MHB Right Triangle: Proving It's a Right Triangle

AI Thread Summary
The discussion centers around proving that a triangle with sides $x, y, z$ satisfies the equation $(x^4+y^4+z^4)^2=2(x^8+y^8+z^8)$ is a right triangle. Participants express confusion about whether negative side lengths are permissible and acknowledge a typo in the original problem statement. Various solutions are proposed, with some members expressing uncertainty about the conclusiveness of their methods. Others appreciate the elegance of different approaches shared in the thread. The conversation highlights the collaborative effort to solve the mathematical challenge.
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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.
 
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A point of confusion:

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.:confused:
 
greg1313 said:
A point of confusion:

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.:confused:

Hi greg1313,

I think I should have mentioned that all three of them $x,\,y,\,z$ aren't equal.
 
Can one of the sides be negative?
 
RLBrown said:
Can one of the sides be negative?

Hi RLBrown,
No, all $x,\,y,\,z$ are supposed to be all different real positive numbers.:)
 
anemone said:
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)
 
Here is my solution. I think there are more elegant approaches.

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.

\[(*). \;\;\; a^4-b^4 = (1-b^2)(1-b^2)-b^4 = 1-2b^2\]
 
lfdahl said:
Here is my solution. I think there are more elegant approaches.

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.

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

Good work, but I'm not sure that proves there are not other values for $x$, $y$ and $z$ for which the equation holds.
 
lfdahl said:
Here is my solution. I think there are more elegant approaches.

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::)

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.
 
  • #10
Hi anemone,

It looks like we both like factoring. :cool: 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.
 
  • #11
Euge said:
Hi anemone,

It looks like we both like factoring. :cool: 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! :)
 

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