A cycloid can be described in parametric form as
<br>x = t − sin(t)
<br>y = 1 − cos(t).
<br>This comes from imagining a circle whose radius is 1 and whose center is at &lt;t, 1> at time t, rolling along the line y = 0.
P is the point on the circumference of the circle leaving the trace.

<p>A half height circle rolling at the same velocity and turning twice as fast yields the half cycloid:
<br>x = t − sin(2t)/2
<br>y = ½ − cos(2t)/2.
<p>The small circle can be seen as rolling inside the large circle
— for all values of t they are mutually tangent at &lt;t, 0>.
The point p is on the small circle that starts at &lt;0, 0> and traverses the small cycloid.
p is always on the diameter of the large circle thru P.
<br>p = &lt; t − sin(2t)/2, ½ − cos(2t)/2>
<br>See <a href=http://www.michaelbach.de/ot/mot-Tusi/index.html>this lovely animation</a>.
<p>P moves as in the first equations.
The other end P' of the diameter moves thus:
<br>P' = &lt; t + sin(t), 1 + cos(t)>
<br>p moves harmonically between P and P' thus:
<br>p = ((1 + cos(t))P + (1 − cos(t))P')/2
<br> =&lt;(1 + cos(t))(t − sin(t)) + (1 − cos(t))( t + sin(t)),
(1 + cos(t))(1 − cos(t)) + (1 − cos(t))(1 + cos(t))>/2
<br>= &lt; t − cos(t)sin(t), 1 − cos<sup>2</sup>(t)>
=  &lt; t − sin(2t)/2, ½ − cos(2t)/2>
<br>which agrees with the description of p as a point on the small circle.
<p>Now if p's velocity is always parallel with the large diameter
then p will be on the envelope of that diameter.
<br>dp/dt = d&lt; t − sin(2t)/2, ½ − cos(2t)/2>/dt
<br>=  &lt; d(2t − sin(2t))/dt, d(1 − cos(2t))/dt>/2
= &lt; 2 − 2cos(2t), 2sin(2t)>/2
= &lt;1 − cos(2t), sin(2t)>.
<p>The diameter as vector is &lt;cos(t), sin(t)> which is parallel to &lt;cos(t), sin(t)>sin(t)
 =  &lt;cos(t)sin(t), sin<sup>2</sup>(t)>

<hr>The claim is that the diameter of the large circle thru P is always tangent to the small cycloid.
Thus the small cycloid is the envelope of the diameter of the big circle.