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anemone
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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.
Prove that it's a right triangle.
greg1313 said:A point of confusion:
I suppose I am probably missing something, but if \(\displaystyle x=y=z=\left(\dfrac23\right)^{\dfrac15}\) then
\(\displaystyle (x^4+y^4+z^4)^2=2(x^3+y^3+z^3)\) and the triangle is equilateral.
RLBrown said:Can one of the sides be negative?
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.
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.
\[(*). \;\;\; a^4-b^4 = (1-b^2)(1-b^2)-b^4 = 1-2b^2\]
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.
\[(*). \;\;\; a^4-b^4 = (1-b^2)(1-b^2)-b^4 = 1-2b^2\]
Euge said:Hi anemone,
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.
A right triangle is a type of triangle that has one angle measuring 90 degrees. This angle is called the right angle, and it is formed by the intersection of the two shorter sides of the triangle.
There are several ways to prove that a triangle is a right triangle. One way is to use the Pythagorean theorem, which states that the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. If this equation holds true for a given triangle, then it is a right triangle.
Yes, there are other methods for proving that a triangle is a right triangle. These include using trigonometric ratios (such as sine, cosine, and tangent) or the properties of special right triangles (such as the 30-60-90 or 45-45-90 triangles).
Right triangles are commonly used in various fields, such as engineering, architecture, and navigation. They are also used in everyday situations, such as measuring the height of a building or calculating the slope of a ramp.
Proving that a triangle is a right triangle not only confirms the properties of the triangle but also allows for the use of various mathematical formulas and relationships that are specific to right triangles. This can be useful in solving various problems and applications involving right triangles.