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 $x = s - a$, $y = s - b$, $z = s - c$:
- $4 \paren {x + y + z} = x y z$
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
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Bachet: $110$