Pythagorean Triangles whose Area equal their Perimeter
Theorem
There exist exactly $2$ Pythagorean triples which define a Pythagorean triangle whose area equals its perimeter:
Proof
From Area of Right Triangle, the area $\AA$ is:
- $\AA = \dfrac {a b} 2$
where $a$ and $b$ are the legs.
$(1): \quad$ The area of the $\tuple {6, 8, 10}$ triangle is $\dfrac {6 \times 8} 2 = 24$.
Its perimeter equals $6 + 8 + 10 = 24$.
$(2): \quad$ The area of the $\tuple {5, 12, 13}$ triangle is $\dfrac {5 \times 12} 2 = 30$.
Its perimeter equals $5 + 12 + 13 = 30$.
It remains to prove that these are the only ones.
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle $T$.
Thus $a, b, c$ form a Pythagorean triple.
By definition of Pythagorean triple, $a, b, c$ are in the form:
- $2 m n, m^2 - n^2, m^2 + n^2$
We have that $m^2 + n^2$ is always the hypotenuse.
Thus the area of $T$ is given by:
- $\AA = m n \paren {m^2 - n^2}$
The perimeter of $T$ is given by:
- $\PP = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$
We need to find all $m$ and $n$ such that $\PP = \AA$.
Thus:
\(\ds 2 m^2 + 2 m n\) | \(=\) | \(\ds m n \paren {m^2 - n^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 m \paren {m + n}\) | \(=\) | \(\ds n \paren {m + n} \paren {m - n}\) | Difference of Two Squares | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds n \paren {m - n}\) | \(=\) | \(\ds 2\) |
As $m$ and $n$ are both (strictly) positive integers, it follows immediately that either:
- $n = 1$
- $m - n = 2$
and so:
- $m = 3, n = 1$
leading to the triangle:
- $a = 6, b = 8, c = 10$
or:
- $n = 2$
- $m - n = 1$
and so:
- $m = 3, n = 2$
leading to the triangle:
- $a = 12, b = 5, c = 13$
and the result follows.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $30$
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Bachet: $110$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $30$