Neat shearing:
You have to make one square out of the three squares (2x2, 3x3 and 6x6) as shown in the figure. How can you do this, cutting the squares into the smallest possible number of pieces?
Attribution: V. Proizvolov
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asked Oct 30 at 21:21
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5 Answers
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Here is a solution where the pieces do not need to be rotated:
answered Oct 31 at 8:33
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1$\begingroup$ This looked somehow familiar. puzzling.stackexchange.com/questions/99590/pythagorean-quilts/… $\endgroup$ Commented Nov 1 at 11:08
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I can do
5 pieces
like so:
The white bits are the two pieces I cut off the yellow 6x6 square.
I suspect this is the minimum, because
We have to cover the 4 corners, and the two smaller pieces cannot cover an entire edge of the 7x7 square. So even if we were to place the small squares in different (adjacent) corners, and cut the biggest possible (6x6) piece to cover the 2 other corners, there would still be empty space on the edge that has the smaller squares; those are more than 6 squares away from both the corners covered by the two pieces from the 6x6.
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1$\begingroup$ Aaaah! I think you got me by a few seconds! $\endgroup$ Commented Oct 30 at 22:13
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2$\begingroup$ I agree with your argument for minimality as far as it goes, but it is incomplete, as it does not handle the case where the two small squares are in diagonally opposite corners. I think that is the only case where improvement is not yet ruled out. Laska's solution has this configuration, but I don't see an easy minimality argument for it. $\endgroup$ Commented Oct 31 at 9:40
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1$\begingroup$ @JaapScherphuis I think one could argue were there only two pieces of the 6x6 then two of the 6-long sides would be preserved and there are only two columns or two rows that could hold them. From there not many configurations are left to rule out. $\endgroup$ Commented Oct 31 at 10:46
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1$\begingroup$ @Albert.Lang Ah, that's clever, I think that actually constitutes a proof already: seeing as the two intact long edges must be the outside edges of the 6x6 square, and they must be placed "back to back" against each other to fit between the smaller squares, there's no way to fill the seventh square in either of those two rows (or, symmetrically, columns). $\endgroup$– BassCommented Oct 31 at 13:12
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3$\begingroup$ Well, in principle, on could cut two "combs", for example with a 5 long backbone and three teeth of lengths 4,4,5 and then not have them back-to-back but have them interlocked by their long teeth. This fills the two long rows (or columns), but, of course, the rest doesn't work out. $\endgroup$ Commented Oct 31 at 13:31
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Here's my answer, just posting it because it's a different approach, putting the two smaller squares into opposite corners.
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You can do it in
5 pieces
by cutting
the red square like this:
and arranging the pieces like this:
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I Can do 3.
"You have to make one square out of the three squares (2x2, 3x3 and 6x6) as shown in the figure. How can you do this, cutting the squares into the smallest possible number of pieces?"
Place the 3x3 on top of the 6x6. Place the 2x2 on top of either of them.
Squares are two dimensional. In two dimensions, we end up with one square.
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$\begingroup$ Lol, I was considering posting this, but I figured it wasn't in the spirit of the question. $\endgroup$ Commented Oct 31 at 22:44
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$\begingroup$ +1 Welcome to PSE (Puzzling Stack Exchange)! Cool answer! $\endgroup$ Commented Nov 1 at 2:38
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$\begingroup$ "on top" isn't meaningful in two dimensions though. $\endgroup$ Commented Nov 3 at 8:04