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        riddles >> medium >> 3 Right Triangles
        
(Message started by: Michael_Dagg on Jan 22nd, 2008, 12:31pm)
Title: 3 Right Triangles
Post by Michael_Dagg on Jan 22nd, 2008, 12:31pm
Up to congruence, show that there are exactly three
right triangles whose side lengths are integers while the
area is twice the perimeter.
Title: Re: 3 Right Triangles
Post by towr on Jan 22nd, 2008, 1:39pm
[hide]
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)
[/hide]

hmm, wait why aren't they all primitive triples... What'd I mess up this time. Oh well, divide as needed.
Title: Re: 3 Right Triangles
Post by Icarus on Jan 22nd, 2008, 7:08pm
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.
Title: Re: 3 Right Triangles
Post by towr on Jan 23rd, 2008, 1:45am

on 01/22/08 at 19:08:47, Icarus wrote:
But some non-primitive triples cannot be expressed this way. For them, your argument needs some minor tweaking.
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 (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1198763501)?
Title: Re: 3 Right Triangles
Post by Icarus on Jan 23rd, 2008, 6:38pm
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.
Title: Re: 3 Right Triangles
Post by Eigenray on Jan 24th, 2008, 5:00am
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 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifm|2n 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif'(m),

where d(n) is the number of divisors of n, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif'(n) is the number of distinct odd prime factors of n.
Title: Re: 3 Right Triangles
Post by towr on Jan 24th, 2008, 6:24am

on 01/23/08 at 18:38:34, 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)).


So, I should have ended up with
[hide]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)
[/hide]
Title: Re: 3 Right Triangles
Post by Joe Fendel on Jan 24th, 2008, 2:10pm

on 01/22/08 at 12:31:46, Michael_Dagg wrote:
Up to congruence


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.
Title: Re: 3 Right Triangles
Post by Grimbal on Jan 24th, 2008, 2:20pm
It covers rotations and reflections.
Title: Re: 3 Right Triangles
Post by Joe Fendel on Jan 24th, 2008, 2:37pm

on 01/24/08 at 14:20:58, Grimbal wrote:
It covers rotations and reflections.

D'oh!  Of course - I was confusing "congruence" with "similarity".

Back to 10th grade for me...   :-[


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