Circumnavigating a unit tetrahedron
Henry
shortest closed route (i.e. returning to its starting point) on the
surface which visits all six edges?
Visiting is a fairly loose concept - it could include crossing an edge
once or more, or just touching it, or going along it, or passing
through an end of the edge. Restricting the problem to properly
crossing each edge once would make it impossible, as each face has an
odd number of edges and so it would not be possible for the number of
times a face is entered to equal the number of times it is exited.
I have the solution and will provide it in due course.
Ted S.
news:3bb788a7...@news.btinternet.com:
[The original problem is being preserved as space for a possible spoiler.]
I can come up with a path of length 3. If the four points of the
tetrahedron are A, B, C, and D, start at the midpoint of AB, go to the
midpoint of BD, then to the midpoint of BC, then to the midpoint of CD,
then to the midpoint of AC, then to the midpoint of AD, and return to the
midpoint of AB.
Can anybody do better than 3? (For some reason, I'm reminded of the spider
and the fly problem.)
--
Ted S.: change .spam to .net to reply by e-mail
When will I learn? The answers to life's problems aren't at the bottom of
a bottle. They're on TV! --Homer Simpson
Paul Underwood
Ted S. wrote in message ...
>Somebody claiming to be se...@btinternet.com (Henry) wrote in
>news:3bb788a7...@news.btinternet.com:
>
>[The original problem is being preserved as space for a possible spoiler.]
>
>> Given a regular tetrahedron with edges of length 1, what is the
>> shortest closed route (i.e. returning to its starting point) on the
>> surface which visits all six edges?
>>
>> Visiting is a fairly loose concept - it could include crossing an edge
>> once or more, or just touching it, or going along it, or passing
>> through an end of the edge. Restricting the problem to properly
>> crossing each edge once would make it impossible, as each face has an
>> odd number of edges and so it would not be possible for the number of
>> times a face is entered to equal the number of times it is exited.
>>
>> I have the solution and will provide it in due course.
>
>I can come up with a path of length 3. If the four points of the
>tetrahedron are A, B, C, and D, start at the midpoint of AB, go to the
>midpoint of BD, then to the midpoint of BC, then to the midpoint of CD,
>then to the midpoint of AC, then to the midpoint of AD, and return to the
>midpoint of AB.
>
>Can anybody do better than 3? (For some reason, I'm reminded of the spider
>and the fly problem.)
>
Start at the vertex a (edges ab,ac,ad covered)
go to vertex b (edges ab,bc,bd visited)
go to mid cd (edges cd)
The union of edges visited is the required six.
The lenth of the path is 1+2*(3^-(1/2))/2 ~ 2.7321
Paul
Nick Wedd
<paulun...@sladroad.fsnet.co.uk> writes
>>> Given a regular tetrahedron with edges of length 1, what is the
>>> shortest closed route (i.e. returning to its starting point) on the
>>> surface which visits all six edges?
>
>Start at the vertex a (edges ab,ac,ad covered)
>go to vertex b (edges ab,bc,bd visited)
>go to mid cd (edges cd)
>The union of edges visited is the required six.
>The lenth of the path is 1+2*(3^-(1/2))/2 ~ 2.7321
You have
(1) misread the question, which requires a closed path
(2) miscalculated the length of your path, which should be 1.7321
I should not scoff - I made both these mistakes myself, but managed to
cancel my response before it got posted!
Nick
--
Nick Wedd ni...@maproom.co.uk
Paul Underwood
Nick Wedd wrote in message ...
oops should have read return to vertex a. Hence path =
1+sqrt(3)/2+sqrt(3)/2 ~ 2.7321
Sorry Paul
Rouben Rostamian
Label the vertices O, A, B, C. Consider O as the origin and A, B,
C as the ends of the vectors OA, OB, OC. For arbitrary r, s, t in
the interval [0,1] define the six vectors a, b, c, d, e, f (each
of which lies on one of the edges of the tetrahedron) by:
a = r * A
b = s * B
c = s * C
d = A + t * (B-A)
e = A + t * (C-A)
f = (B + C) / 2
Then compute the length of the closed path abcdefa. To minimize the
length, set its derivatives with respect to r, s, t to zero and solve
for r, s, t. We get r = 1/4, s = 1/2, t = 3/4 which then yields
length = 3/2*sqrt(3) = 2.5981...
