Area Equals Perimeter: Heronian

Let a,b,c be the sides of the triangle

$a^{2}$ + $b^{2}$ = $c^{2}$

a + b + c = ab/2

c = ab/2 - a - b

$c^{2}$ = $(ab/2 - (a+b))^{2}$

$c^{2}$ = ( $a^{2}$ $b^{2}$ /4) - ab (a + b) + $a^{2}$ + $b^{2}$ + 2ab

= $a^{2}$ + $b^{2}$

$a^{2}$ $b^{2}$ / 4 - ab (a + b) + 2 ab = 0

Divide by ab

ab / 4 - a - b + 2 = 0

ab - 4a - 4b + 8 = 0

a (b - 4) - 4b + 8 = 0

a = (4b - 8) / (b - 4)

Solutions are

6, 8, 10 ... area 24, perimeter 24

5, 12, 13 ... area 30, perimeter 30

After that, as b increases, a gets smaller, approaching 4, so there are no more integer solutions.

Let $x,y,z$ be the sides with hypotenuse $z=\sqrt{x^2+y^2}$ .

$x+y+\sqrt{x^2+y^2}=\dfrac{xy}2$ $\Longrightarrow x+y-\dfrac{xy}2 = -\sqrt{x^2+y^2}$ $\Longrightarrow \dfrac{x^2y^2}4 + 2xy+x^2-xy^2-x^2y=0$ $\Longrightarrow xy-4y-4x+8=0$ $\Longrightarrow (x-4)(y-4)= 8$

$\therefore ~~~ (x,y)=(5,12),(6,8)$

$\therefore ~~~ \sum \dfrac{xy}{2}=\fbox{54}$

Let $a$ , $b$ and $c$ the sides of the right triangle, where $c$ is the hypotenuse.

Now, for some $m$ and $n$ ( $m>n$ ), we have the Pythagorean triple: $(m^{2}-n^{2},2mn,m^{2}+n^{2})$

So, $a=m^{2}-n^{2}$ , $b=2mn$ and $c=m^{2}+n^{2}$ .

The perimeter of the triangle will be: $P=a+b+c=2m(m+n)$ , and the area will be: $A=\frac{ab}{2}=mn(m+n)(m-n)$

We need that $P=A$ , so $2m(m+n)=mn(m+n)(m-n)$ .

Solving for $m$ :

$m=\frac{n^{2}+2}{n}$

And substituting in the area, after simplification, and in the form of function:

$A(n)=\frac{4(n^{2}+1)(n^{2}+2)}{n^{2}}$

We need that the sides be integer numbers, and the formula for the area we got is integer only when $n=1,2$ .

So, the sum of the possible areas is $A(1)+A(2)=\boxed{54}$

Let a <= b <= c be the sides of a right triangle with the same area and perimeter. By Pythagorean theorem, c^2 = a^2+b^2. Using all these facts,

a+b+c = ab/2,

2(a+b+c) = ab,

[2(a+b)-ab]^2 = [2c]^2,

4(a+b)^2+(ab)^2-4ab(a+b) = 4(a^2+b^2),

(ab)(8+ab-4a-4b) = 0,

(a-4)(b-4) = 8,

a-4 = 1 or 2 and b-4 = 8 or 4,

a = 5 or 6 and b = 12 or 8.

The corresponding values of c are 13 and 10. Therefore (a, b, c) = (5, 12, 13) or (6, 8, 10). It follows that the sum of all possible areas of the triangle is 5+12+13+6+8+10 = 54.