# A triangle has area Delta ... ?

## The scalene triangle of area $\Delta$ has lengths $a , b$ and $c$ such that $a \in \mathbb{Q}$ and ${b}^{n} , {c}^{n} \notin \mathbb{Q} , \forall n \in \mathbb{Z}$ Does there exist values of $a , b$ and $c$ such that ${\Delta}^{2 m + 1} \in \mathbb{Q}$ for some $m \in {\mathbb{Z}}^{+}$ If not, give a proof

Jul 10, 2018

Here's a right triangle that does the trick:

$a = 4$

$b = \setminus \sqrt{8 - \setminus \sqrt{60}}$

$c = \setminus \sqrt{8 + \setminus \sqrt{60}}$

$m = 0$

$\setminus \Delta = 1$

#### Explanation:

I'll assume we get to exclude $n = 0$ as an exponent.

Let's see if we can do it with a right triangle ${a}^{2} = {b}^{2} + {c}^{2}$ with irrational legs. I think we can stick with $m = 0.$ If it's true for $m = 0$ it's true for all $m > 0$ as well.

$\Delta = \frac{1}{2} b c$

Let's pick $\setminus \Delta = 1 , a = 4 ,$ abbreviate $A = {a}^{2} , B = {b}^{2} , C = {c}^{2}$ and solve:

$B C = 4$

$B + C = 16 = A$

${x}^{2} - 16 x + 4 = 0$

$B , C = 8 \setminus \pm \sqrt{60}$

That's a solution. We have

$a = 4$

$b = \setminus \sqrt{8 - \setminus \sqrt{60}}$

$c = \setminus \sqrt{8 + \setminus \sqrt{60}}$

$\setminus \Delta = 1$