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A dissection of a polygon $P$ is a partition of $P$ into a finite number of pieces, which can then be rearranged (via planar translations and rotations) and joined (without overlap) to form a new polygon $P'$. Say that a polygon $P$ has an inside-out dissection (my terminology) if $P'$ is congruent to $P$, and the perimeter of $P$ becomes interior to $P'$, and so the perimeter of $P'$ is composed of internal cuts of the dissection of $P$.

I believe every polygon $P$ has an inside-out dissection because (1) $P$ may be triangulated, and (2) every triangle has an inside-out dissection:

      IOTri
One may ask many questions concerning this concept. Here I will confine myself to three:

Q0. Has this notion been explored before, and if so, under what name?

Q1. Is there an inside-out dissection of a generic triangle using fewer than $9$ pieces?

Q2. There is a "$+$" inside-out dissection of any rectangle into $4$ pieces. What is the minimal inside-out dissection of a generic trapezoid?

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  • $\begingroup$ For acute triangles, 7 comes to mind (2 isosceles triangles at each vertex). $\endgroup$
    – The Masked Avenger
    Commented Nov 6, 2014 at 0:47
  • $\begingroup$ It also works for arbitrary polygons (2n+1) if the triangles are skinny enough. $\endgroup$
    – The Masked Avenger
    Commented Nov 6, 2014 at 0:51
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    $\begingroup$ Are you excluding vertices of the dissection that lie on the boundary of both $P$ and $P^\prime$? $\endgroup$
    – Michael Biro
    Commented Nov 6, 2014 at 1:41
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    $\begingroup$ This reminds me of hinged dissections. arxiv.org/abs/0712.2094 The classic hinged dissection of a square into an equilateral triangle is also an inside-out dissection, though not all hinged dissections are. Question: Does the Abbott et al. construction of hinged dissections yield an inside-out dissection? If not, can their result be strengthened to give a dissection that is simultaneously hinged and inside-out? $\endgroup$
    – Timothy Chow
    Commented Nov 6, 2014 at 16:30
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    $\begingroup$ Another thought: Aaron's dissection, and the square/triangle hinged dissection I mentioned above, are what you might call "perfect": Every edge of every component shape appears in the perimeter of exactly one of the two large shapes. Question: When is a perfect inside-out dissection possible? $\endgroup$
    – Timothy Chow
    Commented Nov 7, 2014 at 0:09

4 Answers 4

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Your triangle dissection into $9$ pieces can be modified:enter image description here .

As masterfully noted below, The dissection into $16$ congruent triangles similar to the big triangle can be fused into four pieces. Swap the two triangles and rotate each parallelogram $180^{\circ}.$

enter image description here

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    $\begingroup$ You can do a 4 piece variation that gets rid of pieces 4 and 5, turns 2 and 6 into congruent triangles, and leaves 1 and 3 as parallelograms. $\endgroup$
    – The Masked Avenger
    Commented Nov 6, 2014 at 2:35
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    $\begingroup$ More generally, take a polygon which tiles the plane without reflection. Offsetting this tiling by a carefully chosen epsilon vector should yield an inside out decomposition. $\endgroup$
    – The Masked Avenger
    Commented Nov 6, 2014 at 2:55
  • $\begingroup$ Brilliant! Surely 4 is the minimum for triangles. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 6, 2014 at 12:53
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Do you allow mirror reflection to count as congruent? Because my paper Hinged Kite Mirror Dissection (arXiv preprint, 2001) dissects and reassembles an arbitrary polygon to become its mirror image; all exterior edges of the original polygon become interior edges of the dissection. So this is sort of an answer to Q0. Additionally (unlike your triangulation argument) the dissection is hinged. But the paper doesn't explicitly talk about your inside-out property.

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  • $\begingroup$ Nice, David! I was retricting myself to rotations and translations, so was excluding mirrors as congruent. I had forgotten about your paper. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 6, 2014 at 10:43
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Riffing off the tiling comment to another answer, imagine a square penny packing of circles, and then translate the tiling so that a circle is in the center of four other circles. Now replace each of the five circles with a translate of a sufficiently convex (but not necessarily regular!) polygon, and you will for some polygons get a five piece inside out dissection.

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Here is another view of the @AaronMeyerowitz / @TheMaskedAvenger 4-piece inside-out dissection of a triangle:

      IO4pieces

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    $\begingroup$ Note that the parallelograms do not need to have tha same area. A follow up question: which polygons have such a three piece dkssection? I have only one candidate so far. $\endgroup$
    – The Masked Avenger
    Commented Nov 6, 2014 at 15:08
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    $\begingroup$ Since no one else has remarked on this aspect, I do so: this is not only a hinged dissection, this is a special case of a hinged trapezoidal dissection, answering Q2 of the original post. (I believe 3 pieces does not hide all of the trapezoids corners.) $\endgroup$
    – The Masked Avenger
    Commented Nov 8, 2014 at 18:35

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