# Which of the $43,380$ possible nets for a dodecahedron is the narrowest?

I want to fit multiple regular dodecahedron nets on to an infinitely long roll of paper. I want this to result in the largest possible dodecahedrons, for a roll of a given width.

My hunch is that the longer and narrower the net, the larger the dodecahedron I can produce; proving this one way or the other might be an interesting side-exercise.

My main question for now is simply: for a given size of pentagon, which of the $43,380$ possible nets for a regular dodecahedron fits into the narrowest rectangle?

• In my answer I interpreted your request as "find the rectangle of smallest area". Are you instead looking for the narrowest rectangle, doesn't matter how long it is? Jan 31, 2016 at 22:44
• Yes, the narrowest, not the smallest area. I want to make the biggest dodecahedron possible for a strip of given width. Feb 4, 2016 at 16:40
• See my edited answer. Feb 4, 2016 at 20:51
• Okay, I guess my question was still not totally unambiguous. When I said 'a strip of paper', I meant to suggest that the length of the strip (which is what you've focused on in the edited answer) is irrelevant; I'm only interested in the width, which is to say the shorter dimension of the rectangle. Feb 5, 2016 at 8:55
• I don't understand: 4.02874 is precisely the length of the shortest side of the rectangle. Feb 5, 2016 at 10:45

Just as a starter, I propose the most obvious one. Area of rectangle is 32.89 if every edge of dodecahedron is of unit length.

EDIT.

If one is interested in the narrowest possible net, I think the above disposition is still the best one. Because the central "belt" of six pentagons (yellow in the picture below) cannot be altered without widening the net, and the other surrounding pentagons can be moved to other positions, but this doesn't narrow (at best) the witdth of the net (see possible new positions, in blue, of three pentagons). The width of this net is $$\sqrt{5+2\sqrt5}(3+\sqrt5)/4\approx4.02874$$ times the length of a single edge.

EDIT 2.

Inspired by net n° 9383 in Horiyama's list I could find a strip slightly narrower than the above, at the expense of having its border not parallel to any pentagon side (see picture). Its width is $$\approx 3.93448$$.

EDIT 3.

Oolong discovered the best candidate, up to now: it is n° 43362 in the catalogue, corresponding to a width $$\approx 3.66547$$.

EDIT 4.

Oolong discovered an even narrower net: it is n° 36753 in the catalogue, corresponding to a width $$\approx 3.3166$$.

EDIT 5.

I performed an exhaustive search, using Mathematica and the complete collection of dodecahedron net centers in Mathematica format, which can be found at Horiyama's site. For every net I checked all the lines passing through two vertices: in case all the other vertices lied on the same side of the line, I then computed the distance from the line to the farthest vertex. The shortest of those distances is the "width" of the net.

Here are a few of the best results.

WIDTH     NET NUMBERS
3.07768   41382, 32924, 32920, 32511, 32494, 32492

3.26889   26440, 23967, 23620, 20027, 19706, 19668

3.3166    42665, 42591, 42549, 42546, 39271, 39268, 36753, 36743, 36717,
36716, 36607, 36598, 36581, 36445, 36439, 36408, 36390, 36304,
36298, 36267, 36264, 36263, 29579, 28755, 28742, 28741, 28740,
28734, 28496, 28489, 28488, 28456, 28434, 28433, 28432, 28416,
27807, 27806, 27805, 27729, 27728, 27727, 27674, 27673, 27672


Notice that the narrowest width can be computed exactly: $$3.07768=\sqrt{5+2\sqrt5}$$. Here's a picture of n° 41382, which is one of the "winners":

EDIT 6.

Here's the Mathematica code I used.

(* some definitions *)
lato=2Sin[Pi/5]//Simplify;
sqdist[a_,b_]:=(a-b).(a-b);
rot[a_,b_,t_]:=b+{{Cos[t],-Sin[t]},{Sin[t],Cos[t]}}.(a-b);
cross2[{ax_,ay_},{bx_,by_},{cx_,cy_}]:=(ax*by-ay*bx+bx*cy-cx*by+
cx*ay-cy*ax)/Sqrt[sqdist[{ax,ay},{bx,by}]];

(* main loop; "r04_n.math" are Horiyama's files *)
all={};
Do[
file="/path/r04_"<>ToString[n]<>".math";
Get[file];
net={};

Do[
If[sqdist[p[[i]],p[[k]]]==a^2//Simplify,
cmid=p[[i]]+(p[[k]]-p[[i]])/a;
start=rot[cmid,p[[i]],Pi/5]//Simplify;
pent=Table[rot[start,p[[k]],2j*Pi/5]//FullSimplify,{j,0,5}];
net=Append[net,pent];
pent=Table[rot[start,p[[i]],2j*Pi/5]//FullSimplify,{j,0,5}];
net=Append[net,pent]
],
{i,1,Length[p]-1},
{k,i+1,Length[p]}];

pts=Flatten[net,1]//N;
pts=Union[pts,SameTest -> (sqdist[#1,#2]<0.00001&)];

best=1000;
Do[
wid=-1;
flag=True;
Do[
t=cross2[pts[[i]],pts[[j]],pts[[k]]];
If[Abs[t]<0.0000001,Continue[]];
If[wid<0,wid0=Sign[t];wid=0];
If[t*wid0<-0.0000001,flag=False;Break[]];
If[Abs[t]>Abs[wid],wid=Abs[t]],
{k,1,Length[pts]}];

If[flag && wid/lato<best,best=wid/lato],
{i,1,Length[pts]-1},
{j,i+1,Length[pts]}];

AppendTo[all,{n,best}],

{n,1,43380}];

allsorted=Sort[all,#1[[2]]<#2[[2]]&]

• Oh, sure ... That's great for a rectangle that's narrow top-to-bottom. But what if we want a rectangle that's narrow left-to-right? :) (joke)
– Blue
Jan 31, 2016 at 22:34
• Hmm… Maybe I misinterpreted the request of the OP... Jan 31, 2016 at 23:05
• Just replace your area calculation with a ratio. :)
– Blue
Jan 31, 2016 at 23:51
• Great edit with addition of 9383 and the linked to the exhaustive list… I wonder if that is the narrowest possible? Feb 5, 2016 at 14:17
• Possibly promising: 43362, 43224, 43172, 42665, 42591, 42546. Will investigate properly later, but I think 43,362 has a total width less than three times the length of each side. This is nice: al.ics.saitama-u.ac.jp/horiyama/research/unfolding/catalog/… Feb 5, 2016 at 18:41