| 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 |
Someone asked me if there is an elegant solution to the following
problem;
.circle#1; Radius r1
.Draw a second circle#2 with its center on the circumference
of circle#1
.circle#2; Radius r2
Now find the radius r2 such that area under the intersection of
circle#1 and circle#2 is equal to the remaining area of circle#1
not in the intersection.
If I could come up with an expression for the area in the intersection
of the two circles I would be all set.
Thanks,
Jon
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1469.1 | my attempt, r2 found in term of sector angle | SMAUG::ABBASI | Tue Jul 09 1991 09:32 | 47 | |
2
expression for intersection area is offcourse 1/2 Pi r1
which is half the area of circle 1 (this is given by problem description)
to find expression for r2 from this is not too easy (to me). closest
i get is:
|
|
. . . oa=ob=oe= r1
. | . eh= r2 = eb=ea
. h .
------.--x-o-x--.------------------
ax | xb
x . | . x
x e x
x | x
x | x
x | x
x
draw a line from o (origin of circle 1) to b and a, draw a line ab.
let angle aob be called 2*B
2 2 2
then 1/2 Pi r1 = r2 (Pi-B) + r1 B - 2 * K
-----
2
k = 1/2 r1 Y
2 2 2
Y = 2 r1 - 2 r1 cos (2B)
what i did is find another way of figuring the interstion area, which is
adding sector covered by r2 as it moves inside the intersection area, and
sector covered by r1 as it sweeps also inside the intersection area, and
to compensate for duple addition of area oaeb i subtract this area one time
away and that is given by 2*K.
so r2 is found in terms of internal angle B. not sure if there is direct way
to find anlge B (aob), there is probably a direct way to find B, but i just
ran out of ink (good one, right :-)
/Nasser
| |||||
| 1469.2 | GUESS::DERAMO | duly noted | Tue Jul 09 1991 09:44 | 5 | |
I think that this problem is already in this conference
somewhere, and that the solution was numerical, i.e.,
r2 = <some number to a bunch of decimal places> * r1.
Dan
| |||||
| 1469.3 | a little more clarification on angle | SMAUG::ABBASI | Tue Jul 09 1991 10:07 | 19 | |
ref .1(me)
if there is angle aob (=2B), in terms of r1 only, this could be
solved in closed form.
o
/|\
r1/ | \r1
/ | \
a---|---b oe= r1
\ | /
r2 \|/ r2
e
ref .1 (Dan), I'll definitly would like to see that solution you
mentioned. i'll scan later for it.
| |||||
| 1469.4 | GUESS::DERAMO | duly noted | Tue Jul 09 1991 11:16 | 5 | |
I think it's phrased in terms of how long a rope or cord
must be for the goat tied to the pole to graze half of
the grass in the circle.
Dan
| |||||
| 1469.5 | See 907.* | ELIS::GARSON | V+F = E+2 | Tue Jul 09 1991 14:22 | 0 |
| 1469.7 | search /note=*.* goat | CIVAGE::BUCHANAN | Thu Jul 11 1991 19:09 | 3 | |
And 1284.* as well. This goat just keeps coming back!
Andrew.
| |||||