Math Challenge I could not solve - Printable Version +- HP Forums (https://archived.hpcalc.org/museumforum) +-- Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum-1.html) +--- Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum-2.html) +--- Thread: Math Challenge I could not solve (/thread-236737.html) |
Math Challenge I could not solve - Meindert Kuipers - 01-03-2013 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 ...
The problem is the following: Looking forward to your responses ....
Re: Math Challenge I could not solve - Walter B - 01-03-2013
Quote:Suggest deleting the words between the (added) double parentheses.
d:-)
Re: Math Challenge I could not solve - Steve Hunt - 01-03-2013 I'm no maths expert, but this may help: 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, Re: Math Challenge I could not solve - Kiyoshi Akima - 01-03-2013 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: 3 Jan 2013, 9:33 p.m.
Re: Math Challenge I could not solve - Dieter - 01-03-2013 Simple answer: The Goat Problem.
Dieter
Re: Math Challenge I could not solve - Thomas Klemm - 01-03-2013 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 }
00 { 24-Byte Prgm }
With this program the equation is calculated: 00 { 94-Byte Prgm } 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
Edited: 3 Jan 2013, 7:44 p.m.
Re: Math Challenge I could not solve - Thomas Klemm - 01-03-2013 For those who prefer to use the HP-15C: 001 - 42,21,11 LBL A 020 - 20 x
Solution: 0
Cheers Re: Math Challenge I could not solve - Luiz C. Vieira (Brazil) - 01-03-2013 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)
Re: Math Challenge I could not solve - Thomas Klemm - 01-03-2013 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 Re: Math Challenge I could not solve - Paul Dale - 01-03-2013 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. 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.
Re: Math Challenge I could not solve - Thomas Klemm - 01-03-2013 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
Edited: 3 Jan 2013, 11:35 p.m.
Re: Math Challenge I could not solve - Paul Dale - 01-04-2013 Yes, that is a better way to calculate it -- wasn't really thinking about stability when I wrote my note.
Re: Math Challenge I could not solve - Thomas Klemm - 01-04-2013 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: So it appears HP's numeric integration is well suited for this challenge.
Kind regards Edited: 4 Jan 2013, 1:23 a.m.
Re: Math Challenge I could not solve - Neil Hamilton (Ottawa) - 01-04-2013 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...
Re: Math Challenge I could not solve - Dieter - 01-04-2013 Quote: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
Re: Math Challenge I could not solve - Walter B - 01-04-2013 Build number? Program listing? There must be reasons for the different results, so I'd appreciate getting these data from you and Pauli.
d:-)
Re: Math Challenge I could not solve - Dieter - 01-04-2013 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.
Re: Math Challenge I could not solve - Paul Dale - 01-04-2013 I was running 3.1 3225 -- 100 revisions older. I've been slack and not updated that one for ages.
Re: Math Challenge I could not solve - Dieter - 01-04-2013 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
Re: Math Challenge I could not solve - Paul Dale - 01-04-2013 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
Re: Math Challenge I could not solve - Meindert Kuipers - 01-05-2013 Thank you all for your responses!
Meindert
Re: Math Challenge I could not solve - Thomas Klemm - 01-05-2013 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 Re: Math Challenge I could not solve - PGILLET - 01-07-2013 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...
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