Pythagorean Triangles whose Area equal their Perimeter

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Theorem

There exist exactly $2$ Pythagorean triples which define a Pythagorean triangle whose area equals its perimeter:

$(1): \quad \tuple {6, 8, 10}$, leading to an area and perimeter of $24$
$(2): \quad \tuple {5, 12, 13}$, leading to an area and perimeter of $30$.


Proof

From Area of Right Triangle, the area $\mathcal A$ is:

$\mathcal A = \dfrac {a b} 2$

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:

$\mathcal A = m n \paren {m^2 - n^2}$

The perimeter of $T$ is given by:

$\mathcal P = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$


We need to find all $m$ and $n$ such that $\mathcal P = \mathcal A$.

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