Right triangle where the perimeter = area*k I was doodling on some piece of paper a problem that sprung into my mind. After a few minutes of resultless tries, I advanced to try to solve the problem using computer based means. 
The problem stated is

Does a right angle triangle with integer sides such that
    $$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ exist? 

(Here P is the perimeter of the triangle. Sum of the sides. And A is the area of the triangle. a*b/2 in a triangle with a b c and c is the hypotenuse)
Obviously for the simple cases such triangles exists. For an example 
when
$ A=2P \qquad $ the triangle $12,16,20$ works
$ A=P  \qquad \; \; $ the triangle $6,8,10$ works
$2A=P  \qquad $ the triangle $3,4,5$ works
I tried solving this by hand, first for the special case where $2P=A$. This ended up giving me 
$$ \frac{a \cdot b}{2} = A $$
and
$$ P = \frac{A}{2} = \frac{a \cdot b}{4} \qquad \text{also} \qquad P = a + b + c  $$ 
So 
$$ \frac{ab}{4} = a + b + c $$ 
By knowing that this is a right angle this leads to the equation (Using the Pythagorean theorem)
$$ a^2 + b^2 = c^2 $$
Now we have two equations and three unknowns, which also needs to be integers! Sadly I was not able to continue from here. I have only learned how to solve linear Diophantine equations. Not a system of nonlinear Diophantine equations. 
Just to restate my question below =) 

Is there a right angle triangle with integer sides such that
    $$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ ? 

Regards, Werner

If you choose positive integers $a$ and $b$, you can make a right triangle with sides
$a^2-b^2\text{, } 2ab \text{ and } a^2+b^2,$
and all right triangles can be found this way by varying the values of $a$ and $b$.  
If we have such a triangle, its area is
$$\frac{1}{2}(a^2-b^2)\times 2ab = ab(a^2-b^2)$$
and its perimeter is 
$$a^2-b^2 + 2ab + a^2+b^2 = 2a(a+b).$$
For the first part of your question we want to set
$$ab(a^2-b^2)\times k = 2a(a+b).$$
We can write this as
$$ab(a-b)(a+b) = 2a(a+b)$$
and then cancel the $a$ and the $(a+b)$ on each side (neither are zero so it's ok) and get
$$b(a-b)k = 2\quad\text{, and finally}$$
$$k = \frac{2}{b(a-b)}.$$
We want $k$ to be an integer, so it can only be 1 or 2.
If $k=1$ , $b(a-b)$ must equal $2$ and this can happen in two ways: when $a=3\text{ and }b=2$, which produces the 5-12-13 triangle, and when $a=3$ and $b=1$, which produces the 6-8-10 triangle.
If $k=2$ , $b(a-b)$ must equal $1$ and the only way this can happen is when $a=2\text{ and }b=1$. These values of $a$ and $b$ produce the 3-4-5 triangle.
For the second part of your question we want to set
$$ab(a^2-b^2) = 2a(a+b)\times k.$$
We can reduce this as before and arrive at
$$k = \frac{b(a-b)}{2}.$$
There are lots of different values of $a$ and $b$ that make $k$ an integer.
