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

${I}_{a}^{2}=\frac{bc}{(b+c{)}^{2}}[(b+c{)}^{2}-{a}^{2}].$

I assume that if such a formula exist, it involves areas of the faces of the tetrahedron. Any comments are appreciated.

${I}_{a}^{2}=\frac{bc}{(b+c{)}^{2}}[(b+c{)}^{2}-{a}^{2}].$

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

Expert

2022-08-18Added 14 answers

Write $\hat{P}$ 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 $\overline{DA}$, $\overline{DB}$, $\overline{DC}$.

$X:=B+\frac{\hat{C}}{\hat{B}+\hat{C}}(C-B)=\frac{B\hat{B}+C\hat{C}}{\hat{B}+\hat{C}}\phantom{\rule{2em}{0ex}}Y:=\frac{C\hat{C}+A\hat{A}}{\hat{C}+\hat{A}}\phantom{\rule{2em}{0ex}}Z:=\frac{C\phantom{\rule{thickmathspace}{0ex}}\hat{C}+A\hat{A}}{\hat{C}+\hat{A}}$

The insphere's center, $I$, lies on the line common to all three bisecting planes. $\overleftrightarrow{DI}$ therefore meets $\mathrm{\u25b3}ABC$ at the point, $W$, where the cevians $\overline{AX}$, $\overline{BY}$, $\overline{CZ}$ coincide. One readily shows this point to be

$W:=\frac{A\hat{A}+B\hat{B}+C\hat{C}}{\hat{A}+\hat{B}+\hat{C}}$

How long is $\overline{DW}$? Situate vertex $D$ at the origin. Then

$|\overline{DW}{|}^{2}=W\cdot W=\frac{|A{|}^{2}{\hat{A}}^{2}+|B{|}^{2}{\hat{B}}^{2}+|C{|}^{2}{\hat{C}}^{2}+2(A\cdot B)\hat{A}\hat{B}+2(B\cdot C)\hat{B}\hat{C}+2(C\cdot A)\hat{C}\hat{A}}{(\hat{A}+\hat{B}+\hat{C}{)}^{2}}$

Writing $a$, $b$, $c$ for the lengths of the vectors, and $\alpha $, $\beta $, $\gamma $ for appropriate face angles between them, we have

$|\overline{DW}{|}^{2}=\frac{{\hat{A}}^{2}{a}^{2}+{\hat{B}}^{2}{b}^{2}+{\hat{C}}^{2}{c}^{2}+2\hat{B}\hat{C}bc\mathrm{cos}\alpha +2\hat{C}\hat{A}ca\mathrm{cos}\beta +2\hat{A}\hat{B}ab\mathrm{cos}\gamma}{(\hat{A}+\hat{B}+\hat{C}{)}^{2}}$

We can find points $X$, $Y$, $Z$ on $\overline{BC}$, $\overline{AC}$, $\overline{AB}$ defining bisector planes of dihedral angles along $\overline{DA}$, $\overline{DB}$, $\overline{DC}$.

$X:=B+\frac{\hat{C}}{\hat{B}+\hat{C}}(C-B)=\frac{B\hat{B}+C\hat{C}}{\hat{B}+\hat{C}}\phantom{\rule{2em}{0ex}}Y:=\frac{C\hat{C}+A\hat{A}}{\hat{C}+\hat{A}}\phantom{\rule{2em}{0ex}}Z:=\frac{C\phantom{\rule{thickmathspace}{0ex}}\hat{C}+A\hat{A}}{\hat{C}+\hat{A}}$

The insphere's center, $I$, lies on the line common to all three bisecting planes. $\overleftrightarrow{DI}$ therefore meets $\mathrm{\u25b3}ABC$ at the point, $W$, where the cevians $\overline{AX}$, $\overline{BY}$, $\overline{CZ}$ coincide. One readily shows this point to be

$W:=\frac{A\hat{A}+B\hat{B}+C\hat{C}}{\hat{A}+\hat{B}+\hat{C}}$

How long is $\overline{DW}$? Situate vertex $D$ at the origin. Then

$|\overline{DW}{|}^{2}=W\cdot W=\frac{|A{|}^{2}{\hat{A}}^{2}+|B{|}^{2}{\hat{B}}^{2}+|C{|}^{2}{\hat{C}}^{2}+2(A\cdot B)\hat{A}\hat{B}+2(B\cdot C)\hat{B}\hat{C}+2(C\cdot A)\hat{C}\hat{A}}{(\hat{A}+\hat{B}+\hat{C}{)}^{2}}$

Writing $a$, $b$, $c$ for the lengths of the vectors, and $\alpha $, $\beta $, $\gamma $ for appropriate face angles between them, we have

$|\overline{DW}{|}^{2}=\frac{{\hat{A}}^{2}{a}^{2}+{\hat{B}}^{2}{b}^{2}+{\hat{C}}^{2}{c}^{2}+2\hat{B}\hat{C}bc\mathrm{cos}\alpha +2\hat{C}\hat{A}ca\mathrm{cos}\beta +2\hat{A}\hat{B}ab\mathrm{cos}\gamma}{(\hat{A}+\hat{B}+\hat{C}{)}^{2}}$

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