- 04 Jan '12 10:39It'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

Answer: 25, 100/3, 125/3

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. - 04 Jan '12 11:22Okay, 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.
- 04 Jan '12 11:50

Maybe you can reduce the brute force work using these Pythagorean triples:*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..?

http://en.wikipedia.org/wiki/Pythagorean_triple - 04 Jan '12 14:37That 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. - 05 Jan '12 09:04

Slight mistake, your formule should read:*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.

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

Then choose: k = 25, m = 4, n = 1

--> a = 375, b = 200, c = 425