Insphere of a tetrahedron
Hauke Reddmann
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.)
--
Hauke Reddmann <:-EX8 fc3...@uni-hamburg.de
His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn
Ignacio Larrosa Cañestro
Hauke Reddmann <fc3...@uni-hamburg.de> escribió:
> 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.)
Choose general equations of the planes of the faces that give a positive
value when you susbstitute in them the coordinates of the four vertix. Is
that values is negative, then multiply the equation by -1 (*).
The distance from a point (x0, y0, z0) to a plane A*x + B*y + C*z + D = 0
is
d = |A*x0 + B*y0 + C*z0 + D|/sqrt(A^2 + B^2 + C^2)
In order to find the incentre of the tetrahedron, and not any ex-incentre
(centre of a sphere tangent to one face and the prolongations of the
others), you must eliminate the absolute value, takin in account what is
said above (*).
Then you get a linear system of four equations with four unknowns: the
coordinates x0, y0 and z0 of the incentre an the radius d of the in-sphere.
--
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUIT...@mundo-r.com
Rouben Rostamian
Hauke Reddmann <fc3...@uni-hamburg.de> 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.)
1.
A web search for "tetrahedron incenter" leads to the following
abstract:
http://www.math.uaa.alaska.edu/pnwmaa/abstractlist.pdf
Angie Mai (University of Portland)
On the Incenter of the Triangle and Tetrahedron The incenter
of a triangle (respectively, tetrahedron) is the center of
the inscribed circle (respectively, sphere). The incenter of
a triangle is geometrically characterized by the intersection
of the three angle bisectors and the incenter of a tetrahedron
is the intersection of the four planes bisecting the dihedral
angles formed by adjacent faces. For the triangles and
tetrahedra, the incenters can be expressed as the weighted
averages of the vertices
I haven't seen the contents of that presentation. This is all I know
about it.
2.
You may find the information in the following page useful:
http://www.zebragraph.com/incenter.html
--
Rouben Rostamian
Rob Johnson
Hauke Reddmann <fc3...@uni-hamburg.de> wrote:
>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.)
The perpindiculars from the incenter to each side of the tetrahedron
are all the same length, 6 times the volume of the tetrahedron divided
by the surface area of the tetrahedron. Finding the intersection of 3
of the planes at that distance from the sides should give the incenter
of the tetrahedron.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
Hauke Reddmann
> angles formed by adjacent faces. For the triangles and
> tetrahedra, the incenters can be expressed as the weighted
> averages of the vertices
Too bad the source is so inaccessible, this would be exactly
what I was after.
Rouben Rostamian
Hauke Reddmann <fc3...@uni-hamburg.de> wrote:
>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
Hauke Reddmann
> 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.
And the exspheres (or how they should be called) will be
yielded by a->-a resp. a->-a,b->-b etc, I assume?
As I said, there must be a trick getting this result without
tedious computation - probably vectors. It's always vectors :-)
Thomas Mautsch
> Rouben Rostamian <rou...@pc18.math.umbc.edu> 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.