| 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 |
Here's a quickie:
Draw a chord thru a circle which subtends a small arc. Now draw two more
chords from the endpoints of the arc to its midpoint, forming a rather
flat isosceles triangle.
What is the limiting ratio of the area of the isosceles triangle to the
area bounded by the long chord and the arc as the subtended angle
goes to zero? Can you correctly guess the limit without pencil and
paper?
- Jim
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 489.1 | 1 ? | ROXIE::OSMAN | and silos to fill before I feep, and silos to fill before I feep | Wed May 14 1986 12:30 | 15 |
My guess is 1. For real small angles, the arc itself is hardly longer than the chord, and almost as straight. Gee, it's times like these that I'd just LOVE to be able to draw a sketch in my reply ! Come to think of it, we're in the dark ages in that we can't sketch with the NOTES system. I mean, the technology is definitely here in terms of sketch-resolution tubes, the main problem is the archaic vt100's that so many people have AND the missing mouse or tablet with which sketches would be possible. *sigh*. Especially in a "math" notes file, sketches are badly needed. /Eric | |||||
| 489.2 | LATOUR::JMUNZER | Tue May 20 1986 14:05 | 4 | ||
Three-quarters? Seems like a funny answer, and I need paper to
justify it.
John
| |||||
| 489.3 | ENGINE::ROTH | Tue May 20 1986 20:33 | 6 | ||
Nice guess... did you think of the chord as a flat parabola?
[I guessed either 0 or 1, so much for intuition, and only got the
right answer the pedestrian way.]
- Jim
| |||||
| 489.4 | LATOUR::JMUNZER | Wed May 21 1986 10:34 | 10 | ||
Sorry, .2 was a (pedestrian) calculation, not a (remarkable) guess.
A pleasant way to view the problem is:
[1] Think of the n-sided regular polygon sitting in the circle, P(n)
[2] n triangles total: P(2n) - P(n)
[3] n curved areas total: circle - P(n)
[4] The answer is the limit of: [P(2n) - P(n)] / [circle - P(n)]
John
| |||||