3 Proper Integer Heronian Triangles whose Area and Perimeter are Equal

Theorem
There are exactly $3$ proper integer Heronian triangles whose area and perimeter are equal.

These are the triangles whose sides are:


 * $\tuple {6, 25, 29}$
 * $\tuple {7, 15, 20}$
 * $\tuple {9, 10, 17}$

Proof
First, using Pythagoras's Theorem, we establish that these integer Heronian triangles are indeed proper:

Now we show they have area equal to perimeter.

We use Heron's Formula throughout:


 * $\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

where:
 * $\AA$ denotes the area of the triangle
 * $a$, $b$ and $c$ denote the lengths of the sides of the triangle
 * $s = \dfrac {a + b + c} 2$ denotes the semiperimeter of the triangle.

Thus we take the $3$ triangles in turn:

It remains to be demonstrated that these are indeed the only such proper integer Heronian triangles which match the criterion.

Let $\tuple {a, b, c}$ be the sides of such a triangle.

Using Heron's Formula, we have:

Note that:
 * $\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$

Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:
 * $4 \paren {x + y + z} = x y z$

$s$ is either an integer or a half-integer.

Suppose $s$ is a half-integer.

Then so are $x, y, z$.

But then $4 s = x y z$ is not an integer.

This is a contradiction.

Hence $s, x, y, z \in \N_{>0}$.

By Triple with Product Quadruple the Sum, our equation has solutions:
 * $\tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}, \tuple {2, 3, 10}, \tuple {2, 4, 6}$

Using:
 * $a = s - x = x + y + z - x = y + z$
 * $b = s - y = x + z$
 * $c = s - z = x + y$

the possible sets of side lengths are:
 * $\tuple {29, 25, 6}, \tuple {20, 15, 7}, \tuple {17, 10, 9}, \tuple {13, 12, 5}, \tuple {10, 8, 6}$

of which the final $2$ are Pythagorean Triples, so they are not proper Heronian triangles.

Hence the result.