Re: Insphere of a tetrahedron

In article <dqid7f$7iv$2@xxxxxxxxxxxxxxxxxxxxxxxxxx>,
Hauke Reddmann <fc3a501@xxxxxxxxxxxxxx> wrote:
>Given an irregular (!!) tetrahedron as 4 space point coordinates.
>What are the coordinates of the insphere center?
>(Paper reference suffices :-)
>(Tried to compute that myself, straightforward but the formula
>piggyfies exponentially so a trick would help.)

OK, I worked this out. The results are so simple that I can't
help but to assume that they must be known.

Theorem 1:
Let a, b, c be the lengths of the sides opposite to vertices
A, B, C of the triangle ABC and let O be the triangle's incenter.
Then we have:
O = (a/t) A + (b/t) B + (c/t) C,
where t=a+b+c is the triangle's perimeter.

Note: Here I am viewing A, B, C, O as points in R^2.
The formula above gives the incenter O as a convex
combination of the vertices A, B, C.

Theorem 2:
Let a, b, c, d be the areas of the faces opposite to vertices
A, B, C, D of the tetrahedron $ABCD$ and let O be the
tetrahedron's incenter. Then we have:
O = (a/t) A + (b/t) B + (c/t) C + (d/t) D,
where t = a + b + c + d is the tetrahedron's surface area.

This, too, gives O as a convex combination of the vertices.
It is likely that the formula generalizes to higher dimensions
in the obvious way but I haven't given much thought to that.

--
Rouben Rostamian
.

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