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Topic: 3 Right Triangles (Read 734 times) |
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Michael Dagg
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3 Right Triangles
« on: Jan 22nd, 2008, 12:31pm » |
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Up to congruence, show that there are exactly three
right triangles whose side lengths are integers while the
area is twice the perimeter.
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| « Last Edit: Jan 22nd, 2008, 12:34pm by Michael Dagg » |
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Regards,
Michael Dagg
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towr
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Re: 3 Right Triangles
« Reply #1 on: Jan 22nd, 2008, 1:39pm » |
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combining a pythagorean triple
a=(v^2+u^2)
b=2uv
c=(v^2-u^2)
(where v > u)
with heron's formula
area= sqrt(s(s-a)(s-b)(s-c)), with s=1/2 (a+b+c) = v^2+uv,
then twice the perimeter is 4s, so
4 (v^2+uv) = sqrt( (v^2+uv) (v^2+uv -(v^2-u^2) )(v^2+uv - 2uv)(v^2+uv - (v^2+u^2)) )
4 (v^2+uv) = +/- uv(u^2 - v^2)
4 (v+u) = +/- u(u^2 - v^2)
4 = +/- u (u-v)
v > u > 0, u|4, (u-v)|4
4 = u (v-u)
u=1,2,4
v=4/u+u = 5,4,5
So, (u,v) is (1,5) or (2,4) or (4,5)
and thus (a,b,c) is (26, 10, 24) or (20, 16, 12) or (41, 40, 9)
hmm, wait why aren't they all primitive triples... What'd I mess up this time. Oh well, divide as needed.
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| « Last Edit: Jan 23rd, 2008, 1:46am by towr » |
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Icarus
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Re: 3 Right Triangles
« Reply #2 on: Jan 22nd, 2008, 7:08pm » |
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Primitive pythagorean triples require u and v to be relatively prime, and one of them to be even. This is why the first two results were not primitive. Note, though, that Michael did not restrict to primitive triples. He said that there are only 3 answers even with non-primitives. Note also that multiplying by a constant does not preserve the area = 2*perimeter rule.
The three answers you gave are easily verified. Your argument shows that there are no other solutions that can be so expressed in terms of u and v. This includes all other primitive solutions, and a great many non-primitives as well. But some non-primitive triples cannot be expressed this way. For them, your argument needs some minor tweaking.
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towr
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Re: 3 Right Triangles
« Reply #3 on: Jan 23rd, 2008, 1:45am » |
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on Jan 22nd, 2008, 7:08pm, Icarus wrote:| But some non-primitive triples cannot be expressed this way. For them, your argument needs some minor tweaking. |
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Really? I thought every pythagorean triple (regardless of primitivity, thanks for reminding me) could be expressed as (v2+u2, 2uv, v2-u2)
Wasn't that the issue here?
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Icarus
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Re: 3 Right Triangles
« Reply #4 on: Jan 23rd, 2008, 6:38pm » |
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Perhaps I am misremembering, but I don't think so. The general form for triples is (m(v2+u2) 2muv, m(v2-u2)).
In the thread in your link, the formula was only proved for primitive triples.
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| « Last Edit: Jan 23rd, 2008, 6:39pm by Icarus » |
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Eigenray
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Re: 3 Right Triangles
« Reply #5 on: Jan 24th, 2008, 5:00am » |
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In fact 'most' integers are the hypotenuse of a Pythagorean triple, but are not themselves the sum of two squares. The smallest example would be (9,12,15).
Suppose we asked more generally: how many right triangles are there whose area is n times its perimeter?
By solving this problem in two different ways show that
d(8n2)/2 =  m|2n 2 '(m),
where d(n) is the number of divisors of n, and '(n) is the number of distinct odd prime factors of n.
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towr
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Re: 3 Right Triangles
« Reply #6 on: Jan 24th, 2008, 6:24am » |
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on Jan 23rd, 2008, 6:38pm, Icarus wrote:| Perhaps I am misremembering, but I don't think so. The general form for triples is (m(v2+u2) 2muv, m(v2-u2)). |
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So, I should have ended up with
4 = m u (v-u), v>u>0, m>0, gcd(u,v)=1, v-u is odd
m=1:
u=1: v=5 -> fails, because v-u is even
u=2: v=4 -> fails, because v-u is even
u=4: v=5 -> (41, 40, 9)
m=2:
u=1: v=3 -> fails, because v-u is even
u=2: v=3 -> 2*(5,12,13) = (10,24,26)
m=4:
u=1: v=2 -> 4*(3,4,5) = (12,16,20)
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| « Last Edit: Jan 24th, 2008, 6:25am by towr » |
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Joe Fendel
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Re: 3 Right Triangles
« Reply #7 on: Jan 24th, 2008, 2:10pm » |
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on Jan 22nd, 2008, 12:31pm, Michael_Dagg wrote:
It isn't clear to me what "up to congruence" means. Certainly we can't scale the triangle by an integer, k: we'll multiply the perimeter by k but the area by k2.
I think these words can be removed from the problem statement.
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Grimbal
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Re: 3 Right Triangles
« Reply #8 on: Jan 24th, 2008, 2:20pm » |
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It covers rotations and reflections.
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Joe Fendel
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Re: 3 Right Triangles
« Reply #9 on: Jan 24th, 2008, 2:37pm » |
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on Jan 24th, 2008, 2:20pm, Grimbal wrote:| It covers rotations and reflections. |
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D'oh! Of course - I was confusing "congruence" with "similarity".
Back to 10th grade for me... 
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