Hello Bobbym,
Would you mind if I go to bed now?

Well, are you seeing this link?

Hi Agnishom!

There are other triangles other than right triangles that work.  I was just pointing out that there are
only two that are right triangles.  The proof of that is probably much easier than the proof for triangles
in general.  It will be interesting to see the general proof.

P.S.  I tried your link and found the proof.  Thanks!

Last edited by noelevans (2012-12-12 08:26:34)

Hi!  This is Dan's proof of 4 years ago from your link.

Let a,b,c denote the sides of triangle ∆ABC and P and A its perimeter and area respectively.
Note that Heron’s formula states that the area of ∆ABC is: A = √[s(s-a)(s-b)(s-c)]
Where s denotes the semi-perimeter, s = (a+b+c)/2.
To find all such triangles such that P=A we must have
√(a+b+c)(b+c-a)(a+c-b)(a+b-c) = 4(a+b+c).
Put x=b+c-a, y=a+c-b and z=a+b-c, where x,y,z are positive integers (true by the triangle
inequality). We see that Heron reduces to the Diophantine equation 16(x+y+z)=xyz.

We see from above that x,y,z must all be even integers. So, we may put x=2m, y=2n, z=2k to get,

  mnk=4(m+n+k)
  ⇒ mnk-4m = 4(n+k)
  ⇒ m = 4(n+k)/(nk-4)

Without loss of generality assume m≥n≥k, then we have 2m≥2n≥n+k ⇒ nk≤12.
Since 4<nk≤12 we may test integral values of n and k to find all such triangles.
Finally, we see that there is only 5 such triangles:
(a,b,c) = (5,12,13), (6,8,10), (6,25,29), (7,15,20), (9,10,17)
..................................................................................................................................

I followed pretty easily to the  three little lines starting with mnk=4(m+n+k) but had a little
difficulty figuring out why 4<nk≤12.  So I went back to the line mnk=4(m+n+k) and got it
from there.  First rewrite mnk=4(m+n+k) as  nk = 4(m+n+k)/m. 


Since m≥n≥k then 3m = (m+m+m) ≥ (m+n+k).  Therefore
                              4(3m)          4(m+m+m)          4(m+n+k)
            12 = 3*4 = _____    =     __________    ≥   ________   =  nk  because m≥n≥k.
                             
                                 m                    m                        m
implies  (m+m+m) ≥ (m+n+k).

Also rewriting mnk=4(m+n+k) as  4 = nk*[m/(m+n+k)]  we see that nk is being multiplied
by the quantity [m/(m+n+k)] which is less than one.  Hence nk>4.

So now we have the 4<nk≤12.

Testing all n≥k being bigger than four but less than or equal to 12 gives us the combinations
(n,k) ∈ {(12,1),(6,2),(4,3),(3,2),(4,2),(5,2),(3,3),(6,1),(8,1),(5,1),(7,1), (9,1),(10,1),(11,1)}
to test (14 of them).

Testing means substitute the n and k into mnk=4(m+n+k) and solve for m.  If the result is a
positive integer then double the m, n, and k to get the x, y, and z.  Then put these values
into the equations x=-a+b+c
                           y= a-b+c
                           z= a+b-c    and solve this system for a, b and c (the sides of the triangles).

Example:  n=4 and k=3 gives  m*4*3=4(m+4+3); that is, 12m=4m+28  so  m=28/8 = 7/2
                so this case doesn't give an integral value of m.  Case fails.

Example:  n=6 and k=1 gives  m*6*1=4(m+6+1); that is, 6m=4m+28  so  m=14
                This case gives an integral value for m so x=2m=28, y=2n=12 and z=2k=2.

Substituting these in to the three equations involving a, b, and c we obtain

                           28 = -a + b + c
                           12 =  a -  b  + c
                            2  =  a + b - c

Solving this system by addition/subtraction we obtain a=7, b=15, c=20  which is one of the
triangles that works.  (Adding each pair of equations eliminates two variables leaving the
third which is easy to solve.)

A nice solution to the problem, but Dan leaves out a few steps that one must scratch the
head about to fill in the gaps.     gotta

Hi Agnishom;

The statement 16(x+y+z)=xyz is not generally true. Just look at x = 6, y = 7 and z =8.

He says above, there might something else that implies that they are all even.

Hi Agnishom;

You can but since Dan did not explain you have to consider the possibility that his proof is flawed. I am not saying that it is, just that it is possible.

Do you know Dan?

I don't know him either.

Hi Agnishom;

Thanks for the link! I did not know that one.

Hi Agnishom;

I did a lot of searching for a solution to that. How did you find it?

Hi Agnishom;

Looks like a clever trick. But the sides can be labelled anything as long as the triangle inequality holds. And it does hold,

Hi;

Yes, as long as x, y and z are greater than 0.