Let I be the center of the inscribed sphere of a tetrahedron A B C...

Jaxson Mack

Jaxson Mack

2022-08-17

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.

Answer & Explanation

Marlie Frazier

Marlie Frazier

Expert

2022-08-18Added 14 answers

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'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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