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General Discussion => Other Discussions => Math and Science => Topic started by: Munchor on March 25, 2011, 06:50:42 pm

Title: Pitagoric Triplet
Post by: Munchor on March 25, 2011, 06:50:42 pm

A pitagoric triplet is, as most of you know, a group of numbers a,b,c that solve the following rule:

Code: [Select]

a²+b²=c²

The most common example is 3²+4²=5².

The challenge is the following:

Code: [Select]

a²+b²=c²
a,b,c are all natural numbers
a+b+c=200

You can easily find the answer on the Net, but I want the mathematical (by hand) proof of your work.

Good Luck! If you find the answer like some already did, post it in spoiler tags please.

Title: Re: Pitagoric Triplet
Post by: Goplat on March 25, 2011, 06:59:31 pm

There are actually two solutions (four if you allow a > b), so I'm going to be a smartass and give the unintended one:

Spoiler For Spoiler:

0² + 100² = 100² ;)

Title: Re: Pitagoric Triplet
Post by: Munchor on March 25, 2011, 07:01:39 pm

Quote from: Goplat on March 25, 2011, 06:59:31 pm

There are actually two solutions (four if you allow a > b), so I'm going to be a smartass and give the unintended one:

Spoiler For Spoiler:

0² + 100² = 100² ;)

0 is not a natural number :P

That answer is wrong. The answer is easy, the getting there by hand is hard.

Title: Re: Pitagoric Triplet
Post by: AngelFish on March 25, 2011, 07:04:43 pm

Derivation:

Spoiler For Spoiler:

Let us start with the premises that sqrt(A²+B²)=C and A+B+C=200. From this, we can ascertain that the upper bound on A, B, and C is the same as the upper bound on C. Thus, A<C>B. This suggests that the best course of action is to write the equation 200= sqrt(A²+B²)+A+B, which is equivalent to the second premise.

From that, we get A²+B²=(200-A-B)² or A²+B²=40,000-400A-400B+A²+B²+2AB.

Then, dividing by A²+B², we get 1=(40,000-400A-400B+A²+B²+2AB)/(A²+B²).

Breaking this into partial fractions brings out 1=(40,000)/(A²+B²)-(400A)/(A²+B²)-(400B)/(A²+B²)+(A²+B²)/(A²+B²)+(2AB)/(A²+B²).

Simplifying and subtracting the ones from both sides gives us 0=(40,000-400A-400B+2AB)/(A²+B²). The roots of this equation are the solutions.

I then solved this equation for B, which results in the nice equation B=(200(A-100))/(A-200). The answers should be obvious from this point onwards, so I'll let you do the rest :)

Answer:

Spoiler For Spoiler:

A=40, B=75, C=85. There's another trivial solution where A and B are reversed.

Title: Re: Pitagoric Triplet
Post by: christop on March 25, 2011, 07:21:51 pm

Here's my method and solution:

Spoiler For method:

First I typed this program into my computer (by hand), compiled it (by typing gcc by hand), and then ran it (again, typing the command by hand):

Code: [Select]

#include <stdio.h>

int main()
{
long a, b, c;
long aa, bb, cc;
for (a = 1; a <= 200; ++a) {
aa = a*a;
for (b = 1; b <= 200 - a - 1; ++b) {
bb = b*b;
c = 200 - a - b;
cc = c*c;
if (c > 0 && aa + bb == cc)
printf("a=%ld b=%ld c=%ld\n", a,b,c);
}
}
return 0;
}


Spoiler For solution:

When I ran the program I got this answer:
a=40 b=75 c=85
a=75 b=40 c=85

(one solution and the other trivial solution with A and B reversed)

40^2 + 75^2 = 85^2
40, 75, and 85 all are natural numbers
40 + 75 + 85 = 200

QED. :P

Time between reading the problem and instructing the computer how to find a solution: less than 5 minutes
Time to run the program and obtain a solution: 4 milliseconds
Finding a solution in less time than doing it purely by hand: priceless

Title: Re: Pitagoric Triplet
Post by: Munchor on March 25, 2011, 07:23:04 pm

Quote from: christop on March 25, 2011, 07:21:51 pm

Here's my method and solution:

Spoiler For method:

First I typed this program into my computer (by hand), compiled it (by typing gcc by hand), and then ran it (again, typing the command by hand):

Code: [Select]

#include <stdio.h>

int main()
{
long a, b, c;
long aa, bb, cc;
for (a = 1; a <= 200; ++a) {
aa = a*a;
for (b = 1; b <= 200 - a - 1; ++b) {
bb = b*b;
c = 200 - a - b;
cc = c*c;
if (c > 0 && aa + bb == cc)
printf("a=%ld b=%ld c=%ld\n", a,b,c);
}
}
return 0;
}


Spoiler For solution:

When I ran the program I got this answer:
a=40 b=75 c=85
a=75 b=40 c=85

(one solution and the other trivial solution with A and B reversed)

40^2 + 75^2 = 85^2
40, 75, and 85 all are natural numbers
40 + 75 + 85 = 200

QED. :P

Time between reading the problem and instructing the computer how to find a solution: less than 5 minutes
Time to run the program and obtain a solution: 4 milliseconds
Finding a solution in less time than doing it purely by hand: priceless

Nice Christop, I did the same, but using Python. It's relatively easy to do that (no offense!). I wanna see if someone comes up with a by hand formula of it :O

Title: Re: Pitagoric Triplet
Post by: AngelFish on March 25, 2011, 07:23:53 pm

Actually, that derivation took me about three minutes...

