If a triangle's area equals its perimeter, what special condition must its angles satisfy? If we have a triangle where the Perimeter >0 and the Area >0 , and Area=Perimeter, what special condition must the angles of this triangle satisfy for this to happen?


I've done a bit of research and have found information on Heron's forumla and on integral triangles but I'm at a loss at how to figure this problem out. 

Through research I have found that if the lengths of the sides of the triangle are either {6,8,10),(5,12,13),(9,10,17),(7,15,20), or(6,25,29) we get a triangle that fits the requirements (im not sure if this information is relevant) but the question is about the "special condition" that the angles must meet to create such a triangle not the sides. 

I'm looking for both an answer and an explanation, any help is greatly appreciated. Please let me know if I left anything out (that's all the information I was given) or if something is unclear.
Hint: If you scale the triangle by a factor of $\alpha$, the perimeter is scaled by $\alpha$ and the area is scaled by $\alpha^2$.

Hence, given any triangle, there is a unique triangle that is similar to it, which satisfies perimeter = area.
This a Pythagorean triple treatment of area to perimeter ratios where $\angle AB$ is right and $A^2+B^2=C^2$.

We can find Pythagorean triples, if they exist, for any ratio $R$ of area/perimeter by finding the $m,n$(s) that represent them using the following formula which includes a defined finite search for values of $m$. Whenever the $m$ $R$ combination yields a positive integer for $n$, we have the $m,n$ for a triple.
We begin by solving the area/perimeter equation for $n$ where area=$D$ so as not to confuse it with $A,B,C$.

$$D=\frac{AB}{2}=\frac{(m^2-n^2)*2mn}{2}=mn(m^2-n^2)\quad P=(m^2-n^2)+2mn+(m^2+n^2)=2m^2+2mn$$

$$\frac{D}{P}=\frac{mn(m^2-n^2)}{2m^2+2mn}=\frac{mn(m-n)(m+n)}{2m(m+n)}=\frac{n(m-n)}{2}=R\qquad n^2-mn+2R=0$$

$$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{m\pm\sqrt{m^2-4*1*2R}}{2*1}$$

$$n=\frac{m\pm \sqrt{m^2-8R}}{2}\text{ where }\lceil\sqrt{8R}\space \rceil \le m \le 2R+1$$

Example: For $R=0.5\quad \sqrt{8*0.5}=2\le m \le 2*0.5+1=2$

$$n=\frac{2\pm \sqrt{2^2-8*0.5}}{2}=1\qquad f(2,1)=(3,4,5)$$
But you are asking about $R=1$ and there only two such Pythagorean triples to be found.

$$R=1\rightarrow 3\le m \le 3\quad f(3,2)=(5,12,13)\quad f(3,1)=(8,6,10)$$
For other ratios:
$$R=1.5\rightarrow 4\le m \le 4\quad f(4,3)=(7,24,25)\quad f(4,1)=(15,8,17)$$

$$R=2\rightarrow 4\le m \le 5\quad f(4,2)=(12,16,20)\quad f(5,4)=(9,40,41)\quad f(5,1)=(24,10,26)$$

$$R=2.5\rightarrow 4\le m \le 6\quad f(6,5)=(11,60,61)\quad f(6,1)=(35,12,37)$$

$$R=3\rightarrow 4\le m \le 7\quad f(5,3)=(16,30,34)\quad f(5,2)=(21,20,29)\quad f(7,6)=(13,84,85)\quad f(7,1)=(48,14,50)$$

$$R=18\rightarrow 12\le m \le 37\quad f(12,6)=(108,144,180)\quad f(13,9)=(88,234,250)\quad f(13,4)=(153,104,185)\quad f(15,12)=(81,360,369)\quad f(15,3)=(216,90,234)\quad f(20,18)=(76,720,724)\quad f(20,2)=(396,80,404)\quad f(37,36)=(73,2664,2665)\quad f(37,1)=(1368,74,1370)$$

Aside: you can always find one primitive triple for a given $R$ if you let $(m,n)=(2R+1,2R)$.