# slighty pythagoras

crazyblue
Posers and Puzzles 04 Jan '12 09:54
1. 04 Jan '12 09:54
This one is quite simple to understand, but so far I could only think of trial-error to solve it (which I didn't succeed in).

a²+b²=c², that much we know. Now try to find a set of a, b and c that makes a+b+c=1000.
2. 04 Jan '12 10:39
It's not so difficult. It's actually making a right-angled triangle from a piece of rope with given length.

E.g.:
1) 3²+4²=5² and 3+4+5 = 12
2) 100/12 = 8,33.. --> let's call this S
3) (3S)² + (4S)² = S²(3²+4² ) = S²5² = (5S)² and 3S+4S+5S = 12S = 100

You can do this with any choice of a,b,c for which pythagoras holds.

Your question is more difficult if a,b,c are required to be natural numbers or integers, but then it may not exist.
3. 04 Jan '12 11:22
Okay, I understood your solution. In fact I was instinctively thinking of natural numbers when I tried. So I rechecked the page where I found it and it says indeed natural numbers only. If you try again now, also note that a+b+c = 1000 and not 100. Otherwise it will be difficult for you to find the solution. ðŸ˜‰
4. talzamir
Art, not a Toil
04 Jan '12 11:34
hmm..

a^2 + b^2 = (1000 - a - b)^2
..

I see no natural number solutions for except the trivial where for c = 500.

Proving that by brute force is feasible but uninteresting.. hmm.. for what possible sums of a + b + c is there an all natural number triplet..?
5. 04 Jan '12 11:50
Originally posted by talzamir
hmm..

a^2 + b^2 = (1000 - a - b)^2
..

I see no natural number solutions for except the trivial where for c = 500.

Proving that by brute force is feasible but uninteresting.. hmm.. for what possible sums of a + b + c is there an all natural number triplet..?
Maybe you can reduce the brute force work using these Pythagorean triples:

http://en.wikipedia.org/wiki/Pythagorean_triple
6. 04 Jan '12 14:29
Originally posted by talzamir
Proving that by brute force is feasible but uninteresting.. hmm.. for what possible sums of a + b + c is there an all natural number triplet..?
That's what we have computers for...

200, 375, 425.

Richard
7. talzamir
Art, not a Toil
04 Jan '12 14:37
That is helpful indeed. So, using the buffed-up version of Euclid's method of finding Pythagorean triplets,

a = k (m^2 - n^2)
b = k (2mn)
c = k (m^2 + n^2)

where k, m , n are all integers and m > n, what are the values that a + b + c can get?

a + b + c = k(m^2 - n^2 + 2mn + m^2 + n^2) = 2 k m (m n)

So the possible values are even, and are divisible by a square.

n = 2, m = 5, k = 10 would give 2 x 10 x 5 x 5 x 2 = 1,000.
8. 05 Jan '12 09:04
Originally posted by talzamir
That is helpful indeed. So, using the buffed-up version of Euclid's method of finding Pythagorean triplets,

a = k (m^2 - n^2)
b = k (2mn)
c = k (m^2 + n^2)

where k, m , n are all integers and m > n, what are the values that a + b + c can get?

a + b + c = k(m^2 - n^2 + 2mn + m^2 + n^2) = 2 k m (m n)

So the possible values are even, and are divisible by a square.

n = 2, m = 5, k = 10 would give 2 x 10 x 5 x 5 x 2 = 1,000.