Timeline for On convex polygons that can be cut into convex and mutually congruent pieces in exactly one way
Current License: CC BY-SA 4.0
7 events
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| May 25 at 17:05 | comment | added | Nandakumar R | That seems one more example - a simpler n nicer one than given in question - of a convex polygon that can be cut into 3 congruent pieces only in one way. Thanks. Further, for an arbitrary triangle, there might be only one partition into any perfect square number of congruent pieces; number of pieces =5 could be trickier. | |
| May 25 at 14:53 | comment | added | Gerry Myerson | I think a 30-60-90 triangle can be cut into three congruent pieces by cutting along the bisector of the $60$-dgree angle all the way to the opposite side, and then cutting the resulting $120$-degree angle along its bisector all the way to the hypotenuse of the original triangle. I doubt there's any other way to cut that triangle into three congruent pieces. | |
| May 25 at 14:39 | comment | added | Gerry Myerson | What I had in mind with the right triangle, if you drop perpendiculars to the midpoints of the legs from the midpoint of the hypotenuse, you get two congruent triangles, and a rectangle that can be cut into two further triangles by using either diagonal. But on closer reading of your requirements, I see the two sets of cut lines have a partial intersection with each other, so my construction doesn't count. | |
| May 25 at 9:42 | comment | added | Nandakumar R | i can't readily see how there are two ways to cut a right triangle into 4 pieces in two ways. i suspect your comment belongs better with the post: mathoverflow.net/questions/471894/… . there i had mistakenly written that there are 3 obvious ways to cut any triangle into any perfect square number of pieces (m=perfect square, n =3) and that has been edited out now; i see only n =1. thanks. | |
| May 24 at 2:37 | comment | added | Gerry Myerson | Is there any way to cut a triangle into four convex and mutually congruent pieces, other than by cuts connecting midpoints of the sides? There is another way, if the triangle is a right triangle, but in other cases? | |
| May 23 at 5:03 | history | edited | Nandakumar R | CC BY-SA 4.0 |
added 3 characters in body
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| May 23 at 3:43 | history | asked | Nandakumar R | CC BY-SA 4.0 |