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.
  
  $\blacksquare$
explanation: The document is a detailed mathematical proof that there are exactly three proper integer Heronian triangles whose area and perimeter are equal. It provides the side lengths of these triangles and verifies the conditions using Pythagoras's Theorem and Heron's Formula. The document does not contain a question or an answer to a question; it is a standalone proof.
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distilabel_metadata.content: # 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.
  
  $\blacksquare$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a detailed mathematical proof that there are exactly three proper integer Heronian triangles whose area and perimeter are equal. It provides the side lengths of these triangles and verifies the conditions using Pythagoras's Theorem and Heron's Formula. The document does not contain a question or an answer to a question; it is a standalone proof.
  
  Label: no-question