Area of Integer Heronian Triangle is Multiple of 6

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Theorem

Let $\triangle {ABC}$ be an integer Heronian triangle.

Then the area of $\triangle {ABC}$ is a multiple of $6$.


Proof

Heron's Formula gives us that:

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

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.

We set out to eliminate $s$ and simplify as best possible:

\(\ds \AA\) \(=\) \(\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }\)
\(\ds \AA^2\) \(=\) \(\ds \dfrac {a + b + c} 2 \paren {\dfrac {a + b + c} 2 - a} \paren {\dfrac {a + b + c} 2 - b} \paren {\dfrac {a + b + c} 2 - c}\) substituting for $s$ and squaring
\(\ds \leadsto \ \ \) \(\ds 16 \AA^2\) \(=\) \(\ds \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}\) multiplying through by $16$ and simplifying
\(\ds \) \(=\) \(\ds 2 a^2 b^2 + 2 b^2 c^2 + 2 c^2 a^2 - a^4 - b^4 - c^4\) multiplying out and simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {4 \AA}^2 + \paren {b^2 + c^2 - a^2}\) \(=\) \(\ds \paren {2 b c}^2\) factorising

This is now in the form $p^2 + q^2 = r^2$.

From Solutions of Pythagorean Equation, $\tuple {p, q, r}$ has the parametric solution:

$\tuple {m^2 - n^2, 2 m n, m^2 + n^2}$


There are two steps to showing $6 \divides \AA$:


Step $1$: $2 \divides \AA$

By Euclid's Lemma for Prime Divisors, we just need to show:

$2 \divides \AA^2 = s \paren {s - a} \paren {s - b} \paren {s - b}$

By Semiperimeter of Integer Heronian Triangle is Composite, $s \in \N$.

There are $3$ cases:


Case $1$: There are sides with odd and even lengths

Under this condition, one of $s - x$ will be even.

Then $\AA^2$ is also even.

$\Box$


Case $2$: All sides are of odd length

Suppose all sides are of odd length.

Then the perimeter is also odd.

But then the semiperimeter cannot be an integer.

This is a contradiction.

$\Box$


Case $3$: All sides are of even length

Suppose all sides are of even length.

If the semiperimeter is even, the result follows.


Therefore we consider the case where $s$ is odd.

Then $x := s - a$, $y := s - b$ and $z := s - c$ are also odd.

Note that $x + y + z = 3 s - a - b - c = s$.

Hence $\AA^2 = x y z \paren {x + y + z}$.


Note that each of $x, y, z$ must be equivalent to $\pm 1 \pmod 4$.

If all of $x, y, z \equiv 1 \pmod 4$:

$x + y + z \equiv -1 \pmod 4$

If two of $x, y, z \equiv 1 \pmod 4$:

$x + y + z \equiv 1 \pmod 4$

If one of $x, y, z \equiv 1 \pmod 4$:

$x + y + z \equiv -1 \pmod 4$

If none of $x, y, z \equiv 1 \pmod 4$:

$x + y + z \equiv 1 \pmod 4$

In any of the above cases, we have:

$\AA^2 \equiv -1 \pmod 4$

which is impossible by Square Modulo 4.

This is a contradiction.

$\Box$


In each valid case, we see that $2 \divides \AA^2$.

$\Box$


Step $2$: $3 \divides \AA$

By Euclid's Lemma for Prime Divisors, we just need to show:

$3 \divides 16 \AA^2 = \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}$

We split the problem into $4$ cases:


Case $1$: None of $a, b, c$ are divisible by $3$

If $a \equiv b \equiv c \pmod 3$:

$a + b + c \equiv 3 a \equiv 0 \pmod 3$

Hence:

$3 \divides \paren {a + b + c}$


If two of the lengths have the same remainder when divided by $3$, say $a \equiv b \not \equiv c \pmod 3$:

$a \equiv b \equiv -c \pmod 3$

Hence:

$a + b - c \equiv 3 a \equiv 0 \pmod 3$

Thus:

$3 \divides \paren {a + b - c}$

$\Box$


Case $2$: One of $a, b, c$ is divisible by $3$

Without loss of generality suppose that number is $a$.

If $b \equiv c \pmod 3$:

$a + b - c \equiv a \equiv 0 \pmod 3$

Hence

$3 \divides \paren {a + b - c}$


If $b \not \equiv c \pmod 3$:

$b \equiv -c \pmod 3$

Hence:

$a + b + c \equiv -c + c \equiv 0 \pmod 3$

Thus:

$3 \divides \paren {a + b + c}$

$\Box$


Case $3$: Two of $a, b, c$ is divisible by $3$

Without loss of generality, suppose $3 \divides a, b$.

Then:

\(\ds 16 \AA^2\) \(=\) \(\ds \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}\)
\(\ds \) \(\equiv\) \(\ds c \cdot c \cdot c \cdot \paren {-c}\) \(\ds \pmod 3\)
\(\ds \) \(\equiv\) \(\ds -c^4\) \(\ds \pmod 3\)
\(\ds \) \(\equiv\) \(\ds -1\) \(\ds \pmod 3\) Square Modulo 3

By Square Modulo 3, $16 \AA^2 = \paren {4 \AA}^2$ is not a square.

This is a contradiction.

Therefore this case is not valid.

$\Box$


Case $3$: $a, b, c$ are all divisible by $3$

We have:

$3 \divides \paren {a + b + c}$

$\Box$


We see that in every valid case, $3 \divides 16 \AA^2$.

Hence $3 \divides \AA$, and thus $6 \divides \AA$.

$\blacksquare$


Sources

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