EDIT: Scout, I did come up with a version done by hand.

Title: Re: Pitagoric Triplet
Post by: Munchor on March 25, 2011, 07:27:48 pm

Quote from: Qwerty.55 on March 25, 2011, 07:23:53 pm

Actually, that derivation took me about three minutes...

EDIT: Scout, I did come up with a version done by hand.

NICE QWERTY, show me.

Title: Re: Pitagoric Triplet
Post by: AngelFish on March 25, 2011, 07:28:40 pm

See the spoiler titled "Derivation (http://ourl.ca/9901/189836)"

Title: Re: Pitagoric Triplet
Post by: christop on March 25, 2011, 07:30:00 pm

Spoiler For Spoiler:

I got up to this point:

B=(200(A-100))/(A-200)

but I got stuck when plugging in (200(A-100))/(A-200) for B in the original equation...

Title: Re: Pitagoric Triplet
Post by: AngelFish on March 25, 2011, 07:32:51 pm

I suppose you could do that, but it sounds really painful. At that point, I just recognized that the answer would necessarily be a natural number, so all you have to do is find a number A such that sqrt(A²+((200(A-100))/(A-200))²) is a whole number. That sounds like random guessing, but I already had a good idea of the value of A, so I really only had to deal with thirty values or so, which is easily done in your head.

Title: Re: Pitagoric Triplet
Post by: christop on March 25, 2011, 07:43:49 pm

Quote from: Qwerty.55 on March 25, 2011, 07:32:51 pm

I suppose you could do that, but it sounds really painful. At that point, I just recognized that the answer would necessarily be a natural number, so all you have to do is find a number A such that sqrt(A²+((200(A-100))/(A-200))²) is a whole number. That sounds like random guessing, but I already had a good idea of the value of A, so I really only had to deal with thirty values or so, which is easily done in your head.

It's not easily done in MY head! :)

Title: Re: Pitagoric Triplet
Post by: AngelFish on March 25, 2011, 07:46:22 pm

I must be a freak then, because I still can't count change properly without spending some serious thought :P

Title: Re: Pitagoric Triplet
Post by: Xeda112358 on March 25, 2011, 07:50:19 pm

Aww, I was going in the completely crazy direction... :(

Spoiler For Spoiler:

{a,b,c}={m²-n²,2mn,m²+n²}
From there I did:
200=m²-n²+2mn+m²+n²
Which simplifies to:
200=m²+2mn+m²
200=2m²+2mn
100=m²+mn
100=m(m+n)

That told me that m and n had to be less than 10. I got nowhere. Then I realised that the result was not likely to be a primitive Pythagorean Triple. So I took the prime factors of two hundred:
200=2³*5²
I divided by the highest factor (5) and got 40
20=m(m+n)

So m had to be less than sqrt(20). That got me to m=4, n=1. So:
{a,b,c}={m²-n²,2mn,m²+n²}
{a,b,c}={4²-1²,2*4*1,4²+1²}
{a,b,c}={16-1,8,16+1}
{a,b,c}={15,8,17}

That gives us 15+8+17=40, so multiply those by 5 and you get:
{a,b,c}={75,40,85}

Title: Re: Pitagoric Triplet
Post by: Deep Toaster on March 25, 2011, 08:41:09 pm

Here's a harder one: Find natural numbers a, b, and c where a³+b³=c³ :P

Title: Re: Pitagoric Triplet
Post by: Xeda112358 on March 25, 2011, 08:41:38 pm

:P We discussed this on IRC

Title: Re: Pitagoric Triplet
Post by: AngelFish on March 25, 2011, 10:20:29 pm

Does a proof that it's not possible count?

Title: Re: Pitagoric Triplet
Post by: Munchor on March 26, 2011, 07:14:25 am

I like what Zeda did, at least, I get most of it :)

Title: Re: Pitagoric Triplet
Post by: Deep Toaster on March 26, 2011, 01:26:33 pm

Quote from: Qwerty.55 on March 25, 2011, 10:20:29 pm

Does a proof that it's not possible count?

Sure. Post a full, formal proof here.

Title: Re: Pitagoric Triplet
Post by: Goplat on March 26, 2011, 07:07:35 pm

Quote from: Deep Thought on March 25, 2011, 08:41:09 pm

Here's a harder one: Find natural numbers a, b, and c where a³+b³=c³ :P

0³ + 0³ = 0³

And yes, 0 is a natural number. The standard way to define the natural numbers in modern mathematics is to represent 0 with the empty set, 1 as the set {0}, 2 as the set {0,1}, etc.

Title: Re: Pitagoric Triplet
Post by: AngelFish on March 26, 2011, 07:25:12 pm

That's defining the cardinality as a number, which it does not appear to be. For example, the cardinality of the set of natural numbers is infinity. The cardinality of the set of real numbers is of a greater infinity. If cardinality are real numbers, then would there not be multiple types of natural infinities?