Copyright © University of Cambridge. All rights reserved.
'All Tied Up' printed from https://nrich.maths.org/
A solution based on the one from Richard is
given below. I was not able to use Richard's original diagram but
hope the one submitted is sufficient:
As the path of the ribbon around the box does not change direction
when it reaches each edge, it can be drawn as a single straight
line. As the ribbon crosses two faces of the box twice, a simple
net of the box can show the path as a single straight line.
I therefore numbered the faces 1-6 and checked the order the ribbon
went over them. This led to the diagram submitted, which I used to
work out the length of the ribbon via Pythagoras' theorem.
The length of the ribbon is given by:
$$ \sqrt{(2 \times 5 + 2\times 10)^2 +(2 \times 5 + 2\times
20)^2}$$
By rearranging the above and substituting W, L and H for thelengths
of the sides, the general expression for the length of the ribbon
is therefore:
$$ 2\sqrt{(W+H)^2 + (L+H)^2}$$ or $$ 2\sqrt{W^2 + L^2 + 2H^2 + 2(WH
+ LH)} $$
where $H$ is the length of a shortest edge of a box, $W$ and $L$
the lengths of the longer sets of sides.
A little more from Andrei - many thanks:
The condition of the problem is matched only when the red line
passes through the same $U$, i.e. the dashed blue passes through
the same $A$. So, the length of the ribbon is the distance $A$-$A$.
The blue dashed lines are the extreme positions of the ribbon, and
between these positions the ribbon could glide. Each of these
extreme positions has one of the two parallel lines reduced to a
point.