9
$\begingroup$

A square is halved along a diagonal, creating equal halves A and B. Prove that it is not possible to dissect one half into a finite number of pieces, and through translations only (i.e. no rotations, skews, etc), superimpose the pieces of one half onto the other half.

This was a challenge that was asked as part of Russian elementary school Olympiad, no high-level maths required... had seen the proof a while back, but can no longer remember. Thanks in advance for your help!

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Welcome to Puzzling. Maybe this belong to the math site? $\endgroup$
    – lois6b
    Oct 4 '16 at 10:50
4
$\begingroup$

I don't know how to word this mathematically accurate, but the general idea is the following:

Let's try to dissect the half A to match it with B. Since the number of pieces is finite, the borders of B will be composed of a finite number of pieces, all of which will be straight lines. Now, if we dissect the part B, all each straight cut (even straight parts of a more complex cut) will produce two straight borders for the new pieces, they will be of equal length but of opposite direction (the direction of a straight border part is a vector that's perpendicular to it and points outwards). So the total (added for all pieces) balance of straight borders directions (e.g. 1 cm "north" border, 1 cm "west" border and sqrt(2) "southeast" border) will not change no matter how we cut. When we join the pieces together, it won't change also since these borders will negate each other when joining.

So, with such operations we can only create a figure with the same border balance (opposite directions negate each other when calculating) as the initial figure. But the other half of a square has completely different stats: 1 cm "south" (or -1 "north" for comparison, since these directions are opposite), 1 cm "east" (-1 "west) and sqrt(2) "northwest" (-sqrt2 "southeast"). So we cannot turn one half into another.

The wording is far from being perfect, but I hope that the idea is clear.

Improve this answer
$\endgroup$
5
  • $\begingroup$ The problem I see that it is unclear how to assign three directions uniquely to a piece. But I also think the idea of assigning an additive invariant to pieces should eventually work. $\endgroup$
    – Matsmath
    Oct 4 '16 at 11:04
  • $\begingroup$ @Matsmath You don't assign directions to a piece. You assign directions to each straight part of a border. And then you get a list of directions for each piece. $\endgroup$
    – Verence
    Oct 4 '16 at 11:18
  • $\begingroup$ Basically to any polygon you associate a value in $\oplus_{\mathbb{R}P^1}\mathbb{R}$ which remains invariant unter translative dissections. Since the two triangles have different dissections, you cannot dissect one into the other. $\endgroup$
    – Anon
    Oct 4 '16 at 13:04
  • $\begingroup$ Remark: McFry's way of looking at it is awfully similar to the way that you prove the (much harder) theorem that equal-volume polyhedra, unlike equal-area polygons, aren't interdissectible even if rotations as well as translations are allowed. The term to look up, for anyone curious about this, is "Dehn invariant". $\endgroup$
    – Gareth McCaughan ♦
    Oct 6 '16 at 1:47
  • $\begingroup$ (Er, I should really say "equidecomposable" rather than "interdissectible".) $\endgroup$
    – Gareth McCaughan ♦
    Oct 6 '16 at 1:49
3
$\begingroup$

Suppose we are mapping the top half of the square to the bottom. Consider the top edge.

There is at least one piece on the top edge. If you can't rotate it, how would you place that piece? It can only be matched with a piece with a horizontal bottom edge. But the top half-square doesn't have a such an edge. You can only make a new piece with such an edge, by making a horizontal cut. But when you remove the bottom-edge piece, you are left with a new top-edge piece, and where does that piece go?

And so on until you run out of pieces -- the number of which is finite, so no weird infinity tricks.

Improve this answer
$\endgroup$
2
$\begingroup$

The following assumes that the pieces we dissect into have piecewise smooth boundaries. This is less specific than required by Verence's solution (which he posted while I was writing this) but still more specific than I'd like.

Given any finite set of regions with piecewise-smooth boundaries, consider the set S defined as follows: a direction is in S iff there are an odd number of region-corners from which that direction points "inward" into the region. So, e.g., if your square's sides lie parallel to the coordinate axes and you split it along the NW-SE diagonal, then the set S for each half consists of one quadrant and two octants, and the two halves' direction-sets are complements. (Aside from a little ambiguity about directions pointing along the boundaries.)

Now the point is that dissection doesn't change S (aside, again, from a finite number of directions pointing directly along the boundaries) and that translating regions around obviously doesn't either. But the two halves of the square have completely different choices of S.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What if I make a cut along a curve which is of infinite length (e.g. a snowflake)? $\endgroup$
    – Matsmath
    Oct 4 '16 at 15:36
  • $\begingroup$ Doesn't meet the usual technical definition of a piecewise smooth curve, though admittedly I'm being a bit cheeky here because there are two ways to interpret "piecewise smooth boundary" and this is true only for one of them. It might be clearer to say something like "boundaries are the union of finitely many smooth curves". $\endgroup$
    – Gareth McCaughan ♦
    Oct 4 '16 at 17:56
0
$\begingroup$

This is hard but i will try to explain it as simply as possible.

1-100% of the first half must be put perfectly in the second half.
2-No matter how tiny you cut the pieces, you will need a triangle with the proper angles and direction to fill the corners of the second half.
3-The corners of the first half cannot be used to fill the corners of the second half.
4-Thefore, adequate triangles would have to be cut from the center of the body of the first triangle to fill the corners of the second half.
5-but doing so would create new holes in the same angles and direction that cannot be filled.(no matter how tiny you cut pieces and move them over, the angle and direction of those holes you made will remain and will have to be filled, which can only be done by cutting more identical pieces.)
6-It is therefore impossible to do.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.