Perimeter problem from AMQ

Let $z$ be the hypotenuse of the right triangle, then $A=\dfrac {xy}2$ and by Pythagorean theorem $x^2 + y^2 = z^2$ . Then

$\begin{aligned} x+y+z & = 4A \\ x+y & = 4A-z & \small \blue{\text{Squaring both sides}} \\ x^2 + 2xy + y^2 & = 16A^2 - 8Az + z^2 \\ z^2 + 4A & = 16A^2 - 8Az + z^2 \\ 4A & = 16A^2 - 8Az \\ \implies 4A & = 2z + 1 \end{aligned}$

And

$\begin{aligned} \frac {x^4+y^4+z^4}8 & = 64 - A^2 & \small \blue{\text{Multiply both sides by }8} \\ (x^2+y^2)^2 - 2x^2y^2 + z^4 & = 512 - 8A^2 \\ z^4 - 8A^2 + z^4 & = 512 - 8A^2 \\ z^4 & = 256 \\ \implies z & = 4 \end{aligned}$

Then the perimeter $x+y+z = 4A = 2z+1 = \boxed 9$ .

Since we know that xyz is a right angled triangle: we know that $A=\frac { xy }{ 2 }$ and ${ x }^{ 2 }+{ y }^{ 2 }={ z }^{ 2 }$ are true. Thus ${ x }^{ 4 }+{ y }^{ 4 }+{ z }^{ 4 }=512-2{ x }^{ 2 }{ y }^{ 2 }\quad \Longrightarrow \quad { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 }+{ z }^{ 4 }=512\quad \Longrightarrow \quad 2{ z }^{ 4 }=512\quad \Longrightarrow \quad z=4$ We also know that $x+y+z=4A$ , so $x+y+z=2xy$ would hold. We can then deduce that ${ x }^{ 2 }+{ y }^{ 2 }={ z }^{ 2 }\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+2xy=16+x+y+4\Rightarrow { (x+y) }^{ 2 }-(x+y)-20=0$ . After solving the quadratic, we obtain 5 (we reject the -4), thus the perimeter is equal to 9