Heronian Triangle is Similar to Integer Heronian Triangle

From ProofWiki
Jump to navigation Jump to search


Let $\triangle {ABC}$ be a Heronian triangle.

Then there exists an integer Heronian triangle $\triangle {A'B'C'}$ such that $\triangle {ABC}$ and $\triangle {A'B'C'}$ are similar.


Let $\triangle {ABC}$ have sides whose lengths are $a$, $b$ and $c$.

By definition of Heronian triangle, each of $a$, $b$ and $c$ are rational.

By definition of rational number, we can express:

$a = \dfrac {p_a} {q_a}$, $b = \dfrac {p_b} {q_b}$ and $c = \dfrac {p_c} {q_c}$

where each of $p_a, q_a, p_b, q_b, p_c, q_c$ are integers.

Now let:

\(\ds a'\) \(=\) \(\ds a q_a q_b q_c\)
\(\ds b'\) \(=\) \(\ds b q_a q_b q_c\)
\(\ds c'\) \(=\) \(\ds c q_a q_b q_c\)

Let $\triangle {A'B'C'}$ be the triangle whose sides have lengths $a'$, $b'$ and $c'$.

By definition, $\triangle {ABC}$ and $\triangle {A'B'C'}$ are similar.

Each of $a'$, $b'$ and $c'$ are integers.

Consider the area of triangle $\triangle {A'B'C'}$

Let the area of $\triangle {ABC}$ be $A$.

Then the area $\triangle {A'B'C'}$ is $q_a q_b q_c A$, which is rational.

Hence $\triangle {A'B'C'}$ is an integer Heronian triangle.