MHB Right Triangle: Proving It's a Right Triangle

[anemone]

The sides $x,\,y,\,z$ of a triangle satisfy the equality $(x^4+y^4+z^4)^2=2(x^3+y^3+z^3)$.

Prove that it's a right triangle.

 

[Greg]

A point of confusion:

Spoiler

I suppose I am probably missing something, but if $$x=y=z=\left(\dfrac23\right)^{\dfrac15}$$ then
$$(x^4+y^4+z^4)^2=2(x^3+y^3+z^3)$$ and the triangle is equilateral.

 

[anemone]

A point of confusion:

Spoiler

I suppose I am probably missing something, but if $$x=y=z=\left(\dfrac23\right)^{\dfrac15}$$ then
$$(x^4+y^4+z^4)^2=2(x^3+y^3+z^3)$$ and the triangle is equilateral.

Hi greg1313,

Spoiler

I think I should have mentioned that all three of them $x,\,y,\,z$ aren't equal.

 

[RLBrown]

Can one of the sides be negative?

 

[anemone]

Can one of the sides be negative?

Hi RLBrown,

Spoiler

No, all $x,\,y,\,z$ are supposed to be all different real positive numbers.:)

 

[anemone]

The sides $x,\,y,\,z$ of a triangle satisfy the equality $(x^4+y^4+z^4)^2=2(x^3+y^3+z^3)$.

Prove that it's a right triangle.

Hi MHB,

I want to apologize for the second time in a row today(Tmi) as I also made a typo in this challenge, the problem should read:

The sides $x,\,y,\,z$ of a triangle satisfy the equality $(x^4+y^4+z^4)^2=2(x^8+y^8+z^8)$.

Prove that it's a right triangle.

A special thanks to Opalg for letting me know about the typo that I had made in the original stated problem.(Nod)

 

[lfdahl]

Here is my solution. I think there are more elegant approaches.

Spoiler

The notation is symmetric, so it´s OK to let: $z^2 = x^2+y^2$. I also assume: $z >0$.
I need to prove, that the equation holds.

\[(x^4+y^4+z^4)^2 = 2(x^8+y^2+z^8)
\\\\\Rightarrow \left ( \left ( \frac{x}{z} \right )^4+\left ( \frac{y}{z} \right )^4+1\right )^2=2\left ( \left ( \frac{x}{z} \right )^8+ \left (\frac{y}{z} \right )^8 +1\right )\]

To ease the notation let $a = \frac{x}{z} = cos\alpha$ and $b = \frac{y}{z}=sin\alpha \;\;\; 0<\alpha<\frac{\pi}{2}$ .
I will be using the relation: $a^2+b^2 = 1$ several times:

\[2(a^8+b^8+1)=(a^4+b^4+1)^2 \\\\ \Rightarrow 2(a^8+b^8)+2-(a^8+b^8+2a^4b^4+2(a^4+b^4)+1)=0 \\\\ \Rightarrow a^8+b^8-2a^4b^4-2(a^4+b^4)+1=0 \\\\ \Rightarrow (a^4-b^4)^2-2((a^2+b^2)^2-2a^2b^2)+1=0 \\\\ *\Rightarrow (1-2b^2)^2-2(1-2(1-b^2)b^2)+1=0\\\\ \Rightarrow 1-4b^2+4b^4-2 +4(1-b^2)b^2+1=0\\\\ \Rightarrow 1-4b^2+4b^4+4b^2-4b^4-1 = 0
\\\ \Rightarrow 0=0.\]
This confirms that the sides $x,y$ and $z$ are sides in a right triangle.

\[(*). \;\;\; a^4-b^4 = (1-b^2)(1-b^2)-b^4 = 1-2b^2\]

 

[Greg]

Here is my solution. I think there are more elegant approaches.

Spoiler

The notation is symmetric, so it´s OK to let: $z^2 = x^2+y^2$. I also assume: $z >0$.
I need to prove, that the equation holds.

\[(x^4+y^4+z^4)^2 = 2(x^8+y^2+z^8)
\\\\\Rightarrow \left ( \left ( \frac{x}{z} \right )^4+\left ( \frac{y}{z} \right )^4+1\right )^2=2\left ( \left ( \frac{x}{z} \right )^8+ \left (\frac{y}{z} \right )^8 +1\right )\]

To ease the notation let $a = \frac{x}{z} = cos\alpha$ and $b = \frac{y}{z}=sin\alpha \;\;\; 0<\alpha<\frac{\pi}{2}$ .
I will be using the relation: $a^2+b^2 = 1$ several times:

\[2(a^8+b^8+1)=(a^4+b^4+1)^2 \\\\ \Rightarrow 2(a^8+b^8)+2-(a^8+b^8+2a^4b^4+2(a^4+b^4)+1)=0 \\\\ \Rightarrow a^8+b^8-2a^4b^4-2(a^4+b^4)+1=0 \\\\ \Rightarrow (a^4-b^4)^2-2((a^2+b^2)^2-2a^2b^2)+1=0 \\\\ *\Rightarrow (1-2b^2)^2-2(1-2(1-b^2)b^2)+1=0\\\\ \Rightarrow 1-4b^2+4b^4-2 +4(1-b^2)b^2+1=0\\\\ \Rightarrow 1-4b^2+4b^4+4b^2-4b^4-1 = 0
\\\ \Rightarrow 0=0.\]
This confirms that the sides $x,y$ and $z$ are sides in a right triangle.

\[(*). \;\;\; a^4-b^4 = (1-b^2)(1-b^2)-b^4 = 1-2b^2\]

Spoiler

Good work, but I'm not sure that proves there are not other values for $x$, $y$ and $z$ for which the equation holds.

