New answers tagged geometry
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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, ...
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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.
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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:
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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.
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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
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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!
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My answer:
More explanation as requested:
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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 ...
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Another way, with more geometric, then:
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Another answer, slightly simpler
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Here is the answer drawn to scale.
Analysis:
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Glorfindel has found the answer; here is a uniqueness proof.
Now
And now
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1
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 ...
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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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