Math puzzle - Help with degenerate cases?

Question

Here is the original problem:

Given the following equalities and the values for $v,x,y\in \mathbb{R}$: $$\begin{matrix}v^2 = w(w-2x)(w-2y)(w-2z) & (1) \\ w=x+y+z & (2)\end{matrix}$$

Find all values for $z\in \mathbb{R}$.

I will put my solution in spoiler tags in case you find this problem interesting and want to solve it on your own.

I noticed that this is a modification on Heron's formula. If I had a triangle with sides of lengths $x,y,z$ and area $\dfrac{v}{4}$, it would satisfy Heron's formula with perimeter $w$ and semi-perimeter $\dfrac{w}{2}$. Therefore, I started with the assumption that I had a valid triangle. Thus, if the angle between the sides of length $x$ and $y$ were $\gamma$, then $\dfrac{v}{4} = \dfrac{1}{2}xy\sin \gamma$. Similarly, by the Law of Cosines, I have $z^2 = x^2+y^2-2xy\cos \gamma$. Since $\sin \gamma = \dfrac{v}{2xy} = \dfrac{\text{opp}}{\text{hyp}}$, then $\cos \gamma = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{\sqrt{4x^2y^2-v^2}}{2xy}$. Since $\cos \gamma$ will be negative for any $\gamma > \dfrac{\pi}{2}$, but the RHS will always be nonnegative, I need a $\pm$. Plugging in, this gives: $$z^2 = x^2+y^2\pm \sqrt{4x^2y^2-v^2}$$ Of course, then I just take the square root of both sides, and voila, I have up to four solutions for $z$. The degenerate cases I already found are when $x=0$ or $y=0$ as then $\gamma$ does not exist. If $x=y=0$, then (1) becomes $v^2=-z^4$, which is a contradiction if $v\neq 0$. If only one of them is 0, (say $x$), then the the formula becomes: $$\begin{matrix}v^2 = w^2(w-2y)(w-2z) & (3) \\ w = y+z & (4)\end{matrix}$$ which only has solutions $z = \pm y$. (It is easy to check that if $w\neq 0$ and $v\neq 0$, then $z$ has only complex solutions).

Question: Have I found all degenerate cases? How do I prove this?

Answer 1

After substitution of $w$ and expansion, the equation is biquadratic in $z^2$.

$$z^4-2(x^2+y^2)z^2+(x^2-y^2)^2+v^2=0.$$

It has real roots in $z^2$ when

$$(x^2+y^2)^2-(x^2-y^2)^2-v^2=4x^2y^2-v^2\ge0.$$

Then $$x^2+y^2+\sqrt{4x^2y^2-v^2}$$ is always positive, while

$$x^2+y^2-\sqrt{4x^2y^2-v^2}$$ requires

$$(x^2+y^2)^2\ge4x^2y^2-v^2,$$ $$(x^2-y^2)^2\ge-v^2$$ which is always true.

Hence, under the only condition

$$2|xy|\ge|v|,$$

we have four real roots

$$\pm\sqrt{x^2+y^2\pm\sqrt{4x^2y^2-v^2}}.$$

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Note that we can reduce the variables so that $v=2$ and the locus of $(x,y)$ is made of the four regions disconnected by the equilateral hyperbolas $xy\ge1$ and $-xy\ge1$.

Answer 2

Not an answer, but hints / ideas:

Put $X = w - 2x$, $Y = w - 2y$, $Z = w - 2z$, then you have $z = \frac{X+Y}{2} $ and $$ v^2 = (X+Y+Z)XYZ $$ So for given $v \ne 0$ , cases with no solution arise when $X=0$, or $Y=0$, or $Z=0$, or $X+Y+Z = 0$, or when exactly one or exactly three of the 4 factors are negative.

Conversely, for given $v =0$ , it is necessary that at least one of $X=0$, $Y=0$, $Z=0$, $X+Y+Z = 0$ must hold.

Lots of cases to discuss ....