Heronian Triangle is Similar to Integer Heronian Triangle
Theorem
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.
Proof
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.
$\blacksquare$
Sources
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Bachet