Do we have to work on a flat surface or can we utilize, say, books of arbitrary measurements as ground on which to build our construction?
You have to work on a flat surface.
Damn, that cuts the sphere in half…
…Unless glue can be used.
I think I know possible solution, but I ll post tomorrow and my eyes are closing already 
I mean Can they be piled up/stacked on each other?
I guess the idea is to make a sphere or half (since gravity is a thing) with a diameter of 21cm and then try to make as many pieces of bread touch its outer wall as possible, on all sides.
Yes. This was also possible in Puzzle 22 if you remember.
They cannot be bent or ripped apart though, right?
Correct.
According to my quick and dirty calculations with a cube strategy, as many as 248 gingerbreads can be stacked in a way in which none of them are more than 21cm away from the most distant other bread.
So, yes, Lothar can win the bet.
I arrived at my solution with maths. Explaining it would be a drag. So here are some crappy illegible notes.
I suppose I could elaborate that the basic idea is to take a cube in which the space diagonal is 21cm, then pick two opposing walls of that cube, stack as many gingerbreads as possible in a way that said wall is covered in gingerbreads that touch the wall at least a little (some corner breads only barely touch it with their corners), then do the same for the other two opposing walls, creating an enclosed area which is then filled with gingerbreads, keeping in mind that the breads in the edges need only barely touch the inside of the cube to count as valid breads.
If someone tells me how I can put things inside those nice collapse things I can put that picture inside one such thing.
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Maybe we can do a triangle and place some more inside and then just stack 3 of them
Pictures are not uploading on mobile for some reason
The distance between 3rd stack and 1st stack would be around 20,8
Because height of triangle is around 18,16
and using pythagorean the distance is around 20,8
How far apart are the two areas labled 1?
Just a bit over 12 cm. If you want a precise number, let’s make it 12.02 cm. That allows for areas 2 be 12 cm wide while still having all breads touch the 12.12x12.12x12.12 cube.
Actually it might as well just be 12 cm. No need for any extra space there, what was I thinking.
So for the areas labled 1, you’re stacking 13 together with the 1cm length, and put those on top of each other 4 times with the 7cm length, making for 4 stacks of 13 gingerbread on both sides, correct?
Yes. Areas 2 consist of a similar structure, except there’s 12 of them side-by-side due to space constraints.
Area 3 likewise consists of four stacks, with there being room for 12 side-by-side breads in each layer.
Actually, now that I think about it, areas 1 could fit a 14th bread in each layer, adding 2*4=8 breads into the system, bringing the grand total to 256 breads.
And area 3 could probably fit two more breads in each layer, the extra breads being in a different orientation as the rest to properly use up all available space in a layer, adding an additional 8 breads, bringing the total to 264.
Wait a moment, the breads are 7 cm wide, not 5? I had made my calculations assuming they are 5 cm wide. This drops the number of possible layers from 4 to 3, decreasing the number of breads rather significantly.
Using the updated numbers, it seems this strategy allows for only 198 breads in the system, which is not enough for Lothar to win the bet.
Well then let me do some fun calculations for you. So looking at the lowest level, let’s take the distance between the southernmost of the right stack and the northernmost of the left stack. We’d use Pythagoras here, so that’d be sqrt(12^2+11^2)=sqrt(265). So far so good. But now let’s take the distance between the southernmost gingerbread on the lowest layer to the right and the northernmost gingerbread on the highest layer on the left. Our triangle with a right angle would have two sides with one being the length we just calculated and the other being 14cm. The third is the distance we’re looking for: sqrt(14^2+(sqrt(265))^2)=sqrt(461), which is bigger than 21 since 21^2=441.
But it seems you realized your error yourself.
However, you only know it’s impossible for him winning the bet when using your system.
However, if the sidemost breads of areas 1 in layer 2 are stacked in an orientation that allows 7 breads in a layer of height 7 cm, even if we remove a couple of breads from the middle to account for gravity, I think it would be easy to add in the missing 12 breads. Let’s say that three breads are removed from both sides in layer 2 of area 1. This means we need to place 12+12=24 breads in these holes in our system. We can fit seven breads in each of these holes. Four holes, thus 4*7=28 breads. We arrive at a grand total of 214 breads, and gravity has been taken into account.
I suppose for the sake of beauty, four breads should be removed from each side so that the ultimate result is exactly 210 breads. That also should remove any doubt of gravity screwing things up.
You forgot to decrease the layer for area 3 by 1 I think. That eats 12 more of potential gingerbread you have to place.