Re: Insphere of a tetrahedron

In news:<dqqc58$1fk$1@xxxxxxxxxxxxxxxxxxxxxxxxxx> schrieb Hauke Reddmann:
> Rouben Rostamian <rouben@xxxxxxxxxxxxxxxxxx> wrote:
>
>> 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.
>
> And the exspheres (or how they should be called) will be
> yielded by a->-a resp. a->-a,b->-b etc, I assume?

I think so. - J.H.Conway calls this procedure
in the geometry.research newsgroup "conjugation" in barycentric coordinates.

> As I said, there must be a trick getting this result without
> tedious computation - probably vectors. It's always vectors :-)

Here is an elementary proof without vectors:

It suffices to prove that the point of intersection X
of the bisecting plane of an edge, say CD,
with the opposite edge, AB, satisfies

|AX| : |BX| = a : b.

To show this, call

x := |AX|
y := |BX|
h1 := the length of the altitude of the triangle ABC through C,
h2 := the length of the altitude of the triangle ABD through D,
H := the length of the altitude of the pyramide ABCD through D,
rho := the radius of the insphere of ABCD;

then calculate the volume of the tetrahedron AXCD,
using that the incenter of ABCD is contained in the plane CDX:

1/6 * x * h1 * H2
= 1/3 * a * rho + 1/6 * x * h1 * rho + 1/6 * x * h2 * rho.

This determines the ratio a/x in terms of h1, H2, h2 and rho:

a/x = 1/2 * (x * h1 * H2 / rho - x * h1 - x * h2)

An analogous calculation of the volume of the tetrahedron BXCD
yields:

y/b = 1/2 * (x * h1 * H2 / rho - x * h1 - x * h2),
hence
y/b = x/a.

q.e.d.
.

Relevant Pages

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  • Re: intersection of tetrahedrons
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