A few weeks ago, I explained that two Euclidean polygons are scissor congruent if and only of they have the same area. A scissor congruence is a dissection of the two polygons into smaller polygons (“pieces”) so that the pieces of the first polygon can be translated and rotated into the pieces of the second polygon. Then I asked whether we really need to rotate the pieces, or whether translating them is enough. For instance, can one dissect a square into pieces that can be translated into a second square that is rotated by say 45º?


 Mahlo 01

That this is possible is a consequence of a famous dissection of a square into two squares, shown above, which uses translations only, but tilts the square by an angle which we can make 22.5º. Then we need to repeat this process, tilting the other way, using a mirrored dissection.  This increases the number of pieces needed, and the question arises with how few pieces one can do this. A few years back I ran a contest about this in our department. The best solution with six pieces was found by Seth, one of our (then) undergraduates:

SolpuzzleSince then, I learned that there are solutions with only five pieces. Check them out!

Another cool example is the trans-dissection of a single pentagon into four smaller ones, by Harry Lindgren:

Fourpentagons 01

So, do trans-dissections always exist? Not at all. Let’s try to trans-dissect an equilateral triangle into a copy of it that is rotated by 180º. Suppose we found such a dissection, like the one above. Look at its horizontal edges. There are two types, the ones at the bottom of the pieces, (pointing right), and the ones at the top of the pieces, pointing left. When we add these together (taking the direction into account) for the dissected pieces, the edges in the interior cancel, while the boundary edges add up to the length of the bottom edge of the initial triangle.



Transdiss 01


Thus the oriented length of all horizontal edges of the two tiles that we want to dissect into each other need to be equal. This eliminates the possibility to rotate a triangle at all. The same argument works of course for all directions.

Therefore, in order for two polygons to be trans-scissor-congruent, they not only need to have the same area, but also have the same oriented edge lengths for all edge directions. Remarkably, these conditions are also sufficient, as was proven in 1951 by Paul Glur and Hugo Hadwiger. The argument is a little tricker than the one for general dissections, but not too bad. Maybe I’ll come back to it later.

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