If $\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$ for real $t$ and $0<x<1$, prove that $x=1/2$

Question

$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$ where $0<x<1$ and $t\in\mathbb{R}$. Prove that $x=1/2$.

It is evident that $x=1/2$ satisfies the above equation. Please help.

Answer 1

Well, define $y \in \mathbb{R}$ such that $x=1-y$ (noting that also $0<y<1$), we get $$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$ $$\implies \frac{t^2-(1-y)^2}{(t^2+(1-y)^2)^2}+\frac{y^2-t^2}{(y^2+t^2)^2}=0$$ Note that the first equation implies $$\frac{t^2-x^2}{(t^2+x^2)^2}=\frac{t^2-(1-x)^2}{((1-x)^2+t^2)^2}$$ While the second implies $$\frac{t^2-(1-y)^2}{(t^2+(1-y)^2)^2}=\frac{t^2-y^2}{(y^2+t^2)^2}$$ So, in fact, the two equations are equivalent, which means $y=x \implies 1-x=x \implies x=1/2$

Edit: I figured out that what I proved was if $a$ is a solution (for $x$) then $1-a$ is also a solution. This means that $x=1/2$ for sure is a solution for all $t$.

But when I tried graphing the equation on desmos and added a slider for $t$, I get that as $t$ ranges over all reals (or $t^2$ ranges over positive reals), $x$ and $1-x$ vary with it. For example, $t^2 = \frac{1}{6}\left(\sqrt{601} - 13\right) \implies x \in \{-2,1/2,3\}$

Note that $x=1/2$ is the general solution, and $x=3$ is a specific solution and if $3$ is a solution then $1-3=-2$ is also a solution.

If we want $x=4$ as a solution we have $x=1-4=-3$ is also a solution and $t^2=\frac{1}{6}\left(\sqrt{2353} - 25\right)$ satisfies the equation. So, there's no point of solving for a specific $x$ if $x$ just varies depending on $t^2$ over the reals (and of course $1-x$ accordingly), but for all $t$, $x=1/2$ is certainly a solution.

Answer 2

Seems the fastest way to show this is indeed considering the polynomial left after clearing the denominator and dividing by $2(x-\frac{1}{2})$ $$f(x)=-3 t^4 - 2 t^2 x^2 + 2 t^2 x - t^2 + x^4 - 2 x^3 + x^2$$ Considering $$f'(x)= -2 (2 x - 1) (t^2 - x^2 + x)$$ we see that as $(t^2+x(1-x))>0$ for $0<x<1$ thus the function increases on $\left(0,\frac{1}{2}\right)$ and decreases on $\left(\frac{1}{2},1\right)$ thus we have to check only that $f(0),\,f\left(\frac12\right),\,f(1)$ has the same sign, indeed: $$f(0)=-3t^4-t^2\le 0$$ $$f(1)=-3t^4-t^2\le 0$$ $$f\left(\frac{1}{2}\right)=-3 t^4 - \frac{t^2}{2} + \frac{1}{16}$$ But the latter expression factors as $$-\frac{1}{16} (4 t^2 + 1) (12 t^2 - 1)$$ so for $12t^2-1<0$ the statement is not true. $$$$