 

[anemone]

Here is my solution. I think there are more elegant approaches.

Spoiler

The notation is symmetric, so it´s OK to let: $z^2 = x^2+y^2$. I also assume: $z >0$.
I need to prove, that the equation holds.

\[(x^4+y^4+z^4)^2 = 2(x^8+y^2+z^8)
\\\\\Rightarrow \left ( \left ( \frac{x}{z} \right )^4+\left ( \frac{y}{z} \right )^4+1\right )^2=2\left ( \left ( \frac{x}{z} \right )^8+ \left (\frac{y}{z} \right )^8 +1\right )\]

To ease the notation let $a = \frac{x}{z} = cos\alpha$ and $b = \frac{y}{z}=sin\alpha \;\;\; 0<\alpha<\frac{\pi}{2}$ .
I will be using the relation: $a^2+b^2 = 1$ several times:

\[2(a^8+b^8+1)=(a^4+b^4+1)^2 \\\\ \Rightarrow 2(a^8+b^8)+2-(a^8+b^8+2a^4b^4+2(a^4+b^4)+1)=0 \\\\ \Rightarrow a^8+b^8-2a^4b^4-2(a^4+b^4)+1=0 \\\\ \Rightarrow (a^4-b^4)^2-2((a^2+b^2)^2-2a^2b^2)+1=0 \\\\ *\Rightarrow (1-2b^2)^2-2(1-2(1-b^2)b^2)+1=0\\\\ \Rightarrow 1-4b^2+4b^4-2 +4(1-b^2)b^2+1=0\\\\ \Rightarrow 1-4b^2+4b^4+4b^2-4b^4-1 = 0
\\\ \Rightarrow 0=0.\]
This confirms that the sides $x,y$ and $z$ are sides in a right triangle.

\[(*). \;\;\; a^4-b^4 = (1-b^2)(1-b^2)-b^4 = 1-2b^2\]

Thanks for participating in this challenge, lfdahl, and like greg1313, I'm also less sure that your method is conclusive enough. :(

Here is my solution though::)

Spoiler

Note that

$\begin{align*}(x^2+y^2+z^2)(x^2+y^2-z^2)(x^2-y^2+z^2)(x^2-y^2-z^2)&=(x^4+2x^2y^2+y^4-z^4)(x^4-2x^2y^2+y^4-z^4)\\&=(x^4+y^4-z^4+2x^2y^2)(x^4+y^4-z^4-2x^2y^2)\\&=(x^4+y^4-z^4)^2-4x^4y^4\\&=(x^4+y^4+z^4-2z^4)^2-4x^4y^4\\&=(x^4+y^4+z^4)^2-4x^4y^4-4x^4z^4-4y^4z^4\\&=(x^4+y^4+z^4)^2-2((x^4+y^4+z^4)^2-x^8-y^8-z^8)\\&=2(x^8+y^8+z^8)-(x^4+y^4+z^4)^2\end{align*}$,

This tells us $(x^4+y^4+z^4)^2-2(x^8+y^8+z^8)=(x^2+y^2+z^2)(x^2+y^2-z^2)(x^2-y^2+z^2)(x^2-y^2-z^2)$

Since $x^2+y^2+z^2\ne 0$, we get either $x^2+y^2-z^2=0$, $x^2-y^2+z^2=0$ or $x^2-y^2-z^2=0$, which in turn suggest that $x,\, y,\, z$ can form the sides of a right-angled triangle.

 

[Euge]

Hi anemone,

Spoiler

It looks like we both like factoring. I had the same factoring idea, but I took slightly different steps. Namely, I let $u = x^2, v = y^2, w = z^2$, and write

$$2(u^4 + v^4 + w^4) - (u^2 + v^2 + w^2)^2$$
$$ = (u^4 + v^4 + w^4 + 2u^2v^2 - 2v^2w^2 - 2w^2u^2) - 4u^2v^2$$
$$ = (u^2 + v^2 - w^2)^2 - (2uv)^2$$
$$ = (u^2 + v^2 - w^2 - 2uv)(u^2 + v^2 - w^2 - 2uv)$$
$$ = [(u - v)^2 - w^2][(u + v)^2 - w^2]$$
$$ = (u - v - w)(u - v + w)(u + v - w)(u + v + w)$$

Reverting back to the $x$, $y$, and $z$ variables, we get the same thing.

 

[anemone]

Hi anemone,

Spoiler

It looks like we both like factoring. I had the same factoring idea, but I took slightly different steps. Namely, I let $u = x^2, v = y^2, w = z^2$, and write

$$2(u^4 + v^4 + w^4) - (u^2 + v^2 + w^2)^2$$
$$ = (u^4 + v^4 + w^4 + 2u^2v^2 - 2v^2w^2 - 2w^2u^2) - 4u^2v^2$$
$$ = (u^2 + v^2 - w^2)^2 - (2uv)^2$$
$$ = (u^2 + v^2 - w^2 - 2uv)(u^2 + v^2 - w^2 - 2uv)$$
$$ = [(u - v)^2 - w^2][(u + v)^2 - w^2]$$
$$ = (u - v - w)(u - v + w)(u + v - w)(u + v + w)$$

Reverting back to the $x$, $y$, and $z$ variables, we get the same thing.

Wow, Euge! Your method is definitely neater and more elegant than mine! I therefore have learned something valuable from your solution! :)