How was the explicit closed form for this implicit function derived? The problem comes from reading this [0] paper but I think I can express it in a self contained question.

Consider the implicit function $H(z)$ defined by the relation:
$$F_z(z+H(z))-F_z(z-H(z))=0.5$$
Where $F_z$ is the CDF of an absolutly continuous, unimodal distribution.
The authors point out that when $F'_z=f_z=\max(1-|z|,0)$ (the triangle distribution with parameters $a=-1,c=0$ and $b=1$), 
$$H(z)= \begin{cases}
   1-\sqrt{1/2-z^2} & \text{if } |z| \leq 1/2 \\
   |z|       & \text{if } |z| > 1/2
  \end{cases}$$
(p387, bottom of the article I linked to)
My questions are:


*

*How was the explicit formulation for $H(z)$ derived?

*Can it also be done for other values of $c$, i.e. $c=1$



[0] Ola Hössjer, Peter J. Rousseeuw and Christophe Croux.  Asymptotics of an estimator of a robust spread functional. Statistica Sinica 6(1996), 375-388.
 A: Here is a piecewise formula for $f_z$.
$$
f_z(z) = \begin{cases}
 z+1 & -1\leq z\leq 0 \\
 1-z & 0\leq z\leq 1 \\
 0 & \text{else}.
\end{cases}
$$
Computing a formula for $F_z$ is just a matter of integrating.
$$
F_z(z) = \int_{-1}^z f_z(z) \, dz = 
\begin{cases}
 0 & z\leq -1 \\
 \frac{z^2}{2}+z+\frac{1}{2} &
   -1<z\leq 0 \\
 -\frac{z^2}{2}+z+\frac{1}{2} &
   0<z\leq 1 \\
 1 & z>1.
\end{cases}
$$
Note that the constants are chosen to ensure continuity.  With this, your implicit formula
$$F_z(z+H(z))-F_z(z-H(z))=1/2$$
can be analyzed on certain intervals.  Suppose, for example, that $0\leq z+H(z) \leq 1$ and that $-1\leq z-H(z) \leq 0$, which imples that $-1/2\leq z \leq 1/2$. Then the implicit formula becomes
$$\left(-\frac{1}{2}(z+H(z))^2 + (z+H(z)) + 1/2\right) - \left(\frac{1}{2}(z-H(z))^2 + (z-H(z)) + 1/2)\right) = \frac{1}{2}.$$
Solving this formula for $H(Z)$ yields the formula for $H(z)$ yields the portion of your piecewise expression valid for $-1/2\leq z \leq 1/2$.  The other terms follow in similar fashion.  
I'm certain that a similar thing can be done with different choices of your constant $c$.
