= Solution
{c}
Put $f(\phi)=dx/d\phi$. Since $\sin((2\pi-\phi)/2)=\sin(\phi/2)$, both $f(2\pi-\phi)=f(\phi)$ and $y(2\pi-\phi)=y(\phi)$. Using $x(2\pi)=1$ in the integral for $x$ gives
$$
\boxed{x(2\pi-\phi)=1-x(\phi),\qquad y(2\pi-\phi)=y(\phi).}
$$
Hence the physical track $(x,-y)$ is symmetric about $x=1/2$, with $x(\pi)=1/2$. For $0<\phi<2\pi$, $dx/d\phi>0$, and
$$
\frac{dy}{d\phi}=\frac{\sin(\phi/2)\cos(\phi/2)}{c(1-b\sin(\phi/2))^3}
$$
is positive before $\pi$ and negative after $\pi$. Therefore the \b[physical height $-y$ has its unique minimum at $x=1/2$], with
$$
\boxed{y_{\max}=\frac1{c(1-b)^2},\qquad (x,\text{height})_{\min}=\left(\frac12,-\frac1{c(1-b)^2}\right).}
$$
The depth $y$ has a maximum there. This proves the intended claim for every finite-time member $L>1$; $L=1$ has the degeneracy described in (a).
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2021-constrained-brachistochrone.png]
{title=Stationary brachistochrone arches at several fixed lengths, with vertical endpoint tangents}
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