Theorem: Brachistochrone is Cycloid

The shape of the brachistochrone is a cycloid.

Proof:
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Recall from the Snell-Descartes Law:
:$\dfrac {\sin \alpha_1} {v_1} = \dfrac {\sin \alpha_2} {v_2}$
Here, we invoke a generalization of the Snell-Descartes Law.
This is justified, as we are attempting to demonstrate the curve that takes the smallest time.
Thus we have $\dfrac {\sin \alpha} v = k$, where $k$ is some constant.
By the Principle of Conservation of Energy, the speed of the bead at a particular height is determined by its loss in potential energy in getting there.
Thus, at the point $\tuple {x, y}$, we have:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2 m v^2
      | r = m g y
      | c = Principle of Conservation of Energy
}}
{{eqn | l = v^2
      | r = 2 g y
      | c = multiplying both sides by $\dfrac 2 m$
}}
{{eqn | n = 1
      | l = v
      | r = \sqrt {2 g y}
      | c = taking the square root of both sides
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = \sin \alpha
      | r = \cos \beta
      | c = from the diagram above
}}
{{eqn | r = \frac 1 {\sec \beta}
      | c = {{Defof|Real Secant Function}}
}}
{{eqn | r = \frac 1 {\sqrt {1 + \tan^2 \beta} }
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | n = 2
      | r = \frac 1 {\sqrt {1 + \paren {y'}^2} }
      | c = $y' = \tan \beta$ in the diagram above
}}
{{end-eqn}}
Combining all the above equations, we get:
{{begin-eqn}}
{{eqn | l = \frac {\frac 1 {\sqrt {1 + \paren {y'}^2} } } {\sqrt {2 g y} }
      | r = k
      | c = Putting $(1)$ and $(2)$ into the Snell-Descartes Law
}}
{{eqn | ll= \leadsto
      | l = \frac {\frac 1 {1 + \paren {y'}^2 } } {2 g y }
      | r = k^2
      | c = squaring both sides
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {y \paren {1 + \paren {y'}^2} }
      | r = 2 k^2 g
      | c = multiplying both sides by $2 g$
}}
{{eqn | ll= \leadsto
      | l = y \paren {1 + \paren {y'}^2}
      | r = \frac 1 {2 k^2 g} = c
      | c = taking the reciprocal of both sides
}}
{{end-eqn}}
where $c$ is another (more convenient) constant.
This is the differential equation which defines the brachistochrone.
Now we solve it:
{{begin-eqn}}
{{eqn | l = y \paren {1 + \paren {y'}^2}
      | r = c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1 + \paren {y'}^2
      | r = \dfrac c y
      | c = dividing both sides by $y$
}}
{{eqn | ll= \leadsto
      | l = \paren {y'}^2
      | r = \dfrac c y - 1
      | c = 
}}
{{eqn | n = 3
      | ll= \leadsto
      | l = \frac {\d y} {\d x}
      | r = \sqrt {\frac {c - y} y}
      | c = taking the square root of both sides
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d y}
      | r = \sqrt {\frac y {c - y} }
      | c = taking the reciprocal of both sides
}}
{{eqn | ll= \leadsto
      | l = \rd x
      | r = \sqrt {\frac y {c - y} } \rd y
      | c = 
}}
{{end-eqn}}
Now we introduce a change of variable:
Let $\sqrt {\dfrac y {c - y} } = \tan \phi$
Thus:
{{begin-eqn}}
{{eqn | n = 4
      | l = \sqrt {\frac y {c - y} }
      | r = \tan \phi
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac y {c - y}
      | r = \tan^2 \phi
      | c = squaring both sides
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \paren {c - y} \tan^2 \phi
      | c = multiplying both sides by $\paren {c - y}$
}}
{{eqn | ll= \leadsto
      | l = y
      | r = c \tan^2 \phi - y \tan^2 \phi
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y \paren {1 + \tan^2 \phi}
      | r = c \tan^2 \phi
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {c \tan^2 \phi} {\paren {1 + \tan^2 \phi} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {c \tan^2 \phi} {\sec^2 \phi }
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | n = 5
      | ll= \leadsto
      | l = y
      | r = c \sin^2 \phi
      | c = {{Defof|Real Secant Function}}
}}
{{eqn | n= 6
      | ll= \leadsto
      | l = \frac {\d y} {\d \phi}
      | r = 2 c \sin \phi \cos \phi
      | c = Derivative of Sine Function, Derivative of Square Function
}}
{{end-eqn}}
Also:
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
      | r = \sqrt {\frac {c - y} y}
      | c = from $(3)$ above
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} {\d x}
      | r = \frac 1 {\tan \phi}
      | c = from $(3)$ and $(4)$ above
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} {\d \phi}
      | r = \frac {\d y} {\d x} \frac {\d x} {\d \phi}
      | c = Derivative of Composite Function
}}
{{eqn | r = 2 c \sin \phi \cos \phi
      | c = from $(6)$ above
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d \phi}
      | r = \dfrac {\paren {\frac {\d y} {\d \phi} } } {\paren {\frac {\d y} {\d x} } }
      | c = rearranging Derivative of Composite Function
}}
{{eqn | r = \dfrac {2 c \sin \phi \cos \phi} {\dfrac 1 {\tan \phi} }
      | c = 
}}
{{eqn | r = 2 c \sin^2 \phi
      | c = {{Defof|Real Tangent Function}}
}}
{{eqn | r = c \paren {1 - \cos 2 \phi}
      | c = {{Corollary|Double Angle Formula for Cosine|5}}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d \phi}
      | r = c \paren {1 - \cos 2 \phi}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \rd x
      | r = \int c \paren {1 - \cos 2 \phi} \rd \phi
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac c 2 \paren {2 \phi - \sin 2 \phi} + c_1
      | c = Primitive of $\cos a x$ and Primitive of Constant
}}
{{end-eqn}}
As the curve goes through the origin, we have $x = y = 0$ when $\phi = 0$ and so $c_1 = 0$.
Now we can look again at our expression for $y$:
{{begin-eqn}}
{{eqn | l = y
      | r = c \sin^2 \phi
      | c = from $(5)$ above
}}
{{eqn | r = \frac c 2 \paren {1 - \cos 2 \phi}
      | c = {{Corollary|Double Angle Formula for Cosine|5}}
}}
{{end-eqn}}
To simplify the constants, we can substitute $a = c / 2$ and $\theta = 2 \phi$, and thus we get:
{{begin-eqn}}
{{eqn | l = x
      | r = a \paren {\theta - \sin \theta}
      | c = 
}}
{{eqn | l = y
      | r = a \paren {1 - \cos \theta}
      | c = 
}}
{{end-eqn}}
which are the parametric equations of the cycloid.
{{qed}}