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Triangle ABE is cut from square ABCD (centre E) and placed at one side. Restore the square by a single action different from the exact reversal of this transformation. A solution is to cut triangle BCF from the shape on the right and to use it to fill the gap AEB. Find an additional, completely different, way of cutting the shape on the right into two pieces that can be repositioned to form a square.

enter image description here

Source: https://thatsmaths.com/ where Peter Lynch records that Thane Plambeck got this problem from David Klarner who got it from N G de Bruijn. The ultimate source is unknown.

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    $\begingroup$ hopefully it is not lateral-thinking question, like cutting EC to find a square :) $\endgroup$ – Oray Apr 3 at 10:06
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    $\begingroup$ No lateral-thinking, looking for a bona fide solution. $\endgroup$ – Tom Apr 3 at 10:16
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    $\begingroup$ 'cut triangle BCF' is just 'exact reversal of this transformation' right? $\endgroup$ – newbie Apr 3 at 10:28
  • $\begingroup$ @newbie, yes, instead look for an additional, completely different solution. $\endgroup$ – Tom Apr 3 at 10:33
  • $\begingroup$ What are the constraints on the cut? Can it curve? Can it self-intersect? $\endgroup$ – MooseBoys Apr 3 at 21:23
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Apologies for the crude drawing but I think you need to do something like this

enter image description here
where $G$ is on the extended line of $BE$ such that $|BG| = |BC|$.
$H$ is on $BC$ and $GH$ is parallel to $CD$.
Then move the shape $BGHCF$ so that $G$ goes to $A$, $H$ goes to $E$, $C$ goes to $G$, $F$ goes to $H$. I think it's not too difficult to verify this works as there are only three or four different lengths in play.

Some deduction that went into this answer

To create the original shape, the right angle at $B$ was split in half and moved. I figured that, in the new dissection, we should either preserve the right angle at $B$ and use it as one of the corners of the reassembled square or preserve the newly created right angle at $F$ to use as a square corner.
Because the line $BE$ extends all the way to $D$, I thought it would be easier to make a cut in this direction equal to the side length of the square (so that $B$ could be used as a corner).
With the angle $\angle DCF$ being $\frac{3 \pi}{4}$ it meant that the end point of the cut would have to be at $C$. This means that $F$ would end up part way along one of the sides and the fourth right angle for the square would have to appear from the cut at $C$. There would then have to be another right angle to join to the one at $F$ to form a straight edge. It didn't take too much playing around to see how all of this could be achieved.

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    $\begingroup$ Excellent, and I'm really happy you answered as I didn't know a solution! $\endgroup$ – Tom Apr 3 at 14:23
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    $\begingroup$ Great answer and you've beaten me to it. It's maybe worthwhile to extend the answer with some logical deduction steps that lead to it. f.e. the final square must have the same side-lengths, hence AD-DC must remain as they are.. etc. $\endgroup$ – BmyGuest Apr 3 at 17:04
  • $\begingroup$ @BmyGuest I've tried to add some of my reasoning to get to this answer. Hope it makes sense. $\endgroup$ – hexomino Apr 3 at 18:28
  • $\begingroup$ Ah, thanks Tom, that should have been DCF $\endgroup$ – hexomino Apr 3 at 19:30
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    $\begingroup$ It really is a great answer (and puzzle). The original dissection rotated a triangle through 90 degrees, but here it is done by rotating the triangle in two steps of 45 degrees. The same puzzle would generalise to a regular pentagon or hexagon instead of a square. $\endgroup$ – Jaap Scherphuis Apr 3 at 19:38

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