Theorem: Brachistochrone is Cycloid

The shape of the brachistochrone is a cycloid.

Proof:
Throughout this proof, we use the standard alignment of coordinate axes:
:$X$-axis pointing rightwards
:$Y$-axis is pointing upwards.
Suppose that the curve passes through the point $\tuple {x, y}$ for some value of variable $t$.
Due to smoothness of the curve, one can define velocity $v$ at a point $\tuple {\map x t, \map y t}$:
:$v = \dfrac {\d s} {\d t}$
where $\d s$ is an infinitesimal arc length element.
In Euclidean space the arc length element is:
:$\d s = \sqrt {1 + y'^2} \rd x$
Hence:
:$v = \sqrt {1 + y'^2} \dfrac {\d x} {\d t}$
Due to the symmetries of Euclidean space, the Principle of Conservation of Energy holds:
:$\dfrac {m v^2} 2 + m g y = E$
where $E$ is a constant of motion.
To determine $E$ use the following initial conditions:
:$\tuple {\map x 0, \map y 0} = \mathbf 0$
:$\tuple {\map {\dfrac {\d x} {\d t} } 0, \map {\dfrac {\d y} {\d t} } 0} = \mathbf 0$
Then it follows that:
:$E = 0$
and:
:$v = \sqrt {-2 g y}$
Then the total travel time, integrated {{WRT|Integration}} $x \in \closedint a b$ is:
:$\ds T = \int_a^b \frac {\sqrt {1 + y'^2} } {\sqrt {-2 g y} } \rd x$
Application of Euler's Equation yields:
:$\dfrac {\sqrt {1 + y'^2} } {\sqrt {-2 g y} } - y' \dfrac {2 y'} {2 \sqrt {-2 g y} \sqrt {1 + y'^2} } = c$
or
:$\sqrt C = \sqrt {-y \paren {1 + y'^2} }$
where 
:$C = \dfrac 1 {2 c^2 g}$ 
and $c$ is a real constant.
The aforementioned differential equation can be rearranged to:
:$\dfrac {\d x} {\d y} = \pm \sqrt {\dfrac {-y} {y + C} }$
Since we want to describe a downwards sliding bead, we have:
:$\dfrac {\d y} {\d x} \le 0$
and we choose the minus sign.
This differential equation can be solved for $\map x y$ in the following way:
{{begin-eqn}}
{{eqn | l = x
      | r = -\int \sqrt{\frac {-y} {y + C} } \rd y
}}
{{eqn | r = \int \frac {\sqrt {C u - 1} } {u^2} \rd u
      | c = substituting $u = \dfrac 1 {C + y}$
}}
{{eqn | r = -\frac {\sqrt {C u - 1 } } u + \frac C 2 \int \dfrac {\d u} {u \sqrt {C u - 1} }
      | c = Primitive of $\dfrac {\sqrt {a x + b} } {x^2}$
}}
{{eqn | r = -\frac {\sqrt {C u - 1} } u + \frac C 2 2 \, \map \arctan {C u - 1} + C_1
      | c = Primitive of $\dfrac 1 {x \sqrt {a x + b} }$
}}
{{eqn | r = -\sqrt {-y \paren {C + y} } + C \, \map \arctan {\frac {-y} {C + y} } + C_1
}}
{{end-eqn}}
From the initial condition $\tuple {\map x 0, \map y 0} = \mathbf 0$ it follows that:
:$C_1 = 0$
To bring the solution to parametric form, introduce the following parametric dependence:
:$\sqrt {\dfrac {-y} {C + y} } = \tan \theta$
which can be solved for $y$:
{{begin-eqn}}
{{eqn | l = \sqrt {\frac {-y} {C + y} }
      | r = \tan \theta
}}
{{eqn | ll= \leadsto
      | l = \frac {-y} {C + y}
      | r = \tan^2 \theta
}}
{{eqn | ll= \leadsto
      | l = - 1 - \frac C y
      | r = \frac 1 {\tan^2 \theta}
}}
{{eqn | ll= \leadsto
      | l = y
      | r = -C \sin^2 \theta
}}
{{eqn | r = -\frac C 2 \paren {1 - \map \cos {2 \theta} }
      | c = Square of Sine
}}
{{eqn | ll= \leadsto 
      | r = -\frac C 2 \paren {1 - \map \cos \phi}
      | c = Introduce a new parameter $\phi = 2 \theta$
}}
{{end-eqn}}
Substitution into the expression for $x$ results in:
{{begin-eqn}}
{{eqn | l = x
      | r = C \paren {\theta - \frac {\map \sin {2 \theta} } 2 }
}}
{{eqn | r = C \paren {\frac \phi 2 - \frac {\sin \phi} 2}
      | c = Introduce a new parameter $\phi = 2 \theta$
}}
{{eqn | r = \frac C 2 \paren {\phi - {\sin \phi} }
}}
{{end-eqn}}
To determine $C$ we use the boundary condition for the final point:
:$\intlimits {\dfrac {\d y} {\d x} } {x = b} {} = 0$
For parametric equations we can rewrite this as:
:$\dfrac {\d y} {\d x} = \dfrac {\d y} {\d \phi} \paren {\dfrac {\d x} {\d \phi} }^{-1}$
We need to find to which $\phi$ the point $x = b$ corresponds.
Notice that:
:$\paren {\dfrac {\d y} {\d \phi} = 0 \land \dfrac {\d x} {\d \phi} \ne 0} \implies \paren {\dfrac {\d y} {\d x} = 0}$
Therefore:
:$\dfrac {\d y} {\d \phi} = -\dfrac C 2 \map \sin \phi$
and this derivative vanishes if:
:$\phi = \pi n, n \in \Z$
Similarly:
:$\dfrac {\d x} {\d \phi} = \dfrac C 2 \paren {\map \cos \phi - 1}$
which vanishes if:
:$\phi = 2 \pi n, n \in \Z$
By comparing both conditions on $\phi$ we limit the set of solutions to $\phi = \pi + 2 \pi n, n \in \Z$.
We choose the nearest appropriate value corresponding to $x = b > 0$.
Then substitution into the expression for $x$ results in:
:$b = \dfrac C 2 \pi$
Finally, the parametric form of this curve is described by such a parametric solution:
:$x = \dfrac b \pi \paren {\phi - {\sin \phi} }$
:$y = -\dfrac b \pi \paren {1 - \map \cos \phi}$
This is the form of a cycloid as portrayed upside down.
{{qed}}