Taping a cardboard square is getting me fired!

Question

I was mucking around in the storeroom today when I should have been working. I had a square $20\times20$ cm piece of cardboard, and a roll of masking tape I'd been using (I can't remember the tape's width, but it was an integer number of centimetres). I started mindlessly wrapping the tape around the cardboard...

When I began, I put the tape (sticky side down) with its cut edge entirely on one side of the square and began wrapping. If I hit an edge or a corner, I folded the tape over so as to leave no tape hanging over the edges of square, and none of the sticky side exposed, and continued wrapping. When I finished and cut the tape, the new cut edge was again entirely on one side of the square. On starting and finishing, I cut the strip of tape perpendicular to its long axis.

When my boss entered the room I threw the cardboard into the compactor, but not fast enough, as he caught a glimpse and asked how much tape I had wasted. I didn't know, so he said that unless I can tell him exactly how much tape I used, he will fire me!

Before he walked in, I did take a photo (see below) of one side of the cardboard, but I'm lost - the cardboard and tape were exactly the same colour...

Is it possible to figure out the exact length of tape I used?

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_{This is a purely geometrical puzzle, with a unique solution. Do not assume lengths and angles in the image are to scale. The cardboard is an opaque infinitesimally thin perfect square with two sides (front and back), the tape is an opaque infinitesimally thin foldable but otherwise inflexible rectangle with one sticky side, and the lines within the square represent borders either between tape and tape or tape and cardboard. If lines appear to intersect, they do. If you're confused, it might be worth playing with some real tape to see how the edges and corners work.}

Answer 1

Please tell your boss that you used (I think):

$400/\sqrt{21}\approx 87.287$cm of tape that was 8cm wide

@2012rcampion did a better diagram of the wrapping than mine, so I'm including that here:

Then I built a little spreadsheet with the various geometrical constraints:

Referring to this diagram to do the calculations gives us the following:

$w' = w/\cos(\theta)$ where $w$ is the width of the tape. So $x = 20-2w'$. Then $y = 20\frac{w'-x}{w'}$ and $z = 20\frac{x}{w'}$ (because of similar triangles).
This immediately constrains $w$ to either 7 or 8.
Next I work out $k$ and compare it with $x$. Since I think $k>x$ this leaves $w$ as 8. Woohoo! Now we just need some geometry.
Going back to @2021rcampion's diagram, you'll see that pythag will get us there. The height is 80. The width is $20 + k + w'$.
$w' = 40/\sqrt{21}$.
$k$ is slightly more interesting. But with a bit of fiddling about with trig, it comes to $120\sqrt{21}-20$. This all cancels in a pleasant way, which makes me think there's an easier way to see this. But it all leads to the answer given.

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Edit with logic for why the linking is as it is:

1. The tape must be (at least) $w$ wide, where $w=w'\cos(\theta)$ because of the wide stripe down the middle. 2. Also $\theta<45^{\circ}$ or we wouldn't be able to get that stripe to hit the bottom with some space left over (quite a bit less, obvs, but $45^{\circ}$ is enough to get us going).
3. That means that the little bit in the bottom right must be the start and end of the tape with cardboard between them, since the tape has to be running parallel to the broad stripe in the middle.
4. And it means that the lower of those must the be start because it must be partially covered by the broad stripe in the middle (recall that the ends are fully on the front surface, so the longer one must be the full width (since it's the last to be stuck down it's on top), and the shorter one must be the first and partially covered)).
5. Now work backwards from the end of the tape. Plainly, the last piece wraps around and connects with the broad stripe in the middle. Wrap around again, and there has to be a space between the broad stripe down the middle and the next piece. Hence the middle stripe I've labelled "Cardboard" must, in fact, be cardboard. (Another way to see this: Note that if the cardboard extended to the right, that final piece would wrap around leaving a gap. So a similar gap must exist to the left of the stripe in the middle.)
6. Finally, consider the dotted blue parallel lines in my diagram. The one $k$ away from the left comes from the right hand side of the exposed start of the tape. The one $x$ away comes from the right hand side of the broad stripe down the middle. That means that $k>x$.