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.

Here is the set of all possible $f$.
The algebra gets a little gross, so bear with me.
First, let 
$f_1: \left[1 - \frac{\sqrt{3}}{2}, 1 \right]
\to \mathbb{R}$ be an arbitrary continuous function such that $f_1(x) = 0$ on the closed interval $\left[ \frac{\sqrt{3}}{2}, \frac{7}{4} - \frac{\sqrt{3}}{2} \right]$,
and $f_1(\tfrac12) = 0$,
and finally,
$f_1(\tfrac{\sqrt{3}}{2}) + f_1(\frac{7}{4} - \frac{\sqrt{3}}{2}) = 0$.
Second, define
$f_2 : \left[1 - \frac{\sqrt{3}}{2}, 1 \right] \to \mathbb{R}$
by
$$
f_2(x) =
\begin{cases}
f_1(x)
& \text{if } 1 - \frac{\sqrt{3}}{2} \le x \le \frac{\sqrt{3}}{2} \\
- f_1\left( \frac{1 - \sqrt{4x^2 - 3}}{2} \right)
- f_1\left( \frac{1 + \sqrt{4x^2 - 3}}{2} \right)
& \text{if }
\frac{\sqrt{3}}{2} \le x \le \frac{7}{4} - \frac{\sqrt{3}}{2} \\
f_1(x)
& \text{if }
\frac{7}{4} - \frac{\sqrt{3}}{2} \le x \le 1 \\
\end{cases}
$$
Verify that
$f_2$ is continuous, and satisfies
$f_2(x) + f_2(1-x) + f_2(\sqrt{x^2 - x + 1}) = 0$
on its domain.
This is a bit gross, so I'm skipping the details for now.
Finally, define
$$
f_3(x) =
\begin{cases}
-f_2(1 - x) - f_2(\sqrt{x^2 - x + 1})
&\text{if } 0 \le x \le 1 - \frac{\sqrt{3}}{2} \\
f_2(x)
&\text{if } 1 - \frac{\sqrt{3}}{2} \le x \le 1 \\
\end{cases}
$$
Verify that $f_3$ is continuous,
and satisfies
$f_2(x) + f_2(1-x) + f_2(\sqrt{x^2 - x + 1}) = 0$
on its domain.
This is not too hard given that we already showed it for $f_2$.
Now just set $f = f_3$.

The end result is a function $f$
such that $f_1 = f$ on
$\left[1 - \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2} \right]$
and $\left[ \frac{7}{4} - \frac{\sqrt{3}}{2}, 1 \right]$,
and $f$ satisfies the desired property.
So we get to choose $f$ however we want on
$\left[1 - \frac{\sqrt{3}}{2} , \frac{\sqrt{3}}{2} \right]$
and $\left[ \frac{7}{4} - \frac{\sqrt{3}}{2} , 1 \right]$,
and then $f$ is decided for us everywhere else.
