| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I saw this one long ago: Given ANY triangle, find a way to partition it
(is this the correct English word?) into triangles with "sharp" angles
(LESS than 90 deg.) ONLY. What's the minimum number of triangles?
NITSAN
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 210.1 | HARE::STAN | Thu Jan 17 1985 01:12 | 3 | ||
"partition" is a good word. "acute" would be better for "sharp". | |||||
| 210.2 | METOO::YARBROUGH | Thu Jan 17 1985 14:04 | 7 | ||
I believe the number is seven. It's hard to display here, but maybe I can give a clear verbal description. There is at most one obtuse angle. Contruct a pentagon with one side in the middle of the long edge opposite the obtuse angle, one side on each edge joining at that angle, and the last two sides connecting the loose ends. Join the five corners of the pentagon near the middle. The construction is quite loose and the points can always be chosen so that all angles are <90. | |||||