| The answer is Yes.
First, note that there are two distinct to place the ribbon. Following the
ribbon's path, on each face you either proceed Straight, or turn Right or Left.
The two possible placements are then described by:
R L R L R L and S R L S R L.
The first path can satify the conditions ribbon iff the edges dimensions of the
box can form a triangle!
The second path can always satisfy the conditions.
To analyze the first claim, 'unfold' the box (with dimensions axbxc):
+- b -+- a -+
| |/ |
c c c
| / | |
+- c -+- b -+- a -+
| |/ |
a a a
| / | |
+- a -+- c -+- b -+
| |/ |
b b b
| / | |
*- a -+- c -+
The origin is indicated with a '*', the 'slope' of the ribbon is -1, and it
crosses y=0 at (d,0). Its equation is y(x) = x - d. We require:
0 <= d <= a, and
0 <= y(a) <= b <= y(a+c) <= a+b <= y(a+b+c) <= a+b+c
Equivalently,
max(0,a-b,c-b) <= d <= min(a,a-b+c,c)
For such a d to exist, we require a >= max(0,a-b,c-b), a-b+c >= max(0,a-b,c-b),
and c >= max(0,a-b,c-b). Tackling a >= max(0,a-b,c-b) first, we see that
a >= 0 and a >= a-b are obvious, and we require a >= c-b. Similarly,
a-b+c >= max(0,a-b,c-b) has two obvious inequalities and a-b+c >= 0; finally,
c >= max(0,a-b,c-b) requires c >= a-b. Summarizing: a >= c-b, a-b+c >= 0, and
c >= a-b. Reorganizing: a+b >= c, a+c >= b, b+c >= a. 'Nuff sed.
How wide may the ribbon be? Obviously, min(a,a-b+c,c) - max(0,a-b,c-b).
This 'simplifies' to: min(a+b,a+c,b+c) - max(a,b,c), or a+b+c - 2 max(a,b,c).
How long is the ribbon? (a+b+c)sqrt(2).
To analyze the second claim, 'unfold' the box differently.
+- a -+- b -+- a -+
| | | |
c c c c
| | | |
+- a -+- c -+- a -+- b -+- a -+
| | | |
b b b b
| | | |
*- a -+- c -+- a -+
Etc.
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