Timeline for To cut a triangle into $n$ $p$-sided polygonal regions
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| Dec 10 at 14:17 | history | edited | Nandakumar R | CC BY-SA 4.0 |
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| Dec 10 at 10:56 | history | edited | Nandakumar R | CC BY-SA 4.0 |
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| Nov 21 at 17:40 | history | edited | Nandakumar R | CC BY-SA 4.0 |
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| Jan 14, 2022 at 12:28 | vote | accept | Nandakumar R | ||
| Mar 23, 2021 at 6:17 | history | edited | Nandakumar R | CC BY-SA 4.0 |
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| Feb 1, 2021 at 1:09 | answer | added | Gerry Myerson | timeline score: 5 | |
| Jan 31, 2021 at 13:19 | history | edited | Nandakumar R | CC BY-SA 4.0 |
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| Jan 31, 2021 at 13:13 | history | edited | Nandakumar R | CC BY-SA 4.0 |
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| Jan 31, 2021 at 8:21 | history | edited | Nandakumar R | CC BY-SA 4.0 |
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| Jan 30, 2021 at 22:59 | comment | added | Gerry Myerson | ... (and finishing off with two nonconvex pentagons), which gives you any number of convex pentagons exceeding five. | |
| Jan 30, 2021 at 22:57 | comment | added | Gerry Myerson | You can dissect a triangle into four convex pentagons and a triangle as follows: draw a small, inverted triangle inside the starter triangle, draw line segments connecting two vertices of the small triangle to the nearest sides of the starter, draw a short line segment out from the third remaining vertex of the small triangle, then connect the free end of that short line segment to the same two sides of the starter. Now by iterating this construction or the earlier pentagonal construction on the small triangle you should be able to get any number of convex pentagons of the form $3a+4b$ .... | |
| Jan 30, 2021 at 16:42 | comment | added | Nandakumar R | ...Of course, in the original problem statement, the areas of pieces are not required to be equal but even if such a constraint has to be kept, the earlier quadrilateral and pentagonal pieces constructions appear to go thru - and so does the floor(n/2) convex pieces construction that appears to work for any n and p. | |
| Jan 30, 2021 at 16:42 | comment | added | Nandakumar R | Thanks! I could follow this nifty construction. It does give n-1 convex hexagons but it has the seeming limitation that the number of hexagonal pieces has to be a multiple of 3 (btw, this issue seems to apply to the pentagonal pieces construction as well). Moreover, here, we can't achieve an additional equal area requirement on the pieces, it seems. ...(contd) | |
| Jan 30, 2021 at 11:07 | comment | added | Gerry Myerson | (continued) in the previous step. So now we have three more convex hexagons, and one nonconvex of the sort we just dissected. Now iterate. | |
| Jan 30, 2021 at 11:05 | comment | added | Gerry Myerson | I think you can still do an arbitrarily large number of convex hexagons, with only one nonconvex hexagon. It's a little harder to describe in words. starting with a triangle, draw a short line segment into the triangle from the midpoint of each side. Connect the free end of each of these short line segments to each of the others by a chain of two line segments. We now have three convex hexagons, and one nonconvex,with alternating acute and reflex angles. At each of the reflex angles, draw a short line segment pointing inward, and then connect the ends of those segments as was done (continued) | |
| Jan 30, 2021 at 6:38 | comment | added | Nandakumar R | Thanks again for that neat construction! Still am unable to see such constructions that work from p=6 (hexagonal pieces) upwards. So, for sufficiently large p, we might not be able to get the number of convex pieces close to n... | |
| Jan 29, 2021 at 21:49 | comment | added | Gerry Myerson | Even for pentagons you can get arbitarily large $n$ with only two non-convex. Put a small triangle, upside down, inside your starter triangle, and connect each vertex of the small triangle by a line segment to a point on the nearest edge of the starter triangle. That gives you three convex pentagons and a triangle. Now do the same to the small triangle, and iterate until you get tired. Finish off by cutting the tiny remaining triangle into two non-convex pentagons by drawing a crooked line from a vertex to the opposite side. | |
| Jan 29, 2021 at 14:05 | comment | added | Nandakumar R | Thanks for pointing this out. Guess from pentagons upwards (maybe for sufficiently large p), floor(n/2) might well be a tight upper bound. | |
| Jan 29, 2021 at 14:03 | history | edited | Nandakumar R | CC BY-SA 4.0 |
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| Jan 29, 2021 at 12:10 | comment | added | Gerry Myerson | For any $n$, any triangle can be cut into $n$ quadrilaterals of which $n-3$ are convex. | |
| Jan 29, 2021 at 9:06 | history | edited | YCor | CC BY-SA 4.0 |
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| Jan 29, 2021 at 8:59 | history | asked | Nandakumar R | CC BY-SA 4.0 |