Timeline for Subdivision of pentagon into six congruent pieces
Current License: CC BY-SA 3.0
11 events
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| Dec 16, 2022 at 4:11 | comment | added | user44143 | One could also ask: What is the most regular pentagon that can be divided into six congruent pieces? Eg the pentagon with vertices at $(\pm 1,-2),(\pm 3,1),(0,3)$has angles which range from $90^\circ$ to $125^\circ$, has four equal sides, and can be decomposed into six equal pieces. | |
| Oct 11, 2022 at 20:22 | comment | added | Wlod AA | @AaronMeyerowitz, Gardner used a misdirection. Otherwise, there is no "aha" moment about getting n congruent pieces, it'd be totally and instantly obvious to about everybody. | |
| Oct 11, 2022 at 16:56 | answer | added | Oscar Lanzi | timeline score: 4 | |
| Jun 20, 2018 at 21:43 | comment | added | Aaron Meyerowitz | I think Gardners point was that here are a plethora of ways to get $4$ congruent pieces, for example a path from the center to a side and its rotations by $90$ degrees. Somehow, knowing that makes it hard to see the obvious (and perhaps only) solution for $5.$ The fun is the aha moment. | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 2, 2016 at 14:17 | comment | added | Per Alexandersson | @Steve: No, in "My Best Mathematical and Logic Puzzles", by Garner, problem 57, it is stated verbatim "Your task is to divide the blank square into five parts, all identical in size and shape." The solution is the one you pointed out. I think Gardner is just trolling the reader with this puzzle. | |
| Nov 2, 2016 at 13:32 | comment | added | Steve | @PerAlexandersson, in your post, you say one of MG's favorite problems is to show that "one can divide a square...into five congruent and connected pieces." What I am saying is that there must be some additional constraint (at least to be a favorite problem of MG's), since, as you say, this is easy. | |
| Nov 2, 2016 at 1:47 | comment | added | Per Alexandersson | @Steve I do not follow your solution - it is about subdividing the pentagon, not the square. The square, as you say, is quite easy to divide into any number of congruent pieces. | |
| Nov 2, 2016 at 1:07 | comment | added | Steve | There must be some further condition than congruent and connected--why not just divide one axis of the square evenly into $n$ segments along one axis and cross them with the other axis? Voila, $n$ congruent and connected pieces. | |
| Oct 31, 2016 at 5:56 | comment | added | Aaron Meyerowitz | I'm not sure it is the natural question (although it is a good one, I'd guess no.) The first fact concerns the $1$ dimensional simplex which is also the $1$ dimensional hypercube and cross-polytope. A regular $n$-gon can be dissected in may ways into $n$ congruent pieces. An arbitrary triangle can be naturally dissected into $4$ congruent triangular pieces. Seeing the natural ways to dissect a rectangle into $4$-pieces it is an aha moment to see how to get $5.$ The natural question (which might not be that hard) is about simplices and boxes in $3$ and more dimensions. | |
| Oct 31, 2016 at 2:21 | history | asked | Per Alexandersson | CC BY-SA 3.0 |