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Sep 30, 2001, 5:03:59 PM9/30/01

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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?

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.

Sep 30, 2001, 5:47:42 PM9/30/01

to

Somebody claiming to be se...@btinternet.com (Henry) wrote in

news:3bb788a7...@news.btinternet.com:

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

Sep 30, 2001, 6:05:31 PM9/30/01

to

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

Sep 30, 2001, 6:22:54 PM9/30/01

to

In article <9p84sa$n4e$1...@newsg2.svr.pol.co.uk>, Paul Underwood

<paulun...@sladroad.fsnet.co.uk> writes

<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

Sep 30, 2001, 6:29:31 PM9/30/01

to

hi

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

Sep 30, 2001, 7:43:08 PM9/30/01

to

In article <3bb788a7...@news.btinternet.com>,

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>

Sep 30, 2001, 8:16:03 PM9/30/01

to

Henry <se...@btinternet.com> wrote in message

news:3bb788a7...@news.btinternet.com...

S

P

O

I

L

E

R

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

Oct 1, 2001, 9:38:04 AM10/1/01

to

On 30 Sep 2001 19:43:08 -0400, rou...@math.umbc.edu (Rouben Rostamian)

wrote:

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

Oct 1, 2001, 5:04:47 PM10/1/01

to

On Sun, 30 Sep 2001 21:03:59 GMT, se...@btinternet.com (Henry) wrote:

>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?

>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

Oct 1, 2001, 7:15:58 PM10/1/01

to

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 |

Oct 4, 2001, 5:30:20 PM10/4/01

to

>On Mon, 01 Oct 2001 21:04:47 GMT, se...@btinternet.com (Henry) wrote:

>>For this and other similar routes round a cube or a regular

>>tetrahedron see

>>http://www.btinternet.com/~se16/js/circumnavcubetetra.htm

>>

>>For this and other similar routes round a cube or a regular

>>tetrahedron see

>>http://www.btinternet.com/~se16/js/circumnavcubetetra.htm

>>

Jim Ferry wrote:

>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.

>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.

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