calculus - Continuous functions that satisfies $f(x) + f(1-x) + f(sqrt{x^2+(1-x)}) = 0$ and $f(frac12)=0$

$f:[0,1]\rightarrow \mathbb{R}$ is a continuous function which satisfies

$$f(x) + f(1-x) + f\left(\sqrt{x^2+(1-x)}\right) = 0 \text{ and } f\left(\tfrac12\right)=0.$$

Can someone give explicit examples of $f$, apart from the trivial solution, $f(x)=0$? And are there infinitely many solutions for $f$?

I can derive some properties like $f(\frac{\sqrt3}2)=0$ or $f(0)=-2f(1)$, but I can't generate particular examples. Continuity seems important here, but I can't see how to use it, for if we take $f(x)=0$, for $x\in (0,1)$, and $f(1)=1$, $f(0)=-2$ also works.

Answer

First, notice that $\sqrt{x^2+1-x}\geq\sqrt{3}/2$ for all $x\in[0,1]$. Let $g:[1/2,\sqrt{3}/2]\to\mathbb R$ be an arbitrary continuous function such that $g(1/2)=g(\sqrt{3}/2)=0$. Then, define $f:[0,1]\to\mathbb R$ as follows:
$$\begin{align*} f(x)=\begin{cases} 0&\text{if $x\in[0,1-\sqrt{3}/2]$,}\\ -g(1-x)&\text{if $x\in(1-\sqrt{3}/2,1/2]$,}\\ g(x)&\text{if $x\in(1/2,\sqrt{3}/{2}]$,}\\ 0&\text{if $x\in(\sqrt{3}/2,1]$.} \end{cases} \end{align*}$$

For example: