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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.

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

In accordance with @Tillman's Answer, we can find points $X$, $Y$, $Z$ on $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ defining bisector planes of dihedral angles along $\overline{DA}$, $\overline{DB}$, $\overline{DC}$. $$X := B + \frac{\hat{C}}{\hat{B}+\hat{C}}(C-B) = \frac{B\hat{B}+C\hat{C}}{ \hat{B}+\hat{C}} \qquad Y := \frac{C\hat{C}+A\hat{A}}{\hat{C}+\hat{A}} \qquad Z := \frac{C\;\hat{C}+A\hat{A}}{\hat{C}+\hat{A}}$$

The insphere's center, $I$, lies on the line common to all three bisecting planes. $\overleftrightarrow{DI}$ therefore meets $\triangle ABC$ 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

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

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I'm searching for exact the same formula. I didn't find anything helpful. Maybe there is a generalization of the Theorem of Angel Bisectors on Triangles which says that the angle bisecting line devides the opposite line proportionally to the other lines. But I can't really see what this generalization could say. I hope it makes an assertion on where the angle bisector meets the opposite triangle. Having that the Theorem of Stuart could possibly help.

Best Wishes,

Tillmann

I found a generalization, but it doesn't solve the complete problem. generalization

You can see the whole paper here. I think we are done, if we can verify that the angle bisecting line of the tetrahedron is the angle bisectoring line of the triangle AXD. Can we do this? I think that is wrong, isn't it?

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Ok I think I managed it. The Point is to know where W lies on AX (respectively on XD, what I am using here). But a short calcualtion does show it.

We can calculate AX and XD (Stewart's Theorem - we know where X lies on BC). Now again applying Stewart's Theorem on the triangle AXD - we now know where W lies on XD - we get a formula for the angle bisecting line in the tetrahedron that only depends on knowledge of the edgelengths. I haven't explicitly written down the formula but I expect it to be "nice". Indeed we have to calculate the ratio between the surfaces. There some roots are likely to pop up. But in the use of Stewart's Theorem we use the squares of distances. So it needn't to look to bad in the end.

Hope I can give more than this draft later this day.

I do have the formula now. But it isn't that nice as expected. All the ratios pop up (of course) in terms of squareroots. But maybe that is only bad for my purpose. I did everything like in the sketch above. Here is the formula:

$ ∣w_A ∣^2 = s∣AX∣^2 +(1−s)∣AD∣^2 − s(1−s)∣BC∣^2 $ where $s:=\frac{\hat{B}+\hat{C}}{\hat{B}+\hat{C}+\hat{D}}$.

I texed the complete proof. If your interested contact me.

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