Timeline for Making 7 congruent triangles from the pieces of a triangle dissection
Current License: CC BY-SA 4.0
12 events
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| Aug 20, 2020 at 15:09 | comment | added | Jaap Scherphuis | You are right however that we are not necessarily constrained to using the same number of cuts in each direction. We can use a triangular grid and scale it and skew it to make any three gridpoints match the vertices of the original triangle. By using different numbers of grid line cuts like this, you can divide the original triangle into pieces to make any number of smaller triangles, though those smaller triangles will no longer be similar in shape to the original, and I don't think you can create 7 unit triangles with just 6 cuts in any other way. | |
| Aug 20, 2020 at 14:59 | comment | added | Jaap Scherphuis | @PaulPanzer In the 5-times-smaller-square problem, the larger square splits into a central square and four right-triangles. If the triangles have legs $a$ and $b$ units in length, then the area of the whole square is $(a-b)^2+4(ab/2) = a^2+b^2$ square units. In this problem we have a central triangle surrounded by 3 triangles with legs of $a$ units along one axis and $b$ along another. These each have an area of $ab$ unit triangles, the centre triangle has an area of $(a-b)^2$ unit triangles, for a total of $a^2+ab+b^2$ unit triangles. | |
| Aug 20, 2020 at 14:22 | comment | added | Paul Panzer | @JaapScherphuis i hate to look silly but would you mind explaining your comment? Unless the origami police puts a limit on the number of auxiliary folds can't you just work your way up to any integer in steps of one? And where does the triangle formula come from? | |
| Aug 19, 2020 at 15:52 | comment | added | greenturtle3141 | Nice! Indeed I was reminded of this problem upon seeing the square puzzle. | |
| Aug 19, 2020 at 15:52 | vote | accept | greenturtle3141 | ||
| Aug 19, 2020 at 8:05 | comment | added | Jaap Scherphuis | @PaulPanzer Yes. The 5 times smaller square worked because $5$ can be written as the the sum of two squares $5=a^2+b^2$. For this triangle it works because $7$ can be written in the form $7=a^2+ab+b^2$ with integers $a$, $b$. | |
| Aug 19, 2020 at 7:58 | comment | added | Paul Panzer | So this is a variation of the five times smaller square! I suspected so, but wasn't able to work it out. | |
| Aug 19, 2020 at 7:52 | history | undeleted | Jaap Scherphuis | ||
| Aug 19, 2020 at 7:52 | history | edited | Jaap Scherphuis | CC BY-SA 4.0 |
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| Aug 19, 2020 at 7:38 | history | deleted | Jaap Scherphuis | via Vote | |
| Aug 19, 2020 at 7:29 | history | edited | Jaap Scherphuis | CC BY-SA 4.0 |
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| Aug 19, 2020 at 7:08 | history | answered | Jaap Scherphuis | CC BY-SA 4.0 |