Math Is Fun Forum

Good question.
I want YOU to sweat a little.

Why???

A note: there may exist a figure with arbitary circumfence and constant area.
For the original question:
The figure with the biggest circumfence is the most "angular"(the most different from a circle) figure.

There might be a mistake. Why didn't you use Ex?

Yeah...
dude.

Here's the code (Mathematica, rewritten, but really messy and hard-to-understand):

K[n1_, n2_] := Union[{n1 + n2, n1 - n2, n1*n2, n1/n2}];
KK[list_, num_] := Union[Flatten[Table[K[list[[i]], num], {i, 
      1, Length[list]}]]];
KKK[list1_,
     list2_] := 
      Union[Flatten[Table[KK[list1, list2[[i]]], {i, 1, Length[list2]}]]];
d[a_, b_, c_, d_, f_] := {
    (*abcdfff*)
    f[{a}, f[{b}, f[{c}, {d}]]],
    (*abcfdff*)
    f[{a}, f[f[{b}, {c}], {d}]],
    (*abcffdf*)
    f[f[{a}, f[{b}, {c}]], {d}],
    (*abfcdff*)
    f[f[{a}, {b}], f[{c}, {d}]],
    (*abfcfdf*)
    f[f[f[{a}, {b}], {c}], {d}]
    }
d[l_, f_] := d[l[[1]], l[[2]], l[[3]], l[[4]], f];
dd[l_, f_] := dd[l[[1]], l[[2]], l[[3]], l[[4]], f];
dd[a_, b_, c_, d_, f_] :=
  (
    Print["abcdfff:", f[{a}, f[{b}, f[{c}, {d}]]]];
    Print["abcfdff:", f[{a}, f[f[{b}, {c}], {d}]]];
    Print["abcffdf:", f[f[{a}, f[{b}, {c}]], {d}]];
    Print["abfcdff:", f[f[{a}, {b}], f[{c}, {d}]]];
    Print["abfcfdf:", f[f[f[{a}, {b}], {c}], {d}]];
    )
p = Permutations[{3, 3, 8, 8}];
res = Table[Union[Flatten[d[p[[i]], KKK]]], {i, 1, Length[p]}];
Union[Flatten[res]]

Could explain and rewrite it later.

My program is ready. And guess what-there aren't another solutions except:
8/(3-(8/3))=24!!!

Interesting... Can't you make some program, which gives all possible solutions?

I have to aknowledge i can't really understand the infinitesimal (although I use it).
Here's my question-is it well-defined - does it REALLY exist?
How can a number be greater than infinity?

hi.
me too.

Can't we create our own "objects" as infinitesimals?
Let p be 1 in multiplication and 2 in addition...
What happens???:
for a-real:
pa=a;
p+a=a+2;
p(p+a)=p^2+pa=a+2, so p^2 must be 2 (but only in addition )
(p+a)/p= (a+2)/p=a/p+2/p=a/p+2;
(p+a)/p=p/p+a/p, so p/p=2/p=2.
Others:
(p* means p is used in mult.
p+ means p is used in addit.)
2p-p=2(p*)-(p+)=2-2=0. it is different from:
2p-1p=2(p*)-1(p*)= (2-1)(p*)= (p*)=1.
1p=1=1(p*)=1.1=1.
1(p+0)=1((p+)+0)=1(2+0)=2.
The basic laws won't be true.

Consider a function: {x} - the real part of x. ({2.5}=0.5, {5/3}=1/3, {sqrt(2)}=sqrt(2)-1}).
Let see what's the limit as x goes to 2.
Here we have 2 different limits: left-limit and right-limit: