# The largest side of a right triangle is a^2+b^2 and other side is 2ab. What condition will make the third side to be the smallest side?

Dec 5, 2016

For the third side to be the shortest, we require $\left(1 + \sqrt{2}\right) | b | > \left\mid a \right\mid > \left\mid b \right\mid$ (and that $a$ and $b$ have the same sign).

#### Explanation:

The longest side of a right triangle is always the hypotenuse. So we know the length of the hypotenuse is ${a}^{2} + {b}^{2.}$

Let the unknown side length be $c .$ Then from the Pythagorean theorem, we know

${\left(2 a b\right)}^{2} + {c}^{2} = {\left({a}^{2} + {b}^{2}\right)}^{2}$
or
$c = \sqrt{{\left({a}^{2} + {b}^{2}\right)}^{2} - {\left(2 a b\right)}^{2}}$
$\textcolor{w h i t e}{c} = \sqrt{{a}^{4} + 2 {a}^{2} {b}^{2} + {b}^{4} - 4 {a}^{2} {b}^{2}}$
$\textcolor{w h i t e}{c} = \sqrt{{a}^{4} - 2 {a}^{2} {b}^{2} + {b}^{4}}$
$\textcolor{w h i t e}{c} = \sqrt{{\left({a}^{2} - {b}^{2}\right)}^{2}}$
$\textcolor{w h i t e}{c} = {a}^{2} - {b}^{2}$

We also require that all side lengths be positive, so

• ${a}^{2} + {b}^{2} > 0$
$\implies a \ne 0 \mathmr{and} b \ne 0$
• $2 a b > 0$
$\implies a , b > 0 \mathmr{and} a , b < 0$
• $c = {a}^{2} - {b}^{2} > 0$
$\iff {a}^{2} > {b}^{2}$
$\iff \left\mid a \right\mid > \left\mid b \right\mid$

Now, for any triangle, the longest side must be shorter than the sum of the other two sides. So we have:

$\textcolor{w h i t e}{\implies} 2 a b + \text{ } c \textcolor{w h i t e}{X X} > {a}^{2} + {b}^{2}$
$\implies 2 a b + \left({a}^{2} - {b}^{2}\right) > {a}^{2} + {b}^{2}$
$\implies 2 a b \textcolor{w h i t e}{X X X X X X} > 2 {b}^{2}$

$\implies \left\{\begin{matrix}a > b \text{ & " if b > 0 \\ a < b" & } \mathmr{if} b < 0\end{matrix}\right.$

Further, for third side to be smallest, ${a}^{2} - {b}^{2} < 2 a b$
or ${a}^{2} - 2 a b + {b}^{2} < 2 {b}^{2}$ or $a - b < \sqrt{2} b$ or $a < b \left(1 + \sqrt{2}\right)$

Combining all of these restrictions, we can deduce that in order for the third side to be the shortest, we must have $\left(1 + \sqrt{2}\right) | b | > \left\mid a \right\mid > \left\mid b \right\mid \mathmr{and} \left(a , b < 0 \mathmr{and} a , b > 0\right) .$