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Some time ago a colleague gave me a math challenge that he could not solve, and according to him others had huge problems getting it solved. I tried, and could not really come to a satisfying solution, so I am calling the forum's help. Due to lack of regular exercise own math shape is really bad ...
I would like to end up with a formula or part of a program (for HP41 or WP43s preferably) to find a solution.
The problem is the following:
A farmer has a perfectly circular piece of land with a give radius R. He wants to sell half of it to his neighbour, by putting a pole on the edge of the piece of land, and, using a wire of a certain length L tied to the pole, marks off a part of the farmers land. The pole is the center of a new circle segment. The result is a lensshaped piece of land wich has half the area of this lensshaped part is half of the original piece of land. The big question: how long is the rope L depending on the original radius R.
Looking forward to your responses ....
Meindert
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Quote:
The result is a lensshaped piece of land wich has half the area ((of this lensshaped part is half)) of the original piece of land.
Suggest deleting the words between the (added) double parentheses.
d:)
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I'm no maths expert, but this may help:
http://mathworld.wolfram.com/CircleCircleIntersection.html
Equation 14 gives the expression for the lens area. It simplifies when you place the centre of the second circle on the perimeter of the first circle by setting d=R.
Hope that helps,
Steve
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I get L = about 1.37R .
Yes, I could give ten digits as calculated by my 15C, but first I want to know whether I'm in the right ballpark. Incidentally, it didn't require a calculator until the final step to numerically evaluate an expression involving roots.
Edited: If I'm in a ballpark, it's the wrong one.
Edited: 3 Jan 2013, 9:33 p.m.
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Simple answer: The Goat Problem.
Dieter
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You could use the HP42S and combine the solver with the numeric integration.
I'm cutting the lensshaped piece vertically into two circular segments that are integrated.
Then I use the solver to make sure that the area is half of the circle with radius 1.
Using the following definitions I end up with this equation:
Now I define the two functions f(t) and g(t):
00 { 16Byte Prgm }
01 LBL "f(t)"
02 MVAR "t"
03 RCL "t"
04 ACOS
05 SIN
06 END
00 { 24Byte Prgm }
01 LBL "g(t)"
02 MVAR "r"
03 MVAR "t"
04 RCL "r"
05 X^2
06 RCL "t"
07 X^2
08 
09 SQRT
10 END
With this program the equation is calculated:
00 { 94Byte Prgm }
01 LBL "GOAT"
02 MVAR "x"
03 RCL "x"
04 2
05 *
06 COS
07 STO "LLIM"
08 1
09 STO "ULIM"
10 PRGMINT "f(t)"
11 INTEG "t"
12 STO "F"
13 1
14 RCL  "LLIM"
15 STO "LLIM"
16 RCL "x"
17 SIN
18 2
19 *
20 STO "r"
21 STO "ULIM"
22 PGMINT "g(t)"
23 INTEG "t"
24 RCL + "F"
25 2
26 *
27 PI
28 2
29 /
30 
31 END
Solve this equation for the variable x and you end up with x = 0.617948.
Use r = 2 sin(x) and you will get r = 1.158728.
This result is in accordance with the aforementioned page of the Goat Problem.
Kind regards
Thomas
Edited: 3 Jan 2013, 7:44 p.m.
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First of all, congratulations! It is a very clever reduction/synthesis.
Just for the records: after running the program, how long did it take to find the answer? Did you use a regular HP42S or an emulator?
Thanks!
Luiz (Brazil)
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I used Free42 on an iPhone and the answer comes almost immediately. As for the HP15C I used ClassicRPN on the iPhone as well. It took about 37 seconds, but I have no idea how this relates to the original calculator. In both cases the result is correct to 6 decimal places.
And then I tried to use WolframAlpha but ran out of time ...
Cheers
Thomas
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That's how I saw the problem:
In the lensshaped piece you can see an isosceles triangle with sides (1,1,r) and angles (a,a,pi2*a).
Divided in two rightangled triangles, it leads to 1*cos(pi2*a)=d and r*cos(a)=1d => d=1r*r/2.
The grazed surface can then be expressed as
S=2*(integ(sqrt(1x*x),x,1r*r/2,1)+integ(sqrt(r*rx*x),x,r*r/2,r)).
But we know that integ(sqrt(r*rx*x),x)=x/2*sqrt(r*rx*x)+r*r/2*arcsin(x/r).
