Right Triangle Question?

Question

The right triangle with sides of length 5, 12, and 13 has the property that its area is equal to its perimeter. What is a another triangle with sides that are integers that also has this property?

Answer 1

a triangle with sides 6, 8, and 10 also has this property....
6 + 8 + 10 = 24
(8 x 6)/2 = 24

Answer 2

As already posted, a 6, 8, 10 triangle has this
property. It turns out that the 5, 12, 13 and 6, 8, 10
triangles are the only right triangles with integer
sides with this property.
Here is the proof:
We know that a² + b² = c² (1)
and ab/2 = a + b + c.
So c = ab/2 - a - b.
Plug this into (1):
a² + b² = (ab/2 -a-b)² = a²b²/4 - a²b + a² - 2b(ab/2-a) + b².
Cancelling a² + b² from both sides gives
a²b²/4 -a²b - ab² + 2ab=0.
Now a and b are positive, so we can divide out ab:
ab/4 -a - b + 2 = 0.
Now solve for b
b(a/4-1) = a-2
b = 4( a-2)/(a-4). (2)
Since the fraction on the rhs of (2 ) > 1 we see
b > 4
Now consider the function
y = (x-2)/(x-4).
Its derivative is -2/(x-4)², which is always negative,
so y is a decreasing function of x. Consequently
the right side of (2) decreases as a increases.
Now if a= 18, b = 4(18-2)/(18-4) <4.58
and b will be even smaller for larger values of a.
But if 4 < b < 4.58, there is no integer b that
can satisfy (2).
Consequently, there are no solutions for a ≥ 18.
If we now try all the possible values of a,
we find that 5, 12, 13 and 6, 8, 10 (and the
triples with a and b reversed)
are the only triples that satisfy the conditions of
the problem.

Answer 3

6, 8, 10

Answer 4

yes! Only 2 triangles, not more. But what if s=10p, how many?