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Florian F's 2nd pattern is far and away my favorite, but if anyone was curious, I'll post my answers. First I wanted to show an example of something that doesn't work but really seems like it should: It's just like the third example from the question, but it uses joinery such that the pieces slide together at an angle. It comes apart in the same way, ...


Although I was too lazy to get the angles exact, the idea should hold in principle if the picture isn't quite right. They interlock.


To a real-life problem I had to give a real-life answer: But you asked for an actual tiling, without gaps, so here it is. PS: there is a simpler pattern where pairs disassemble with a single translation:


How about the Any two adjacent pieces can be slid apart, but I don't think the whole thing can be split by sliding from any direction.


This is just for the teaser question (do not know the answer for the real one). I think you can pull the puzzle apart by


Maybe this works? Note: I am using MS Excel to finish this. The grids are not perfect squares, so they, when analysed using this image, may not be accurately identical. Yet I hope you all get the idea. Thanks!


My answer: More explanation as requested:


Also, $6^2 = a^2 + c^2 $


No. The rectangle has an area of $75$ square units. The only other rectangle with integer sides that would have the same area would be $3*25$ (or $1*75$ technically, I guess). The rectangles you have are $3*6, 2*3, 3*2, 2*4, 4*2, 10*2$ and $9*1$. The $9*1$ needs a $9*2$ paired with it to fill those rows and prevent leaving a $1*x$ space that cannot be ...


Another way, with more geometric, then:


Another answer, slightly simpler


Here is the answer drawn to scale. Analysis:



Glorfindel has found the answer; here is a uniqueness proof. Now And now


It looks like Explanation:


Triangle case: If the triangle has three weights, we can subtract the smallest value from each side. That reduces to the case of only two unknown weights and a zero in the third pan. The original problem defines the maximum weight as 40. Rather than examine this exact case, I assume an arbitrarily large MAX weight and solved for that. This will give an ...


Suppose we label the corner on the table like this: Now we want to move from $A$ to $D$. Now, imagine the table like this: Here, Now, to hits all $4$ edges, that means To get the shortest path, Thus, Which is like this:

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