Maria's Puzzle Book Reloaded [Puzzle 26]

Thank you :stuck_out_tongue: And I realized my embarassing typo. I meant connecting. :stuck_out_tongue:

ok i need to sleep im writing stupid shit

Also, now that we’re in the topic of simple and silly things, this isn’t really even a puzzle of any kind, just a surprising little fact.

I take a very long rope and loop it around the entire Earth by the equator. Then, I stretch the rope so that everywhere, the rope is one meter away from the surface of the Earth. How much does the length of the rope increase?

(we assume here that the Earth is a perfect sphere)

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4 pi meters, no?

Just 2 pi, actually, but yeah.

It’s surprisingly little, and the calculation makes it apparent that the increase doesn’t even depend on the radius of the object it was draped around in the first place. If the same thing was done with the moon, the result would still be roughly 6.3 meters.

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Is it possible for Lothar to win the bet or was he overconfident this time?

He is fat, overconfident and lazy. But that’s Mrs. Sugarhoney’s fall. Instead of proper parenting she spends all her time in the kitchen doing gingerbreads, maybe she should spend more time with her kid! Learn something from this!

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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.

Step 1)

Step 2)
batter

Step 3)
crepecooking

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I think I know possible solution, but I ll post tomorrow and my eyes are closing already :kinzo:

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