text: # 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:
  
   $\ds 6^2 + 25^2$ $=$ $\ds 661$ $\ds$ $\ne$ $\ds 29^2$ so not right-angled
   $\ds 7^2 + 15^2$ $=$ $\ds 274$ $\ds$ $\ne$ $\ds 20^2$ so not right-angled
   $\ds 9^2 + 10^2$ $=$ $\ds 181$ $\ds$ $\ne$ $\ds 17^2$ so not right-angled
  
  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:
  
   $\text {(6, 25, 29)}: \quad$ $\ds s$ $=$ $\ds \frac {6 + 25 + 29} 2$ $\ds$ $=$ $\ds 30$ $\ds \leadsto \ \$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ $=$ $\ds \sqrt {30 \paren {30 - 6} \paren {30 - 25} \paren {30 - 29} }$ $\ds$ $=$ $\ds \sqrt {30 \times 24 \times 5 \times 1}$ $\ds$ $=$ $\ds \sqrt {3600}$ $\ds$ $=$ $\ds 60$ $\ds$ $=$ $\ds 6 + 25 + 29$
  
   $\text {(7, 15, 20)}: \quad$ $\ds s$ $=$ $\ds \frac {7 + 15 + 20} 2$ $\ds$ $=$ $\ds 21$ $\ds \leadsto \ \$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ $=$ $\ds \sqrt {21 \paren {21 - 7} \paren {21 - 15} \paren {21 - 20} }$ $\ds$ $=$ $\ds \sqrt {21 \times 14 \times 6 \times 1}$ $\ds$ $=$ $\ds \sqrt {1764}$ $\ds$ $=$ $\ds 42$ $\ds$ $=$ $\ds 7 + 15 + 20$
  
   $\text {(9, 10, 17)}: \quad$ $\ds s$ $=$ $\ds \frac {9 + 10 + 17} 2$ $\ds$ $=$ $\ds 18$ $\ds \leadsto \ \$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ $=$ $\ds \sqrt {18 \paren {18 - 9} \paren {18 - 10} \paren {18 - 17} }$ $\ds$ $=$ $\ds \sqrt {18 \times 9 \times 8 \times 1}$ $\ds$ $=$ $\ds \sqrt {1296}$ $\ds$ $=$ $\ds 36$ $\ds$ $=$ $\ds 9 + 10 + 17$
  
  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:
  
   $\ds 2 s$ $=$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ $\ds \leadsto \ \$ $\ds 4 s^2$ $=$ $\ds s \paren {s - a} \paren {s - b} \paren {s - c}$ $\ds \leadsto \ \$ $\ds 4 s$ $=$ $\ds \paren {s - a} \paren {s - b} \paren {s - c}$
  
  Note that:
  
  $\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$
  
  Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:
  
  $4 \paren {x + y + z} = x y z$
  
  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.
  
  $\blacksquare$