cube + cube + cube = cube - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:59:47Z http://mathoverflow.net/feeds/question/53048 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53048/cube-cube-cube-cube cube + cube + cube = cube Denis Serre 2011-01-24T12:07:20Z 2013-05-07T21:50:50Z <p>The following identity is a bit isolated in the arithmetics of natural integers $$3^3+4^3+5^3=6^3.$$ Let $K_6$ be a cube whose side has length $6$. We view it as the union of $216$ elementary unit cubes. We wish to cut it into $N$ connected components, each one being a union of elementary unit cubes, such that these components can be assembled so as to form three cubes of sizes $3,4$ and $5$. Of course, the latter are made simultaneously: a component may not be used in two cubes. There is a solution with $9$ pieces. </p> <blockquote> <p>What is the minimal number $N$ of pieces into which to cut $K_6$ ?</p> </blockquote> <p>About connectedness: a piece is connected if it is a union of elementary cubes whose centers are the nodes of a connected graph with arrows of unit length parallel to the coordinate axes.</p> <p><strong>Edit</strong>. Several comments ask for a reference for the $8$-pieces puzzle, mentionned at first in the question. Actually, $8$ was a mistake, as the solution I know constists of $9$ pieces. The only one that I have is the photograph in François's answer below. Yet it is not very informative, so let me give you additional informations (I manipulated the puzzle a couple weeks ago). There is a $2$-cube (middle) and a $3$-cube (right). At left, the $4$-cube is not complete, as two elementary cubes are missing at the end of an edge. Of course, one could not have both a $3$-cube and a $4$-cube in a $6$-cube. So you can imagine how the $3$-cube and the imperfect $4$-cube match (two possibilities). Other rather symmetric pieces are a $1\times1\times2$ (it fills the imperfect $4$-cube when you build the $3$-, $4$- and $5$-cubes) and a $1\times2\times3$. Two other pieces have only a planar symmetry, whereas the last one has no symmetry at all.</p> <p>Here is a <a href="http://www.umpa.ens-lyon.fr/~serre/DPF/1ere_MataLy.tiff" rel="nofollow">photography</a> of the cut mentionned above. </p> http://mathoverflow.net/questions/53048/cube-cube-cube-cube/53806#53806 Answer by JHI for cube + cube + cube = cube JHI 2011-01-30T17:43:36Z 2013-05-07T21:50:50Z <p>8 is the least.</p> <p>You can't have a piece of length 6, thus no two corners of the 6x6x6 cube can be part of the same piece, the cube has 8 corners, we need then a minimum of 8 pieces.</p> http://mathoverflow.net/questions/53048/cube-cube-cube-cube/127710#127710 Answer by François Brunault for cube + cube + cube = cube François Brunault 2013-04-16T15:11:55Z 2013-04-16T15:11:55Z <p>The solution mentioned in the question consists in fact of $9$ pieces, as shown in the following picture I took yesterday. I don't know whether a solution with $8$ pieces exists.</p> <p><img src="http://perso.ens-lyon.fr/francois.brunault/P1000848b.jpg" alt="Solution"></p> http://mathoverflow.net/questions/53048/cube-cube-cube-cube/129754#129754 Answer by Noam D. Elkies for cube + cube + cube = cube Noam D. Elkies 2013-05-05T18:39:17Z 2013-05-06T20:36:02Z <p>JHI's elegant lower bound of $8$ on $N$ is achieved by an explicit dissection. I show my construction below; you might want to try to find a solution yourself before proceeding $-$ it makes for a neat puzzle. There may well be other ways to do it.</p> <p>If somebody can make a "$3$-dimensional" graphic or picture of the $8$-piece dissection, you're welcome to add it by editing my answer. My diagrams are two-dimensional, labeling each piece with its height. Fortunately the dissection is simple enough for this to be possible; in particular, the eight pieces comprise four boxes and four L-shaped prisms. This also made it possible to find the solution using just pencil and paper on an otherwise uneventful international flight.</p> <p>Begin by cutting the $6 \times 6 \times 6$ cube top to bottom into three pieces, as shown in top view in the first square diagram. Then cut each piece horizontally in two, dividing AB into $3+3$,$\phantom.$ C into $4+2$, and D into $5+1$. Each AB piece is then further subdivided into a box B and an L-shaped prism A. The second diagram shows (say) the bottom layer of four pieces, and the third diagram shows the top. Note that the AB subdivisions are not quite the same.</p> <p><img src="http://math.harvard.edu/~elkies/cube3456.png"></p> <p>Pieces with the same color will come together to form a smaller cube. The larger C piece is a $4$-cube, and the two A pieces form a $3$-cube as shown. It remains to construct the $5$-cube from the remaining five pieces. The last two diagrams show the bottom and top of the $5$-cube.</p> <p><img src="http://math.harvard.edu/~elkies/cube3456a.png"></p> <p>The two $5$'s are the larger B piece, rotated to span the entire height of the cube, and the thick D piece. The thin D piece completes the bottom, with width $1$. The top is filled by the thinner C piece and the smaller B, both rotated to height 4. <strong>QEF</strong></p> <p>I guess that a physical model won't make for a hard puzzle to reconstitute into either one or three cubes (e.g. the AB, C, and D parts of the $6$-cube are independent) but would still make a nice physical model of the identity $3^3 + 4^3 + 5^3 = 6^3$.</p> <p>This dissection is specific to the solution $(a,b,c;d)=(3,4,5;6)$ of the Diophantine equation $a^3+b^3+c^3=d^3$; I don't know whether an $8$-piece dissection is possible for any other solution. JHI's analysis shows that one can never get below $8$, and in some cases even that's not possible: if <code>$a&lt;b&lt;c$</code> and <code>$a+c&lt;d$</code> then there's at least one corner of the $d$-cube, say $(1,1,1)$, that contributes to the $a$-cube, but then any cell $(x,y,z)$ with $\max(x,y,z) = a+1$ cannot connect to any corner. This first happens for $(a,b,c;d) = (6,32,33;41)$.</p> <p>What's the minimal dissection for the "taxicab" identity $1^3 + 12^3 = 9^3 + 10^3$? JHI's corner-cutting argument shows that at least nine pieces are needed.</p>