ¯  ,             A      B        ¯   defining bisector planes of dihedral angles along             D      A        ¯  ,             D      B        ¯  ,             D      C        ¯  .  X  :=  B  +                    C        ^                                      B          ^                    +                        C          ^                      (  C  −  B  )  =            B                        B          ^                    +      C                        C          ^                                              B          ^                    +                        C          ^                        Y  :=            C                        C          ^                    +      A                        A          ^                                              C          ^                    +                        A          ^                        Z  :=            C                              C          ^                    +      A                        A          ^                                              C          ^                    +                        A          ^                    The insphere&#039;s center,   I, lies on the line common to all three bisecting planes.             D      I        ↔   therefore meets   △  A  B  C at the point,   W, where the cevians             A      X        ¯  ,             B      Y        ¯  ,             C      Z        ¯   coincide. One readily shows this point to be  W  :=            A                        A          ^                    +      B                        B          ^                    +      C                        C          ^                                              A          ^                    +                        B          ^                    +                        C          ^                    How long is             D      W        ¯  ? Situate vertex   D at the origin. Then      |              D      W        ¯              |        2    =  W  ⋅  W  =                    |            A                        |                2                                          A            ^                          2            +              |            B                        |                2                                          B            ^                          2            +              |            C                        |                2                                          C            ^                          2            +      2      (      A      ⋅      B      )                        A          ^                                      B          ^                    +      2      (      B      ⋅      C      )                        B          ^                                      C          ^                    +      2      (      C      ⋅      A      )                        C          ^                                      A          ^                            (                        A          ^                    +                        B          ^                    +                        C          ^                            )        2            Writing   a,   b,   c for the lengths of the vectors, and   α,   β,   γ for appropriate face angles between them, we have      |              D      W        ¯              |        2    =                                          A            ^                          2                    a        2            +                                    B            ^                          2                    b        2            +                                    C            ^                          2                    c        2            +      2                        B          ^                                      C          ^                    b      c      cos      ⁡      α      +      2                        C          ^                                      A          ^                    c      a      cos      ⁡      β      +      2                        A          ^                                      B          ^                    a      b      cos      ⁡      γ              (                        A          ^                    +                        B          ^                    +                        C          ^                            )        2

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