3 Proper Integer Heronian Triangles whose Area and Perimeter are Equal

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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:

\(\displaystyle 6^2 + 25^2\) \(=\) \(\displaystyle 661\)
\(\displaystyle \) \(\ne\) \(\displaystyle 29^2\) so not right-angled
\(\displaystyle 7^2 + 15^2\) \(=\) \(\displaystyle 274\)
\(\displaystyle \) \(\ne\) \(\displaystyle 20^2\) so not right-angled
\(\displaystyle 9^2 + 10^2\) \(=\) \(\displaystyle 181\)
\(\displaystyle \) \(\ne\) \(\displaystyle 17^2\) 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:

\(\text {(6, 25, 29)}: \quad\) \(\displaystyle s\) \(=\) \(\displaystyle \frac {6 + 25 + 29} 2\)
\(\displaystyle \) \(=\) \(\displaystyle 30\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }\) \(=\) \(\displaystyle \sqrt {30 \paren {30 - 6} \paren {30 - 25} \paren {30 - 29} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {30 \times 24 \times 5 \times 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {3600}\)
\(\displaystyle \) \(=\) \(\displaystyle 60\)
\(\displaystyle \) \(=\) \(\displaystyle 6 + 25 + 29\)


\(\text {(7, 15, 20)}: \quad\) \(\displaystyle s\) \(=\) \(\displaystyle \frac {7 + 15 + 20} 2\)
\(\displaystyle \) \(=\) \(\displaystyle 21\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }\) \(=\) \(\displaystyle \sqrt {21 \paren {21 - 7} \paren {21 - 15} \paren {21 - 20} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {21 \times 14 \times 6 \times 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {1764}\)
\(\displaystyle \) \(=\) \(\displaystyle 42\)
\(\displaystyle \) \(=\) \(\displaystyle 7 + 15 + 20\)


\(\text {(9, 10, 17)}: \quad\) \(\displaystyle s\) \(=\) \(\displaystyle \frac {9 + 10 + 17} 2\)
\(\displaystyle \) \(=\) \(\displaystyle 18\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }\) \(=\) \(\displaystyle \sqrt {18 \paren {18 - 9} \paren {18 - 10} \paren {18 - 17} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {18 \times 9 \times 8 \times 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {1296}\)
\(\displaystyle \) \(=\) \(\displaystyle 36\)
\(\displaystyle \) \(=\) \(\displaystyle 9 + 10 + 17\)

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:

\(\displaystyle 2 s\) \(=\) \(\displaystyle \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 4 s^2\) \(=\) \(\displaystyle s \paren {s - a} \paren {s - b} \paren {s - c}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 4 s\) \(=\) \(\displaystyle \paren {s - a} \paren {s - b} \paren {s - c}\)

Note that:

$\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$

Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:

$4 \paren {x + y + z} = x y z$


By Semiperimeter of Integer Heronian Triangle is Composite, $s$ is an integer.

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.

$\blacksquare$


Sources