Math Challenge I could not solve

[Meindert Kuipers]

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 lens-shaped piece of land wich has half the area of this lens-shaped 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

[Walter B]

Quote:

---

The result is a lens-shaped piece of land wich has half the area ((of this lens-shaped part is half)) of the original piece of land.

---

Suggest deleting the words between the (added) double parentheses.

d:-)

[Steve Hunt]

I'm no maths expert, but this may help:
http://mathworld.wolfram.com/Circle-CircleIntersection.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

[Neil Hamilton (Ottawa)]

This would be link below for those of you who, like me, primarily use a handheld device to read this site and cannot easily cut and paste:

http://mathworld.wolfram.com/Circle-CircleIntersection.html

All the best...

[Kiyoshi Akima]

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.

[Dieter]

Simple answer: The Goat Problem.

Dieter

[Thomas Klemm]

You could use the HP-42S and combine the solver with the numeric integration.

I'm cutting the lens-shaped 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 { 16-Byte Prgm }

01 LBL "f(t)"

02 MVAR "t"

03 RCL "t"

04 ACOS

05 SIN

06 END

00 { 24-Byte 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 { 94-Byte 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.

[Luiz C. Vieira (Brazil)]

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)

[Thomas Klemm]

I used Free42 on an iPhone and the answer comes almost immediately. As for the HP-15C 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

[PGILLET]

That's how I saw the problem:

In the lens-shaped piece you can see an isosceles triangle with sides (1,1,r) and angles (a,a,pi-2*a).

Divided in two right-angled triangles, it leads to 1*cos(pi-2*a)=d and r*cos(a)=1-d => d=1-r*r/2.

The grazed surface can then be expressed as

S=2*(integ(sqrt(1-x*x),x,1-r*r/2,1)+integ(sqrt(r*r-x*x),x,r*r/2,r)).

But we know that integ(sqrt(r*r-x*x),x)=x/2*sqrt(r*r-x*x)+r*r/2*arcsin(x/r).

So developing and simplifying S naturally leads to

S=arccos(1-r*r/2)+r*r*arccos(r/2)-r/2*sqrt(4-r*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...

[Thomas Klemm]

For those who prefer to use the HP-15C:

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

[Paul Dale]

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.

[Thomas Klemm]

Or to avoid extinction when x is close to 1:

On the Free42 you can set the variable "ACC" to 1e-20 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.

[Paul Dale]

Yes, that is a better way to calculate it -- wasn't really thinking about stability when I wrote my note.

- Pauli

[Thomas Klemm]

As you may know the HP-34C (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:

What Method Underlies the Integrate Key?

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.

[Dieter]

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

[Walter B]

Build number? Program listing? There must be reasons for the different results, so I'd appreciate getting these data from you and Pauli.

d:-)

[Dieter]

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.

[Paul Dale]

I was running 3.1 3225 -- 100 revisions older. I've been slack and not updated that one for ages.

- Pauli

[Dieter]

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

[Paul Dale]

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

[Meindert Kuipers]

Thank you all for your responses!

Meindert

[Thomas Klemm]

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