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If you take a cube, and grow a new cube out from each of its six faces, you will get a "hyper plus sign":

enter image description here

This 3D solid has an interesting property. It can be sliced along its edges and unfolded into a single 2D shape that can then be re-folded to perfectly cover the surface of a cube. Your challenge is to show how this can be done.

You must:

  1. Show how to cut the surface of the above 3D solid along its edges so the entire surface can be unfolded into a single 2D shape.
  2. Then show how to re-fold this 2D shape onto the surface of a cube.

Some rules and clarifications:

  • The original 3D solid may only be cut along its edges.
  • You must unfold the entire surface of the original solid, in one continuous piece.
  • The 2D shape must cover the entire surface of the cube, with no gaps and no overlaps.
  • The edges of the 2D shape do not need to line up with the edges of the final cube. (If this was a requirement, the puzzle would be unsolvable.)
  • I am aware of one two basic solutions, but there are endless trivial variations of these solutions possible. Your solution does not need to look identical to mine. It just needs to meet all of the requirements above.
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  • $\begingroup$ The first step is to calculate the length of the side of the final cube. After that it is pretty clear how it would look like. $\endgroup$
    – WhatsUp
    May 26 at 13:11
  • $\begingroup$ Side of $\sqrt 5$ which is equal to length (over surface) between two opposite corners of a small 1 sized cube. $\endgroup$
    – z100
    May 26 at 15:02

2 Answers 2

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Phew! To start,

we're blowing up each cube face into five identical faces, which multiplies the area by 5 and the side length by $\sqrt{5}$. A good way to start would be to take a grid and superimpose it on another one scaled by $\sqrt{5}$ and rotated to line up nicely, so that's what I did. (There's another puzzle on here that illustrates how to do this, but I can't find it...) Initially I tried nets that folded outward from one of the smaller cubes' centers, but after some unsalvageable close calls I got the bright idea to start from an edge joining two subcubes and look for rotationally-symmetric solutions.

After much trial and error, I got a net that looks like this:

enter image description here

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    $\begingroup$ Well done! This is a surprisingly different solution from the one I found. (Yours is a bit nicer than mine, honestly.) When time permits I'll try and post mine as a supplementary self-answer. (but, of course, the check mark is yours, and is well-earned) $\endgroup$ May 26 at 22:00
  • $\begingroup$ I think that in the raw version, you accidentally cut the topmost tip of the brown side. Also, the bottom third of the "Cut Flat" drawing is missing the horizontal black lines. Further, making the "Cut Flat" drawing touching the "Folded into large cube" makes it a bit confusing. $\endgroup$ May 29 at 1:40
  • $\begingroup$ All consequences of working in Paint; not much I can do without remaking the entire image from scratch. $\endgroup$ May 29 at 18:47
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I'm posting my own solution as a self-answer here because it's quite different from the solution posted by AxiomaticSystem. It seems worthwhile to have both of them recorded for posterity. (But honestly, of the two of them, I like AxiomaticSystem's solution better.)

Solution as flattened 2D surface. Solution as 3D before/after solids. Before/after 3D solids from below.

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  • $\begingroup$ This was actually my initial idea, but I used T-nets instead, which meant that extending into the "bottom cube" grabbed the wrong squares. $\endgroup$ Jun 1 at 3:28
  • $\begingroup$ @AxiomaticSystem rot13(V unq frireny snyfr fgnegf, jurer V hajenccrq gur ulcre cyhf fvta naq gura jbhaq hc jvgu n funcr gung pbhyqa'g or jenccrq nebhaq gur phor. Jung svanyyl jbexrq sbe zr jnf gb jbex ba obgu fbyvqf va cnenyyry, nqqvat rdhvinyrag phg yvarf gb obgu, naq gura onpxvat hc jurarire rvgure bs gurz jnfa'g jbexvat. V zbfgyl yvxr gur sbhe-nezrq ybbx urer, ohg V nqzvg gung bar rkgen flzzrgel-oernxvat fdhner xvaq bs ohtf zr.) $\endgroup$ Jun 2 at 3:09

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