The figure above represents the starting region of our track. Our posited optimal track is S-P-R where P is the point at which the cycloid part of the track meets the floor. 

Hitting the  floor left of P:  If our track were to hit the floor to the left of P and then rolled horizontally to P, we would necessarily take more time than our cycloid track S-P-R. This is because the cycloid is the optimal track to get to P as was shown 3 centuries ago!

Hitting the floor to the right of P (or never hitting the floor):

Consider a curve such as the one in black above that hits the floor at point R. Let’s say that this curve passes directly above our posited optimum at point Q. The marble will inevitably arrive at Q at a slower speed than our path at P, because it has fallen through a smaller vertical distance. But this marble may also arrive in less time at Q. This suggests there might be a path from Q to R such that in total we arrive faster at R than through S-P-R leading to a better solution! We have to show this is not possible.

Now consider this point Q. The original Brachistochrone solution tells us that the optimal path to Q has to be a cycloid (shown in purple) and NOT the curve in black. So we’d presumably use the purple cycloid from S to Q and then switch to the black.

Now let’s calculate the time it takes to get from the start S to the point Q along the cycloid. Let the height of Q above P, i.e. \(PQ = a \). Using the original parametric cycloid equations along with the condition that it pass through the point Q \( (5\pi, a) \) gives us:

\begin{align*}
5\pi &= r(\theta – \sin \theta) \\
a &= 10 – r(1- \cos \theta) \\
t &= \theta \sqrt{\frac{r}{g}}
\end{align*}

Remember that \(r\) is the radius of the cycloid, and \(t\) is the time from the start. Using \(r\) from the first equation and substituting in the third equation (for time) gives us:

\[
t = \theta \sqrt{\frac{5\pi}{g(\theta- \sin \theta)}}
\]

Now, keep in mind that \(\theta\) necessarily has to be greater than \(\pi\) since \(r\) has to be less than the 5 cms for our optimal path for the cycloid to get to Q. This is also seen in the figure above. 

Now recollect that the time to get to P in our optimal path was:

\[
t_{P} = \pi \sqrt{\frac{5}{g}}
\]

Comparing the time \( t\) to get to Q with \(t_P\) the time to get to P we can actually show that the former is greater than the latter. i.e. it takes more time to get to any Q than to get to P!

\begin{align*}
t &\stackrel{?}{\geq} t_p \\
\theta \sqrt{\frac{5\pi}{g(\theta- \sin \theta)}} \ &\stackrel{?}{\geq} \ \pi \sqrt{\frac{5}{g}} \\
\Rightarrow \qquad \theta \sqrt{\frac{\pi}{\theta- \sin \theta}} \ &\stackrel{?}{\geq} \ \pi \\
\Rightarrow \qquad \frac{{\theta}^2}{\pi(\theta – \sin \theta)} \ &\stackrel{?}{\geq} \ 1 \\
\Rightarrow \qquad \theta (\theta – \pi) + \pi \sin \theta &\stackrel{?}{\geq} \ 0 \\ \\
\textit{Since } \theta > \pi \ \textit{ Let } \theta &= \pi + \phi \qquad 0 < \phi \leq \pi \\
\Rightarrow \qquad (\pi + \phi)\phi +\pi \sin (\pi + \phi) &\stackrel{?}{\geq} \ 0 \\
\Rightarrow \qquad \phi^2 + \pi (\phi – \sin \phi) &\stackrel{?}{\geq} \ 0 \\ \\
\textit{always true since} \quad \phi &\geq \sin \phi \qquad 0 < \phi \leq \pi \\
\end{align*}

This shows you can never get to any Q above P faster than getting to P. So there can be no better path that meets the floor after P. Also there can never be one that never meets the floor. If this is non obvious, consider P’, Q’ symmetrically on the other (right) side of the cycloid. Going from this Q’ to the finish will take longer than going from P’. Also the PP’ journey necessarily has to be faster than any QQ’ since the speed at P is greater than at Q and any QQ’ cannot be shorter than QQ’.

Q.E.D.