Jaxson Mack

2022-08-17

Let $I$ be the center of the inscribed sphere of a tetrahedron $ABCD$ 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
${I}_{a}^{2}=\frac{bc}{\left(b+c{\right)}^{2}}\left[\left(b+c{\right)}^{2}-{a}^{2}\right].$
I assume that if such a formula exist, it involves areas of the faces of the tetrahedron. Any comments are appreciated.

Marlie Frazier

Expert

Write $\stackrel{^}{P}$ for the area of the face opposite vertex $P$.

We can find points $X$, $Y$, $Z$ on $\overline{BC}$, $\overline{AC}$, $\overline{AB}$ defining bisector planes of dihedral angles along $\overline{DA}$, $\overline{DB}$, $\overline{DC}$.
$X:=B+\frac{\stackrel{^}{C}}{\stackrel{^}{B}+\stackrel{^}{C}}\left(C-B\right)=\frac{B\stackrel{^}{B}+C\stackrel{^}{C}}{\stackrel{^}{B}+\stackrel{^}{C}}\phantom{\rule{2em}{0ex}}Y:=\frac{C\stackrel{^}{C}+A\stackrel{^}{A}}{\stackrel{^}{C}+\stackrel{^}{A}}\phantom{\rule{2em}{0ex}}Z:=\frac{C\phantom{\rule{thickmathspace}{0ex}}\stackrel{^}{C}+A\stackrel{^}{A}}{\stackrel{^}{C}+\stackrel{^}{A}}$
The insphere's center, $I$, lies on the line common to all three bisecting planes. $\stackrel{↔}{DI}$ therefore meets $\mathrm{△}ABC$ at the point, $W$, where the cevians $\overline{AX}$, $\overline{BY}$, $\overline{CZ}$ coincide. One readily shows this point to be
$W:=\frac{A\stackrel{^}{A}+B\stackrel{^}{B}+C\stackrel{^}{C}}{\stackrel{^}{A}+\stackrel{^}{B}+\stackrel{^}{C}}$
How long is $\overline{DW}$? Situate vertex $D$ at the origin. Then
$|\overline{DW}{|}^{2}=W\cdot W=\frac{|A{|}^{2}{\stackrel{^}{A}}^{2}+|B{|}^{2}{\stackrel{^}{B}}^{2}+|C{|}^{2}{\stackrel{^}{C}}^{2}+2\left(A\cdot B\right)\stackrel{^}{A}\stackrel{^}{B}+2\left(B\cdot C\right)\stackrel{^}{B}\stackrel{^}{C}+2\left(C\cdot A\right)\stackrel{^}{C}\stackrel{^}{A}}{\left(\stackrel{^}{A}+\stackrel{^}{B}+\stackrel{^}{C}{\right)}^{2}}$
Writing $a$, $b$, $c$ for the lengths of the vectors, and $\alpha$, $\beta$, $\gamma$ for appropriate face angles between them, we have
$|\overline{DW}{|}^{2}=\frac{{\stackrel{^}{A}}^{2}{a}^{2}+{\stackrel{^}{B}}^{2}{b}^{2}+{\stackrel{^}{C}}^{2}{c}^{2}+2\stackrel{^}{B}\stackrel{^}{C}bc\mathrm{cos}\alpha +2\stackrel{^}{C}\stackrel{^}{A}ca\mathrm{cos}\beta +2\stackrel{^}{A}\stackrel{^}{B}ab\mathrm{cos}\gamma }{\left(\stackrel{^}{A}+\stackrel{^}{B}+\stackrel{^}{C}{\right)}^{2}}$

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