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| Author |
Message |
ric
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Math Problem
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Dec 12 14:13 UTC 2000 |
A farmer has a circular pen for his donkey. The donkey's rope is tied to one
of the posts on the perimeter of the pen. The rope allows the donkey to reach
exactly 50% of the area of the pen. Assume that the donkey's reach extends
only as far as the rope does.
How long, as a percentage of the diameter of the circular pen, is the rope?
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| 20 responses total. |
flem
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response 1 of 20:
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Dec 12 15:43 UTC 2000 |
I set it up, but I'm too lazy to do the integral right now. :)
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carson
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response 2 of 20:
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Dec 12 17:08 UTC 2000 |
View hidden response.
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wjw
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response 3 of 20:
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Dec 12 20:23 UTC 2000 |
The rope is *about* 0.58 times the diameter. Looking at this number
and trying to recognize something that looks familiar, I'm going to
guess that the rope is 4/7ths of the diameter (which would be 0.571428
times the diameter). Just a guess.
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flem
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response 4 of 20:
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Dec 13 16:57 UTC 2000 |
I've got a meeting coming up in a few minutes, so maybe I'll set it up and
solve it there. I'll need something to do. ;)
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nspeed
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response 5 of 20:
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Dec 15 22:30 UTC 2000 |
I get the following monstrous equation:
4arccos(P)*P^2 + 2(pi - arccos(P)) = SQRT[(4P^2 - 1)(1 - cos[2(pi -
arccos(P)))])^2]
, where P = rope length / pen diameter and all the trigonometry is to be
carried out in radians. I suppose this could be solved graphically, or by
other means.
Problems like this are fun until you actually try to get a number..
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flem
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response 6 of 20:
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Dec 15 23:30 UTC 2000 |
Yeah. I tried the integral during that meeting, which was fun, but I ended
up with something that I needed tables to reduce, and didn't have any tables
with me. Lack of forethought, I suppose. Then I tried something clever, and
came up with something that was harder to set up, but if I'd kept to it, might
have led to a readable answer. But I didn't.
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carson
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response 7 of 20:
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Dec 15 23:40 UTC 2000 |
(it sounds as if you guys are overanalyzing the problem. try starting
with area being equal to the value of pi, multiplied by the square of
the radius. one area's half the size of the other.)
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nspeed
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response 8 of 20:
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Dec 16 02:06 UTC 2000 |
but the area the donkey can reach is not a circle, right?
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remmers
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response 9 of 20:
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Dec 16 02:23 UTC 2000 |
Right. The donkey is tied to the perimeter of the pen, so the
area that the donkey can reach will be the intersection of two
overlapping circles.
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carson
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response 10 of 20:
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Dec 16 12:10 UTC 2000 |
(right, but the area the donkey *can't* reach is a circle, and just
happens to have the same area as what the donkey *can* reach.)
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remmers
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response 11 of 20:
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Dec 16 12:38 UTC 2000 |
No, the area the donkey can't reach is a kind of crescent moon shape.
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ric
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response 12 of 20:
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Dec 16 15:22 UTC 2000 |
re 10 - that's definately not true..
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wjw
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response 13 of 20:
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Dec 16 16:34 UTC 2000 |
Well, I cheated. The answer is available on the web. In order to spare
you all the extreme temptation I will not post the exact url, but, I
will tell you that my original estimate (rope = .58 times the diameter
of the pen) is *very* close. And the sloution is not trivial or easy.
nspeed's equation looks vaguely similar to the solution I saw, but I did
not do the math to see if they are equivalent.
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ric
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response 14 of 20:
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Dec 16 18:31 UTC 2000 |
(This math problem, by the way, was posted on a BBS that I used to use back
in the mid 80's - perhaps even before I started m-netting).
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rcurl
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response 15 of 20:
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Dec 16 19:45 UTC 2000 |
I saw that it wasn't hard to set up by using the formulae for areas of
segments and sectors of circles. These give simultaneous transcendental
equations, but my need for the answer wasn't great enough to do the work.
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carson
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response 16 of 20:
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Dec 16 20:31 UTC 2000 |
resp:11 (oops, you're right. that'll teach me.)
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flem
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response 17 of 20:
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Dec 17 19:33 UTC 2000 |
re 15: Right. If I'd had the formula handy for the area of a chord, this
would have been a little simpler. Well, it still would have involved arctan,
or something equivalent. But no calc.
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janc
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response 18 of 20:
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Dec 18 02:58 UTC 2000 |
"simultaneous transcendental equations". Sometimes I love mathematical
language.
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rcurl
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response 19 of 20:
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Dec 18 03:14 UTC 2000 |
Pretty transcendental, eh?
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aruba
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response 20 of 20:
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Dec 18 04:45 UTC 2000 |
The best I can do is to say that if p is the desired ratio, and
u = arccos(2p^2-1), then sin(u) - u*cos(u) = pi/2. Or, to put it another
way,
p*sqrt(1-p^2) - (2p^2-1)arccos(p) = pi/4.
If there's a closed form for the solution, I'm interested to see it.
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