So developing and simplifying S naturally leads to
S=arccos(1r*r/2)+r*r*arccos(r/2)r/2*sqrt(4r*r).
(It's just another way to the same formula as in the http://mathworld.wolfram.com/GoatProblem.html web site).
Solving S=pi/2 on r also :) leads to r=1.15872847...
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For those who prefer to use the HP15C:
001  42,21,11 LBL A 020  20 x
002  44 0 STO 0 021  43 26 PI
003  2 2 022  2 2
004  20 x 023  10 /
005  24 COS 024  30 
006  44 2 STO 2 025  43 32 RTN
007  1 1 026  42,21, 0 LBL 0
008  42,20, 0 INTEGRATE 0 027  43 24 ACOS
009  44 3 STO 3 028  23 SIN
010  1 1 029  43 32 RTN
011  45,30, 2 RCL 2 030  42,21, 1 LBL 1
012  45 0 RCL 0 031  45 1 RCL 1
013  23 SIN 032  43 11 x^2
014  2 2 033  34 x<>y
015  20 x 034  43 11 x^2
016  44 1 STO 1 035  30 
017  42,20, 1 INTEGRATE 1 036  11 SQRT
018  45,40, 3 RCL+ 3 037  43 32 RTN
019  2 2
 LBL A: GOAT
 LBL 0: f(x)
 LBL 1: g(x)
 REG 0: x
 REG 1: r
 REG 2: d
 REG 3: F
Solution:
0
ENTER
1
SOLVE A
running
0.617948
SIN
2
*
1.158728
Cheers
Thomas
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This same program and procedure works on the 34S, although slowly (probably the solver again). It gives 1.158728473020422 and is incorrect by one in the 12^{th} digit. The correct answer being 1.1587284730181215...
I also suspect that calculating would be faster done directly instead of as SIN(COS^{1}(t)).
 Pauli
Edited: 3 Jan 2013, 10:33 p.m.
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Or to avoid extinction when x is close to 1:
On the Free42 you can set the variable "ACC" to 1e20 and you will end up with 25 correct places. Though I wonder what we could do with these.
Cheers
Thomas
Edited: 3 Jan 2013, 11:35 p.m.
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Yes, that is a better way to calculate it  wasn't really thinking about stability when I wrote my note.
 Pauli
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As you may know the HP34C (and all successors) uses a Romberg method with nodes that are spaced nonuniformly using a substitution:
Details can be found in this section of William M. Kahan's article:
Quote:
Second, can be calculated efficiently when
or where g(u) is
everywhere a smooth function, without any of the expedients
that would otherwise be required to cope with the infinite
values taken by the derivative f'(u) at u = x or u = y. Such integrals are encountered often during calculations of areas enclosed by smooth closed curves. For example, the area of a circle of radius 1 is
which consumes only 60 seconds when evaluated in SCI 5
and only 110 seconds to get 3.141592654±1.4x10^{9}
in SCI 9.
So it appears HP's numeric integration is well suited for this challenge.
Kind regards
Thomas
Edited: 4 Jan 2013, 1:23 a.m.
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Quote:
This same program and procedure works on the 34S, although slowly (probably the solver again). It gives 1.158728473020422 and is incorrect by one in the 12th digit.
Strange. I just tried the same on a (hardware) 34s and I finally got 1,158728473018121  which is exact within 1 ULP. The calculator was set to ALL mode, of course.
Dieter
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Build number? Program listing? There must be reasons for the different results, so I'd appreciate getting these data from you and Pauli.
d:)
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Version = 3.1 3325, and the program was the same as the 15C version posted in this thread. OK, I used LBL B, 55 and 66 instead. ;)
Dieter
Edited: 4 Jan 2013, 1:57 p.m.
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I was running 3.1 3225  100 revisions older. I've been slack and not updated that one for ages.
 Pauli
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Now, that's even stranger. I still have an emulator with version 3.1 3225 here. Entered the same program, got the same (exact) result: 1,158728473018121. So something else must be different. Which display mode did you set?
Dieter
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Unknown anymore, I've been making changes to the display code this morning. It was probably ALL something. I was running in single precision.
 Pauli
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Thank you all for your responses!
Meindert
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Let me thank you for the challenge. And Dieter for pointing out that this is the "Goat Problem". I'm always amazed by this forum. The last time I posted a challenge: zing and the answer was that this is "Kaprekar's constant". Never heard of that before.
Kind regards
Thomas