--
Rouben Rostamian <rost...@umbc.edu>
Mark Sproson
Henry <se...@btinternet.com> wrote in message
news:3bb788a7...@news.btinternet.com...
S
P
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I can make it around in 2.598 units.
Consider the tetrahedron 'sat' on a surface on one of its faces. Start at
the mid-point of one edge of the bottom face. Follow a line perpendicular to
an adjacent edge - this has length sqrt(3)/4. Now continue to the midpoint
of the next edge of the bottom face, and so on around the three 'upper'
faces - six lines in all. Total length 6/4*sqrt(3)=2.598.
Can't quite prove this is minimal - but I believe it is.
Mark
Wim Benthem
wrote:
But there seems to be more than 1 way to order the vertices on this
path.
abcdefa can't be right, as c isn't on the same face as d, but you could
have abcefda or abcfdea or abcfeda. I'm not sure you'd get the same
minimum in all cases.
--
Wim Benthem
Henry
>Given a regular tetrahedron with edges of length 1, what is the
>shortest closed route (i.e. returning to its starting point) on the
>surface which visits all six edges?
Gold stars to Mark Sproson and Rouben Rostamian who have the correct
values. For this and other similar routes round a cube or a regular
tetrahedron see
http://www.btinternet.com/~se16/js/circumnavcubetetra.htm
Jim Ferry
It seems that you can generalize the tetrahedron result to the
octahedron and icosahedron: the minimal circumnavigation is
sqrt(3) E / 4, where E is the number of edges (12 and 30).
Referring to the graph at the bottom of the above web page,
let a, b, and c be the number of crossings of each family of
parallel lines. The total length of the corresponding path
is sqrt((a^2 + b^2 + c^2) / 2), and a, b, and c must satisfy:
a + b + c = 0, and
|a| + |b| + |c| = E.
Let a be the largest in absolute value. Then |a| = |b| + |c| = E/2.
The distance is minimized for |b| = |c| = E/4, which is allowed
provided E is even. This yields sqrt(3) E / 4.
The minimum is achieved if the edge graph has a Hamiltonian. I think
it does in each case.
I haven't checked this argument too carefully, but I think something
like it should establish the minimum.
| Jim Ferry | Center for Simulation |
+------------------------------------+ of Advanced Rockets |
| http://www.uiuc.edu/ph/www/jferry/ +------------------------+
| jferry@[delete_this]uiuc.edu | University of Illinois |
Henry
>>For this and other similar routes round a cube or a regular
>>tetrahedron see
>>http://www.btinternet.com/~se16/js/circumnavcubetetra.htm
>>
>It seems that you can generalize the tetrahedron result to the
>octahedron and icosahedron: the minimal circumnavigation is
>sqrt(3) E / 4, where E is the number of edges (12 and 30).
>
>Referring to the graph at the bottom of the above web page,
>let a, b, and c be the number of crossings of each family of
>parallel lines. The total length of the corresponding path
>is sqrt((a^2 + b^2 + c^2) / 2), and a, b, and c must satisfy:
> a + b + c = 0, and
>|a| + |b| + |c| = E.
>Let a be the largest in absolute value. Then |a| = |b| + |c| = E/2.
>The distance is minimized for |b| = |c| = E/4, which is allowed
>provided E is even. This yields sqrt(3) E / 4.
>The minimum is achieved if the edge graph has a Hamiltonian. I think
>it does in each case.
>I haven't checked this argument too carefully, but I think something
>like it should establish the minimum.
I don't quite follow this - but it would seem to suggest a minimum for
a unit octahedron of sqrt(3)*12/4=5.196... and a unit icosahedron of
sqrt(3)*30/4=12.990... for closed surface routes which meet all the
edges.
There is certainly a shorter closed route visiting all 12 octahedron
edges of length 4, namely the four sides of a square which divides the
octahedron into two square pyramids.
For a unit icosahedron I think there is a route made up of twelve
altitudes of triangles for a length of 12*sqrt(3)/2=10.392...
I do not know whether either of these are the shortest, but I would
expect the octahedral case to be optimal.