center of tetrahedron
STUARTe
vertices. what is the distance from the center to each vertex? all 4
edge lines have length 1.
Robert Israel
Easy to figure out using coordinates, though a bit tedious.
This looks more like a homework problem than a puzzle. For extra
credit try an icosidodecahedron.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Philippe 92
> In article <1113789714.7...@g14g2000cwa.googlegroups.com>,
> STUARTe <jablu...@hotmail.com> wrote:
>> a regular tetrahedron has a center point that is equidistant to all 4
>> vertices. what is the distance from the center to each vertex? all 4
>> edge lines have length 1.
>
> Easy to figure out using coordinates, though a bit tedious.
As the coordinates of vertices can't be easily deduced from
just the length of edges, this is not only tedious but seems
unnecessarily complicated.
Cut the tetrahedron by a suitable plane, and use
Pythagorean theorem to find altitude of tetrahedron, then
distance of center to vertex.
A more direct method uses the centroid formula :
sum(ai * MAi^2) = sum(ai * GAi^2) + (sum(ai) * MG^2)
With a suitable choice of point M, this gives GA immediately.
> This looks more like a homework problem than a puzzle.
shure...
> For extra credit try an icosidodecahedron.
--
philippe
mail : chephip+misc at free dot fr
replace misc by news otherwise message will go to trash
Philippe 92
(sorry, clicked the wrong button)
Robert Israel wrote :
> In article <1113789714.7...@g14g2000cwa.googlegroups.com>,
> STUARTe <jablu...@hotmail.com> wrote:
>> a regular tetrahedron has a center point that is equidistant to all 4
>> vertices. what is the distance from the center to each vertex? all 4
>> edge lines have length 1.
>
> Easy to figure out using coordinates, though a bit tedious.
As the coordinates of vertices can't be easily deduced from
just the length of edges, this is not only tedious but seems
unnecessarily complicated.
Cut the tetrahedron by a suitable plane, and use
Pythagorean theorem to find altitude of tetrahedron, then
distance of center to vertex.
A more direct method uses the centroid formula :
sum(ai * MAi^2) = sum(ai * GAi^2) + (sum(ai) * MG^2)
With a suitable choice of point M, this gives GA immediately.
> This looks more like a homework problem than a puzzle.
shure...
> For extra credit try an icosidodecahedron.
The cutting plane method works fine.
But altitude of pentagon is less well known as altitude of
equilateral triangle !
Regards.
Martin DeMello
>
> As the coordinates of vertices can't be easily deduced from
> just the length of edges, this is not only tedious but seems
> unnecessarily complicated.
Use the fact that a regular tetrahedron can be obtained by joining
vertices of a cube (each of the tetrahedron edges is a face diagonal of
the cube).
martin
Philippe 92
Yes of course, then the trick is to show that the center of tetrahedron
is
also the center of the cube, as midpoint of a median of the
tetrahedron.
(median being line joining midpoints of two opposite edges of
tetrahedron)
This avoids a lot of calculations.
STUARTe
hint: it's a ratio of 2 square roots
Philippe 92
> can someone give me a number
> hint: it's a ratio of 2 square roots
or not... as sqrt(a) / sqrt(b) = sqrt(ab) / b
for instance sqrt(3) / sqrt(2) = sqrt(6) / 2
(not a real value just an example)
Back to the tetrahedron problem.
what is the face diagonal of a unit cube ?
what is the distance from center to vertex in a unit cube ?
isn't it half the big diagonal of cube ?
what is the ratio ?
Cheers.
The Last Danish Pastry
news:1113789714.7...@g14g2000cwa.googlegroups.com...
> a regular tetrahedron has a center point that is equidistant to all 4
> vertices. what is the distance from the center to each vertex? all 4
> edge lines have length 1.
Consider a tetrahedron of side sqrt(2) embedded in 4-dimensional space so
that its vertices are at (1,0,0,0) (0,1,0,0) (0,0,1,0) and (0,0,0,1). [The
tetrahedron lies in the 3-space w+x+y+z=1] Its center is clearly at
(1/4,1/4,1/4,1/4). The distance from the center to (1,0,0,0) is
sqrt(9/16+1/16+1/16+1/16)=sqrt(12/16)=sqrt(3/4). Hence, in the case where
the edge length is 1, the required distance is sqrt(3/4)/sqrt(2)=sqrt(3/8).
--
Clive Tooth
http://www.clivetooth.dk
STUARTe
the solution of a tetrahedron with vertices on a cube also works.
i solved via simultaneous equations for spheres with centers at the 4
vertices. the simultaneous solution is the coordinates of the
tetrahedron's center. it was an interesting exercise.
clive, plese reply with the theorem that states the vertex to center
distance of a regular 3d n-hedron is invariant when embedded in 4d.
The Last Danish Pastry
Just take the 3-space that the tetrahedron is in, and construct some 4-space
which contains that 3-space. No metrical properties of the tetrahedron are
affected.
Imagine a square (for example) embedded in a 2-dimensional space (a plane).
Now imagine that plane as part of some 3-dimensional space. No measurements
change when we imagine an object embedded in spaces of different
dimensionality.
Martin DeMello
Marvellous.
martin
The Last Danish Pastry
news:6_G9e.1086888$Xk.643471@pd7tw3no...
Thanks. This method can easily be generalized to compute a rather simple
formula for the radius of the n-sphere which touches the j-dimensional
components of an n-simplex (the n-dimensional generalization of the
equilateral triangle, the tetrahedron, etc).
Consider an n-simplex, with edge sqrt(2), embedded in (n+1)-space with its
n+1 vertices at
(1,0,...,0) [n+1 coordinates]
...
(0,...,0,1,0,...,0)
...
(0,...,0,1)
The center of this n-simplex is the point
A=(1/(n+1),1/(n+1),...,1/(n+1))
The first j+1 vertices in the above sequence are the vertices of a j-simplex
with center
B=(1/(j+1),1/(j+1),...,1/(j+1),0,...,0) [j+1 non-zero terms and n-j zero
terms]
The distance AB is thus
r
=sqrt((j+1)*(1/(j+1)-1/(n+1))^2 + (n-j)*(1/(n+1))^2)
=sqrt(1/(j+1) - 2/(n+1) + (j+1)/(n+1)^2 + (n-j)/(n+1)^2)
=sqrt(1/(j+1) - 2/(n+1) + (n+1)/(n+1)^2)
=sqrt(1/(j+1) - 2/(n+1) + 1/(n+1))
=sqrt(1/(j+1) - 1/(n+1))
This is the radius of the n-sphere centered at A and touching the j-simplex
at its center.
Putting n=3 and j=0 gives r=sqrt(1/1-1/4)=sqrt(3/4), as noted in my first
post.
Putting n=3 and j=1 gives r=sqrt(1/2-1/4)=1/2, this is the radius of the
sphere which touches all 6 edges of the tetrahedron.
Putting n=3 and j=2 gives r=sqrt(1/3-1/4)=sqrt(1/12), this is the radius of
the sphere which touches the 4 faces of the tetrahedron.
This is all in Coxeter, "Regular Polytopes", 8.8 Metrical properties.