| 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've not looked at this conference before, and I have to say that most
of the discussions here are a bit complex for me. In addition, I
apologise if this kind of problem has been entered before. It's
something I thought up a few years ago, and I've never really worked
out how to solve it. Any ideas?
Take a circle, and then draw an arc within it which touches the
circumference at each end and whose centre is on the circumference such
that the area within the circle is the same on each side of the arc.
What is the radius of the arc expressed in terms of the radius of the
circle?
[If that description above is not clear enough let me know and I'll try
to make it clearer]/
Thanks,
Keith
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1284.1 | lucky guess | HERON::BUCHANAN | combinatorial bomb squad | Thu Aug 09 1990 15:48 | 13 |
Well, dir/title=<anything sensible> wouldn't get you anything, but my second guess of dir /title=goat hits the jackpot. Note 907 treats this problem. So far as I can tell, they didn't got an answer though. I have just a moment before I head off to dinner, but it seems clear that integration is not necessary. You can get the area munched by the goat by adding together the area of two arcs, and subtracting the intersection (a kite) to avoid counting it twice. Then equate this to �pi*r�. Thus you have a transcendental equation for the length of the tether which one can only solve numerically. regards, Andrew. | |||||